Polar coordinates (disk, sector, annulus)

Solve the heat equation in polar coordinates for $u(r,t)$ \[ u_t = k (u_{rr} + \frac {1}{r} u_r) \] For $0<r<L$ and $t>0$. The boundary conditions are such it is insulated \begin {align*} u_r(L,t) &= 0 \end {align*}

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[r, t], t] == k*(D[u[r, t], {r, 2}] + 1/r*D[u[r, t], r]); 
ic  = u[r, 0] == f[r]; 
bc  = Derivative[1,0][u][L, t] == 0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[r, t], {r, t}, Assumptions->{r<L,L>0,k>0,t>0}], 60*10]];

\[\left \{\left \{u(r,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {\sqrt {2} e^{-\frac {k t \left (j_{1,K[1]}\right ){}^2}{L^2}} J_0\left (\frac {r j_{1,K[1]}}{L}\right ) \int _0^L \frac {\sqrt {2} r J_0\left (\frac {r j_{1,K[1]}}{L}\right ) f(r)}{L J_0\left (j_{1,K[1]}\right )} \, dr}{L J_0\left (j_{1,K[1]}\right )}+\frac {\sqrt {2} \int _0^L \frac {\sqrt {2} r f(r)}{L} \, dr}{L}\right \}\right \}\]

Maple ✓

restart; 
pde := diff(u(r,t),t)= k*(diff(u(r,t),r$2)+ 1/r*diff(u(r,t),r)); 
ic  := u(r,0)=f(r); 
bc  := D[1](u)(L,t)=0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],u(r,t)) assuming k>0,r>0,L>0,r<L,t>0 ),output='realtime'));

\[u \left (r , t\right ) = \sqrt {k}\, \left (\int _{-L +r}^{0}\mathcal {L}^{-1}\left (\frac {\BesselJ \left (0, \frac {\sqrt {-s}\, \left (r -\tau \right )}{\sqrt {k}}\right ) \mathcal {L}\left (\frac {-D\left (f \right )\left (L +\tau \right )+\left (-L -\tau \right ) D^{\left (2\right )}\left (f \right )\left (L +\tau \right )}{L +\tau }, t , s\right )}{\sqrt {-s}\, \BesselJ \left (1, \frac {\sqrt {-s}\, L}{\sqrt {k}}\right )}, s , t\right )d \tau \right )-\sqrt {k}\, \mathcal {L}^{-1}\left (\frac {\BesselJ \left (0, \frac {\sqrt {-s}\, r}{\sqrt {k}}\right ) \mathcal {L}\left (-D\left (f \right )\left (L \right ), t , s\right )}{\sqrt {-s}\, \BesselJ \left (1, \frac {\sqrt {-s}\, L}{\sqrt {k}}\right )}, s , t\right )+f \left (r \right )\] Cant get series solution

Solve for $u\left ( r,t\right ) $\begin {align*} u_{t} & =k\left ( u_{rr}+\frac {1}{r}u_{r}\right ) \\ \left \vert u\left ( 0,t\right ) \right \vert & <\infty \\ u_{r}\left ( L,t\right ) & =0\\ u\left ( r,0\right ) & =f\left ( r\right ) \end {align*}

for $0<r<L\,$. Let $u=RT$. Substituting in the PDE gives\begin {align*} \frac {T^{\prime }R}{k} & =R^{\prime \prime }T+\frac {1}{r}R^{\prime }T\\ \frac {T^{\prime }}{kT} & =\frac {R^{\prime \prime }}{R}+\frac {1}{r}\frac {R^{\prime }}{R}=-\lambda \end {align*}

The above gives the following ODE’s to solve\[ T^{\prime }+\lambda kT=0 \] And\begin {align} R^{\prime \prime }+\frac {1}{r}R^{\prime }+\lambda R & =0\tag {1}\\ R^{\prime }\left ( L\right ) & =0\nonumber \\ R\left ( 0\right ) & <\infty \nonumber \end {align}

Let $\lambda =-n^{2}$ for positive $n$. (1) becomes\begin {align*} R^{\prime \prime }+\frac {1}{r}R^{\prime }-n^{2}R & =0\\ r^{2}R^{\prime \prime }+rR^{\prime }-n^{2}r^{2}R & =0 \end {align*}

This is a Bessel ODE whose solution is $R\left ( r\right ) =C_{1}\operatorname {BesselI}\left ( 0,nr\right ) +C_{2}\operatorname {Besselk}\left ( 0,nr\right ) $. Since $R\left ( r\right ) $ is bounded at $r=0$ then $C_{2}=0$ since $\operatorname {Besselk}$ blows up at $r=0$. The solution becomes $R\left ( r\right ) =C_{1}\operatorname {BesselI}\left ( 0,nr\right ) $. Then $R^{\prime }\left ( r\right ) =C_{1}\operatorname {BesselI}\left ( 1,nr\right ) $. Hence we need to solve for $n$ in the following $R^{\prime }\left ( L\right ) =0=\operatorname {BesselI}\left ( 1,nL\right ) $. But $\operatorname {BesselI}$ has zero only at zero. Therefore $\lambda <0$ is not possible eigenvalue.

Equation (1) becomes $R^{\prime \prime }+\frac {1}{r}R^{\prime }=0$. The solution is $R\left ( r\right ) =C_{1}\ln \left ( r\right ) +C_{2}$. Since bounded at $r=0$ then $C_{1}=0$. The solution becomes $R\left ( r\right ) =C_{2}$ and $R^{\prime }\left ( r\right ) =0$. Which satisﬁes the boundary condition for any constant $C_{2}$. Therefore $\lambda =0$ is an eigenvalue with eigenfunction $R_{0}\left ( r\right ) =1$ as the choice of the constant.

Equation (1) becomes\[ r^{2}R^{\prime \prime }+rR^{\prime }+\lambda r^{2}R=0 \] This is a Bessel ODE whose solution is $R\left ( r\right ) =C_{1}\operatorname {BesselJ}\left ( 0,\sqrt {\lambda }r\right ) +C_{2}\operatorname {BesselY}\left ( 0,\sqrt {\lambda }r\right ) $. Since $R\left ( r\right ) $ is bounded at $r=0$ then $C_{2}=0$ since $\operatorname {BesselY}$ blows up at $r=0$. The solution becomes $R\left ( r\right ) =C_{1}\operatorname {BesselJ}\left ( 0,\sqrt {\lambda }r\right ) $. Then $R^{\prime }\left ( r\right ) =-C_{1}\operatorname {BesselJ}\left ( 1,\sqrt {\lambda }r\right ) $. Absorbing the minus sign into the constant, hence $R^{\prime }\left ( r\right ) =C_{1}\operatorname {BesselJ}\left ( 1,\sqrt {\lambda }r\right ) $. At $r=L$ we want

\[ 0=C_{1}\operatorname {BesselJ}\left ( 1,\sqrt {\lambda }L\right ) \] For non-trivial solution, $\sqrt {\lambda }L$ are the zeros of the $\operatorname {BesselJ}\left ( 1,x\right ) $. Let these zeros be $\Lambda _{n}$ where $n=1,2,\cdots $. Hence \begin {align*} \sqrt {\lambda _{n}}L & =\Lambda _{n}\qquad n=1,2,\cdots \\ \lambda _{n} & =\left ( \frac {\Lambda _{n}}{L}\right ) ^{2} \end {align*}

The corresponding eigenfunctions are\[ R_{n}\left ( r\right ) =\operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{L}r\right ) \qquad n=1,2,\cdots \] Now that we found the eigenvalues and eigenfunctions, we can solve the time ODE. For the zero eigenvalue $\lambda =0$ the ODE $T^{\prime }+\lambda kT=0$ becomes $T^{\prime }=0$, hence $T_{0}\left ( t\right ) $ is a constant. For $\lambda >0$ the time ODE has solution $T_{n}\left ( t\right ) =e^{-k\lambda _{n}t}=e^{-k\left ( \frac {\Lambda _{n}}{L}\right ) ^{2}t}$. Therefore the complete solution is\[ u\left ( r,t\right ) =C_{0}+\sum _{n=1}^{\infty }C_{n}e^{-k\left ( \frac {\Lambda _{n}}{L}\right ) ^{2}t}\operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{L}r\right ) \] Now we ﬁnd the constants from the initial conditions. At $t=0$\[ f\left ( r\right ) =C_{0}+\sum _{n=1}^{\infty }C_{n}\operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{L}r\right ) \] For $n=0$, applying orthogonality gives\begin {align*} \int _{0}^{L}f\left ( r\right ) rdr & =\int _{0}^{L}C_{0}rdr\\ & =C_{0}\frac {L^{2}}{2}\\ C_{0} & =\frac {2}{L^{2}}\int _{0}^{L}f\left ( r\right ) rdr \end {align*}

For $n>0$\begin {align*} \int _{0}^{L}f\left ( r\right ) \operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{L}r\right ) rdr & =\int _{0}^{L}C_{n}\operatorname {BesselJ}^{2}\left ( 0,\frac {\Lambda _{n}}{L}r\right ) rdr\\ C_{n} & =\frac {\int _{0}^{L}f\left ( r\right ) \operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{L}r\right ) rdr}{\int _{0}^{L}\operatorname {BesselJ}^{2}\left ( 0,\frac {\Lambda _{n}}{L}r\right ) rdr} \end {align*}

Hence the ﬁnal solution is\[ u\left ( r,t\right ) =\frac {2}{L^{2}}\int _{0}^{L}f\left ( r\right ) rdr+\sum _{n=1}^{\infty }\left ( \frac {\int _{0}^{L}f\left ( r\right ) \operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{L}r\right ) rdr}{\int _{0}^{L}\operatorname {BesselJ}^{2}\left ( 0,\frac {\Lambda _{n}}{L}r\right ) rdr}\right ) e^{-k\left ( \frac {\Lambda _{n}}{L}\right ) ^{2}t}\operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{L}r\right ) \] Where $\Lambda _{n}$ are the zeros of the $\operatorname {BesselJ}\left ( 1,x\right ) $.

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3.2.2.2 [259] no $\theta $ dependency, insulated (Speciﬁc solution)

Solve the heat equation in polar coordinates for $u(r,t)$ \[ u_t = k (u_{rr} + \frac {1}{r} u_r) \] For $0<r<L$ and $t>0$ and $k=1,L=1$. The boundary conditions are such it is insulated \begin {align*} u_r(L,t) &= 0 \end {align*}

Mathematica ✓

ClearAll["Global`*"]; 
k=1; 
 L=1; 
pde =  D[u[r, t], t] == k*(D[u[r, t], {r, 2}] + 1/r*D[u[r, t], r]); 
ic  = u[r, 0] == 2*L*r-r^2; 
bc  = Derivative[1,0][u][L, t] == 0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[r, t], {r, t}], 60*10]];

\[\left \{\left \{u(r,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}-\frac {2 e^{-t \left (j_{1,K[1]}\right ){}^2} J_0\left (r j_{1,K[1]}\right ) \left (\frac {2 J_2\left (j_{1,K[1]}\right )}{\left (j_{1,K[1]}\right ){}^2}-\frac {2}{3} \, _1F_2\left (\frac {3}{2};1,\frac {5}{2};-\frac {1}{4} \left (j_{1,K[1]}\right ){}^2\right )-\frac {J_3\left (j_{1,K[1]}\right )}{j_{1,K[1]}}\right )}{J_0\left (j_{1,K[1]}\right ){}^2}+\frac {5}{6}\right \}\right \}\]

Maple ✓

restart; 
k:=1; 
L:=1; 
pde := diff(u(r,t),t)= k*(diff(u(r,t),r$2)+ 1/r*diff(u(r,t),r)); 
ic  := u(r,0)= 2*L*r-r^2; 
bc  := D[1](u)(L,t)=0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],u(r,t))),output='realtime'));

\[u \left (r , t\right ) = -r^{2}+2 r -\left (\int _{0}^{r -1}\mathcal {L}^{-1}\left (\frac {\BesselJ \left (0, \sqrt {-s}\, \left (r -\tau \right )\right ) \mathcal {L}\left (\frac {4 \tau +2}{\tau +1}, t , s\right )}{\sqrt {-s}\, \BesselJ \left (1, \sqrt {-s}\right )}, s , t\right )d \tau \right )\] Do not understand Maple solution

Solve for $u\left ( r,t\right ) $\begin {align*} u_{t} & =k\left ( u_{rr}+\frac {1}{r}u_{r}\right ) \\ \left \vert u\left ( 0,t\right ) \right \vert & <\infty \\ u_{r}\left ( 1,t\right ) & =0\\ u\left ( r,0\right ) & =2Lr-r^{2} \end {align*}

The basic solution for this type of PDE was already given in problem 3.2.2.1 on page 873 as\[ u\left ( r,t\right ) =\frac {2}{L^{2}}\int _{0}^{L}f\left ( r\right ) rdr+\sum _{n=1}^{\infty }\left ( \frac {\int _{0}^{L}f\left ( r\right ) \operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{L}r\right ) rdr}{\int _{0}^{L}\operatorname {BesselJ}^{2}\left ( 0,\frac {\Lambda _{n}}{L}r\right ) rdr}\right ) e^{-k\left ( \frac {\Lambda _{n}}{L}\right ) ^{2}t}\operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{L}r\right ) \] Where $\Lambda _{n}$ are the zeros of the $\operatorname {BesselJ}\left ( 1,x\right ) $. In this problem $L=1$ and $k=1,f\left ( r\right ) =2Lr-r^{2}$. This is animation of the above solution using these speciﬁc values for for $0.2$ seconds. (Animation will only show in the HTML version)

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3.2.2.3 [260] no $\theta $ dependency

Solve the heat equation in polar coordinates for $u(r,t)$ \[ u_t = k (u_{rr} + \frac {1}{r} u_r) \] For $0<r<L$ and $t>0$. The boundary conditions are Newton Law of cooling, where $h>0$ \begin {align*} L u(L,t) &= -h u(L,t) \end {align*}

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[r, t], t] == k*(D[u[r, t], {r, 2}] + 1/r*D[u[r, t], r]); 
ic  = u[r, 0] == f[r]; 
bc  = L*Derivative[1,0][u][L, t] == -h*u[L,t]; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[r, t], {r, t}, Assumptions->{r<L,L>0,k>0,t>0,h>0}], 60*10]];

\[\left \{\left \{u(r,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {\sqrt {2} e^{-k t K[2,K[1]]^2} J_0(r K[2,K[1]]) \left (\int _0^L \frac {\sqrt {2} r J_0(r K[2,K[1]]) f(r) K[2,K[1]]}{J_0(L K[2,K[1]]) \sqrt {h^2+L^2 K[2,K[1]]^2}} \, dr\right ) K[2,K[1]]}{L J_0(L K[2,K[1]]) \sqrt {\frac {h^2}{L^2}+K[2,K[1]]^2}} & h J_0(L K[2,K[1]])=L J_1(L K[2,K[1]]) K[2,K[1]]\land K[1]\in \mathbb {Z}\land K[1]\geq 1\land K[2,K[1]]>0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]

Maple ✓

restart; 
pde := diff(u(r,t),t)= k*(diff(u(r,t),r$2)+ 1/r*diff(u(r,t),r)); 
ic  := u(r,0)=f(r); 
bc  := L*D[1](u)(L,t)=-h*u(L,t); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],u(r,t)) assuming k>0,r>0,L>0,r<L,t>0,h>0 ),output='realtime'));

\[u \left (r , t\right ) = h f \left (L \right ) \ln \left (L \right )-h f \left (L \right ) \ln \left (r \right )-2 \left (\sum _{n=0}^{\infty }\frac {\left (-\cosh \left (\frac {k t \lambda _{n}^{2}}{L^{2}}\right )+\sinh \left (\frac {k t \lambda _{n}^{2}}{L^{2}}\right )\right ) \BesselJ \left (0, \frac {r \lambda _{n}}{L}\right ) \left (\int _{0}^{L}\left (-h f \left (L \right ) \ln \left (L \right )+h f \left (L \right ) \ln \left (r \right )-f \left (L \right )+f \left (r \right )\right ) r \BesselJ \left (0, \frac {r \lambda _{n}}{L}\right )d r \right )}{\left (\BesselJ \left (0, \lambda _{n}\right )^{2}+\BesselJ \left (1, \lambda _{n}\right )^{2}\right ) L^{2}}\right )+f \left (L \right )\boldsymbol {\mathrm {where}}\left \{-h \BesselJ \left (0, \lambda _{n}\right )+\lambda _{n} \BesselJ \left (1, \lambda _{n}\right )=0\wedge -\infty \le \lambda _{n}\le \infty \right \}\]

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3.2.2.4 [261] no $\theta $ dependency

Solve the heat equation in polar coordinates for $u(r,t)$ \[ \frac { \partial u}{\partial t}= \frac { \partial ^2 u}{\partial r^2} + \frac {1}{r} \frac { \partial u}{\partial r} \] For $0<r<1$ and $t>0$. The boundary conditions are \begin {align*} u(1,t) &= 0 \end {align*}

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[r, t], t] == D[u[r, t], {r, 2}] + (1*D[u[r, t], r])/r; 
ic  = u[r, 0] == 1 - r; 
bc  = u[1, t] == 0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[r, t], {r, t}], 60*10]]; 
sol =  sol /. K[1] -> n;

\[\left \{\left \{u(r,t)\to \underset {n=1}{\overset {\infty }{\sum }}\frac {e^{-t \left (j_{0,n}\right ){}^2} \pi J_0\left (r j_{0,n}\right ) \left (J_1\left (j_{0,n}\right ) \pmb {H}_0\left (j_{0,n}\right )-J_0\left (j_{0,n}\right ) \pmb {H}_1\left (j_{0,n}\right )\right )}{J_1\left (j_{0,n}\right ){}^2 \left (j_{0,n}\right ){}^2}\right \}\right \}\]

Maple ✓

restart; 
pde := diff(u(r,t),t)= diff(u(r,t),r$2)+ 1/r*diff(u(r,t),r); 
ic  := u(r,0)=1-r; 
bc  := u(1,t) =0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],u(r,t),HINT =boundedseries(r=0))),output='realtime'));

\[u \left (r , t\right ) = \sum _{n=1}^{\infty }\left (-\frac {\pi \left (\BesselJ \left (0, \lambda _{n}\right ) \StruveH \left (1, \lambda _{n}\right )-\BesselJ \left (1, \lambda _{n}\right ) \StruveH \left (0, \lambda _{n}\right )\right ) \left (\cosh \left (t \lambda _{n}^{2}\right )-\sinh \left (t \lambda _{n}^{2}\right )\right ) \BesselJ \left (0, r \lambda _{n}\right )}{\left (\BesselJ \left (0, \lambda _{n}\right )^{2}+\BesselJ \left (1, \lambda _{n}\right )^{2}\right ) \lambda _{n}^{2}}\right )\boldsymbol {\mathrm {where}}\left \{\lambda _{n}=\mathit {BesselJZeros}\left (0, n\right )\wedge 0<\lambda _{n}\right \}\]

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3.2.2.5 [262] Haberman 8.3.5 (General solution)

Problem 8.3.5 from Richard Haberman applied partial diﬀerential equations book, 5th edition

Solve for $u(r,t)$ \[ u_t= k (u_rr+1/r u_r) + f(r,t) \] Inside the circle ($r<a$) with $u=0$ at $r=a$ and initially $u=0$.

One of the problems here, is how to tell CAS the implicit condition when solving this which is that $u(0,t)<\infty $.

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[r, t], t] == k*(D[u[r, t], {r, 2}] + D[u[r, t], r]/r) + f[r, t]; 
ic  = u[r, 0] == 0; 
bc  = u[a, t] == 0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[r, t], {r, t}, Assumptions -> r < a], 60*10]];

\[\left \{\left \{u(r,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {\sqrt {2} J_0\left (\frac {r j_{0,K[1]}}{a}\right ) \int _0^t e^{-\frac {k \left (j_{0,K[1]}\right ){}^2 (t-K[2])}{a^2}} \text {Integrate}\left [\frac {\sqrt {2} r J_0\left (\frac {r j_{0,K[1]}}{a}\right ) f(r,K[2])}{a J_1\left (j_{0,K[1]}\right )},\{r,0,a\},\text {Assumptions}\to a\geq 0\land r<a\right ] \, dK[2]}{a J_1\left (j_{0,K[1]}\right )}\right \}\right \}\]

Maple ✓

restart; 
pde := diff(u(r,t),t)= k*(diff(u(r,t),r$2)+ 1/r*diff(u(r,t),r)) + f(r,t); 
ic  := u(r,0)=0; 
bc  := u(a,t) =0; 
#do not use HINT=boundedseries below, Maple will  not solve it then 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],u(r,t)) assuming r<a),output='realtime'));

\[u \left (r , t\right ) = -\left (\int _{-a +r}^{0}\mathcal {L}^{-1}\left (\frac {\BesselJ \left (0, \sqrt {-\frac {s}{k}}\, \left (r -\tau \right )\right ) \mathcal {L}\left (-\frac {f \left (a +\tau , t\right )}{k}, t , s\right )}{\BesselJ \left (0, \sqrt {-\frac {s}{k}}\, a \right )}, s , t\right )d \tau \right )\]

Since this problem has homogeneous B.C. but has time dependent source (i.e. non-homogenous in the PDE itself), then we will use the method of eigenfunction expansion. In this method, we ﬁrst ﬁnd the eigenfunctions $\phi _{n}\left ( x\right ) $ of the associated homogenous PDE without the source being present. Then use these $\phi _{n}\left ( x\right ) $ to expand the source $f\left ( x,t\right ) $ as generalized Fourier series. We now switch to the associated homogenous PDE in order to ﬁnd the eigenfunctions. $u\equiv u\left ( r,t\right ) $. There is no $\theta $. Hence\begin {align} \frac {\partial u\left ( r,t\right ) }{\partial t} & =k\left ( \frac {\partial ^{2}u}{\partial r^{2}}+\frac {1}{r}\frac {\partial u}{\partial r}\right ) \tag {1}\\ u\left ( a,t\right ) & =0\nonumber \\ \left \vert u\left ( 0,t\right ) \right \vert & <\infty \nonumber \\ u\left ( r,0\right ) & =0\nonumber \end {align}

We need to solve the above in order to ﬁnd the eigenfunctions $\phi _{n}\left ( r\right ) $. Let $u=R\left ( r\right ) T\left ( t\right ) $. Substituting this back into (1) gives\[ T^{\prime }R=k\left ( R^{\prime \prime }T+\frac {1}{r}R^{\prime }T\right ) \] Dividing by $RT$\[ \frac {1}{k}\frac {T^{\prime }}{T}=\frac {R^{\prime \prime }}{R}+\frac {1}{r}\frac {R^{\prime }}{R}=-\lambda \] Where $\lambda $ is the separation constant. The above gives\[ T^{\prime }+k\lambda T=0 \] And\[ rR^{\prime \prime }+R^{\prime }+\lambda rR=0 \] This is a singular Sturm-Liouville ODE. Standard form is\[ \left ( rR^{\prime }\right ) ^{\prime }=-\lambda rR \] Hence\begin {align*} p & =r\\ q & =0\\ \sigma & =r \end {align*}

The ODE$\ rR^{\prime \prime }+R^{\prime }+\lambda rR=0$ is Bessel ODE whose solution is\[ R\left ( r\right ) =C_{1}\operatorname {BesselJ}\left ( 0,\sqrt {\lambda }r\right ) +C_{2}\operatorname {BesselY}\left ( 0,\sqrt {\lambda }r\right ) \] Since $\operatorname {BesselY}\left ( 0,\sqrt {\lambda }r\right ) $ blows up at $r=0$, then $C_{2}=0$ and the solution becomes \[ R\left ( r\right ) =C_{1}\operatorname {BesselJ}\left ( 0,\sqrt {\lambda }r\right ) \] At $r=a$ the above becomes $0=C_{1}\operatorname {BesselJ}\left ( 0,\sqrt {\lambda }a\right ) $. Non trivial solution requires that $\sqrt {\lambda }a$ are the zeros of $\operatorname {BesselJ}\left ( 0,x\right ) $. Let the zeros be called $\Lambda _{n},n=1,2,3,\cdots $. Therefore $\sqrt {\lambda _{n}}a=\Lambda _{n}$ or\[ \lambda _{n}=\left ( \frac {\Lambda _{n}}{a}\right ) ^{2}\qquad n=1,2,3,\cdots \] The corresponding eigenfunctions are $R_{n}\left ( r\right ) =\operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) $. Now that the eigenfunctions for the homogeneous PDE are found, eigenfunction expansion is used to ﬁnd the general solution. Let \begin {equation} u\left ( r,t\right ) =\sum _{n=1}^{\infty }a_{n}\left ( t\right ) \operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) \tag {2} \end {equation} Where $a_{n}\left ( t\right ) $ is function of time since it includes the time solution in it. Substituting the above back into the original nonhomogeneous PDE \begin {align} u_{t} & =k\nabla ^{2}u+f\left ( r,t\right ) \tag {3}\\ & =k\left ( \frac {\partial ^{2}u}{\partial r^{2}}+\frac {1}{r}\frac {\partial u}{\partial r}\right ) +f\left ( r,t\right ) \nonumber \end {align}

Where $\nabla ^{2}u=-\lambda ru$. Substituting (2) into (3), and using $f\left ( r,t\right ) =\sum _{n=1}^{\infty }b_{n}\left ( t\right ) \operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) $ gives

\[ \sum _{n=1}^{\infty }a_{n}^{\prime }\left ( t\right ) \operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) =ka_{n}\left ( t\right ) \left ( \sum _{n=1}^{\infty }\operatorname {BesselJ}^{\prime \prime }\left ( 0,\frac {\Lambda _{n}}{a}r\right ) +\frac {1}{r}\operatorname {BesselJ}^{^{\prime }}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) \right ) +\sum _{n=1}^{\infty }b_{n}\left ( t\right ) \operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) \]

But $\sum _{n=1}^{\infty }\operatorname {BesselJ}^{\prime \prime }\left ( 0,\frac {\Lambda _{n}}{a}r\right ) +\frac {1}{r}\operatorname {BesselJ}^{^{\prime }}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) =-\lambda _{n}\operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) $. The above becomes

\begin {align*} \sum _{n=1}^{\infty }a_{n}^{\prime }\left ( t\right ) \operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) & =-ka_{n}\left ( t\right ) \sum _{n=1}^{\infty }\left ( \frac {\Lambda _{n}}{a}\right ) ^{2}\operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) +\sum _{n=1}^{\infty }b_{n}\left ( t\right ) \operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) \\ \sum _{n=1}^{\infty }\left ( a_{n}^{\prime }\left ( t\right ) +k\left ( \frac {\Lambda _{n}}{a}\right ) ^{2}a_{n}\left ( t\right ) \right ) \operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) & =\sum _{n=1}^{\infty }b_{n}\left ( t\right ) \operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) \end {align*}

The above simpliﬁes to\[ a_{n}^{\prime }\left ( t\right ) +k\left ( \frac {\Lambda _{n}}{a}\right ) ^{2}a_{n}\left ( t\right ) =b_{n}\left ( t\right ) \] The solution is\[ a_{n}\left ( t\right ) =e^{-k\left ( \frac {\Lambda _{n}}{a}\right ) ^{2}t}\int _{0}^{t}b_{n}\left ( \tau \right ) e^{k\left ( \frac {\Lambda _{n}}{a}\right ) ^{2}\tau }d\tau +a_{n}\left ( 0\right ) e^{-k\left ( \frac {\Lambda _{n}}{a}\right ) ^{2}t}\] Hence the solution (2) becomes\[ u\left ( r,t\right ) =\sum _{n=1}^{\infty }\left ( e^{-k\left ( \frac {\Lambda _{n}}{a}\right ) ^{2}t}\left ( \int _{0}^{t}b_{n}\left ( \tau \right ) e^{k\left ( \frac {\Lambda _{n}}{a}\right ) ^{2}\tau }d\tau \right ) +a_{n}\left ( 0\right ) e^{-k\left ( \frac {\Lambda _{n}}{a}\right ) ^{2}t}\right ) \operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) \] To ﬁnd $a_{n}\left ( 0\right ) $, putting $t=0$ in the above gives\[ 0=\sum _{n=1}^{\infty }a_{n}\left ( 0\right ) \operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) \] Hence $a_{n}\left ( 0\right ) =0$. Therefore $a_{n}\left ( t\right ) $ becomes.\[ a_{n}\left ( t\right ) =e^{-k\left ( \frac {\Lambda _{n}}{a}\right ) ^{2}t}\int _{0}^{t}b_{n}\left ( \tau \right ) e^{k\left ( \frac {\Lambda _{n}}{a}\right ) ^{2}\tau }d\tau \] Hence the solution from (2) now becomes\[ u\left ( r,t\right ) =\sum _{n=1}^{\infty }\left ( e^{-k\left ( \frac {\Lambda _{n}}{a}\right ) ^{2}t}\int _{0}^{t}b_{n}\left ( \tau \right ) e^{k\left ( \frac {\Lambda _{n}}{a}\right ) ^{2}\tau }d\tau \right ) \operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) \] And ﬁnally, to ﬁnd $b_{n}\left ( t\right ) $, which is the generalized Fourier coeﬃcient of the expansion of the source in (3) above, orthogonality is used as follows\begin {align*} \int _{0}^{a}f\left ( r,t\right ) \operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) rdr & =b_{n}\left ( t\right ) \int _{0}^{a}\operatorname {BesselJ}^{2}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) rdr\\ b_{n}\left ( t\right ) & =\frac {\int _{0}^{a}f\left ( r,t\right ) \operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) rdr}{\int _{0}^{a}\operatorname {BesselJ}^{2}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) rdr} \end {align*}

Summary of solution \begin {align*} u\left ( r,t\right ) & =\sum _{n=1}^{\infty }a_{n}\left ( t\right ) \operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) \\ & =\sum _{n=1}^{\infty }\left ( \int _{0}^{t}b_{n}\left ( \tau \right ) e^{k\left ( \frac {\Lambda _{n}}{a}\right ) ^{2}\tau }d\tau \right ) e^{-k\left ( \frac {\Lambda _{n}}{a}\right ) ^{2}t}\operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) \end {align*}

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3.2.2.6 [263] Speciﬁc example of the above

Problem 8.3.5 from Richard Haberman applied partial diﬀerential equations book, 5th edition, but added speciﬁc values for parameters in order to do animations

Solve for $u(r,t)$ \[ u_t= k (u_rr+1/r u_r) + f(r,t) \] Inside the circle ($r<a$) with $u=0$ at $r=a$ and initially $u=0$. Let $k=\frac {1}{100},a=2$. Let $f(r,t)=sin(t)$

Mathematica ✓

ClearAll["Global`*"]; 
a=2; 
 k=1/100; 
 f=Sin[t]; 
pde =  D[u[r, t], t] == k*(D[u[r, t], {r, 2}] + D[u[r, t], r]/r) + f; 
ic  = u[r, 0] == 0; 
bc  = u[a, t] == 0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[r, t], {r, t}, Assumptions -> r < a], 60*10]];

\[\left \{\left \{u(r,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {800 J_0\left (\frac {1}{2} r j_{0,K[1]}\right ) \left (\sin (t) \left (j_{0,K[1]}\right ){}^2+400 \left (e^{-\frac {1}{400} t \left (j_{0,K[1]}\right ){}^2}-\cos (t)\right )\right )}{J_1\left (j_{0,K[1]}\right ) j_{0,K[1]} \left (\left (j_{0,K[1]}\right ){}^4+160000\right )}\right \}\right \}\]

Maple ✓

restart; 
a:=2; 
k:=1/100; 
f:=sin(t); 
pde := diff(u(r,t),t)= k*(diff(u(r,t),r$2)+ 1/r*diff(u(r,t),r)) + f; 
ic  := u(r,0)=0; 
bc  := u(a,t)=0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],u(r,t)) assuming r<a),output='realtime'));

\[u \left (r , t\right ) = -\cos \left (t \right )-\mathcal {L}^{-1}\left (\frac {\BesselJ \left (0, 10 \sqrt {-s}\, r \right )}{s \BesselJ \left (0, 20 \sqrt {-s}\right )}, s , t\right )+\mathcal {L}^{-1}\left (\frac {s \BesselJ \left (0, 10 \sqrt {-s}\, r \right )}{\left (s^{2}+1\right ) \BesselJ \left (0, 20 \sqrt {-s}\right )}, s , t\right )+1\]

The basic solution for this type of PDE was already given in problem 3.2.2.5 on page 892 as\begin {align*} u\left ( r,t\right ) & =\sum _{n=1}^{\infty }a_{n}\left ( t\right ) \operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) \\ & =\sum _{n=1}^{\infty }\left ( \int _{0}^{t}b_{n}\left ( \tau \right ) e^{k\left ( \frac {\Lambda _{n}}{a}\right ) ^{2}\tau }d\tau \right ) e^{-k\left ( \frac {\Lambda _{n}}{a}\right ) ^{2}t}\operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) \end {align*}

Where\[ b_{n}\left ( t\right ) =\frac {\int _{0}^{a}f\left ( r,t\right ) \operatorname {BesselJ}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) rdr}{\int _{0}^{a}\operatorname {BesselJ}^{2}\left ( 0,\frac {\Lambda _{n}}{a}r\right ) rdr}\] Where $\Lambda _{n}$ are the $n^{th}$ zeros of $\operatorname {BesselJ}\left ( 0,x\right ) $. In this problem \begin {align*} a & =2\\ k & =\frac {1}{100}\\ f\left ( r,t\right ) & =\sin \left ( t\right ) \end {align*}

This is animation of the above solution using these speciﬁc values for for $50$ seconds. (Animation will only show in the HTML version)

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3.2.2.7 [264] Speciﬁc example of the above

Problem 8.3.5 from Richard Haberman applied partial diﬀerential equations book, 5th edition, but added speciﬁc values for the parameters in order to do animations

Solve for $u(r,t)$ \[ u_t= k (u_rr+1/r u_r) + f(r,t) \] Inside the circle ($r<a$) with $u=0$ at $r=a$ and initially $u=0$. Let $k=\frac {1}{100},a=2$. Let $f(r,t)=r t e^{-t}$

Mathematica ✓

ClearAll["Global`*"]; 
a=2; 
 k=1/100; 
 f=r*t*Exp[-t]; 
pde =  D[u[r, t], t] == k*(D[u[r, t], {r, 2}] + D[u[r, t], r]/r) + f; 
ic  = u[r, 0] == 0; 
bc  = u[a, t] == 0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[r, t], {r, t}, Assumptions -> r < a], 60*10]];

\[\left \{\left \{u(r,t)\to \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {1600 e^{-\frac {1}{400} t \left (j_{0,K[1]}\right ){}^2} J_0\left (\frac {1}{2} r j_{0,K[1]}\right ) \left (e^{\frac {1}{400} t \left (j_{0,K[1]}\right ){}^2-t} \left (t \left (j_{0,K[1]}\right ){}^2-400 (t+1)\right )+400\right ) \, _1F_2\left (\frac {3}{2};1,\frac {5}{2};-\frac {1}{4} \left (j_{0,K[1]}\right ){}^2\right )}{3 J_1\left (j_{0,K[1]}\right ){}^2 \left (\left (j_{0,K[1]}\right ){}^2-400\right ){}^2}\right \}\right \}\]

Maple ✗

restart; 
a:=2; 
k:=1/100; 
f:=r*t*exp(-t); 
pde := diff(u(r,t),t)= k*(diff(u(r,t),r$2)+ 1/r*diff(u(r,t),r)) + f; 
ic  := u(r,0)=0; 
bc  := u(a,t)=0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],u(r,t)) assuming r<a),output='realtime'));

sol=()

This is animation of the above solution using these speciﬁc values for for $100$ seconds. (Animation will only show in the HTML version)

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3.2.2.8 [265] Articolo 6.9.1

Example 6.9.1 from Partial diﬀerential equations and boundary value problems with Maple/George A. Articolo, 2nd ed :

We seek the temperature distribution $u(r,\theta ,t)$ in a thin circular plate over the two-dimensional domain $D=\{(r,\theta ) | 0<r<1, 0< \theta < \frac {\pi }{2}\}$.

The lateral surfaces of the plate are insulated. The edges $r = 1$ and $\theta = 0$ are at a ﬁxed temperature of $0$, and the edge $\theta = \frac {\pi }{2}$ is insulated. The initial temperature distribution $u(r, \theta , 0) = f(r, \theta )$ is $u(r,\theta ,0)=(r-r^3)\sin (\theta )$.

Solve for $u(r,\theta ,t)$ the heat PDE \[ u_t= k \left ( u_{rr} + \frac {1}{r} u_r +\frac {1}{r^2} u_{\theta \theta } \right ) \] With boundary conditions \begin {align*} |u(0,\theta ,t)| &< \infty \\ u(1,\theta ,t) &= 0\\ u(r,0,t) &=0\\ \frac {\partial u}{\partial \theta }(1,\frac {\pi }{2},t) &= 0 \end {align*}

Mathematica ✓

ClearAll["Global`*"]; 
k = 1/50; 
pde =  D[u[r, theta, t], t] == k*(D[u[r, theta, t], {r, 2}] + D[u[r, theta, t], r]/r + (1*D[u[r, theta, t], {theta, 2}])/r^2); 
 bcOnR = u[1, theta, t] == 0; 
 bcOnTheta = {u[r, 0, t] == 0, Derivative[0, 1, 0][u][r, Pi/2, t] == 0}; 
ic  = u[r, theta, 0] == (r - (1*r^3)/3)*Sin[theta]; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, bcOnR, bcOnTheta, ic}, u[r, theta, t], {r, theta, t},Assumptions -> t > 0], 60*10]];

\[\left \{\left \{u(r,\theta ,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[3]=1}{\overset {\infty }{\sum }}\frac {4 e^{-\frac {1}{50} t \left (j_{1,K[3]}\right ){}^2} J_1\left (r j_{1,K[3]}\right ) \left (J_3\left (j_{1,K[3]}\right )+J_2\left (j_{1,K[3]}\right ) j_{1,K[3]}\right ) \sin (\theta )}{3 J_0\left (j_{1,K[3]}\right ){}^2 \left (j_{1,K[3]}\right ){}^2} & (K[1]|K[3])\in \mathbb {Z}\land K[1]\geq 1\land K[3]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]

Maple ✗

restart; 
k:=1/50; 
pde := diff(u(r, theta, t), t) = k*( diff(u(r, theta, t), r$2) + 1/r* diff(u(r, theta, t), r) +  1/r^2 * diff(u(r ,theta, t), theta$2) ); 
bc_on_r:= u(1,theta,t)=0; 
bc_on_theta:= u(r,0,t)=0, eval(diff(u(r,theta,t),theta),theta=Pi/2)=0; 
ic  := u(r,theta,0)=(r-1/3*r^3)*sin(theta); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc_on_r, bc_on_theta, ic], u(r, theta, t),HINT = boundedseries(r = [0])) assuming r<a,t>0),output='realtime'));

sol=()

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3.2.2.9 [266] Articolo 6.9.2

Example 6.9.2 from Partial diﬀerential equations and boundary value problems with Maple/George A. Articolo, 2nd ed :

We seek the temperature distribution in a thin circular plate over the two-dimensional domain $ D = \{(r,\theta ) | 0 < r < 1, 0 < \theta < \pi \}$. The lateral surfaces of the plate are insulated. The sides $\theta = 0$ and $\theta = \pi $ are at a ﬁxed temperature of $0$, and the edge $r = 1$ is insulated. The initial temperature distribution is $u(r, \theta , 0) = \left ( r - \frac {r^3}{3} \right )\sin \theta $.

The thermal diﬀusivity is $k = \frac {1}{25}$. Solve for $u(r,\theta ,t)$ the heat PDE \[ u_t= k \left ( u_{rr} + \frac {1}{r} u_r +\frac {1}{r^2} u_{\theta \theta } \right ) \] With boundary conditions \begin {align*} |u(0,\theta ,t) &< \infty \\ u(1,\theta ,t) &= 0\\ u(r,0,t) &=0\\ u(r,\pi ,t) & 0 \end {align*}

Mathematica ✓

ClearAll["Global`*"]; 
k = 1/25; 
pde =  D[u[r, theta, t], t] == k*(D[u[r, theta, t], {r, 2}] + D[u[r, theta, t], r]/r + (1*D[u[r, theta, t], {theta, 2}])/r^2); 
 bcOnR = Derivative[1, 0, 0][u][1, theta, t] == 0; 
 bcOnTheta = {u[r, 0, t] == 0, u[r, Pi, t] == 0}; 
ic  = u[r, theta, 0] == (r - (1*r^3)/3)*Sin[theta]; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, bcOnR, bcOnTheta, ic}, u[r, theta, t], {r, theta, t}], 60*10]];

\[\left \{\left \{u(r,\theta ,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[3]=1}{\overset {\infty }{\sum }}\frac {4 e^{-\frac {1}{25} t K[2,1,K[3]]^2} J_1(r K[2,1,K[3]]) (J_3(K[2,1,K[3]])+J_2(K[2,1,K[3]]) K[2,1,K[3]]) \sin (\theta )}{3 K[2,1,K[3]] \left (\left (J_0(K[2,1,K[3]]){}^2+J_1(K[2,1,K[3]]){}^2\right ) K[2,1,K[3]]-2 J_0(K[2,1,K[3]]) J_1(K[2,1,K[3]])\right )} & (K[1]|K[3])\in \mathbb {Z}\land J_{K[1]-1}(K[2,K[1],K[3]])=J_{K[1]+1}(K[2,K[1],K[3]])\land K[1]\geq 1\land K[3]\geq 1\land K[2,1,K[3]]>0\land K[2,K[1],K[3]]>0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]

Maple ✓

restart; 
k:=1/25; 
pde := diff(u(r, theta, t), t) = k*( diff(u(r, theta, t), r$2) + 1/r* diff(u(r, theta, t), r) +  1/r^2 * diff(u(r, theta, t), theta$2) ); 
bc_on_r:= eval(diff(u(r,theta,t),r),r=1)=0; 
bc_on_theta:= u(r,0,t)=0, u(r,Pi,t)=0; 
ic  := u(r,theta,0)=(r-1/3*r^3)*sin(theta); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc_on_r, bc_on_theta, ic], u(r, theta, t),HINT = boundedseries(r = [0]))),output='realtime'));

\[u \left (r , \theta , t\right ) = \sum _{n=0}^{\infty }-\frac {4 \left (\lambda _{n}^{3} \BesselJ \left (0, \lambda _{n}\right )-\lambda _{n}^{2} \BesselJ \left (1, \lambda _{n}\right )+4 \lambda _{n} \BesselJ \left (0, \lambda _{n}\right )-8 \BesselJ \left (1, \lambda _{n}\right )\right ) \BesselJ \left (1, r \lambda _{n}\right ) {\mathrm e}^{-\frac {t \lambda _{n}^{2}}{25}} \sin \left (\theta \right )}{3 \left (\lambda _{n} \BesselJ \left (0, \lambda _{n}\right )^{2}+\lambda _{n} \BesselJ \left (1, \lambda _{n}\right )^{2}-2 \BesselJ \left (0, \lambda _{n}\right ) \BesselJ \left (1, \lambda _{n}\right )\right ) \lambda _{n}^{3}}\boldsymbol {\mathrm {where}}\left \{\lambda _{n} \BesselJ \left (2, \lambda _{n}\right )-\BesselJ \left (1, \lambda _{n}\right )=0\wedge 0<\lambda _{n}\right \}\]

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3.2.2.10 [267] Haberman 8.2.5 with $\theta $ dependency (General case)

Problem 8.2.5 from from Richard Haberman applied partial diﬀerential equations book, 5th edition.

Solve the initial value problem for a two-dimensional heat equation inside a circle (of radius $a$) $u_t = k \nabla ^2 u$ with time-independent boundary conditions: \begin {align*} u(a,\theta ,t) &= g(\theta ) \end {align*}

And initial conditions $u(r,\theta ,0) =f(r,\theta )$. There is an implied periodic boundary conditions on $\theta $ \begin {align*} u(r,-\pi ,t) &= u(r,\pi ,t)\\ \frac {\partial u}{\partial \theta }(r,-\pi ,t) &= \frac {\partial u}{\partial \theta }(r,\pi ,t) \end {align*}

Mathematica ✓

ClearAll["Global`*"]; 
pde = D[u[r, theta, t], t] == k*(D[u[r, theta, t], {r, 2}] + D[u[r, theta, t], r]/r + D[u[r, theta, t], {theta, 2}]/r^2); 
bcOnR = u[a, theta, t] == g[theta]; 
bcOnTheta = {u[r, -Pi, t] == u[r, Pi, t], Derivative[0, 1, 0][u][r, -Pi, t] == Derivative[0, 1, 0][u][r, Pi, t]}; 
ic  = u[r, theta, 0] == f[r, theta]; 
sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bcOnR, bcOnTheta, ic}, u[r, theta, t], {r, theta, t}, Assumptions -> {a > 0, r<a, k > 0,t>0}], 60*10]];

\[\left \{\left \{u(r,\theta ,t)\to \left (\begin {array}{cc} \{ & \begin {array}{cc} \underset {K[3]=1}{\overset {\infty }{\sum }}\frac {e^{-\frac {k t \left (j_{0,K[3]}\right ){}^2}{a^2}} J_0\left (\frac {r j_{0,K[3]}}{a}\right ) \int _0^a\int _{-\pi }^{\pi }r J_0\left (\frac {r j_{0,K[3]}}{a}\right ) \left (2 \pi f(r,\theta )-\text {Integrate}[g(\theta ),\{\theta ,-\pi ,\pi \},\text {Assumptions}\to a>0\land k>0\land t>0\land r<a]-2 \pi \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {a^{-K[1]} r^{K[1]} (\cos (\theta K[1]) \text {Integrate}[\cos (\theta K[1]) g(\theta ),\{\theta ,-\pi ,\pi \},\text {Assumptions}\to a>0\land k>0\land t>0\land r<a]+\text {Integrate}[g(\theta ) \sin (\theta K[1]),\{\theta ,-\pi ,\pi \},\text {Assumptions}\to a>0\land k>0\land t>0\land r<a] \sin (\theta K[1]))}{\pi }\right )d\theta dr}{2 a^2 \pi ^2 J_1\left (j_{0,K[3]}\right ){}^2}+\underset {K[3]=1}{\overset {\infty }{\sum }}\left (\underset {K[1]=1}{\overset {\infty }{\sum }}e^{-\frac {k t \left (j_{K[1],K[3]}\right ){}^2}{a^2}} \left (\frac {J_{K[1]}\left (\frac {r j_{K[1],K[3]}}{a}\right ) \cos (\theta K[1]) \int _0^a\int _{-\pi }^{\pi }r J_{K[1]}\left (\frac {r j_{K[1],K[3]}}{a}\right ) \cos (\theta K[1]) \left (2 \pi f(r,\theta )-\text {Integrate}[g(\theta ),\{\theta ,-\pi ,\pi \},\text {Assumptions}\to a>0\land k>0\land t>0\land r<a]-2 \pi \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {a^{-K[1]} r^{K[1]} (\cos (\theta K[1]) \text {Integrate}[\cos (\theta K[1]) g(\theta ),\{\theta ,-\pi ,\pi \},\text {Assumptions}\to a>0\land k>0\land t>0\land r<a]+\text {Integrate}[g(\theta ) \sin (\theta K[1]),\{\theta ,-\pi ,\pi \},\text {Assumptions}\to a>0\land k>0\land t>0\land r<a] \sin (\theta K[1]))}{\pi }\right )d\theta dr}{a^2 \pi ^2 J_{K[1]-1}\left (j_{K[1],K[3]}\right ){}^2}+\frac {J_{K[1]}\left (\frac {r j_{K[1],K[3]}}{a}\right ) \left (\int _0^a\int _{-\pi }^{\pi }r J_{K[1]}\left (\frac {r j_{K[1],K[3]}}{a}\right ) \sin (\theta K[1]) \left (2 \pi f(r,\theta )-\text {Integrate}[g(\theta ),\{\theta ,-\pi ,\pi \},\text {Assumptions}\to a>0\land k>0\land t>0\land r<a]-2 \pi \underset {K[1]=1}{\overset {\infty }{\sum }}\frac {a^{-K[1]} r^{K[1]} (\cos (\theta K[1]) \text {Integrate}[\cos (\theta K[1]) g(\theta ),\{\theta ,-\pi ,\pi \},\text {Assumptions}\to a>0\land k>0\land t>0\land r<a]+\text {Integrate}[g(\theta ) \sin (\theta K[1]),\{\theta ,-\pi ,\pi \},\text {Assumptions}\to a>0\land k>0\land t>0\land r<a] \sin (\theta K[1]))}{\pi }\right )d\theta dr\right ) \sin (\theta K[1])}{a^2 \pi ^2 J_{K[1]-1}\left (j_{K[1],K[3]}\right ){}^2}\right )\right ) & (K[1]|K[3])\in \mathbb {Z}\land K[1]\geq 1\land K[3]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right )+\underset {K[1]=1}{\overset {\infty }{\sum }}\frac {a^{-K[1]} r^{K[1]} \left (\cos (\theta K[1]) \int _{-\pi }^{\pi } \cos (\theta K[1]) g(\theta ) \, d\theta +\left (\int _{-\pi }^{\pi } g(\theta ) \sin (\theta K[1]) \, d\theta \right ) \sin (\theta K[1])\right )}{\pi }+\frac {\int _{-\pi }^{\pi } g(\theta ) \, d\theta }{2 \pi }\right \}\right \}\]

Maple ✗

restart; 
pde := diff(u(r,theta,t),t)=k*(diff(u(r,theta,t),r$2) + 1/r*diff(u(r,theta,t),r)+1/r^2*diff(u(r,theta,t),theta$2)); 
bcOnR:= u(a,theta,t)=g(theta); 
bcOnTheta:= u(r,-Pi,t)=u(r,Pi,t),eval(diff(u(r,theta,t),theta),theta=-Pi)=eval(diff(u(r,theta,t),theta),theta=Pi); 
ic  := u(r,theta,0)=f(r,theta); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bcOnR,bcOnTheta, ic], u(r, theta, t), HINT = boundedseries(r = [0])) assuming a>0,r<a,k>0,t>0),output='realtime'));

sol=()

\begin {align*} \frac {\partial u\left ( r,\theta ,t\right ) }{\partial t} & =k\left ( \frac {\partial ^{2}u}{\partial r^{2}}+\frac {1}{r}\frac {\partial u}{\partial r}+\frac {1}{r^{2}}\frac {\partial ^{2}u}{\partial \theta ^{2}}\right ) \\ \left \vert u\left ( 0,\theta ,t\right ) \right \vert & <\infty \\ u\left ( a,\theta ,t\right ) & =g\left ( \theta \right ) \\ u\left ( r,-\pi ,t\right ) & =u\left ( r,\pi ,t\right ) \\ \frac {\partial u}{\partial \theta }\left ( r,-\pi ,t\right ) & =\frac {\partial u}{\partial \theta }\left ( r,\pi ,t\right ) \end {align*}

With initial conditions $u\left ( r,\theta ,0\right ) =f\left ( r,\theta \right ) $.

Since the boundary conditions are not homogenous, and since there are no time dependent sources, then in this case we look for $u_{E}\left ( r,\theta \right ) $ which is solution at steady state which needs to satisfy the nonhomogeneous B.C., where $u\left ( r,\theta ,t\right ) =\overset {\text {transient}}{\overbrace {v\left ( r,\theta ,t\right ) }}+\overset {\text {steady state}}{\overbrace {u_{E}\left ( r,\theta \right ) }}$ and $v\left ( r,\theta ,t\right ) $ solves the PDE but with homogenous B.C. Therefore, we need to ﬁnd equilibrium (steady state) solution for Laplace PDE on disk, that only needs to satisfy the nonhomogeneous B.C.\begin {align*} \nabla ^{2}u_{E} & =0\\ \frac {\partial ^{2}u_{E}}{\partial r^{2}}+\frac {1}{r}\frac {\partial u_{E}}{\partial r}+\frac {1}{r^{2}}\frac {\partial ^{2}u_{E}}{\partial \theta ^{2}} & =0 \end {align*}

With boundary condition\begin {align*} \left \vert u_{E}\left ( 0,\theta \right ) \right \vert & <\theta \\ u_{E}\left ( a,\theta \right ) & =g\left ( \theta \right ) \\ u_{E}\left ( r,-\pi \right ) & =u_{E}\left ( r,\pi \right ) \\ \frac {\partial u_{E}}{\partial \theta }\left ( r,-\pi \right ) & =\frac {\partial u_{E}}{\partial \theta }\left ( r,\pi \right ) \end {align*}

Let \[ u_{E}\left ( r,\theta \right ) =R\left ( r\right ) \Theta \left ( \theta \right ) \] Where $R\left ( r\right ) $ is the solution in radial dimension and $\Theta \left ( \theta \right ) $ is solution in angular dimension. Substituting $u_{E}\left ( r,\theta \right ) $ in the PDE gives\[ R^{\prime \prime }\Theta +\frac {1}{r}R^{\prime }\Theta +\frac {1}{r^{2}}\Theta ^{\prime \prime }R=0 \] Dividing by $R\left ( r\right ) \Phi \left ( \theta \right ) $\begin {align*} \frac {R^{\prime \prime }}{R}+\frac {1}{r}\frac {R^{\prime }}{R}+\frac {1}{r^{2}}\frac {\Theta ^{\prime \prime }}{\Theta } & =0\\ r^{2}\frac {R^{\prime \prime }}{R}+r\frac {R^{\prime }}{R} & =-\frac {\Theta ^{\prime \prime }}{\Theta } \end {align*}

Hence each side is equal to constant, say $\lambda $ and we obtain\begin {align*} r^{2}\frac {R^{\prime \prime }}{R}+r\frac {R^{\prime }}{R} & =\lambda \\ -\frac {\Theta ^{\prime \prime }}{\Theta } & =\lambda \end {align*}

Or\begin {align} r^{2}R^{\prime \prime }+rR^{\prime }-\lambda R & =0\tag {1}\\ \Theta ^{\prime \prime }+\lambda \Theta & =0 \tag {2} \end {align}

We start with $\Phi $ ODE. The boundary conditions on (3) are \begin {align*} \Theta \left ( -\pi \right ) & =\Theta \left ( \pi \right ) \\ \frac {\partial \Theta }{\partial \theta }\left ( -\pi \right ) & =\frac {\partial \Theta }{\partial \theta }\left ( \pi \right ) \end {align*}

case $\lambda =0$ The solution is $\Phi =c_{1}\theta +c_{2}$. Hence we obtain, from ﬁrst initial conditions\begin {align*} -\pi c_{1}+c_{2} & =\pi c_{1}+c_{2}\\ c_{1} & =0 \end {align*}

Second boundary conditions just says that $c_{2}=c_{2}$, so any constant will do. Hence $\lambda =0$ is an eigenvalue with constant being eigenfunction.

case $\lambda >0$ The solution is \[ \Theta \left ( \theta \right ) =c_{1}\cos \sqrt {\lambda }\theta +c_{2}\sin \sqrt {\lambda }\theta \] The ﬁrst boundary conditions gives\begin {align} c_{1}\cos \left ( -\sqrt {\lambda }\pi \right ) +c_{2}\sin \left ( -\sqrt {\lambda }\pi \right ) & =c_{1}\cos \left ( \sqrt {\lambda }\pi \right ) +c_{2}\sin \left ( \sqrt {\lambda }\pi \right ) \nonumber \\ c_{1}\cos \left ( \sqrt {\lambda }\pi \right ) -c_{2}\sin \left ( \sqrt {\lambda }\pi \right ) & =c_{1}\cos \left ( \sqrt {\lambda }\pi \right ) +c_{2}\sin \left ( \sqrt {\lambda }\pi \right ) \nonumber \\ 2c_{2}\sin \left ( \sqrt {\lambda }\pi \right ) & =0\tag {3} \end {align}

From second boundary conditions we obtain\[ \Theta ^{\prime }\left ( \theta \right ) =-\sqrt {\lambda }c_{1}\sin \sqrt {\lambda }\theta +c_{2}\sqrt {\lambda }\cos \sqrt {\lambda }\theta \] Therefore\begin {align} -\sqrt {\lambda }c_{1}\sin \left ( -\sqrt {\lambda }\pi \right ) +c_{2}\sqrt {\lambda }\cos \left ( -\sqrt {\lambda }\pi \right ) & =-\sqrt {\lambda }c_{1}\sin \left ( \sqrt {\lambda }\pi \right ) +c_{2}\sqrt {\lambda }\cos \left ( \sqrt {\lambda }\pi \right ) \nonumber \\ \sqrt {\lambda }c_{1}\sin \left ( \sqrt {\lambda }\pi \right ) +c_{2}\sqrt {\lambda }\cos \left ( \sqrt {\lambda }\pi \right ) & =-\sqrt {\lambda }c_{1}\sin \left ( \sqrt {\lambda }\pi \right ) +c_{2}\sqrt {\lambda }\cos \left ( \sqrt {\lambda }\pi \right ) \nonumber \\ \sqrt {\lambda }c_{1}\sin \left ( \sqrt {\lambda }\pi \right ) & =-\sqrt {\lambda }c_{1}\sin \left ( \sqrt {\lambda }\pi \right ) \nonumber \\ 2c_{1}\sin \left ( \sqrt {\lambda }\pi \right ) & =0\tag {4} \end {align}

Both (3) and (4) are satisﬁed if\begin {align*} \sqrt {\lambda }\pi & =n\pi \qquad n=1,2,3,\cdots \\ \lambda & =n^{2}\qquad n=1,2,3,\cdots \end {align*}

Therefore\begin {equation} \Theta _{n}\left ( \theta \right ) =\overset {\lambda =0}{\overbrace {A_{0}}}+\sum _{n=1}^{\infty }A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \tag {5} \end {equation} Now we go back to the $R$ ODE (1) given by $r^{2}R^{\prime \prime }+rR^{\prime }-\lambda _{n}R=0$ and solve it. This is Euler ODE whose solution is found by substituting $R\left ( r\right ) =r^{\alpha }$. The solution comes out to be\begin {equation} R_{n}\left ( r\right ) =c_{0}+\sum _{n=1}^{\infty }c_{n}r^{n}\tag {6} \end {equation} Combining (5,6) we now ﬁnd $u_{E}$ as\begin {align} u_{E_{n}}\left ( r,\theta \right ) & =R_{n}\left ( r\right ) \Theta _{n}\left ( \theta \right ) \nonumber \\ u_{E}\left ( r,\theta \right ) & =A_{0}+\sum _{n=1}^{\infty }r^{n}\left ( A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \right ) \tag {7} \end {align}

Where $c_{0}$ was combined with $A_{0}$. Now the above equilibrium solution needs to satisfy the non-homogenous B.C. $u_{E}\left ( a,\theta \right ) =g\left ( \theta \right ) $. Using orthogonality on (7) to ﬁnd $A_{n},B_{n}$ gives\[ g\left ( \theta \right ) =A_{0}+\sum _{n=1}^{\infty }a^{n}\left ( A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \right ) \] For $n=0$\begin {align*} \int _{0}^{2\pi }g\left ( \theta \right ) d\theta & =A_{0}\int _{0}^{2\pi }d\theta \\ A_{0} & =\frac {1}{2\pi }\int _{0}^{2\pi }g\left ( \theta \right ) d\theta \end {align*}

For $n>0$, applying orthogonality using cosine to ﬁnd $A_{n}$ gives\begin {align*} \int _{0}^{2\pi }g\left ( \theta \right ) \cos \left ( n\theta \right ) d\theta & =A_{n}\int _{0}^{2\pi }\cos ^{2}\left ( n\theta \right ) a^{n}d\theta \\ A_{n} & =\frac {1}{\pi }\int _{0}^{2\pi }g\left ( \theta \right ) \cos \left ( n\theta \right ) d\theta \end {align*}

Similarly, applying orthogonality using sin to ﬁnd $B_{n}$ gives\[ B_{n}=\frac {1}{\pi }\int _{0}^{2\pi }g\left ( \theta \right ) \sin \left ( n\theta \right ) d\theta \] Therefore, we have found $u_{E}\left ( r,\theta \right ) $ completely now. It is given by (7)\begin {align} u_{E}\left ( r,\theta \right ) & =A_{0}+\sum _{n=1}^{\infty }r^{n}\left ( A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \right ) \tag {7A}\\ A_{0} & =\frac {1}{2\pi }\int _{0}^{2\pi }g\left ( \theta \right ) d\theta \nonumber \\ A_{n} & =\frac {1}{\pi }\int _{0}^{2\pi }g\left ( \theta \right ) \cos \left ( n\theta \right ) d\theta \nonumber \\ B_{n} & =\frac {1}{\pi }\int _{0}^{2\pi }g\left ( \theta \right ) \sin \left ( n\theta \right ) d\theta \nonumber \end {align}

Now, since $u\left ( r,\theta ,t\right ) =v\left ( r,\theta ,t\right ) +u_{E}\left ( r,\theta \right ) $, then we need to solve now for $v\left ( r,\theta ,t\right ) $ with homogeneous boundary conditions \begin {align} v_{t}\left ( r,\theta ,t\right ) & =k\left ( \frac {\partial ^{2}v}{\partial r^{2}}+\frac {1}{r}\frac {\partial v}{\partial r}+\frac {1}{r^{2}}\frac {\partial ^{2}v}{\partial \theta ^{2}}\right ) \tag {8}\\ \left \vert v\left ( 0,\theta ,t\right ) \right \vert & <\theta \nonumber \\ v\left ( a,\theta ,t\right ) & =0\nonumber \\ v\left ( r,-\pi ,t\right ) & =v\left ( r,\pi ,t\right ) \nonumber \\ \frac {\partial v}{\partial \theta }\left ( r,-\pi ,t\right ) & =\frac {\partial v}{\partial \theta }\left ( r,\pi ,t\right ) \nonumber \end {align}

Let $v\left ( r,\theta ,t\right ) =R\left ( r\right ) \Theta \left ( \theta \right ) T\left ( t\right ) $. Substituting into (8) gives\[ T^{\prime }R\Theta =k\left ( R^{\prime \prime }T\Theta +\frac {1}{r}R^{\prime }T\Theta +\frac {1}{r^{2}}\Theta ^{\prime \prime }RT\right ) \] Dividing by $R\left ( r\right ) \Theta \left ( \theta \right ) T\left ( t\right ) \neq 0$ gives\[ \frac {1}{k}\frac {T^{\prime }}{T}=\frac {R^{\prime \prime }}{R}+\frac {1}{r}\frac {R^{\prime }}{R}+\frac {1}{r^{2}}\frac {\Theta ^{\prime \prime }}{\Theta }\] Let ﬁrst separation constant be $-\lambda $, hence the above becomes\begin {align*} \frac {1}{k}\frac {T^{\prime }}{T} & =-\lambda \\ \frac {R^{\prime \prime }}{R}+\frac {1}{r}\frac {R^{\prime }}{R}+\frac {1}{r^{2}}\frac {\Theta ^{\prime \prime }}{\Theta } & =-\lambda \end {align*}

Or\begin {align*} T^{\prime }+\lambda kT & =0\\ r^{2}\frac {R^{\prime \prime }}{R}+r\frac {R^{\prime }}{R}+r^{2}\lambda & =-\frac {\Theta ^{\prime \prime }}{\Theta } \end {align*}

We now separate the second equation above using $\mu $ giving\begin {align*} r^{2}\frac {R^{\prime \prime }}{R}+r\frac {R^{\prime }}{R}+r^{2}\lambda & =\mu \\ -\frac {\Theta ^{\prime \prime }}{\Theta } & =\mu \end {align*}

Or\begin {align} R^{\prime \prime }+\frac {1}{r}R^{\prime }+R\left ( \lambda -\frac {\mu }{r^{2}}\right ) & =0\tag {9}\\ \Theta ^{\prime \prime }+\mu \Theta & =0\tag {10} \end {align}

Equation (9) is Sturm-Liouville ODE with boundary conditions $R\left ( a\right ) =0$ and bounded at $r=0$ and (10) has periodic boundary conditions as was solved above. The solution to (10) is given in (5) above, no change for this part.\begin {equation} \Theta _{n}\left ( \theta \right ) =\overset {\lambda =0}{\overbrace {\alpha _{0}}}+\sum _{n=1}^{\infty }\alpha _{n}\cos \left ( n\theta \right ) +\beta _{n}\sin \left ( n\theta \right ) \tag {11} \end {equation} Therefore (9) becomes $R^{\prime \prime }+\frac {1}{r}R^{\prime }+R\left ( \lambda -\frac {n^{2}}{r^{2}}\right ) =0$ with $n=0,1,2,\cdots $. We found the solution to this Sturm-Liouville before, it is given by\begin {equation} R_{nm}\left ( r\right ) =J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \qquad n=0,1,2,\cdots ,m=1,2,3,\cdots \tag {12} \end {equation} Where $\sqrt {\lambda _{nm}}=\frac {a}{z_{nm}}$ where $a$ is the radius of the disk and $z_{nm}$ is the $m^{th}$ zero of the Bessel function of order $n$. This is found numerically. We now just need to ﬁnd the time solution from $T^{\prime }+\lambda _{nm}kT=0$. For This has solution \begin {equation} T_{nm}\left ( t\right ) =e^{-k\lambda _{nm}t}\tag {13} \end {equation} Now we combine (11,12,13) to ﬁnd solution for $v\left ( r,\theta ,t\right ) $, and combining constants gives\begin {align} v_{nm}\left ( r,\theta ,t\right ) & =\Theta _{n}\left ( \theta \right ) R_{nm}\left ( r\right ) T_{nm}\left ( t\right ) \nonumber \\ v\left ( r,\theta ,t\right ) & =\alpha _{0,1}J_{0}\left ( \sqrt {\lambda _{0,1}}r\right ) +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) e^{-k\lambda _{nm}t}\left ( \alpha _{nm}\cos \left ( n\theta \right ) +\beta _{nm}\sin \left ( n\theta \right ) \right ) \nonumber \\ & =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \left ( \alpha _{nm}\cos \left ( n\theta \right ) +\beta _{nm}\sin \left ( n\theta \right ) \right ) \tag {14} \end {align}

We now need to ﬁnd $\alpha _{0},\alpha _{nn},\beta _{nm}$, which are found from initial conditions on $v\left ( r,\theta ,0\right ) $ which is given by\begin {align*} v\left ( r,\theta ,0\right ) & =u\left ( r,\theta ,0\right ) -u_{E}\left ( r,\theta \right ) \\ & =f\left ( r,\theta \right ) -u_{E}\left ( r,\theta \right ) \end {align*}

Hence from (14), at $t=0$\begin {equation} f\left ( r,\theta \right ) -u_{E}\left ( r,\theta \right ) =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \left ( \alpha _{nm}\cos \left ( n\theta \right ) +\beta _{nm}\sin \left ( n\theta \right ) \right ) \tag {15} \end {equation}

For each $n$, inside the $m$ sum, $\cos \left ( n\theta \right ) $ and $\sin \left ( n\theta \right ) $ will be constant. So we need to apply orthogonality twice in order to remove both sums. Multiplying (15) by $\cos \left ( n^{\prime }\theta \right ) $ and integrating gives\begin {align*} \int _{-\pi }^{\pi }\left ( f\left ( r,\theta \right ) -u_{E}\left ( r,\theta \right ) \right ) \cos \left ( n^{\prime }\theta \right ) d\theta & =\int _{-\pi }^{\pi }\sum _{n=0}^{\infty }\left ( \sum _{m=1}^{\infty }\alpha _{nn}J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \right ) \cos \left ( n\theta \right ) \cos \left ( n^{\prime }\theta \right ) d\theta \\ & +\int _{-\pi }^{\pi }\sum _{n=1}^{\infty }\left ( \sum _{m=1}^{\infty }\beta _{nm}J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) \right ) \sin \left ( n\theta \right ) \cos \left ( n^{\prime }\theta \right ) \end {align*}

The second sum in the RHS above goes to zero due to $\int _{-\pi }^{\pi }\sin \left ( n\theta \right ) \cos \left ( n^{\prime }\theta \right ) d\theta $ and we end up with\[ \int _{-\pi }^{\pi }\left ( f\left ( r,\theta \right ) -u_{E}\left ( r,\theta \right ) \right ) \cos \left ( n\theta \right ) d\theta =\alpha _{nn}\int _{-\pi }^{\pi }\cos ^{2}\left ( n\theta \right ) \sum _{m=1}^{\infty }J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) d\theta \] We now apply orthogonality again, but on Bessel functions and remembering to add the weight $r$, the above becomes \begin {align*} \int _{0}^{a}\int _{-\pi }^{\pi }\left ( f\left ( r,\theta \right ) -u_{E}\left ( r,\theta \right ) \right ) \cos \left ( n\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm^{\prime }}}r\right ) rd\theta dr & =\alpha _{nn}\int _{0}^{a}\int _{-\pi }^{\pi }\cos ^{2}\left ( n\theta \right ) \sum _{m=1}^{\infty }J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) J_{n}\left ( \sqrt {\lambda _{nm^{\prime }}}r\right ) rd\theta dr\\ & =\alpha _{nn}\int _{0}^{a}\int _{-\pi }^{\pi }\cos ^{2}\left ( n\theta \right ) J_{n}^{2}\left ( \sqrt {\lambda _{nm^{\prime }}}r\right ) rd\theta dr \end {align*}

Therefore\[ \alpha _{nn}=\frac {\int _{0}^{a}\int _{-\pi }^{\pi }\left ( f\left ( r,\theta \right ) -u_{E}\left ( r,\theta \right ) \right ) \cos \left ( n\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) rd\theta dr}{\int _{0}^{a}\int _{-\pi }^{\pi }\cos ^{2}\left ( n\theta \right ) J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rd\theta dr}\qquad n=0,1,2,\cdots ,m=1,2,3,\cdots \] We will repeat the same thing to ﬁnd $\beta _{nm}$. The only diﬀerence now is to use $\sin n\theta $. repeating these steps gives\[ \beta _{nm}=\frac {\int _{0}^{a}\int _{-\pi }^{\pi }\left ( f\left ( r,\theta \right ) -u_{E}\left ( r,\theta \right ) \right ) \sin \left ( n\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) rd\theta dr}{\int _{0}^{a}\int _{-\pi }^{\pi }\sin ^{2}\left ( n\theta \right ) J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rd\theta dr}\qquad n=0,1,2,\cdots ,m=1,2,3,\cdots \] This complete the solution.

Summary of solution\begin {align*} u\left ( r,\theta ,t\right ) & =v\left ( r,\theta ,t\right ) +u_{E}\left ( r,\theta \right ) \\ & =u_{E}\left ( r,\theta \right ) +\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) e^{-k\lambda _{nm}t}\left ( \alpha _{nn}\cos \left ( n\theta \right ) +\beta _{nm}\sin \left ( n\theta \right ) \right ) \end {align*}

Where\begin {align*} u_{E}\left ( r,\theta \right ) & =A_{0}+\sum _{n=1}^{\infty }r^{n}\left ( A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \right ) \\ A_{0} & =\frac {1}{2\pi }\int _{0}^{2\pi }g\left ( \theta \right ) d\theta \\ A_{n} & =\frac {1}{\pi }\int _{0}^{2\pi }g\left ( \theta \right ) \cos \left ( n\theta \right ) d\theta \\ B_{n} & =\frac {1}{\pi }\int _{0}^{2\pi }g\left ( \theta \right ) \sin \left ( n\theta \right ) d\theta \end {align*}

And\[ \alpha _{nn}=\frac {\int _{0}^{a}\int _{-\pi }^{\pi }\left ( f\left ( r,\theta \right ) -u_{E}\left ( r,\theta \right ) \right ) \cos \left ( n\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) rd\theta dr}{\int _{0}^{a}\int _{-\pi }^{\pi }\cos ^{2}\left ( n\theta \right ) J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rd\theta dr}\qquad n=0,1,2,\cdots ,m=1,2,3,\cdots \] And\[ \beta _{nm}=\frac {\int _{0}^{a}\int _{-\pi }^{\pi }\left ( f\left ( r,\theta \right ) -u_{E}\left ( r,\theta \right ) \right ) \sin \left ( n\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) rd\theta dr}{\int _{0}^{a}\int _{-\pi }^{\pi }\sin ^{2}\left ( n\theta \right ) J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rd\theta dr}\qquad n=0,1,2,\cdots ,m=1,2,3,\cdots \] Where $\sqrt {\lambda _{nm}}=\frac {a}{z_{nm}}$ where $a$ is the radius of the disk and $z_{nm}$ is the $m^{th}$ zero of the Bessel function of order $n$.

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3.2.2.11 [268] With $\theta $ dependency (Speciﬁc example)

Solve the initial value problem for a two-dimensional heat equation inside a circle (of radius $a=1$) $u_t = k \nabla ^2 u$ with $k=1$ with time-independent boundary conditions: \begin {align*} u(1,\theta ,t) &= 0 \end {align*}

And initial conditions $u(r,\theta ,0) =1-r^2$. There is an implied periodic boundary conditions on $\theta $ \begin {align*} u(r,-\pi ,t) &= u(r,\pi ,t)\\ \frac {\partial u}{\partial \theta }(r,-\pi ,t) &= \frac {\partial u}{\partial \theta }(r,\pi ,t) \end {align*}

Mathematica ✓

ClearAll["Global`*"]; 
k=1; 
a=1; 
pde = D[u[r, theta, t], t] == k*(D[u[r, theta, t], {r, 2}] + D[u[r, theta, t], r]/r + D[u[r, theta, t], {theta, 2}]/r^2); 
bcOnR = u[a, theta, t] == 0; 
bcOnTheta = {u[r, -Pi, t] == u[r, Pi, t], Derivative[0, 1, 0][u][r, -Pi, t] == Derivative[0, 1, 0][u][r, Pi, t]}; 
ic  = u[r, theta, 0] == 1-r^2; 
sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bcOnR, bcOnTheta, ic}, u[r, theta, t], {r, theta, t}], 60*10]];

\[\left \{\left \{u(r,\theta ,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[3]=1}{\overset {\infty }{\sum }}\frac {4 e^{-t \left (j_{0,K[3]}\right ){}^2} J_0\left (r j_{0,K[3]}\right ) J_2\left (j_{0,K[3]}\right )}{J_1\left (j_{0,K[3]}\right ){}^2 \left (j_{0,K[3]}\right ){}^2} & (K[1]|K[3])\in \mathbb {Z}\land K[1]\geq 1\land K[3]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]

Maple ✗

restart; 
a:=1; 
k:=1; 
pde := diff(u(r,theta,t),t)=k*(diff(u(r,theta,t),r$2) + 1/r*diff(u(r,theta,t),r)+1/r^2*diff(u(r,theta,t),theta$2)); 
bcOnR:= u(a,theta,t)=0; 
bcOnTheta:= u(r,-Pi,t)=u(r,Pi,t),eval(diff(u(r,theta,t),theta),theta=-Pi)=eval(diff(u(r,theta,t),theta),theta=Pi); 
ic  := u(r,theta,0)=1-r^2; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bcOnR,bcOnTheta, ic], u(r, theta, t), HINT = boundedseries(r = [0]))),output='realtime'));

sol=()

\begin {align*} \frac {\partial u\left ( r,\theta ,t\right ) }{\partial t} & =k\left ( \frac {\partial ^{2}u}{\partial r^{2}}+\frac {1}{r}\frac {\partial u}{\partial r}+\frac {1}{r^{2}}\frac {\partial ^{2}u}{\partial \theta ^{2}}\right ) \\ \left \vert u\left ( 0,\theta ,t\right ) \right \vert & <\infty \\ u\left ( a,\theta ,t\right ) & =0\\ u\left ( r,-\pi ,t\right ) & =u\left ( r,\pi ,t\right ) \\ \frac {\partial u}{\partial \theta }\left ( r,-\pi ,t\right ) & =\frac {\partial u}{\partial \theta }\left ( r,\pi ,t\right ) \\ u\left ( r,\theta ,0\right ) & =1-r^{2} \end {align*}

This problem was solved in problem 3.2.2.10 on page 920 (since B.C. is zero, we just need to use $v\left ( r,\theta ,t\right ) $ in the solution given in the above problem). Hence \[ u\left ( r,\theta ,t\right ) =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) e^{-k\lambda _{nm}t}\left ( A_{nm}\cos \left ( n\theta \right ) +B_{nm}\sin \left ( n\theta \right ) \right ) \] Where\[ A_{nm}=\frac {\int _{0}^{a}\int _{-\pi }^{\pi }f\left ( r,\theta \right ) \cos \left ( n\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) rd\theta dr}{\int _{0}^{a}\int _{-\pi }^{\pi }\cos ^{2}\left ( n\theta \right ) J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rd\theta dr}\qquad n=0,1,2,\cdots ,m=1,2,3,\cdots \] And\[ B_{nm}=\frac {\int _{0}^{a}\int _{-\pi }^{\pi }f\left ( r,\theta \right ) \sin \left ( n\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) rd\theta dr}{\int _{0}^{a}\int _{-\pi }^{\pi }\sin ^{2}\left ( n\theta \right ) J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rd\theta dr}\qquad n=0,1,2,\cdots ,m=1,2,3,\cdots \] Where $\sqrt {\lambda _{nm}}=\frac {a}{z_{nm}}$ and $z_{nm}$ is the $m^{th}$ zero of the Bessel function of order $n$. where $a$ is the radius of the disk and $z_{nm}$ is the $m^{th}$ zero of the Bessel function of order $n$. Since in this problem $k=1,a=1,f\left ( r,\theta \right ) =1-r^{2}$ then

\[ u\left ( r,\theta ,t\right ) =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) e^{-\frac {t}{z_{nm}^{2}}}\left ( A_{nm}\cos \left ( n\theta \right ) +B_{nm}\sin \left ( n\theta \right ) \right ) \]\begin {align*} A_{nm} & =\frac {\int _{0}^{1}\int _{-\pi }^{\pi }\left ( 1-r^{2}\right ) \cos \left ( n\theta \right ) J_{n}\left ( \frac {r}{z_{nm}}\right ) rd\theta dr}{\int _{0}^{1}\int _{-\pi }^{\pi }\cos ^{2}\left ( n\theta \right ) J_{n}^{2}\left ( \frac {r}{z_{nm}}\right ) rd\theta dr}\qquad n=0,1,2,\cdots ,m=1,2,3,\cdots \\ B_{nm} & =\frac {\int _{0}^{1}\int _{-\pi }^{\pi }\left ( 1-r^{2}\right ) \sin \left ( n\theta \right ) J_{n}\left ( \frac {r}{z_{nm}}\right ) rd\theta dr}{\int _{0}^{1}\int _{-\pi }^{\pi }\sin ^{2}\left ( n\theta \right ) J_{n}^{2}\left ( \frac {r}{z_{nm}}\right ) rd\theta dr}\qquad n=0,1,2,\cdots ,m=1,2,3,\cdots \end {align*}

Where $\sqrt {\lambda _{nm}}=\frac {a}{z_{nm}}$ where $a$ is the radius of the disk and $z_{nm}$ is the $m^{th}$ zero of the Bessel function of order $n$.

This is animation of the solution for 0.4 seconds. (Animation will only show in the HTML version)

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3.2.2.12 [269] With $\theta $ dependency (speciﬁc example)

And initial conditions $u(r,\theta ,0) =(r-r^3)\sin \theta $. There is an implied periodic boundary conditions on $\theta $ \begin {align*} u(r,-\pi ,t) &= u(r,\pi ,t)\\ \frac {\partial u}{\partial \theta }(r,-\pi ,t) &= \frac {\partial u}{\partial \theta }(r,\pi ,t) \end {align*}

Mathematica ✓

ClearAll["Global`*"]; 
k=1; 
a=1; 
pde = D[u[r, theta, t], t] == k*(D[u[r, theta, t], {r, 2}] + D[u[r, theta, t], r]/r + D[u[r, theta, t], {theta, 2}]/r^2); 
bcOnR = u[a, theta, t] == 0; 
bcOnTheta = {u[r, -Pi, t] == u[r, Pi, t], Derivative[0, 1, 0][u][r, -Pi, t] == Derivative[0, 1, 0][u][r, Pi, t]}; 
ic  = u[r, theta, 0] == (r-r^3)*Sin[theta]; 
sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bcOnR, bcOnTheta, ic}, u[r, theta, t], {r, theta, t}], 60*10]];

\[\left \{\left \{u(r,\theta ,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[3]=1}{\overset {\infty }{\sum }}\frac {4 e^{-t \left (j_{1,K[3]}\right ){}^2} J_1\left (r j_{1,K[3]}\right ) J_3\left (j_{1,K[3]}\right ) \sin (\theta )}{J_0\left (j_{1,K[3]}\right ){}^2 \left (j_{1,K[3]}\right ){}^2} & (K[1]|K[3])\in \mathbb {Z}\land K[1]\geq 1\land K[3]\geq 1 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]

Maple ✗

restart; 
a:=1; 
k:=1; 
pde := diff(u(r,theta,t),t)=k*(diff(u(r,theta,t),r$2) + 1/r*diff(u(r,theta,t),r)+1/r^2*diff(u(r,theta,t),theta$2)); 
bcOnR:= u(a,theta,t)=0; 
bcOnTheta:= u(r,-Pi,t)=u(r,Pi,t),eval(diff(u(r,theta,t),theta),theta=-Pi)=eval(diff(u(r,theta,t),theta),theta=Pi); 
ic  := u(r,theta,0)=(r-r^3)*sin(theta); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bcOnR,bcOnTheta, ic], u(r, theta, t), HINT = boundedseries(r = [0]))),output='realtime'));

sol=()

Solve\begin {align*} \frac {\partial u\left ( r,\theta ,t\right ) }{\partial t} & =k\left ( \frac {\partial ^{2}u}{\partial r^{2}}+\frac {1}{r}\frac {\partial u}{\partial r}+\frac {1}{r^{2}}\frac {\partial ^{2}u}{\partial \theta ^{2}}\right ) \\ \left \vert u\left ( 0,\theta ,t\right ) \right \vert & <\infty \\ u\left ( a,\theta ,t\right ) & =0\\ u\left ( r,-\pi ,t\right ) & =u\left ( r,\pi ,t\right ) \\ \frac {\partial u}{\partial \theta }\left ( r,-\pi ,t\right ) & =\frac {\partial u}{\partial \theta }\left ( r,\pi ,t\right ) \\ u\left ( r,\theta ,0\right ) & =\left ( r-r^{3}\right ) \sin \theta \end {align*}

This problem was solved in problem 3.2.2.10 on page 920 (since B.C. is zero, we just need to use $v\left ( r,\theta ,t\right ) $ in the solution given in the above problem). Hence \[ u\left ( r,\theta ,t\right ) =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) e^{-k\lambda _{nm}t}\left ( A_{nm}\cos \left ( n\theta \right ) +B_{nm}\sin \left ( n\theta \right ) \right ) \] Where\[ A_{nm}=\frac {\int _{0}^{a}\int _{-\pi }^{\pi }f\left ( r,\theta \right ) \cos \left ( n\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) rd\theta dr}{\int _{0}^{a}\int _{-\pi }^{\pi }\cos ^{2}\left ( n\theta \right ) J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rd\theta dr}\qquad n=0,1,2,\cdots ,m=1,2,3,\cdots \] And\[ B_{nm}=\frac {\int _{0}^{a}\int _{-\pi }^{\pi }f\left ( r,\theta \right ) \sin \left ( n\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) rd\theta dr}{\int _{0}^{a}\int _{-\pi }^{\pi }\sin ^{2}\left ( n\theta \right ) J_{n}^{2}\left ( \sqrt {\lambda _{nm}}r\right ) rd\theta dr}\qquad n=0,1,2,\cdots ,m=1,2,3,\cdots \] Where $\sqrt {\lambda _{nm}}=\frac {a}{z_{nm}}$ and $z_{nm}$ is the $m^{th}$ zero of the Bessel function of order $n$ and $a$ is the radius of the disk and $z_{nm}$ is the $m^{th}$ zero of the Bessel function of order $n$. Since in this problem $k=1,a=1,f\left ( r,\theta \right ) =\left ( r-r^{3}\right ) \sin \theta $ then the above solution becomes

\[ u\left ( r,\theta ,t\right ) =\sum _{n=0}^{\infty }\sum _{m=1}^{\infty }A_{n}\cos \left ( n\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) e^{-\frac {t}{z_{nm}^{2}}}+\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }B_{n}\sin \left ( n\theta \right ) J_{n}\left ( \sqrt {\lambda _{nm}}r\right ) e^{-\frac {t}{z_{nm}^{2}}}\]\begin {align*} A_{n} & =\frac {\int _{0}^{1}\int _{-\pi }^{\pi }\left ( r-r^{3}\right ) \sin \theta \cos \left ( n\theta \right ) J_{n}\left ( \frac {r}{z_{nm}}\right ) rd\theta dr}{\int _{0}^{1}\int _{-\pi }^{\pi }\cos ^{2}\left ( n\theta \right ) J_{n}^{2}\left ( \frac {r}{z_{nm}}\right ) rd\theta dr}\\ B_{n} & =\frac {\int _{0}^{1}\int _{-\pi }^{\pi }\left ( r-r^{3}\right ) \sin \theta \sin \left ( n\theta \right ) J_{n}\left ( \frac {r}{z_{nm}}\right ) rd\theta dr}{\int _{0}^{1}\int _{-\pi }^{\pi }\sin ^{2}\left ( n\theta \right ) J_{n}^{2}\left ( \frac {r}{z_{nm}}\right ) rd\theta dr} \end {align*}

This is animation of the solution for 0.11 seconds. (Animation will only show in the HTML version)

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3.2.2.13 [270] Insulated with $\theta $ dependency (General solution)

Solve for $u(r,\theta ,t)$, the initial value problem for a two-dimensional heat equation inside a circle of radius $a$ \[ u_t = k \nabla ^2 u \] with time-independent boundary conditions: \begin {align*} u_r(a,\theta ,t) &= 0 \end {align*}

Mathematica ✓

ClearAll["Global`*"]; 
pde = D[u[r, theta, t], t] == k*(D[u[r, theta, t], {r, 2}] + D[u[r, theta, t], r]/r + D[u[r, theta, t], {theta, 2}]/r^2); 
bcOnR = Derivative[1,0,0][u][L, theta, t] == 0; 
bcOnTheta = {u[r, -Pi, t] == u[r, Pi, t], Derivative[0, 1, 0][u][r, -Pi, t] == Derivative[0, 1, 0][u][r, Pi, t]}; 
ic  = u[r, theta, 0] == f[r,theta]; 
sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bcOnR, bcOnTheta, ic}, u[r, theta, t], {r, theta, t},Assumptions->{L>0,r>0,r<L,t>0,k>0}], 60*10]];

\[\left \{\left \{u(r,\theta ,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {\int _0^L\int _{-\pi }^{\pi }r f(r,\theta )d\theta dr}{L^2 \pi }+\underset {K[3]=1}{\overset {\infty }{\sum }}\frac {e^{-k t K[2,0,K[3]]^2} J_0(r K[2,0,K[3]]) \left (\int _0^L\int _{-\pi }^{\pi }r J_0(r K[2,0,K[3]]) f(r,\theta )d\theta dr\right ) \sqrt {K[2,0,K[3]]}}{L^2 \pi \sqrt {J_0(L K[2,0,K[3]]){}^2+J_1(L K[2,0,K[3]]){}^2} \sqrt {\left (J_0(L K[2,0,K[3]]){}^2+J_1(L K[2,0,K[3]]){}^2\right ) K[2,0,K[3]]}}+\underset {K[3]=1}{\overset {\infty }{\sum }}\left (\underset {K[1]=1}{\overset {\infty }{\sum }}e^{-k t K[2,K[1],K[3]]^2} \left (\frac {2 J_{K[1]}(r K[2,K[1],K[3]]) \left (\int _0^L\int _{-\pi }^{\pi }r J_{K[1]}(r K[2,K[1],K[3]]) f(r,\theta ) \sin (\theta K[1])d\theta dr\right ) \sin (\theta K[1]) K[2,K[1],K[3]]^{3/2}}{L \pi \sqrt {L \left (J_{K[1]-1}(L K[2,K[1],K[3]]){}^2+J_{K[1]}(L K[2,K[1],K[3]]){}^2\right ) K[2,K[1],K[3]]-2 J_{K[1]-1}(L K[2,K[1],K[3]]) J_{K[1]}(L K[2,K[1],K[3]]) K[1]} \sqrt {K[2,K[1],K[3]] \left (L \left (J_{K[1]-1}(L K[2,K[1],K[3]]){}^2+J_{K[1]}(L K[2,K[1],K[3]]){}^2\right ) K[2,K[1],K[3]]-2 J_{K[1]-1}(L K[2,K[1],K[3]]) J_{K[1]}(L K[2,K[1],K[3]]) K[1]\right )}}+\frac {2 J_{K[1]}(r K[2,K[1],K[3]]) \cos (\theta K[1]) \left (\int _0^L\int _{-\pi }^{\pi }r J_{K[1]}(r K[2,K[1],K[3]]) \cos (\theta K[1]) f(r,\theta )d\theta dr\right ) K[2,K[1],K[3]]^{3/2}}{L \pi \sqrt {L \left (J_{K[1]-1}(L K[2,K[1],K[3]]){}^2+J_{K[1]}(L K[2,K[1],K[3]]){}^2\right ) K[2,K[1],K[3]]-2 J_{K[1]-1}(L K[2,K[1],K[3]]) J_{K[1]}(L K[2,K[1],K[3]]) K[1]} \sqrt {K[2,K[1],K[3]] \left (L \left (J_{K[1]-1}(L K[2,K[1],K[3]]){}^2+J_{K[1]}(L K[2,K[1],K[3]]){}^2\right ) K[2,K[1],K[3]]-2 J_{K[1]-1}(L K[2,K[1],K[3]]) J_{K[1]}(L K[2,K[1],K[3]]) K[1]\right )}}\right )\right ) & J_{K[1]-1}(L K[2,K[1],K[3]])=J_{K[1]+1}(L K[2,K[1],K[3]])\land (K[1]|K[3])\in \mathbb {Z}\land K[1]\geq 1\land K[3]\geq 1\land K[2,K[1],K[3]]>0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]

Maple ✗

restart; 
pde := diff(u(r,theta,t),t)=k*(diff(u(r,theta,t),r$2) + 1/r*diff(u(r,theta,t),r)+1/r^2*diff(u(r,theta,t),theta$2)); 
bcOnR:= D[1](u)(L,theta,t)=0; 
bcOnTheta:= u(r,-Pi,t)=u(r,Pi,t),eval(diff(u(r,theta,t),theta),theta=-Pi)=eval(diff(u(r,theta,t),theta),theta=Pi); 
ic  := u(r,theta,0)=f(r,theta); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bcOnR,bcOnTheta, ic], u(r, theta, t)) assuming L>0,r>0,r<L,t>0,k>0),output='realtime'));

sol=()

Solve\[ \frac {\partial u\left ( r,\theta ,t\right ) }{\partial t}=k\left ( \frac {\partial ^{2}u}{\partial r^{2}}+\frac {1}{r}\frac {\partial u}{\partial r}+\frac {1}{r^{2}}\frac {\partial ^{2}u}{\partial \theta ^{2}}\right ) \] With boundary conditions\begin {align*} \left \vert u\left ( 0,\theta ,t\right ) \right \vert & <\infty \\ u_{t}\left ( L,\theta ,t\right ) & =g\left ( \theta \right ) \\ u\left ( r,-\pi ,t\right ) & =u\left ( r,\pi ,t\right ) \\ \frac {\partial u}{\partial \theta }\left ( r,-\pi ,t\right ) & =\frac {\partial u}{\partial \theta }\left ( r,\pi ,t\right ) \end {align*}

And initial conditions $u\left ( r,\theta ,0\right ) =f\left ( r,\theta \right ) $. Let $y\left ( r,\theta ,t\right ) =R\left ( r\right ) \Theta \left ( \theta \right ) T\left ( t\right ) $. Substituting into (1) gives\[ T^{\prime }R\Theta =k\left ( R^{\prime \prime }T\Theta +\frac {1}{r}R^{\prime }T\Theta +\frac {1}{r^{2}}\Theta ^{\prime \prime }RT\right ) \] Dividing by $R\left ( r\right ) \Theta \left ( \theta \right ) T\left ( t\right ) \neq 0$ gives\[ \frac {1}{k}\frac {T^{\prime }}{T}=\frac {R^{\prime \prime }}{R}+\frac {1}{r}\frac {R^{\prime }}{R}+\frac {1}{r^{2}}\frac {\Theta ^{\prime \prime }}{\Theta }\] Let ﬁrst separation constant be $-\lambda $, hence the above becomes\begin {align*} \frac {1}{k}\frac {T^{\prime }}{T} & =-\lambda \\ \frac {R^{\prime \prime }}{R}+\frac {1}{r}\frac {R^{\prime }}{R}+\frac {1}{r^{2}}\frac {\Theta ^{\prime \prime }}{\Theta } & =-\lambda \end {align*}

Or\begin {align*} T^{\prime }+\lambda kT & =0\\ r^{2}\frac {R^{\prime \prime }}{R}+r\frac {R^{\prime }}{R}+r^{2}\lambda & =-\frac {\Theta ^{\prime \prime }}{\Theta }=\mu \end {align*}

Hence \begin {align*} r^{2}\frac {R^{\prime \prime }}{R}+r\frac {R^{\prime }}{R}+r^{2}\lambda & =\mu \\ -\frac {\Theta ^{\prime \prime }}{\Theta } & =\mu \end {align*}

Or\begin {align} r^{2}R^{\prime \prime }\left ( r\right ) +rR^{\prime }\left ( r\right ) +R\left ( r\right ) \left ( r^{2}\lambda -\mu \right ) & =0\tag {2}\\ \Theta ^{\prime \prime }\left ( \theta \right ) +\mu \Theta \left ( \theta \right ) & =0 \tag {3} \end {align}

Equation (2) is Sturm-Liouville ODE with boundary conditions $R^{\prime }\left ( a\right ) =0$ and bounded at $r=0$ and (3) has periodic boundary conditions.

We start with $\Theta $ ODE. The boundary conditions on (3) are \begin {align*} \Theta \left ( -\pi \right ) & =\Theta \left ( \pi \right ) \\ \frac {\partial \Theta }{\partial \theta }\left ( -\pi \right ) & =\frac {\partial \Theta }{\partial \theta }\left ( \pi \right ) \end {align*}

Negative $\mu $ is not possible, since this gives exponential and the boundary conditions on $\Theta \left ( \theta \right ) $ are periodic.

case $\mu =0$ The solution is $\Phi =c_{1}\theta +c_{2}$. Hence we obtain, from ﬁrst initial conditions\begin {align*} -\pi c_{1}+c_{2} & =\pi c_{1}+c_{2}\\ c_{1} & =0 \end {align*}

Second boundary conditions just says that $c_{2}=c_{2}$, so any constant will do. Hence $\mu =0$ is an eigenvalue with constant being eigenfunction $\Theta _{0}\left ( \theta \right ) =1$.

case $\mu >0$ The solution is \[ \Theta \left ( \theta \right ) =c_{1}\cos \sqrt {\mu }\theta +c_{2}\sin \sqrt {\mu }\theta \] The ﬁrst boundary conditions gives\begin {align} c_{1}\cos \left ( -\sqrt {\mu }\pi \right ) +c_{2}\sin \left ( -\sqrt {\mu }\pi \right ) & =c_{1}\cos \left ( \sqrt {\mu }\pi \right ) +c_{2}\sin \left ( \sqrt {\mu }\pi \right ) \nonumber \\ c_{1}\cos \left ( \sqrt {\mu }\pi \right ) -c_{2}\sin \left ( \sqrt {\mu }\pi \right ) & =c_{1}\cos \left ( \sqrt {\mu }\pi \right ) +c_{2}\sin \left ( \sqrt {\mu }\pi \right ) \nonumber \\ 2c_{2}\sin \left ( \sqrt {\mu }\pi \right ) & =0 \tag {4} \end {align}

From second boundary conditions we obtain\[ \Theta ^{\prime }\left ( \theta \right ) =-\sqrt {\mu }c_{1}\sin \sqrt {\mu }\theta +c_{2}\sqrt {\mu }\cos \sqrt {\mu }\theta \] Therefore\begin {align} -\sqrt {\mu }c_{1}\sin \left ( -\sqrt {\mu }\pi \right ) +c_{2}\sqrt {\mu }\cos \left ( -\sqrt {\mu }\pi \right ) & =-\sqrt {\mu }c_{1}\sin \left ( \sqrt {\mu }\pi \right ) +c_{2}\sqrt {\lambda }\cos \left ( \sqrt {\mu }\pi \right ) \nonumber \\ \sqrt {\mu }c_{1}\sin \left ( \sqrt {\mu }\pi \right ) +c_{2}\sqrt {\mu }\cos \left ( \sqrt {\mu }\pi \right ) & =-\sqrt {\mu }c_{1}\sin \left ( \sqrt {\mu }\pi \right ) +c_{2}\sqrt {\mu }\cos \left ( \sqrt {\mu }\pi \right ) \nonumber \\ \sqrt {\mu }c_{1}\sin \left ( \sqrt {\mu }\pi \right ) & =-\sqrt {\mu }c_{1}\sin \left ( \sqrt {\mu }\pi \right ) \nonumber \\ 2c_{1}\sin \left ( \sqrt {\mu }\pi \right ) & =0 \tag {5} \end {align}

Both (4) and (5) are satisﬁed if\begin {align} \sqrt {\mu }\pi & =n\pi \qquad n=1,2,3,\cdots \tag {5A}\\ \mu & =n^{2}\qquad n=1,2,3,\cdots \nonumber \end {align}

Therefore $\mu _{n}=1,4,9,16,\cdots $. Or $\sqrt {\mu _{n}}=n=1,2,3,\cdots $ \begin {align} \Theta _{n}\left ( \theta \right ) & =\overset {n=0}{\overbrace {A_{0}}}+\sum _{n=1}^{\infty }A_{n}\cos \left ( \sqrt {\mu _{n}}\theta \right ) +B_{n}\sin \left ( \sqrt {\mu _{n}}\theta \right ) \tag {6}\\ & =A_{0}+\sum _{n=1}^{\infty }A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \end {align}

Now we go back to the $R\left ( r\right ) $ ODE (2) given by $r^{2}R^{\prime \prime }+rR^{\prime }+R\left ( r^{2}\lambda -\mu \right ) =0$ or since $\mu _{n}=n^{2}$, we write it as $r^{2}R^{\prime \prime }+rR^{\prime }+R\left ( r^{2}\lambda -n^{2}\right ) =0$ for $n=0,1,2,\cdots $.

The $R\left ( r\right ) $ ODE becomes $r^{2}R^{\prime \prime }+rR^{\prime }+\lambda r^{2}R=0$. This is a Bessel ODE. If $\lambda <0$, this leads to solution (contains $\operatorname {Besselk}$ and $\operatorname {BesselI}$) that can not satisfy the boundary conditions. If $\lambda =0$ then the ODE becomes $r^{2}R^{\prime \prime }+rR^{\prime }=0$ whose solution is $R\left ( r\right ) =C_{1}\ln \left ( r\right ) +C_{2}$. Since bounded at $r=0$. Then $R\left ( r\right ) =C_{2}$. Hence $R^{\prime }\left ( r\right ) =0$. Which satisﬁes the boundary condition at $r=L$ for any constant $C_{2}$. Therefore $\lambda =0$ is possible eigenvalue with constant as its eigenfunction $R_{0}\left ( r\right ) =1$.

If $\lambda >0$ the ODE is a Bessel ODE whose solution is \[ R\left ( r\right ) =C_{1}\operatorname {BesselJ}\left ( 0,\sqrt {\lambda }r\right ) +C_{2}\operatorname {BesselY}\left ( 0,\sqrt {\lambda }r\right ) \] We set $C_{2}=0$ since $\operatorname {BesselY}\left ( 0,\sqrt {\lambda }r\right ) $ blows up at $r=0$. Hence $R\left ( r\right ) =C_{1}\operatorname {BesselJ}\left ( 0,\sqrt {\lambda }r\right ) $ and $R^{\prime }\left ( r\right ) =-C_{1}\sqrt {\lambda }\operatorname {BesselJ}\left ( 1,\sqrt {\lambda }r\right ) $. At $r=L$ we obtain

$R^{\prime }\left ( L\right ) =-C_{1}\sqrt {\lambda }\operatorname {BesselJ}\left ( 1,\sqrt {\lambda }L\right ) =0$. Therefore for non-trivial solution we want $\sqrt {\lambda }L$ to be the zeros of $\operatorname {BesselJ}\left ( 1,x\right ) $. Let these zeros be $\Lambda _{m}$, $m=1,2,3,\cdots $. Hence \begin {align*} \sqrt {\lambda _{0,m}}L & =\Lambda _{m}\\ \lambda _{0,m} & =\left ( \frac {\Lambda _{m}}{L}\right ) ^{2}\qquad m=1,2,3,\cdots \end {align*}

Where we added zero as subscript to $\lambda _{0,m}$ to indicate that this is associated with Case $n=0$. Hence the eigenfunctions for $\mu =0$ are\[ R_{0,m}\left ( r\right ) =\operatorname {BesselJ}\left ( 0,\frac {\Lambda _{m}}{L}r\right ) \qquad n=0,m=1,2,3,\cdots \] And $\Theta _{0}\left ( \theta \right ) =1$.

The $R\left ( r\right ) $ ODE becomes $r^{2}R^{\prime \prime }+rR^{\prime }+R\left ( r^{2}\lambda -n^{2}\right ) =0$.

This is a Bessel ODE. If $\lambda <0$, this leads to solution that can not satisfy the boundary conditions (contains $\operatorname {Besselk}$ and $\operatorname {BesselI}$). If $\lambda =0$ then the ODE becomes $r^{2}R^{\prime \prime }+rR^{\prime }-n^{2}R=0$. This is Euler ODE whose solution is $R\left ( r\right ) =C_{1}r^{n}+C_{2}r^{-n}$. Since bounded at $r=0$. Then $C_{2}=0$ and the solution becomes $R\left ( r\right ) =C_{1}r^{n}$. Hence $R^{\prime }\left ( r\right ) =C_{1}nr^{n-1}$. At $r=L$ this becomes $C_{1}nL^{n-1}=0$. Since $L\neq 0$, then this gives $C_{1}=0$, trivial solution. Hence $\lambda =0$ is not possible eigenvalue when $\mu >0$.

If $\lambda >0$ the ODE becomes $r^{2}R^{\prime \prime }+rR^{\prime }+R\left ( r^{2}\lambda -n^{2}\right ) =0$ with now $\lambda $ positive. This is a Bessel ODE whose solution is

\[ R_{n}\left ( r\right ) =C_{1}\operatorname {BesselJ}\left ( n,\sqrt {\lambda }r\right ) +C_{2}\operatorname {BesselY}\left ( n,\sqrt {\lambda }r\right ) \qquad n=1,2,3,\cdots \] Since $R\left ( r\right ) $ is bounded at $r=0$ then $C_{2}=0$ and the above becomes\[ R_{n}\left ( r\right ) =C_{1}\operatorname {BesselJ}\left ( n,\sqrt {\lambda }r\right ) \qquad n=1,2,3,\cdots \] Therefore \[ R_{n}^{\prime }\left ( r\right ) =C_{1}\frac {d}{dr}\operatorname {BesselJ}\left ( n,\sqrt {\lambda }r\right ) \] At $r=L$\[ R_{n}^{\prime }\left ( L\right ) =C_{1}\frac {d}{dr}\operatorname {BesselJ}\left ( n,\sqrt {\lambda }L\right ) \] For non-trivial solution we want\begin {equation} \frac {d}{dr}\operatorname {BesselJ}\left ( n,\sqrt {\lambda }L\right ) =0\tag {7} \end {equation} The eigenvalues $\lambda $ are solved for numerically by ﬁnding all zeros of $\frac {d}{dx}\operatorname {BesselJ}\left ( n,x\right ) $. Let these zeros be called $\Gamma _{nm}$ for $n=1,2,3,\cdots ,m=1,2,3,\cdots $. Hence\begin {align*} \sqrt {\lambda _{nm}}L & =\Gamma _{nm}\\ \lambda _{nm} & =\left ( \frac {\Gamma _{nm}}{L}\right ) ^{2}\qquad n=1,2,3,\cdots ,m=1,2,3,\cdots \end {align*}

With associated eigenfunctions $R_{n,m}\left ( r\right ) =\operatorname {BesselJ}\left ( n,\frac {\Gamma _{nm}}{L}r\right ) $. Now that we solved for $R_{n,m}\left ( r\right ) $ and $\Theta _{n,m}\left ( \theta \right ) $, what is left is to solve $\frac {1}{k}\frac {T^{\prime }}{T}=-\lambda _{nm}$.

For $\lambda =0$ the solution is constant. Say $T_{00}\left ( t\right ) =1$. For $\lambda >0$ the solution is $T_{0m}\left ( t\right ) =e^{-k\lambda _{0m}t}$. Where $\lambda _{0,m}=\left ( \frac {\Lambda _{m}}{L}\right ) ^{2},m=1,2,3,\cdots $ and $\Lambda _{m}$ are the zeros of $\operatorname {BesselJ}\left ( 1,x\right ) $.

For $n>0$ In this case $\lambda >0$ only. The solution is $T_{nm}\left ( t\right ) =e^{-k\lambda _{nm}t}$. Where $\lambda _{n,m}$ are now the solutions of (7). Hence $T_{nm}\left ( t\right ) =e^{-k\left ( \frac {\Gamma _{nm}}{L}\right ) ^{2}t}$

The solution now becomes\[ u\left ( r,\theta ,t\right ) =\overset {\text {arbitrary constant}}{\overbrace {R_{0,0}\left ( r\right ) \Theta _{0,0}\left ( \theta \right ) T_{0,0}\left ( t\right ) }}+\overset {n=0,m>0}{\overbrace {R_{0,m}\left ( r\right ) \Theta _{0,m}\left ( \theta \right ) T_{0.m}\left ( t\right ) }}+\overset {n>0,m>0}{\overbrace {R_{n,m}\left ( r\right ) \Theta _{n,m}\left ( \theta \right ) T_{n.m}\left ( t\right ) }}\] Or\begin {align*} u\left ( r,\theta ,t\right ) & =\beta _{0}\\ & +\sum _{m=1}^{\infty }\alpha _{0,m}\operatorname {BesselJ}\left ( 0,\frac {\Lambda _{m}}{L}r\right ) e^{-k\left ( \frac {\Lambda _{m}}{L}\right ) ^{2}t}\\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }e^{-k\left ( \frac {\Gamma _{nm}}{L}\right ) ^{2}t}\operatorname {BesselJ}\left ( n,\frac {\Gamma _{nm}}{L}r\right ) \left ( A_{nm}\cos \left ( n\theta \right ) +B_{nm}\sin \left ( n\theta \right ) \right ) \end {align*}

Where $\Gamma _{nm}$ are the $m^{th}$ zeros $\frac {d}{dx}\operatorname {BesselJ}\left ( n,x\right ) $ and $\Lambda _{m}$ are the $m^{th}$ zeros of $\operatorname {BesselJ}\left ( 1,x\right ) $. These have to be found numerically.

Now we ﬁnd the constants $\beta _{0},\alpha _{m},A_{nm},B_{nm}$. For $n=0,m=0$, and at initial conditions at $t=0$, and applying orthogonality gives (all eigenfunctions are constants, assumed $1$, in this case)

\begin {align*} \int _{-\pi }^{\pi }\int _{0}^{L}f\left ( r,\theta \right ) rdrd\theta & =\beta _{0}\int _{-\pi }^{\pi }\int _{0}^{L}rdrd\theta \\ \beta _{0} & =\frac {\int _{-\pi }^{\pi }\int _{0}^{L}f\left ( r,\theta \right ) rdrd\theta }{\pi L^{2}} \end {align*}

For $n=0,m>0$\begin {align*} \int _{-\pi }^{\pi }\int _{0}^{L}f\left ( r,\theta \right ) \operatorname {BesselJ}\left ( 0,\frac {\Lambda _{m}}{L}r\right ) rdrd\theta & =\alpha _{0,m}\int _{-\pi }^{\pi }\int _{0}^{L}\operatorname {BesselJ}^{2}\left ( 0,\frac {\Lambda _{m}}{L}r\right ) rdrd\theta \\ & =2\pi \alpha _{0,m}\int _{0}^{L}\operatorname {BesselJ}^{2}\left ( 0,\frac {\Lambda _{m}}{L}r\right ) rdr\\ \alpha _{0,m} & =\frac {\int _{-\pi }^{\pi }\int _{0}^{L}f\left ( r,\theta \right ) \operatorname {BesselJ}\left ( 0,\frac {\Lambda _{m}}{L}r\right ) rdrd\theta }{2\pi \int _{0}^{L}\operatorname {BesselJ}^{2}\left ( 0,\frac {\Lambda _{m}}{L}r\right ) rdr} \end {align*}

For $n>0,m>0$, applying orthogonality on $\operatorname {BesselJ}\left ( n,\frac {\Gamma _{nm}}{L}r\right ) $ gives\[ \int _{0}^{L}f\left ( r,\theta \right ) \operatorname {BesselJ}\left ( n,\frac {\Gamma _{nm}}{L}r\right ) rdr=\sum _{n=1}^{\infty }\int _{0}^{L}\operatorname {BesselJ}^{2}\left ( n,\frac {\Gamma _{nm}}{L}r\right ) rdr\left ( A_{nm}\cos \left ( n\theta \right ) +B_{nm}\sin \left ( n\theta \right ) \right ) \] applying orthogonality on $\cos \left ( n\theta \right ) $ the above becomes\begin {align*} \int _{-\pi }^{\pi }\int _{0}^{L}f\left ( r,\theta \right ) \operatorname {BesselJ}\left ( n,\frac {\Gamma _{nm}}{L}r\right ) \cos \left ( n\theta \right ) rdrd\theta & =A_{nm}\int _{-\pi }^{\pi }\int _{0}^{L}\operatorname {BesselJ}^{2}\left ( n,\frac {\Gamma _{nm}}{L}r\right ) \cos ^{2}\left ( n\theta \right ) rdrd\theta \\ A_{nm} & =\frac {\int _{-\pi }^{\pi }\int _{0}^{L}f\left ( r,\theta \right ) \operatorname {BesselJ}\left ( n,\frac {\Gamma _{nm}}{L}r\right ) \cos \left ( n\theta \right ) rdrd\theta }{\int _{-\pi }^{\pi }\int _{0}^{L}\operatorname {BesselJ}^{2}\left ( n,\frac {\Gamma _{nm}}{L}r\right ) \cos ^{2}\left ( n\theta \right ) rdrd\theta } \end {align*}

Similarly\[ B_{nm}=\frac {\int _{-\pi }^{\pi }\int _{0}^{L}f\left ( r,\theta \right ) \operatorname {BesselJ}\left ( n,\frac {\Gamma _{nm}}{L}r\right ) \sin \left ( n\theta \right ) rdrd\theta }{\int _{-\pi }^{\pi }\int _{0}^{L}\operatorname {BesselJ}^{2}\left ( n,\frac {\Gamma _{nm}}{L}r\right ) \sin ^{2}\left ( n\theta \right ) rdrd\theta }\] This complete the solution. In summary, the solution is\begin {align*} u\left ( r,\theta ,t\right ) & =\beta _{0}\\ & +\sum _{m=1}^{\infty }\alpha _{0,m}\operatorname {BesselJ}\left ( 0,\frac {\Lambda _{m}}{L}r\right ) e^{-k\left ( \frac {\Lambda _{m}}{L}\right ) ^{2}t}\\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }e^{-k\left ( \frac {\Gamma _{nm}}{L}\right ) ^{2}t}\operatorname {BesselJ}\left ( n,\frac {\Gamma _{nm}}{L}r\right ) \left ( A_{nm}\cos \left ( n\theta \right ) +B_{nm}\sin \left ( n\theta \right ) \right ) \end {align*}

Where\begin {align*} \beta _{0} & =\frac {\int _{-\pi }^{\pi }\int _{0}^{L}f\left ( r,\theta \right ) rdrd\theta }{\pi L^{2}}\\ \alpha _{0,m} & =\frac {\int _{-\pi }^{\pi }\int _{0}^{L}f\left ( r,\theta \right ) \operatorname {BesselJ}\left ( 0,\frac {\Lambda _{m}}{L}r\right ) rdrd\theta }{2\pi \int _{0}^{L}\operatorname {BesselJ}^{2}\left ( 0,\frac {\Lambda _{m}}{L}r\right ) rdr}\\ A_{nm} & =\frac {\int _{-\pi }^{\pi }\int _{0}^{L}f\left ( r,\theta \right ) \operatorname {BesselJ}\left ( n,\frac {\Gamma _{nm}}{L}r\right ) \cos \left ( n\theta \right ) rdrd\theta }{\int _{-\pi }^{\pi }\int _{0}^{L}\operatorname {BesselJ}^{2}\left ( n,\frac {\Gamma _{nm}}{L}r\right ) \cos ^{2}\left ( n\theta \right ) rdrd\theta }\\ B_{nm} & =\frac {\int _{-\pi }^{\pi }\int _{0}^{L}f\left ( r,\theta \right ) \operatorname {BesselJ}\left ( n,\frac {\Gamma _{nm}}{L}r\right ) \sin \left ( n\theta \right ) rdrd\theta }{\int _{-\pi }^{\pi }\int _{0}^{L}\operatorname {BesselJ}^{2}\left ( n,\frac {\Gamma _{nm}}{L}r\right ) \sin ^{2}\left ( n\theta \right ) rdrd\theta } \end {align*}

And $\Gamma _{nm}$ are the $m^{th}$ zeros of $\frac {d}{dx}\operatorname {BesselJ}\left ( n,x\right ) $ and $\Lambda _{m}$ are the $m^{th}$ zeros of $\operatorname {BesselJ}\left ( 1,x\right ) $. These have to be found numerically.

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3.2.2.14 [271] Insulated with $\theta $ dependency (Speciﬁc example)

Solve for $u(r,\theta ,t)$, the initial value problem for a two-dimensional heat equation inside a circle of radius $a$ \[ u_t = k \nabla ^2 u \] For $t>0$ and $0<r<L$ with $L=1,k=1$ and time-independent boundary conditions: \begin {align*} u_r(L,\theta ,t) &= 0 \end {align*}

And initial conditions $u(r,\theta ,0) =(2 L r-r^2)\cos \theta \sin \theta $. There is an implied periodic boundary conditions on $\theta $ \begin {align*} u(r,-\pi ,t) &= u(r,\pi ,t)\\ \frac {\partial u}{\partial \theta }(r,-\pi ,t) &= \frac {\partial u}{\partial \theta }(r,\pi ,t) \end {align*}

Mathematica ✓

ClearAll["Global`*"]; 
L=1; 
k=1; 
pde = D[u[r, theta, t], t] == k*(D[u[r, theta, t], {r, 2}] + D[u[r, theta, t], r]/r + D[u[r, theta, t], {theta, 2}]/r^2); 
bcOnR = Derivative[1,0,0][u][L, theta, t] == 0; 
bcOnTheta = {u[r, -Pi, t] == u[r, Pi, t], Derivative[0, 1, 0][u][r, -Pi, t] == Derivative[0, 1, 0][u][r, Pi, t]}; 
ic  = u[r, theta, 0] == (2*L*r-r^2)*Cos[theta]*Sin[theta]; 
sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bcOnR, bcOnTheta, ic}, u[r, theta, t], {r, theta, t},Assumptions->{r>0,r<L,t>0}], 60*10]];

\[\left \{\left \{u(r,\theta ,t)\to \begin {array}{cc} \{ & \begin {array}{cc} \underset {K[3]=1}{\overset {\infty }{\sum }}\frac {e^{-t K[2,2,K[3]]^2} J_2(r K[2,2,K[3]]) \left (2 \, _1F_2\left (\frac {5}{2};3,\frac {7}{2};-\frac {1}{4} K[2,2,K[3]]^2\right )-5 \, _0\tilde {F}_1\left (;4;-\frac {1}{4} K[2,2,K[3]]^2\right )\right ) K[2,2,K[3]]^3 \sin (2 \theta )}{40 \left (\left (J_1(K[2,2,K[3]]){}^2+J_2(K[2,2,K[3]]){}^2\right ) K[2,2,K[3]]-4 J_1(K[2,2,K[3]]) J_2(K[2,2,K[3]])\right )} & J_{K[1]-1}(K[2,K[1],K[3]])=J_{K[1]+1}(K[2,K[1],K[3]])\land (K[1]|K[3])\in \mathbb {Z}\land K[1]\geq 1\land K[3]\geq 1\land K[2,K[1],K[3]]>0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \}\]

Maple ✗

restart; 
k:=1; 
L:=1; 
pde := diff(u(r,theta,t),t)=k*(diff(u(r,theta,t),r$2) + 1/r*diff(u(r,theta,t),r)+1/r^2*diff(u(r,theta,t),theta$2)); 
bcOnR:= D[1](u)(L,theta,t)=0; 
bcOnTheta:= u(r,-Pi,t)=u(r,Pi,t),eval(diff(u(r,theta,t),theta),theta=-Pi)=eval(diff(u(r,theta,t),theta),theta=Pi); 
ic  := u(r,theta,0)= (2*L*r-r^2)*cos(theta)*sin(theta); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bcOnR,bcOnTheta, ic], u(r, theta, t)) assuming r>0,r<L,t>0),output='realtime'));

sol=()

Solve for $u\left ( r,\theta ,t\right ) $\[ u_{t}=k\left ( \frac {\partial ^{2}u}{\partial r^{2}}+\frac {1}{r}\frac {\partial u}{\partial r}+\frac {1}{r^{2}}\frac {\partial ^{2}u}{\partial \theta ^{2}}\right ) \] With boundary conditions\begin {align*} \left \vert u\left ( 0,\theta ,t\right ) \right \vert & <\infty \\ u_{t}\left ( L,\theta ,t\right ) & =0\\ u\left ( r,-\pi ,t\right ) & =u\left ( r,\pi ,t\right ) \\ \frac {\partial u}{\partial \theta }\left ( r,-\pi ,t\right ) & =\frac {\partial u}{\partial \theta }\left ( r,\pi ,t\right ) \end {align*}

And initial conditions \begin {align*} u\left ( r,\theta ,0\right ) & =f\left ( r,\theta \right ) =\left ( 2rL-r^{2}\right ) \cos \theta \sin \theta \\ L & =1\\ k & =1 \end {align*}

The basic solution for this type of PDE was already given in problem 3.2.2.13 on page 944 as\begin {align*} u\left ( r,\theta ,t\right ) & =\beta _{0}\\ & +\sum _{m=1}^{\infty }\alpha _{0,m}\operatorname {BesselJ}\left ( 0,\frac {\Lambda _{m}}{L}r\right ) e^{-k\left ( \frac {\Lambda _{m}}{L}\right ) ^{2}t}\\ & +\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }e^{-k\left ( \frac {\Gamma _{nm}}{L}\right ) ^{2}t}\operatorname {BesselJ}\left ( n,\frac {\Gamma _{nm}}{L}r\right ) \left ( A_{nm}\cos \left ( n\theta \right ) +B_{nm}\sin \left ( n\theta \right ) \right ) \end {align*}

____________________________________________________________________________________

3.2.2.15 [272] Insulated with $\theta $ dependency (Speciﬁc example)

And initial conditions $u(r,\theta ,0) =(2 L r - r^2) \theta \sin \theta e^{\cos \theta }$. There is an implied periodic boundary conditions on $\theta $ \begin {align*} u(r,-\pi ,t) &= u(r,\pi ,t)\\ \frac {\partial u}{\partial \theta }(r,-\pi ,t) &= \frac {\partial u}{\partial \theta }(r,\pi ,t) \end {align*}

Mathematica ✗

ClearAll["Global`*"]; 
L=1; 
k=1; 
pde = D[u[r, theta, t], t] == k*(D[u[r, theta, t], {r, 2}] + D[u[r, theta, t], r]/r + D[u[r, theta, t], {theta, 2}]/r^2); 
bcOnR = Derivative[1,0,0][u][L, theta, t] == 0; 
bcOnTheta = {u[r, -Pi, t] == u[r, Pi, t], Derivative[0, 1, 0][u][r, -Pi, t] == Derivative[0, 1, 0][u][r, Pi, t]}; 
ic  = u[r, theta, 0] == (2*L*r - r^2)*theta*Sin[theta]*Exp[Cos[theta]]; 
sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bcOnR, bcOnTheta, ic}, u[r, theta, t], {r, theta, t},Assumptions->{r>0,r<L,t>0}], 60*10]];

$Aborted

Maple ✗

restart; 
k:=1; 
L:=1; 
pde := diff(u(r,theta,t),t)=k*(diff(u(r,theta,t),r$2) + 1/r*diff(u(r,theta,t),r)+1/r^2*diff(u(r,theta,t),theta$2)); 
bcOnR:= D[1](u)(L,theta,t)=0; 
bcOnTheta:= u(r,-Pi,t)=u(r,Pi,t),eval(diff(u(r,theta,t),theta),theta=-Pi)=eval(diff(u(r,theta,t),theta),theta=Pi); 
ic  := u(r,theta,0)= (2*L*r -r^2)*theta*sin(theta)*exp(cos(theta)); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bcOnR,bcOnTheta, ic], u(r, theta, t)) assuming r>0,r<L,t>0),output='realtime'));

sol=()

Solve for $u\left ( r,\theta ,t\right ) $\[ u_{t}=k\left ( \frac {\partial ^{2}u}{\partial r^{2}}+\frac {1}{r}\frac {\partial u}{\partial r}+\frac {1}{r^{2}}\frac {\partial ^{2}u}{\partial \theta ^{2}}\right ) \] With boundary conditions\begin {align*} \left \vert u\left ( 0,\theta ,t\right ) \right \vert & <\infty \\ u_{t}\left ( L,\theta ,t\right ) & =0\\ u\left ( r,-\pi ,t\right ) & =u\left ( r,\pi ,t\right ) \\ u_{\theta }\left ( r,-\pi ,t\right ) & =u_{\theta }\left ( r,\pi ,t\right ) \end {align*}

And initial conditions \begin {align*} u\left ( r,\theta ,0\right ) & =f\left ( r,\theta \right ) =\left ( 2rL-r^{2}\right ) \theta \sin \theta e^{\cos \theta }\\ L & =1\\ k & =1 \end {align*}

The basic solution for this type of PDE was already given in problem 3.2.2.13 on page 944 as

\[ u\left ( r,\theta ,t\right ) =\beta _{0}+\sum _{m=1}^{\infty }\alpha _{0,m}\operatorname {BesselJ}\left ( 0,\frac {\Lambda _{m}}{L}r\right ) e^{-k\left ( \frac {\Lambda _{m}}{L}\right ) ^{2}t}+\sum _{n=1}^{\infty }\sum _{m=1}^{\infty }e^{-k\left ( \frac {\Gamma _{nm}}{L}\right ) ^{2}t}\operatorname {BesselJ}\left ( n,\frac {\Gamma _{nm}}{L}r\right ) \left ( A_{nm}\cos \left ( n\theta \right ) +B_{nm}\sin \left ( n\theta \right ) \right ) \] Where\begin {align*} \beta _{0} & =\frac {\int _{-\pi }^{\pi }\int _{0}^{L}f\left ( r,\theta \right ) rdrd\theta }{\pi L^{2}}\\ \alpha _{0,m} & =\frac {\int _{-\pi }^{\pi }\int _{0}^{L}f\left ( r,\theta \right ) \operatorname {BesselJ}\left ( 0,\frac {\Lambda _{m}}{L}r\right ) rdrd\theta }{2\pi \int _{0}^{L}\operatorname {BesselJ}^{2}\left ( 0,\frac {\Lambda _{m}}{L}r\right ) rdr}\\ A_{nm} & =\frac {\int _{-\pi }^{\pi }\int _{0}^{L}f\left ( r,\theta \right ) \operatorname {BesselJ}\left ( n,\frac {\Gamma _{nm}}{L}r\right ) \cos \left ( n\theta \right ) rdrd\theta }{\int _{-\pi }^{\pi }\int _{0}^{L}\operatorname {BesselJ}^{2}\left ( n,\frac {\Gamma _{nm}}{L}r\right ) \cos ^{2}\left ( n\theta \right ) rdrd\theta }\\ B_{nm} & =\frac {\int _{-\pi }^{\pi }\int _{0}^{L}f\left ( r,\theta \right ) \operatorname {BesselJ}\left ( n,\frac {\Gamma _{nm}}{L}r\right ) \sin \left ( n\theta \right ) rdrd\theta }{\int _{-\pi }^{\pi }\int _{0}^{L}\operatorname {BesselJ}^{2}\left ( n,\frac {\Gamma _{nm}}{L}r\right ) \sin ^{2}\left ( n\theta \right ) rdrd\theta } \end {align*}

3.2.2 Polar coordinates (disk, sector, annulus)

3.2.2.1 [258] no \(\theta \) dependency, insulated (General solution)

3.2.2.2 [259] no \(\theta \) dependency, insulated (Speciﬁc solution)

3.2.2.3 [260] no \(\theta \) dependency

3.2.2.4 [261] no \(\theta \) dependency

3.2.2.5 [262] Haberman 8.3.5 (General solution)

3.2.2.6 [263] Speciﬁc example of the above

3.2.2.7 [264] Speciﬁc example of the above

3.2.2.8 [265] Articolo 6.9.1

3.2.2.9 [266] Articolo 6.9.2

3.2.2.10 [267] Haberman 8.2.5 with \(\theta \) dependency (General case)

3.2.2.11 [268] With \(\theta \) dependency (Speciﬁc example)

3.2.2.12 [269] With \(\theta \) dependency (speciﬁc example)

3.2.2.13 [270] Insulated with \(\theta \) dependency (General solution)

3.2.2.14 [271] Insulated with \(\theta \) dependency (Speciﬁc example)

3.2.2.15 [272] Insulated with \(\theta \) dependency (Speciﬁc example)