5.2.2 Polar coordinates

5.2.2.1 [404] no $\theta$ dependency, fixed boundary, general case

problem number 404

Added January 12, 2020

Circular disk. fixed edge of disk, no $\theta$ dependency, with initial position and velocity given

Solve for $u(r,t)$ with $0<r<a$ and $t>0$. $$u_{tt}=c^2(u_{rr}+\frac{1}{r}{u}_r)$$ With boundary conditions $$\begin{array}{rl}u(a,t)& =0\end{array}$$

With initial conditions $$\begin{array}{rl}u(r,0)& =f(r)\\ \frac{\partial u}{\partial t}(r,0)& =g(r)\end{array}$$

Figure 5.101: PDE specification

Mathematica ✓

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

$$\left\{\{u(r,t)\to \underset{n=1}{\sum ^{\mathrm{\infty }}}\boxed{\frac{2J_0\left(\frac{r{j}_{0,n}}{a}\right)(\sqrt{c^2}{j}_{0,n}\cos \left(\frac{\sqrt{c^2}t{j}_{0,n}}{a}\right){\int }_0^{a}r{J}_0\left(\frac{r{j}_{0,n}}{a}\right)f(r)dr+a({\int }_0^{a}r{J}_0\left(\frac{r{j}_{0,n}}{a}\right)g(r)dr)\sin \left(\frac{\sqrt{c^2}t{j}_{0,n}}{a}\right))}{a^2\sqrt{c^2}(J_0\left(j_{0,n}\right){}^2+J_1\left(j_{0,n}\right){}^2)j_{0,n}}\text{ if }{j}_{0,n}\ge 0}\}\right\}$$

Maple ✓

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

$$u(r,t)=\frac{1}{c^2}(\mathit{i}\mathit{n}\mathit{v}\mathit{l}\mathit{a}\mathit{p}\mathit{l}\mathit{a}\mathit{c}\mathit{e}(\mathrm{BesselK}(0,\frac{sr}{c})\int \mathrm{BesselK}(0,\frac{sa}{c})a(f\left(a\right)s+g\left(a\right))\mathrm{d}a\mathrm{BesselI}(0,\frac{sa}{c}){\left(\mathrm{BesselK}(0,\frac{sa}{c})\right)}^{-1},s,t)-\mathit{i}\mathit{n}\mathit{v}\mathit{l}\mathit{a}\mathit{p}\mathit{l}\mathit{a}\mathit{c}\mathit{e}(\mathrm{BesselK}(0,\frac{sr}{c})\int \mathrm{BesselI}(0,\frac{sa}{c})a(f\left(a\right)s+g\left(a\right))\mathrm{d}a,s,t)-\mathit{i}\mathit{n}\mathit{v}\mathit{l}\mathit{a}\mathit{p}\mathit{l}\mathit{a}\mathit{c}\mathit{e}(\int \mathrm{BesselK}(0,\frac{sr}{c})(g\left(r\right)+sf\left(r\right))r\mathrm{d}r\mathrm{BesselI}(0,\frac{sr}{c}),s,t)+\mathit{i}\mathit{n}\mathit{v}\mathit{l}\mathit{a}\mathit{p}\mathit{l}\mathit{a}\mathit{c}\mathit{e}(\int \mathrm{BesselI}(0,\frac{sr}{c})(g\left(r\right)+sf\left(r\right))r\mathrm{d}r\mathrm{BesselK}(0,\frac{sr}{c}),s,t))$$ Has unresolved Invlaplace calls

Hand solution

Assuming $u=T\left(t\right)R\left(r\right)$. Substituting in the PDE gives$$\frac{1}{c^2}{T}^{\prime \prime }R=R^{\prime \prime }T+\frac{1}{r}{R}^{\prime }T$$ Dividing by $RT$$$\frac{1}{c^2}\frac{T^{\prime \prime }}{T}=\frac{R^{\prime \prime }}{R}+\frac{1}{r}\frac{R^{\prime }}{R}$$ Hence$$\begin{array}{rl}\frac{1}{c^2}\frac{T^{\prime \prime }}{T}& =-\lambda \\ \frac{R^{\prime \prime }}{R}+\frac{1}{r}\frac{R^{\prime }}{R}& =-\lambda \end{array}$$

The time ODE is$$T^{\prime \prime }+c^2\lambda T=0$$ And the $r$ ODE is (Sturm-Liouville)

$$r{R}^{\prime \prime }+R^{\prime }+\lambda rR=0$$ Where $p=r,q=0,\sigma =r$. This is singular SL.  The solution turns out to be  $$R_n\left(r\right)=A_n{J}_0\left(\sqrt{{\lambda }_n}r\right)n=1,2,3,\cdots$$ Where ${\lambda }_n$ is found from roots of $0=J_n\left(\sqrt{{\lambda }_n}a\right)$ giving the eigenvalues. Now the time ODE is solved$$\begin{array}{rl}{T}_n^{\prime \prime }+c^2{\lambda }_n{T}_n& =0\\ T_n& =B_n\cos \left(c\sqrt{{\lambda }_n}t\right)+C_n\sin \left(c\sqrt{{\lambda }_n}t\right)n=1,2,3,\dots ,\end{array}$$

Hence the solution is$$\begin{array}{rl}u(r,t)& =\sum _{n=1}^{\mathrm{\infty }}{T}_n{R}_n\\ \text{(1)}& & =\sum _{n=1}^{\mathrm{\infty }}{A}_n\cos \left(c\sqrt{{\lambda }_n}t\right)J_0\left(\sqrt{{\lambda }_n}r\right)+B_n\sin \left(c\sqrt{{\lambda }_n}t\right)J_0\left(\sqrt{{\lambda }_n}r\right)\end{array}$$

Now initial conditions $u(r,0)=f\left(r\right)$ is used to find $A_n$using orthogonality. At $t=0$ the solution simplifies to $$u(r,0)=\sum _{n=1}^{\mathrm{\infty }}{A}_n{J}_0\left(\sqrt{{\lambda }_n}r\right)$$ Hence$$\begin{array}{rl}f\left(r\right)& =\sum _{n=1}^{\mathrm{\infty }}{A}_n{J}_0\left(\sqrt{{\lambda }_n}r\right)\\ {\int }_0^{a}f\left(r\right)J_0\left(\sqrt{{\lambda }_n}r\right)rdr& =A_n{\int }_0^a{J}_0^2\left(\sqrt{{\lambda }_n}r\right)rdr\\ A_n& =\frac{{\int }_0^{a}f\left(r\right)J_0\left(\sqrt{{\lambda }_n}r\right)rdr}{{\int }_0^a{J}_0^2\left(\sqrt{{\lambda }_n}r\right)rdr}\end{array}$$

Now we will look at the second initial conditions $\frac{\partial u}{\partial t}(r,0)=g\left(r\right).$ Taking derivative w.r.t. time $t$ of the solution in (1) gives$$\frac{\partial u}{\partial t}(r,t)=\sum _{n=1}^{\mathrm{\infty }}-c\sqrt{{\lambda }_n}{A}_n\sin \left(c\sqrt{{\lambda }_n}t\right)J_0\left(\sqrt{{\lambda }_n}r\right)+B_{n}c\sqrt{{\lambda }_n}\cos \left(c\sqrt{{\lambda }_n}t\right)J_0\left(\sqrt{{\lambda }_n}r\right)$$ At time $t=0$ the above becomes $$g\left(r\right)=\sum _{n=1}^{\mathrm{\infty }}{B}_{n}c\sqrt{{\lambda }_n}{J}_0\left(\sqrt{{\lambda }_n}r\right)$$ Now orthogonality is used. The above becomes$$B_n=\frac{{\int }_0^{a}g\left(r\right)J_0\left(\sqrt{{\lambda }_n}r\right)rdr}{c\sqrt{{\lambda }_n}{\int }_0^a{J}_0^2\left(\sqrt{{\lambda }_n}r\right)rdr}$$ Summary of solution$$u(r,t)=\sum _{n=1}^{\mathrm{\infty }}{A}_n\cos \left(c\sqrt{{\lambda }_n}t\right)J_0\left(\sqrt{{\lambda }_n}r\right)+B_n\sin \left(c\sqrt{{\lambda }_n}t\right)J_0\left(\sqrt{{\lambda }_n}r\right)$$$$A_n=\frac{{\int }_0^{a}f\left(r\right)J_0\left(\sqrt{{\lambda }_n}r\right)rdr}{{\int }_0^a{J}_0^2\left(\sqrt{{\lambda }_n}r\right)rdr}$$$$B_n=\frac{{\int }_0^{a}g\left(r\right)J_0\left(\sqrt{{\lambda }_n}r\right)rdr}{c\sqrt{{\lambda }_n}{\int }_0^a{J}_0^2\left(\sqrt{{\lambda }_n}r\right)rdr}$$ With ${\lambda }_n$ being the solutions for $0=J_0\left(\sqrt{{\lambda }_n}a\right)$. We have infinite number of zeros. This generates all the needed ${\lambda }_n$. Hence $\sqrt{{\lambda }_n}a=BesselJZero(0,n)$, therefore $\sqrt{{\lambda }_n}=\frac{a}{BesselJZero(0,n)}$

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5.2.2.2 [405] no $\theta$ dependency. Specific example. Both initial conditions not zero

problem number 405

Taken from Mathematica helps pages on DSolve

In circular disk. fixed edge of disk, no $\theta$ dependency, with initial position and velocity given

Solve for $u(r,t)$ with $0<r<1$ and $t>0$. $$u_{tt}=c^2(u_{rr}+\frac{1}{r}{u}_r)$$ With boundary conditions $$\begin{array}{rl}u(1,t)& =0\end{array}$$

With initial conditions $$\begin{array}{rl}u(r,0)& =1\\ \frac{\partial u}{\partial t}(r,0)& =\frac{r}{3}\end{array}$$

Figure 5.102: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[r, t], {t, 2}] == c^2*(D[u[r, t], {r, 2}] + 1/r*D[u[r, t], r]); 
ic  = {u[r, 0] == 1, Derivative[0, 1][u][r, 0] == r/3}; 
bc  = u[1, t] == 0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[r, t], {r, t}], 60*10]]; 
sol =  sol /. K[1] -> n; 
sol =  FullSimplify[sol];
 

$$\left\{\{u(r,t)\to \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2J_0\left(r{j}_{0,n}\right)(9\sqrt{c^2}{J}_1\left(j_{0,n}\right)\cos \left(ct{j}_{0,n}\right)+{}_1{F}_2(\frac{3}{2};1,\frac{5}{2};-\frac{1}{4}\left(j_{0,n}\right){}^2)\sin \left(\sqrt{c^2}t{j}_{0,n}\right))}{9\sqrt{c^2}(J_0\left(j_{0,n}\right){}^2+J_1\left(j_{0,n}\right){}^2)j_{0,n}}\}\right\}$$

Maple ✓

restart; 
pde := diff(u(r, t), t$2) = c^2*( diff(u(r,t), r$2)+ (1/r)* diff(u(r,t),r)  ); 
ic  := u(r,0)=1, eval( diff(u(r,t),t),t=0)=r/3; 
bc  := u(1,t)=0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic,bc], u(r, t)) assuming t>0,r>0,r<1),output='realtime'));
 

$$u(r,t)=-\mathit{i}\mathit{n}\mathit{v}\mathit{l}\mathit{a}\mathit{p}\mathit{l}\mathit{a}\mathit{c}\mathit{e}(\frac{1}{s}\mathrm{BesselI}(0,\frac{sr}{c}){\left(\mathrm{BesselI}(0,\frac{s}{c})\right)}^{-1},s,t)-1/3\mathit{i}\mathit{n}\mathit{v}\mathit{l}\mathit{a}\mathit{p}\mathit{l}\mathit{a}\mathit{c}\mathit{e}(\frac{1}{s^2}\mathrm{BesselI}(0,\frac{sr}{c}){\left(\mathrm{BesselI}(0,\frac{s}{c})\right)}^{-1},s,t)+1/6\pi (\mathit{i}\mathit{n}\mathit{v}\mathit{l}\mathit{a}\mathit{p}\mathit{l}\mathit{a}\mathit{c}\mathit{e}(\frac{1}{s^3}\mathrm{BesselI}(0,\frac{sr}{c})\mathrm{StruveL}(0,\frac{s}{c}){\left(\mathrm{BesselI}(0,\frac{s}{c})\right)}^{-1},s,t)-\mathit{i}\mathit{n}\mathit{v}\mathit{l}\mathit{a}\mathit{p}\mathit{l}\mathit{a}\mathit{c}\mathit{e}(\frac{1}{s^3}\mathrm{StruveL}(0,\frac{sr}{c}),s,t))c+1+1/3tr$$ Has unresolved Invlaplace calls

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5.2.2.3 [406] no $\theta$ dependency. Specific example. Both initial conditions not zero

problem number 406

Added January 12, 2020.

In circular disk. fixed edge of disk, no $\theta$ dependency, with initial position and velocity given

Solve for $u(r,t)$ with $0<r<a$ and $t>0$. $$u_{tt}=c^2(u_{rr}+\frac{1}{r}{u}_r)$$ With boundary conditions $$\begin{array}{rl}u(a,t)& =0\end{array}$$

With initial conditions $$\begin{array}{rl}u(r,0)& =f(r)\\ \frac{\partial u}{\partial t}(r,0)& =g(r)\end{array}$$

Using $a=1,c=\frac{2}{10},g(r)=0,f(r)=r$.

Figure 5.103: PDE specification

Mathematica ✓

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

$$\left\{\{u(r,t)\to \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{2J_0\left(r{j}_{0,n}\right)\cos \left(\frac{t{j}_{0,n}}{5}\right){}_1{F}_2(\frac{3}{2};1,\frac{5}{2};-\frac{1}{4}\left(j_{0,n}\right){}^2)}{3(J_0\left(j_{0,n}\right){}^2+J_1\left(j_{0,n}\right){}^2)}\}\right\}$$

Maple ✓

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

$$u(r,t)=\mathit{i}\mathit{n}\mathit{v}\mathit{l}\mathit{a}\mathit{p}\mathit{l}\mathit{a}\mathit{c}\mathit{e}(\frac{\mathit{l}\mathit{a}\mathit{p}\mathit{l}\mathit{a}\mathit{c}\mathit{e}(-1,t,s)\mathrm{BesselI}(0,5sr)}{\mathrm{BesselI}(0,5s)},s,t)+{\int }_0^{r-1}\mathit{i}\mathit{n}\mathit{v}\mathit{l}\mathit{a}\mathit{p}\mathit{l}\mathit{a}\mathit{c}\mathit{e}(\frac{\mathrm{BesselI}(0,5s(r-\tau ))}{\mathrm{BesselI}(0,5s)}\mathit{l}\mathit{a}\mathit{p}\mathit{l}\mathit{a}\mathit{c}\mathit{e}(-{(\tau +1)}^{-1},t,s),s,t)\mathrm{d}\tau +r$$ Has unresolved Invlaplace calls. How to get series solution?

Hand solution

The basic solution for this type of PDE was already given in problem 5.2.2.1 on page 1545 as

$$u(r,t)=\sum _{n=1}^{\mathrm{\infty }}{A}_n\cos \left(c\sqrt{{\lambda }_n}t\right)J_0\left(\sqrt{{\lambda }_n}r\right)+B_n\sin \left(c\sqrt{{\lambda }_n}t\right)J_0\left(\sqrt{{\lambda }_n}r\right)$$$$A_n=\frac{{\int }_0^{a}f\left(r\right)J_0\left(\sqrt{{\lambda }_n}r\right)rdr}{{\int }_0^a{J}_0^2\left(\sqrt{{\lambda }_n}r\right)rdr}$$$$B_n=\frac{{\int }_0^{a}g\left(r\right)J_0\left(\sqrt{{\lambda }_n}r\right)rdr}{c\sqrt{{\lambda }_n}{\int }_0^a{J}_0^2\left(\sqrt{{\lambda }_n}r\right)rdr}$$ With ${\lambda }_n$ being the solutions for $0=J_0\left(\sqrt{{\lambda }_n}a\right)$. We have infinite number of zeros. This generates all the needed ${\lambda }_n$. Hence $\sqrt{{\lambda }_n}a=BesselJZero(0,n)$, therefore $\sqrt{{\lambda }_n}=\frac{a}{BesselJZero(0,n)}$.

In this problem $c=\frac{2}{10},a=1,g\left(r\right)=0$ and $f\left(r\right)=r$, hence the solution becomes

$$u(r,t)=\sum _{n=1}^{\mathrm{\infty }}{A}_n\cos \left(\frac{2}{10}\sqrt{{\lambda }_n}t\right)J_0\left(\sqrt{{\lambda }_n}r\right)$$

Where $\sqrt{{\lambda }_n}=\frac{1}{BesselJZero(0,n)}$.

This animation runs for 40 seconds.

Source code for all the above animation

(*By Nasser M Abbasi*) 
SetDirectory[NotebookDirectory[]] 
 
(*Axis symmetric*) 
(*definitions*) 
ClearAll[a,c,n,m,r,theta,f,g,u] 
A0[n_,a_,lam0_]:= Module[{num,den,theta,r,f}, 
   f=r; 
   num=N[(BesselJ[1,lam0] (2 lam0-Pi StruveH[0,lam0])+ 
        Pi BesselJ[0,lam0] StruveH[1,lam0])/(2 lam0^2)]; 
 
    den=0.5 (BesselJ[0,lam0]^2+BesselJ[1,lam0]^2); 
    num/den 
]; 
 
u[r_,theta_,t_]:=Sum[A0tbl[[n]]*Cos[c lamtbl[[n]] t] *BesselJ[0,lamtbl[[n]]*r],{n,1,maxN}]; 
 
maxN=6; 
a=1; 
c=.2; 
 
lam[n_,a_]:=Module[{x}, 
   x=BesselJZero[0,n]; 
   N[(x/a)] 
]; 
 
 
lamtbl=Table[lam[n,a],{n,1,maxN}]; 
A0tbl=Table[A0[n,a,lamtbl[[n]]],{n,1,maxN}]; 
 
 
t=.1 
ParametricPlot3D[{r Cos[theta],r Sin[theta],Evaluate[u[r,theta,t]]},{r,0,1}, 
      {theta,0,2 Pi},AxesLabel->{"r","theta","u"},ImageMargins->5, 
      PerformanceGoal->"Quality",BoxRatios->{1,1,1}, 
      PlotRange->{Automatic,Automatic,{-5,5}},Mesh->10,MaxRecursion->1] 
 
Animate[ParametricPlot3D[{r Cos[theta],r Sin[theta], 
        Evaluate[u[r,theta,t]]},{r,0,1},{theta,0,2 Pi}, 
        AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, 
        PerformanceGoal->"Speed",BoxRatios->{1,1,1},PlotRange->{Automatic,Automatic,{-5,5}}, 
        Mesh->10,MaxRecursion->1],{t,0,50,.01}] 
 
r=Table[ 
      Labeled[ParametricPlot3D[{r Cos[theta],r Sin[theta], 
           Evaluate[u[r,theta,t]]},{r,0,1},{theta,0,2 Pi},AxesLabel->{"r","theta","u"}, 
           BaseStyle->15,ImageMargins->5,PerformanceGoal->"Speed",BoxRatios->{1,1,1}, 
           PlotRange->{Automatic,Automatic,{-5,5}},Mesh->10,MaxRecursion->1], 
           Row[{"time (sec)",Round@t}]],{t,0,40,.1}]; 
 
Export["anim_axis.gif",r,"DisplayDurations"->Table[.05,{Length[r]}]]

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5.2.2.4 [407] no $\theta$ dependency. Using integral transforms. Source present. Specific example

problem number 407

Added Oct 6, 2019.

Taken from https://www.mapleprimes.com/posts/211274-Integral-Transforms-revamped-And-PDE

Solve $$\frac{{\partial }^2}{\partial r^2}u(r,t)+\frac{\frac{\partial }{\partial r}u(r,t)}{r}+\frac{{\partial }^2}{\partial t^2}u(r,t)=-Q_{0}q\left(r\right)$$ With initial conditions $$\begin{array}{rl}u(r,0)& =0\end{array}$$

Mathematica ✗

ClearAll["Global`*"]; 
pde = D[u[r, t], {r, 2}] + D[u[r, t], r]/r + D[u[r, t], {t, 2}] == -Q0*q[r]; 
ic = u[r, 0] == 0; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic}, u[r, t], {r, t}], 60*10]];
 

Failed

Maple ✓

restart; 
pde := diff(u(r, t), r$2) + diff(u(r, t), r)/r + diff(u(r, t), t$2) = -Q__0*q(r); 
iv := u(r, 0) = 0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,iv],u(r,t))),output='realtime')); 
sol:=convert(sol,Int,only = hankel);
 

$$u(r,t)=Q_0(-{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-st}{\int }_0^{\mathrm{\infty }}q\left(r\right)r\mathrm{BesselJ}(0,sr)\mathrm{d}r\mathrm{BesselJ}(0,sr)}{s}\mathrm{d}s+{\int }_0^{\mathrm{\infty }}\frac{{\int }_0^{\mathrm{\infty }}q\left(r\right)r\mathrm{BesselJ}(0,sr)\mathrm{d}r\mathrm{BesselJ}(0,sr)}{s}\mathrm{d}s)$$

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5.2.2.5 [408] no $\theta$ dependency. Using integral transforms. Source present. Specific example

problem number 408

Added Oct 6, 2019.

Taken from https://www.mapleprimes.com/posts/211274-Integral-Transforms-revamped-And-PDE

Solve $$c^2(\frac{{\partial }^2}{\partial r^2}u(r,t)+\frac{\frac{\partial }{\partial r}u(r,t)}{r})=\frac{{\partial }^2}{\partial t^2}u(r,t)$$ With initial conditions $$\begin{array}{rl}u(r,0)& =\frac{Aa}{\sqrt{a^2+r^2}}\\ \frac{\partial u(r,0)}{\partial t}& =0\end{array}$$

Mathematica ✗

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

Failed

Maple ✓

restart; 
pde := c^2*(diff(u(r, t), r, r) + diff(u(r, t), r)/r) = diff(u(r, t), t, t); 
iv := u(r, 0) = A*a*(a^2 + r^2)^(-1/2), D[2](u)(r, 0) = 0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,iv],u(r,t),method = Hankel) assuming (0 < r, 0 < t, 0 < a) ),output='realtime'));
 

$$u(r,t)=1/2\frac{Aa(\sqrt{-c^2{t}^2+2iact+a^2+r^2}+\sqrt{-c^2{t}^2-2iact+a^2+r^2})}{\sqrt{-c^2{t}^2-2iact+a^2+r^2}\sqrt{-c^2{t}^2+2iact+a^2+r^2}}$$

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5.2.2.6 [409] $\theta$ dependency, fixed on edges, general solution

problem number 409

Added January 11, 2020

Solve for $u(r,\theta ,t)$ with $0<r<a$ and $t>0$ and $-\pi <\theta <\pi$ $$\frac{{\partial }^{2}u}{\partial t^2}=c^2(\frac{{\partial }^{2}u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{{\partial }^{2}u}{\partial {\theta }^2})$$ With boundary conditions $$\begin{array}{rl}u(a,\theta ,t)& =0\\ |u(0,\theta ,t)|<\mathrm{\infty }\\ u(r,-\pi ,t)& =u(r,\pi ,t)\\ \frac{\partial u}{\partial \theta }(r,-\pi ,t)& =\frac{\partial u}{\partial \theta }(r,\pi ,t)\end{array}$$

With initial conditions $$\begin{array}{rl}u(r,\theta ,0)& =f(r,\theta )\\ \frac{\partial u}{\partial t}(r,\theta ,0)& =g(r,\theta )\end{array}$$

Figure 5.104: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[r, theta, t], {t, 2}] == c^2*Laplacian[u[r,theta,t],{r,theta},"Polar"]; 
ic  = {u[r, theta, 0] == f[r, theta], Derivative[0, 0, 1][u][r, theta, 0] == g[r,theta]}; 
bc  = {u[a, theta, t] == 0, u[r, -Pi, t] == u[r, Pi, t], Derivative[0, 1, 0][u][r, -Pi, t] == Derivative[0, 1, 0][u][r, Pi, t]}; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[r, theta, t], {r, theta, t}, Assumptions -> {0 < r < a, a > 0, t > 0, -Pi < theta < Pi}], 60*10]];
 

$$\left\{\{u(r,\theta ,t)\to \begin{array}{cc}\{& \begin{array}{cc}\underset{K[3]=1}{\sum ^{\mathrm{\infty }}}-\frac{\sqrt{\frac{2}{\pi }}{J}_0\left(\frac{r{j}_{0,K[3]}}{a}\right)(-\frac{\sqrt{\frac{2}{\pi }}\cos \left(\frac{\sqrt{c^2}t{j}_{0,K[3]}}{a}\right){\int }_0^a{\int }_{-\pi }^{\pi }r{J}_0\left(\frac{r{j}_{0,K[3]}}{a}\right)f(r,\theta )d\theta dr}{a{J}_1\left(j_{0,K[3]}\right)}-\frac{\sqrt{\frac{2}{\pi }}({\int }_0^a{\int }_{-\pi }^{\pi }r{J}_0\left(\frac{r{j}_{0,K[3]}}{a}\right)g(r,\theta )d\theta dr)\sin \left(\frac{\sqrt{c^2}t{j}_{0,K[3]}}{a}\right)}{|c|J_1\left(j_{0,K[3]}\right)j_{0,K[3]}})}{a{J}_1\left(j_{0,K[3]}\right)}+\underset{K[3]=1}{\sum ^{\mathrm{\infty }}}\left(\underset{K[1]=1}{\sum ^{\mathrm{\infty }}}(\frac{\sqrt{\frac{2}{\pi }}{J}_{K[1]}\left(\frac{r{j}_{K[1],K[3]}}{a}\right)\cos (\theta K[1])(\frac{\sqrt{\frac{2}{\pi }}\cos \left(\frac{\sqrt{c^2}t{j}_{K[1],K[3]}}{a}\right){\int }_0^a{\int }_{-\pi }^{\pi }r{J}_{K[1]}\left(\frac{r{j}_{K[1],K[3]}}{a}\right)\cos (\theta K[1])f(r,\theta )d\theta dr}{a{J}_{K[1]-1}\left(j_{K[1],K[3]}\right)}+\frac{\sqrt{\frac{2}{\pi }}({\int }_0^a{\int }_{-\pi }^{\pi }r{J}_{K[1]}\left(\frac{r{j}_{K[1],K[3]}}{a}\right)\cos (\theta K[1])g(r,\theta )d\theta dr)\sin \left(\frac{\sqrt{c^2}t{j}_{K[1],K[3]}}{a}\right)}{|c|J_{K[1]-1}\left(j_{K[1],K[3]}\right)j_{K[1],K[3]}})}{a{J}_{K[1]-1}\left(j_{K[1],K[3]}\right)}+\frac{\sqrt{\frac{2}{\pi }}{J}_{K[1]}\left(\frac{r{j}_{K[1],K[3]}}{a}\right)(\frac{\sqrt{\frac{2}{\pi }}\cos \left(\frac{\sqrt{c^2}t{j}_{K[1],K[3]}}{a}\right){\int }_0^a{\int }_{-\pi }^{\pi }r{J}_{K[1]}\left(\frac{r{j}_{K[1],K[3]}}{a}\right)f(r,\theta )\sin (\theta K[1])d\theta dr}{a{J}_{K[1]-1}\left(j_{K[1],K[3]}\right)}+\frac{\sqrt{\frac{2}{\pi }}({\int }_0^a{\int }_{-\pi }^{\pi }r{J}_{K[1]}\left(\frac{r{j}_{K[1],K[3]}}{a}\right)g(r,\theta )\sin (\theta K[1])d\theta dr)\sin \left(\frac{\sqrt{c^2}t{j}_{K[1],K[3]}}{a}\right)}{|c|J_{K[1]-1}\left(j_{K[1],K[3]}\right)j_{K[1],K[3]}})\sin (\theta K[1])}{a{J}_{K[1]-1}\left(j_{K[1],K[3]}\right)})\right)& (K[1]|K[3])\in \mathbb{Z}\wedge K[1]\ge 1\wedge K[3]\ge 1\wedge c^2\left(j_{K[1],K[3]}\right){}^2>0\\ \text{Indeterminate}& \text{True}\end{array}\end{array}\}\right\}$$

Maple ✗

restart; 
pde := diff(u(r, theta, t), t$2) = c^2*VectorCalculus:-Laplacian(u(r,theta,t),'polar'[r,theta]); 
ic  := u(r, theta, 0) = f(r, theta) , (D[3](u))(r, theta, 0) = g(r,theta); 
bc  := u(a, theta, t) = 0, 
       u(r, -Pi, t) = u(r, Pi, t), 
       (D[2](u))(r, -Pi, t) = (D[2](u))(r, Pi, t); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic,bc], u(r, theta ,t),HINT = boundedseries(r=0))),output='realtime'));
 

sol=()

Hand solution

Assuming $u=T\left(t\right)R\left(r\right)\mathrm{\Theta }\left(\theta \right)$ and substituting in the PDE gives$$\frac{1}{c^2}{T}^{\prime \prime }R\mathrm{\Theta }=R^{\prime \prime }T\mathrm{\Theta }+\frac{1}{r}{R}^{\prime }T\mathrm{\Theta }+\frac{1}{r^2}{\mathrm{\Theta }}^{\prime \prime }RT$$ Dividing by $RT\mathrm{\Theta }$$$\frac{1}{c^2}\frac{T^{\prime \prime }}{T}=\frac{R^{\prime \prime }}{R}+\frac{1}{r}\frac{R^{\prime }}{R}+\frac{1}{r^2}\frac{{\mathrm{\Theta }}^{\prime \prime }}{\mathrm{\Theta }}$$ Hence$$\begin{array}{rl}\frac{1}{c^2}\frac{T^{\prime \prime }}{T}& =-\lambda \\ \frac{R^{\prime \prime }}{R}+\frac{1}{r}\frac{R^{\prime }}{R}+\frac{1}{r^2}\frac{{\mathrm{\Theta }}^{\prime \prime }}{\mathrm{\Theta }}& =-\lambda \end{array}$$

The time ODE is$$T^{\prime \prime }+c^2\lambda T=0$$ Now we separate again the space ODE’s (remember to move the $\lambda$ with the $R$ and not the $\mathrm{\Theta }$)$$\begin{array}{rl}\frac{R^{\prime \prime }}{R}+\frac{1}{r}\frac{R^{\prime }}{R}+\lambda & =-\frac{1}{r^2}\frac{{\mathrm{\Theta }}^{\prime \prime }}{\mathrm{\Theta }}\\ r^2\frac{R^{\prime \prime }}{R}+r\frac{R^{\prime }}{R}+r^2\lambda & =-\frac{{\mathrm{\Theta }}^{\prime \prime }}{\mathrm{\Theta }}\end{array}$$

Let the new separation constant be $\mu$, therefore$$\begin{array}{rl}-\frac{{\mathrm{\Theta }}^{\prime \prime }}{\mathrm{\Theta }}& =\mu \\ {\mathrm{\Theta }}^{\prime \prime }+\mu \mathrm{\Theta }& =0\end{array}$$

With periodic boundary conditions and$$\begin{array}{rl}{r}^2\frac{R^{\prime \prime }}{R}+r\frac{R^{\prime }}{R}+r^2\lambda & =\mu \\ r^2{R}^{\prime \prime }+r{R}^{\prime }+\lambda r^{2}R-\mu R& =0\\ r{R}^{\prime \prime }+R^{\prime }-\frac{\mu }{r}R& =-\lambda rR\end{array}$$

Now it is in Sturm Liouville form, where $p=r,q=-\frac{\mu }{r},\sigma =r$. This is singular SL. Can be written as$$R^{\prime \prime }+\frac{1}{r}{R}^{\prime }+(\lambda -\frac{\mu }{r^2})R=0$$ Before we solve the above $R$ ODE, we solve the ${\mathrm{\Theta }}^{\prime \prime }+\mu \mathrm{\Theta }=0$ to find $\mu$ Eigenvalues. The solution is$$\mathrm{\Theta }=A\cos \left(\sqrt{\mu }\theta \right)+B\sin \left(\sqrt{\mu }\theta \right)$$ With B.C $\mathrm{\Theta }(-\pi )=\mathrm{\Theta }\left(\pi \right)$ and ${\mathrm{\Theta }}^{\prime }(-\pi )={\mathrm{\Theta }}^{\prime }\left(\pi \right)$. From first B.C. we obtain$$\begin{array}{rl}A\cos \left(\sqrt{\mu }\pi \right)-B\sin \left(\sqrt{\mu }\pi \right)& =A\cos \left(\sqrt{\mu }\pi \right)+B\sin \left(\sqrt{\mu }\pi \right)\\ \text{(1)}& 2B\sin \left(\sqrt{\mu }\pi \right)& =0\end{array}$$

Looking at second B.C. ${\mathrm{\Theta }}^{\prime }(-\pi )={\mathrm{\Theta }}^{\prime }\left(\pi \right)$$${\mathrm{\Theta }}^{\prime }\left(\theta \right)=-A\sqrt{\mu }\sin \left(\sqrt{\mu }\theta \right)+\sqrt{\mu }B\cos \left(\sqrt{\mu }\theta \right)$$ Hence$$\begin{array}{rl}A\sqrt{\mu }\sin \left(\sqrt{\mu }\pi \right)+\sqrt{\mu }B\cos \left(\sqrt{\mu }\pi \right)& =-A\sqrt{\mu }\sin \left(\sqrt{\mu }\pi \right)+\sqrt{\mu }B\cos \left(\sqrt{\mu }\pi \right)\\ A\sqrt{\mu }\sin \left(\sqrt{\mu }\pi \right)& =-A\sqrt{\mu }\sin \left(\sqrt{\mu }\pi \right)\\ \text{(2)}& 2A\sin \left(\sqrt{\mu }\pi \right)& =0\end{array}$$

From (1,2), we see that both are satisfied if $$\begin{array}{rl}\sqrt{\mu }\pi & =n\pi n=1,2,3,\dots \\ \mu & =n^2\end{array}$$

Hence $${\mathrm{\Theta }}_n=A_n\cos \left(n\theta \right)+B_n\sin \left(n\theta \right)$$ There is another solution for $\mu =0$ which is constant (that is why one of the sums below starts from $n=0$). We can combine the zero eigenvalue with the above and write$${\mathrm{\Theta }}_n=A_n\cos \left(n\theta \right)+B_n\sin \left(n\theta \right)n=0,1,2,3,\dots$$ Since at $n=0$ the above reduces to constant $A_0$.

Now that we know ${\mu }_n=n^2$, from solving the $\theta$ part, we go and solve the $r$ ODE. For each $n$, the solution to the $r$ (Bessel) ode

$$R^{\prime \prime }+\frac{1}{r}{R}^{\prime }+(\lambda -\frac{n^2}{r^2})R=0$$ The solution turns out to be  $$R_{nm}\left(r\right)=J_n\left(\sqrt{{\lambda }_{nm}}r\right)m=1,2,3,\cdots$$ Where ${\lambda }_{nm}$ is found from roots of $0=J_n\left(\sqrt{{\lambda }_{nm}}a\right)$ giving the eigenvalues. Now the time ODE is solved$$\begin{array}{rl}{T}_{nm}^{\prime \prime }+c^2{\lambda }_{nm}{T}_{nm}& =0\\ T_{nm}& =C_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)+D_{nm}\sin \left(c\sqrt{{\lambda }_{nm}}t\right)n=0,1,2,3,\dots ,m=1,2,3,\cdots \end{array}$$

Hence the solution is$$\begin{array}{rl}u(r,\theta ,t)& =\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{T}_{nm}{R}_{nm}{\mathrm{\Theta }}_n\\ & =\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}(C_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)+D_{nm}\sin \left(c\sqrt{{\lambda }_{nm}}t\right))J_n\left(\sqrt{{\lambda }_{nm}}r\right)(A_n\cos \left(n\theta \right)+B_n\sin \left(n\theta \right))\end{array}$$

We now break this sum as follows$$\begin{array}{rl}u(r,\theta ,t)& =\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}(C_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)+D_{nm}\sin \left(c\sqrt{{\lambda }_{nm}}t\right))J_n\left(\sqrt{{\lambda }_{nm}}r\right)A_n\cos \left(n\theta \right)\\ & +\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}(C_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)+D_{nm}\sin \left(c\sqrt{{\lambda }_{nm}}t\right))J_n\left(\sqrt{{\lambda }_{nm}}r\right)B_n\sin \left(n\theta \right)\end{array}$$

Or$$\begin{array}{rl}u(r,\theta ,t)& =\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{C}_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)A_n\cos \left(n\theta \right)+D_{nm}\sin \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)A_n\cos \left(n\theta \right)\\ & +\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{C}_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)B_n\sin \left(n\theta \right)+D_{nm}\sin \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)B_n\sin \left(n\theta \right)\end{array}$$

Then we break the above into 4 sums$$\begin{array}{rl}u(r,\theta ,t)& =\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{C}_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)A_n\cos \left(n\theta \right)\\ & +\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{D}_{nm}\sin \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)A_n\cos \left(n\theta \right)\\ & +\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{C}_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)B_n\sin \left(n\theta \right)\\ & +\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{D}_{nm}\sin \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)B_n\sin \left(n\theta \right)\end{array}$$

Finally, we merge constants in the above as follows$$\begin{array}{rl}{A}_n{C}_{nm}& \equiv A_{nm}\\ A_n{D}_{nm}& \equiv B_{nm}\\ B_n{C}_{nm}& \equiv C_{nm}\\ B_n{D}_{nm}& \equiv D_{nm}\end{array}$$

Hence the final solution now becomes$$\begin{array}{rl}u(r,\theta ,t)& =\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{A}_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\cos \left(n\theta \right)\\ & +\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{B}_{nm}\sin \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\cos \left(n\theta \right)\\ & +\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{C}_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\sin \left(n\theta \right)\\ \text{(3)}& & +\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{D}_{nm}\sin \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\sin \left(n\theta \right)\end{array}$$

Now initial conditions $u(r,\theta ,0)=f(r,\theta )$ is used to find $A_{nm},C_{nm}$using orthogonality. At $t=0$ the solution simplifies to (all terms with $\sin \left(c\sqrt{{\lambda }_{nm}}t\right)$ vanish giving$$\begin{array}{rl}u(r,\theta ,t)& =\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{A}_{nm}{J}_n\left(\sqrt{{\lambda }_{nm}}r\right)\cos \left(n\theta \right)\\ & +\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{C}_{nm}{J}_n\left(\sqrt{{\lambda }_{nm}}r\right)\sin \left(n\theta \right)\end{array}$$

Hence$$\begin{array}{}\text{(4)}& f(r,\theta )=\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{A}_{nm}{J}_n\left(\sqrt{{\lambda }_{nm}}r\right)\cos \left(n\theta \right)+\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{C}_{nm}{J}_n\left(\sqrt{{\lambda }_{nm}}r\right)\sin \left(n\theta \right)\end{array}$$ When iterating over $m$ index, the terms $\cos \left(n\theta \right)$ and $\sin \left(n\theta \right)$ will be constant. So for each $n$, we have $\sum _{m=1}^{\mathrm{\infty }}{A}_{nm}{J}_n\left(\sqrt{{\lambda }_{nm}}r\right)$ and $\sum _{m=1}^{\mathrm{\infty }}{C}_{nm}{J}_n\left(\sqrt{{\lambda }_{nm}}r\right)$. So orthogonality is carried out on the $m$ index on the Bessel functions. Multiplying (4) by $J_n\left(\sqrt{{\lambda }_{nk}}r\right)$ and integrating$$\begin{array}{rl}{\int }_0^{a}f(r,\theta )J_n\left(\sqrt{{\lambda }_{nk}}r\right)rdr& =\sum _{n=0}^{\mathrm{\infty }}\left({\int }_0^a\sum _{m=1}^{\mathrm{\infty }}{A}_{nm}{J}_n\left(\sqrt{{\lambda }_{nm}}r\right)J_n\left(\sqrt{{\lambda }_{nk}}r\right)rdr\right)\cos \left(n\theta \right)\\ & +\sum _{n=1}^{\mathrm{\infty }}\left({\int }_0^a\sum _{m=1}^{\mathrm{\infty }}{C}_{nm}{J}_n\left(\sqrt{{\lambda }_{nm}}r\right)J_n\left(\sqrt{{\lambda }_{nk}}r\right)rdr\right)\sin \left(n\theta \right)\end{array}$$

Or$$\begin{array}{rl}{\int }_0^{a}f(r,\theta )J_n\left(\sqrt{{\lambda }_{nk}}r\right)rdr& =\sum _{n=0}^{\mathrm{\infty }}{A}_{nk}\left({\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nk}}r\right)rdr\right)\cos \left(n\theta \right)\\ & +\sum _{n=1}^{\mathrm{\infty }}{C}_{nk}\left({\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nk}}r\right)rdr\right)\sin \left(n\theta \right)\end{array}$$

Replacing $k$ back with $m$, the above becomes$$\begin{array}{rl}{\int }_0^{a}f(r,\theta )J_n\left(\sqrt{{\lambda }_{nm}}r\right)rdr& =\sum _{n=0}^{\mathrm{\infty }}{A}_{nm}\left({\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr\right)\cos \left(n\theta \right)\\ \text{(5)}& & +\sum _{n=1}^{\mathrm{\infty }}{C}_{nm}\left({\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr\right)\sin \left(n\theta \right)\end{array}$$

We now apply orthogonality on $n$ using the $\cos \left(n\theta \right)$ results in$${\int }_{-\pi }^{\pi }\left({\int }_0^{a}f(r,\theta )J_n\left(\sqrt{{\lambda }_{nm}}r\right)rdr\right)\cos \left(n\theta \right)d\theta =A_{nm}{\int }_{-\pi }^{\pi }\left({\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr\right){\cos }^2\left(n\theta \right)d\theta$$ But ${\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr$ does not depend on $\theta$, therefore the above becomes$$\begin{array}{rl}{\int }_{-\pi }^{\pi }\left({\int }_0^{a}f(r,\theta )J_n\left(\sqrt{{\lambda }_{nm}}r\right)rdr\right)\cos \left(n\theta \right)d\theta & =A_{nm}\left({\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr\right){\int }_{-\pi }^{\pi }{\cos }^2\left(n\theta \right)d\theta \\ & =A_{nm}\pi {\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr\end{array}$$

Therefore$$A_{nm}=\frac{{\int }_{-\pi }^{\pi }\left({\int }_0^{a}f(r,\theta )J_n\left(\sqrt{{\lambda }_{nm}}r\right)rdr\right)\cos \left(n\theta \right)d\theta }{\pi {\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr}$$ Similarly for $\sin \left(n\theta \right)$, which gives$$C_{nm}=\frac{{\int }_{-\pi }^{\pi }\left({\int }_0^{a}f(r,\theta )J_n\left(\sqrt{{\lambda }_{nm}}r\right)rdr\right)\sin \left(n\theta \right)d\theta }{\pi {\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr}$$ Now we will look at the second initial conditions $\frac{\partial u}{\partial t}(r,\theta ,0)=g(r,\theta ).$ Taking derivative w.r.t. time $t$ of the solution in (3) gives$$\begin{array}{rl}\frac{\partial u}{\partial t}(r,\theta ,t)& =\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}-c\sqrt{{\lambda }_{nm}}{A}_{nm}\sin \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\cos \left(n\theta \right)\\ & +\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}c\sqrt{{\lambda }_{nm}}{B}_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\cos \left(n\theta \right)\\ & +\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}-c\sqrt{{\lambda }_{nm}}{C}_{nm}\sin \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\sin \left(n\theta \right)\\ & +\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}c\sqrt{{\lambda }_{nm}}{D}_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\sin \left(n\theta \right)\end{array}$$

At time $t=0$ the above becomes (all terms with $\sin \left(c\sqrt{{\lambda }_{nm}}t\right)$ vanish).$$\begin{array}{rl}g(r,\theta )=& \sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}c\sqrt{{\lambda }_{nm}}{B}_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\cos \left(n\theta \right)\\ & +\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}c\sqrt{{\lambda }_{nm}}{D}_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\sin \left(n\theta \right)\end{array}$$

Now orthogonality is used. At $t=0$ the above becomes$$\begin{array}{rl}g(r,\theta )& =\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}c\sqrt{{\lambda }_{nm}}{B}_{nm}{J}_n\left(\sqrt{{\lambda }_{nm}}r\right)\cos \left(n\theta \right)\\ & +\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}c\sqrt{{\lambda }_{nm}}{D}_{nm}{J}_n\left(\sqrt{{\lambda }_{nm}}r\right)\sin \left(n\theta \right)\end{array}$$

Similarly to the above we now find $B_{nm}$ and $D_{nm}$. The only difference, is that now we have extra $c\sqrt{{\lambda }_{nm}}$ terms that show up. The final result will be$$B_{nm}=\frac{{\int }_{-\pi }^{\pi }{\int }_0^{a}g(r,\theta )J_n\left(\sqrt{{\lambda }_{nm}}r\right)\cos \left(n\theta \right)rd\theta dr}{c\pi \sqrt{{\lambda }_{nm}}{\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr}$$ And$$D_{nm}=\frac{{\int }_{-\pi }^{\pi }{\int }_0^{a}g(r,\theta )J_n\left(\sqrt{{\lambda }_{nm}}r\right)\sin \left(n\theta \right)rd\theta dr}{c\pi \sqrt{{\lambda }_{nm}}{\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr}$$ Summary of solution$$\begin{array}{rl}u(r,\theta ,t)& =\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{A}_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\cos \left(n\theta \right)\\ & +\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{B}_{nm}\sin \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\cos \left(n\theta \right)\\ & +\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{C}_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\sin \left(n\theta \right)\\ & +\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{D}_{nm}\sin \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\sin \left(n\theta \right)\end{array}$$

$$A_{nm}=\frac{{\int }_{-\pi }^{\pi }\left({\int }_0^{a}f(r,\theta )J_n\left(\sqrt{{\lambda }_{nm}}r\right)rdr\right)\cos \left(n\theta \right)d\theta }{\pi {\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr}$$$$C_{nm}=\frac{{\int }_{-\pi }^{\pi }\left({\int }_0^{a}f(r,\theta )J_n\left(\sqrt{{\lambda }_{nm}}r\right)rdr\right)\sin \left(n\theta \right)d\theta }{\pi {\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr}$$$$B_{nm}=\frac{{\int }_{-\pi }^{\pi }{\int }_0^{a}g(r,\theta )J_n\left(\sqrt{{\lambda }_{nm}}r\right)\cos \left(n\theta \right)rd\theta dr}{c\pi \sqrt{{\lambda }_{nm}}{\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr}$$$$D_{nm}=\frac{{\int }_{-\pi }^{\pi }{\int }_0^{a}g(r,\theta )J_n\left(\sqrt{{\lambda }_{nm}}r\right)\sin \left(n\theta \right)rd\theta dr}{c\pi \sqrt{{\lambda }_{nm}}{\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr}$$ With ${\lambda }_{nm}$ being the solutions for $0=J_n\left(\sqrt{{\lambda }_{nm}}a\right)$. For each $n$, we find ${\lambda }_{n,1},{\lambda }_{n,2},{\lambda }_{n,3},\cdots$, which are the zeros of the Bessel $J_n\left(x\right)$ function. So for each $n$, there are infinite number of zeros. This generates all eigenvalues ${\lambda }_{nm}$. Hence $\sqrt{{\lambda }_{nm}}a=BesselJZero(n,m)$, therefore $\sqrt{{\lambda }_{nm}}=\frac{a}{BesselJZero(n,m)}$

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5.2.2.7 [410] $\theta$ dependency, fixed on edges, zero initial velocity, general solution

problem number 410

Solve for $u(r,\theta ,t)$ with $0<r<a$ and $t>0$ and $-\pi <\theta <\pi$ $$\frac{{\partial }^{2}u}{\partial t^2}=c^2(\frac{{\partial }^{2}u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{{\partial }^{2}u}{\partial {\theta }^2})$$ With boundary conditions $$\begin{array}{rl}u(a,\theta ,t)& =0\\ |u(0,\theta ,t)|<\mathrm{\infty }\\ u(r,-\pi ,t)& =u(r,\pi ,t)\\ \frac{\partial u}{\partial \theta }(r,-\pi ,t)& =\frac{\partial u}{\partial \theta }(r,\pi ,t)\end{array}$$

With initial conditions $$\begin{array}{rl}u(r,\theta ,0)& =f(r,\theta )\\ \frac{\partial u}{\partial t}(r,\theta ,0)& =0\end{array}$$

Figure 5.105: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
pde =  D[u[r, theta, t], {t, 2}] == c^2*Laplacian[u[r,theta,t],{r,theta},"Polar"]; 
ic  = {u[r, theta, 0] == f[r, theta], Derivative[0, 0, 1][u][r, theta, 0] == 0}; 
bc  = {u[a, theta, t] == 0, u[r, -Pi, t] == u[r, Pi, t], Derivative[0, 1, 0][u][r, -Pi, t] == Derivative[0, 1, 0][u][r, Pi, t]}; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[r, theta, t], {r, theta, t}, Assumptions -> {0 < r < a, a > 0, t > 0, -Pi < theta < Pi}], 60*10]];
 

$$\left\{\{u(r,\theta ,t)\to \begin{array}{cc}\{& \begin{array}{cc}\underset{K[3]=1}{\sum ^{\mathrm{\infty }}}\frac{2J_0\left(\frac{r{j}_{0,K[3]}}{a}\right)\cos \left(\frac{\sqrt{c^2}t{j}_{0,K[3]}}{a}\right){\int }_0^a{\int }_{-\pi }^{\pi }r{J}_0\left(\frac{r{j}_{0,K[3]}}{a}\right)f(r,\theta )d\theta dr}{a^2\pi J_1\left(j_{0,K[3]}\right){}^2}+\underset{K[3]=1}{\sum ^{\mathrm{\infty }}}\left(\underset{K[1]=1}{\sum ^{\mathrm{\infty }}}(\frac{2J_{K[1]}\left(\frac{r{j}_{K[1],K[3]}}{a}\right)\cos \left(\frac{\sqrt{c^2}t{j}_{K[1],K[3]}}{a}\right)\cos (\theta K[1]){\int }_0^a{\int }_{-\pi }^{\pi }r{J}_{K[1]}\left(\frac{r{j}_{K[1],K[3]}}{a}\right)\cos (\theta K[1])f(r,\theta )d\theta dr}{a^2\pi J_{K[1]-1}\left(j_{K[1],K[3]}\right){}^2}+\frac{2J_{K[1]}\left(\frac{r{j}_{K[1],K[3]}}{a}\right)\cos \left(\frac{\sqrt{c^2}t{j}_{K[1],K[3]}}{a}\right)({\int }_0^a{\int }_{-\pi }^{\pi }r{J}_{K[1]}\left(\frac{r{j}_{K[1],K[3]}}{a}\right)f(r,\theta )\sin (\theta K[1])d\theta dr)\sin (\theta K[1])}{a^2\pi J_{K[1]-1}\left(j_{K[1],K[3]}\right){}^2})\right)& (K[1]|K[3])\in \mathbb{Z}\wedge K[1]\ge 1\wedge K[3]\ge 1\wedge c^2\left(j_{K[1],K[3]}\right){}^2>0\\ \text{Indeterminate}& \text{True}\end{array}\end{array}\}\right\}$$

Maple ✗

restart; 
pde := diff(u(r, theta, t), t$2) =c^2*VectorCalculus:-Laplacian(u(r,theta,t),'polar'[r,theta]); 
ic  := u(r, theta, 0) = f(r, theta) , (D[3](u))(r, theta, 0) = 0; 
bc  := u(a, theta, t) = 0, 
       u(r, -Pi, t) = u(r, Pi, t), 
       (D[2](u))(r, -Pi, t) = (D[2](u))(r, Pi, t); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic,bc], u(r, theta ,t),HINT = boundedseries(r=0))),output='realtime'));
 

sol=()

Hand solution

The basic solution for this type of PDE was already given in problem 5.2.2.6 on page 1557 as$$\begin{array}{rl}u(r,\theta ,t)& =\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{A}_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\cos \left(n\theta \right)\\ & +\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{B}_{nm}\sin \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\cos \left(n\theta \right)\\ & +\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{C}_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\sin \left(n\theta \right)\\ & +\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{D}_{nm}\sin \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\sin \left(n\theta \right)\end{array}$$

$$A_{nm}=\frac{{\int }_{-\pi }^{\pi }\left({\int }_0^{a}f(r,\theta )J_n\left(\sqrt{{\lambda }_{nm}}r\right)rdr\right)\cos \left(n\theta \right)d\theta }{\pi {\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr}$$$$C_{nm}=\frac{{\int }_{-\pi }^{\pi }\left({\int }_0^{a}f(r,\theta )J_n\left(\sqrt{{\lambda }_{nm}}r\right)rdr\right)\sin \left(n\theta \right)d\theta }{\pi {\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr}$$$$B_{nm}=\frac{{\int }_{-\pi }^{\pi }{\int }_0^{a}g(r,\theta )J_n\left(\sqrt{{\lambda }_{nm}}r\right)\cos \left(n\theta \right)rd\theta dr}{c\pi \sqrt{{\lambda }_{nm}}{\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr}$$$$D_{nm}=\frac{{\int }_{-\pi }^{\pi }{\int }_0^{a}g(r,\theta )J_n\left(\sqrt{{\lambda }_{nm}}r\right)\sin \left(n\theta \right)rd\theta dr}{c\pi \sqrt{{\lambda }_{nm}}{\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr}$$ With ${\lambda }_{nm}$ being the solutions for $0=J_n\left(\sqrt{{\lambda }_{nm}}a\right)$. For each $n$, we find ${\lambda }_{n,1},{\lambda }_{n,2},{\lambda }_{n,3},\cdots$, which are the zeros of the Bessel $J_n\left(x\right)$ function. So for each $n$, there are infinite number of zeros. This generates all eigenvalues ${\lambda }_{nm}$. Hence $\sqrt{{\lambda }_{nm}}a=BesselJZero(n,m)$, therefore $\sqrt{{\lambda }_{nm}}=\frac{a}{BesselJZero(n,m)}$. Since $g(r,\theta )=0$ in this case, then $B_{nm}=0,D_{nm}=0$ and the solution simplifies to

$$u(r,\theta ,t)=\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{A}_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\cos \left(n\theta \right)+\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{C}_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\sin \left(n\theta \right)$$

____________________________________________________________________________________

5.2.2.8 [411] $\theta$ dependency, fixed on edges, zero initial velocity, specific example

problem number 411

Added January 11 2020.

Solve for $u(r,\theta ,t)$ with $0<r<a$ and $t>0$ and $-\pi <\theta <\pi$ $$\frac{{\partial }^{2}u}{\partial t^2}=c^2(\frac{{\partial }^{2}u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{{\partial }^{2}u}{\partial {\theta }^2})$$ With boundary conditions $$\begin{array}{rl}u(a,\theta ,t)& =0\\ |u(0,\theta ,t)|<\mathrm{\infty }\\ u(r,-\pi ,t)& =u(r,\pi ,t)\\ \frac{\partial u}{\partial \theta }(r,-\pi ,t)& =\frac{\partial u}{\partial \theta }(r,\pi ,t)\end{array}$$

With initial conditions $$\begin{array}{rl}u(r,\theta ,0)& =f(r,\theta )\\ \frac{\partial u}{\partial t}(r,\theta ,0)& =0\end{array}$$

Using $a=1,c=0.2,f(r,\theta )=r\theta$.

Figure 5.106: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
f[r_,theta_]:=r*theta; 
c=2/10; a=1; 
pde =  D[u[r, theta, t], {t, 2}] == c^2*Laplacian[u[r,theta,t],{r,theta},"Polar"]; 
ic  = {u[r, theta, 0] == f[r, theta], Derivative[0, 0, 1][u][r, theta, 0] == 0}; 
bc  = {u[a, theta, t] == 0, u[r, -Pi, t] == u[r, Pi, t], Derivative[0, 1, 0][u][r, -Pi, t] == Derivative[0, 1, 0][u][r, Pi, t]}; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[r, theta, t], {r, theta, t}, Assumptions -> {0 < r < a,t > 0, -Pi < theta < Pi}], 60*10]];
 

$$\left\{\{u(r,\theta ,t)\to \begin{array}{cc}\{& \begin{array}{cc}\underset{K[3]=1}{\sum ^{\mathrm{\infty }}}\left(\underset{K[1]=1}{\sum ^{\mathrm{\infty }}}\frac{(-1{)}^{K[1]+1}{J}_{K[1]}\left(r{j}_{K[1],K[3]}\right)j_{K[1],K[3]}\cos \left(\frac{1}{5}t{j}_{K[1],K[3]}\right)\text{Gamma}(\frac{1}{2}(K[1]+3)){}_1{\tilde{F}}_2(\frac{1}{2}(K[1]+3);\frac{1}{2}(K[1]+5),K[1]+1;-\frac{1}{4}\left(j_{K[1],K[3]}\right){}^2)\sin (\theta K[1])}{J_{K[1]-1}\left(j_{K[1],K[3]}\right){}_0{\tilde{F}}_1(;K[1];-\frac{1}{4}\left(j_{K[1],K[3]}\right){}^2)K[1]}\right)& (K[1]|K[3])\in \mathbb{Z}\wedge K[1]\ge 1\wedge K[3]\ge 1\\ \text{Indeterminate}& \text{True}\end{array}\end{array}\}\right\}$$

Maple ✗

restart; 
f:=(r,theta)->r*theta; 
c:=2/10; 
a:=1; 
pde := diff(u(r, theta, t), t$2) = c^2*VectorCalculus:-Laplacian(u(r,theta,t),'polar'[r,theta]); 
ic  := u(r, theta, 0) = f(r, theta) , (D[3](u))(r, theta, 0) = 0; 
bc  := u(a, theta, t) = 0, 
       u(r, -Pi, t) = u(r, Pi, t), 
       (D[2](u))(r, -Pi, t) = (D[2](u))(r, Pi, t); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic,bc], u(r, theta ,t),HINT = boundedseries(r=0))),output='realtime'));
 

sol=()

Hand solution

The basic solution for this type of PDE was already given in problem 5.2.2.6 on page 1557 as$$\begin{array}{rl}u(r,\theta ,t)& =\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{A}_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\cos \left(n\theta \right)\\ & +\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{B}_{nm}\sin \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\cos \left(n\theta \right)\\ & +\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{C}_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\sin \left(n\theta \right)\\ & +\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{D}_{nm}\sin \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\sin \left(n\theta \right)\end{array}$$

$$A_{nm}=\frac{{\int }_{-\pi }^{\pi }\left({\int }_0^{a}f(r,\theta )J_n\left(\sqrt{{\lambda }_{nm}}r\right)rdr\right)\cos \left(n\theta \right)d\theta }{\pi {\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr}$$$$C_{nm}=\frac{{\int }_{-\pi }^{\pi }\left({\int }_0^{a}f(r,\theta )J_n\left(\sqrt{{\lambda }_{nm}}r\right)rdr\right)\sin \left(n\theta \right)d\theta }{\pi {\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr}$$$$B_{nm}=\frac{{\int }_{-\pi }^{\pi }{\int }_0^{a}g(r,\theta )J_n\left(\sqrt{{\lambda }_{nm}}r\right)\cos \left(n\theta \right)rd\theta dr}{c\pi \sqrt{{\lambda }_{nm}}{\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr}$$$$D_{nm}=\frac{{\int }_{-\pi }^{\pi }{\int }_0^{a}g(r,\theta )J_n\left(\sqrt{{\lambda }_{nm}}r\right)\sin \left(n\theta \right)rd\theta dr}{c\pi \sqrt{{\lambda }_{nm}}{\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr}$$ With ${\lambda }_{nm}$ being the solutions for $0=J_n\left(\sqrt{{\lambda }_{nm}}a\right)$. For each $n$, we find ${\lambda }_{n,1},{\lambda }_{n,2},{\lambda }_{n,3},\cdots$, which are the zeros of the Bessel $J_n\left(x\right)$ function. So for each $n$, there are infinite number of zeros. This generates all eigenvalues ${\lambda }_{nm}$. Hence $\sqrt{{\lambda }_{nm}}a=BesselJZero(n,m)$, therefore $\sqrt{{\lambda }_{nm}}=\frac{a}{BesselJZero(n,m)}$.

In this problem $g(r,\theta )=0,f(r,\theta )=r\theta ,a=1,c=\frac{2}{10}$, then $B_{nm}=0,D_{nm}=0$ and the solution simplifies to

$$u(r,\theta ,t)=\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{A}_{nm}\cos \left(\frac{2}{10}\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\cos \left(n\theta \right)+\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{C}_{nm}\cos \left(\frac{2}{10}\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\sin \left(n\theta \right)$$

Where

$$\begin{array}{rl}{A}_{nm}& =\frac{{\int }_{-\pi }^{\pi }\left({\int }_0^{a}r\theta J_n\left(\sqrt{{\lambda }_{nm}}r\right)rdr\right)\cos \left(n\theta \right)d\theta }{\pi {\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr}\\ C_{nm}& =\frac{{\int }_{-\pi }^{\pi }\left({\int }_0^{a}r\theta J_n\left(\sqrt{{\lambda }_{nm}}r\right)rdr\right)\sin \left(n\theta \right)d\theta }{\pi {\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr}\end{array}$$

The following animations run for 80 seconds. They are for different $n,m$modes. (This only show in the HTML version)

5.2.2.8.1 Cases for $n=0$

5.2.2.8.2 Cases for $n=1$

5.2.2.8.3 Cases for $n=2$

5.2.2.8.4 Cases for $n=3$

Source code for all the above animations

(*By Nasser M. Abbasi*) 
SetDirectory[NotebookDirectory[]] 
 X:\data\public_html\my_notes\PDE_animations\problems\4 
 
(*definitions*) 
ClearAll[a,c,n,m,r,theta,f,g,u,maxM,maxN,t] 
maxN=4; 
maxM=4; 
a=1; 
c=.2; 
minZ=-8; 
maxZ=10; 
A0[n_,m_]:= Module[{num,den,r,theta,f}, 
f=r*theta; 
num=Integrate[f* BesselJ[n,lamtbl[[n+1,m]]  r] Cos[n theta] r,{r,0,a},{theta,0,2Pi}]; 
den=Integrate[(BesselJ[n,lamtbl[[n+1,m]] r])^2  (Cos[n theta])^2 r,{r,0,a},{theta,0,2Pi}]; 
num/den 
]; 
 
C0[n_,m_]:= Module[{num,den,f,r,theta}, 
   f=r*theta; 
   num=Integrate[f* BesselJ[n,lamtbl[[n+1,m]]  r] Sin[n theta] r,{r,0,a},{theta,0,2Pi}]; 
   den=Integrate[(BesselJ[n,lamtbl[[n+1,m]] r])^2  (Sin[n theta])^2 r,{r,0,a},{theta,0,2Pi}]; 
   num/den 
   ]; 
 
lam[n_,m_]:=Module[{x}, 
    x=BesselJZero[n,m]; 
    N[(x/a)] 
   ]; 
 
u[r_,theta_,t_,maxN_,maxM_]:=Module[{tmp,n,m}, 
     tmp=Sum[A0tbl[[n+1,m]]*Cos[c lamtbl[[n+1,m]] t] *BesselJ[n,lamtbl[[n+1,m]]*r]* 
         Cos[n*theta],{n,0,maxN},{m,1,maxM}]; 
 
     tmp=tmp+Sum[C0tbl[[n+1,m]]*Cos[c lamtbl[[n+1,m]] t] *BesselJ[n,lamtbl[[n+1,m]]*r]* 
        Sin[n*theta],{n,1,maxN},{m,1,maxM}] 
   ]; 
 
lamtbl=Table[lam[n,m],{n,0,maxN},{m,1,maxN}]; 
 
A0tbl=Table[A0[n,m],{n,0,maxN},{m,1,maxM}] 
C0tbl=Table[C0[n,m],{n,1,maxN},{m,1,maxM}] 
 
(*n=0,m=1*) 
ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0],Evaluate[u[r0,theta0,12,0,1]]}, 
   {r0,0,1},{theta0,0,2 Pi},AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, 
   PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1}, 
   PlotRange->{Automatic,Automatic,{minZ,maxZ}}] 
 
r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], 
     Evaluate[u[r0,theta0,t,0,1]]},{r0,0,1},{theta0,0,2 Pi},AxesLabel->{"r","theta","u"}, 
     BaseStyle->15,ImageMargins->5,PerformanceGoal->"Quality",Mesh->10, 
     MaxRecursion->1, BoxRatios->{1,1,1},PlotRange->{Automatic,Automatic,{minZ,maxZ}}], 
     Row[{"time (sec)",Round@t,", N = ", 0, ", M = ",1}]],{t,0,80,.5}]; 
 
Export["anim_n_0_m_1.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] 
 
 
(*n=0,m=2*) 
ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0],Evaluate[u[r0,theta0,0,0,2]]},{r0,0,1}, 
   {theta0,0,2 Pi},AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, 
   PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1}, 
   PlotRange->{Automatic,Automatic,{minZ,maxZ}}] 
 
(*to speed it up, make Z in {t,0,100,Z} larger, and then make Z in Table[Z,{Length[r]}] smaller.*) 
r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], 
     Evaluate[u[r0,theta0,t,0,2]]},{r0,0,1},{theta0,0,2 Pi},AxesLabel->{"r","theta","u"}, 
     BaseStyle->15,ImageMargins->5,PerformanceGoal->"Quality",Mesh->10, 
     MaxRecursion->1, BoxRatios->{1,1,1},PlotRange->{Automatic,Automatic,{minZ,maxZ}}], 
     Row[{"time (sec)",Round@t,", N = ", 0, ", M = ",2}]],{t,0,80,.5}]; 
 
Export["anim_n_0_m_2.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] 
 
(*n=0,m=3*) 
r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], 
          Evaluate[u[r0,theta0,t,0,3]]},{r0,0,1},{theta0,0,2 Pi}, 
          AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, 
          PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1}, 
          PlotRange->{Automatic,Automatic,{minZ,maxZ}}], 
          Row[{"time (sec)",Round@t,", N = ", 0, ", M = ",3}]],{t,0,80,.5}]; 
 
Export["anim_n_0_m_3.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] 
 
 
(*n=0,m=4*) 
t=1; 
ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0],Evaluate[u[r0,theta0,t,0,4]]}, 
        {r0,0,1},{theta0,0,2 Pi},AxesLabel->{"r","theta","u"},BaseStyle->15, 
        ImageMargins->5,PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, 
        BoxRatios->{1,1,1},PlotRange->{Automatic,Automatic,{minZ,maxZ}}] 
 
r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], 
          Evaluate[u[r0,theta0,t,0,4]]},{r0,0,1},{theta0,0,2 Pi},AxesLabel->{"r","theta","u"}, 
          BaseStyle->15,ImageMargins->5,PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, 
          BoxRatios->{1,1,1},PlotRange->{Automatic,Automatic,{minZ,maxZ}}], 
          Row[{"time (sec)",Round@t,", N = ", 0, ", M = ",4}]],{t,0,80,.5}]; 
 
Export["anim_n_0_m_4.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] 
 
(*n=1,m=1*) 
t=1; 
ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], 
       Evaluate[u[r0,theta0,t,1,1]]},{r0,0,1},{theta0,0,2 Pi}, 
       AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, 
       PerformanceGoal->"Speed",Mesh->10, BoxRatios->{1,1,1}, 
       PlotRange->{Automatic,Automatic,{minZ,maxZ}}] 
 
r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], 
            Evaluate[u[r0,theta0,t,1,1]]},{r0,0,1},{theta0,0,2 Pi}, 
            AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, 
            PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, 
            BoxRatios->{1,1,1},PlotRange->{Automatic,Automatic,{minZ,maxZ}}], 
            Row[{"time (sec)",Round@t,", N = ", 1, ", M = ",1}]],{t,0,80,.5}]; 
 
Export["anim_n_1_m_1.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] 
 
(*n=1,m=2*) 
 
r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], 
         Evaluate[u[r0,theta0,t,1,2]]},{r0,0,1},{theta0,0,2 Pi}, 
         AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, 
         PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, 
         BoxRatios->{1,1,1},PlotRange->{Automatic,Automatic,{minZ,maxZ}}], 
         Row[{"time (sec)",Round@t,", N = ", 1, ", M = ",2}]],{t,0,80,.5}]; 
 
Export["anim_n_1_m_2.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] 
 
(*n=1,m=3*) 
r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], 
    Evaluate[u[r0,theta0,t,1,3]]},{r0,0,1},{theta0,0,2 Pi}, 
    AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, 
    PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1}, 
    PlotRange->{Automatic,Automatic,{minZ,maxZ}}], 
    Row[{"time (sec)",Round@t,", N = ", 1, ", M = ",3}]],{t,0,80,.5}]; 
 
Export["anim_n_1_m_3.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] 
 
(*n=1,m=4*) 
r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], 
           Evaluate[u[r0,theta0,t,1,4]]},{r0,0,1},{theta0,0,2 Pi}, 
           AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, 
           PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, 
           BoxRatios->{1,1,1},PlotRange->{Automatic,Automatic,{minZ,maxZ}}], 
           Row[{"time (sec)",Round@t,", N = ", 1, ", M = ",4}]],{t,0,80,.5}]; 
 
Export["anim_n_1_m_4.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] 
 
(*n=2,m=1*) 
r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], 
          Evaluate[u[r0,theta0,t,2,1]]},{r0,0,1},{theta0,0,2 Pi}, 
          AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, 
          PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1}, 
          PlotRange->{Automatic,Automatic,{minZ,maxZ}}], 
          Row[{"time (sec)",Round@t,", N = ", 2 ", M = ",1}]] 
          ,{t,0,80,.5}]; 
 
Export["anim_n_2_m_1.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] 
 
(*n=2,m=2*) 
r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], 
         Evaluate[u[r0,theta0,t,2,2]]},{r0,0,1},{theta0,0,2 Pi}, 
         AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, 
         PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1}, 
         PlotRange->{Automatic,Automatic,{minZ,maxZ}}], 
         Row[{"time (sec)",Round@t,", N = ", 2 ", M = ",2}]],{t,0,80,.5}]; 
 
Export["anim_n_2_m_2.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] 
 
(*n=2,m=3*) 
r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], 
       Evaluate[u[r0,theta0,t,2,3]]},{r0,0,1},{theta0,0,2 Pi},AxesLabel->{"r","theta","u"}, 
       BaseStyle->15,ImageMargins->5,PerformanceGoal->"Quality",Mesh->10, 
       MaxRecursion->1, BoxRatios->{1,1,1},PlotRange->{Automatic,Automatic,{minZ,maxZ}}], 
       Row[{"time (sec)",Round@t,", N = ", 2 ", M = ",3}]],{t,0,80,.5}]; 
 
Export["anim_n_2_m_3.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] 
 
(*n=2,m=4*) 
r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], 
         Evaluate[u[r0,theta0,t,2,4]]},{r0,0,1},{theta0,0,2 Pi}, 
         AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, 
         PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1}, 
         PlotRange->{Automatic,Automatic,{minZ,maxZ}}], 
         Row[{"time (sec)",Round@t,", N = ", 2 ", M = ",4}]],{t,0,80,.5}]; 
 
Export["anim_n_2_m_4.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] 
 
(*n=3,m=1*) 
r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], 
      Evaluate[u[r0,theta0,t,3,1]]},{r0,0,1},{theta0,0,2 Pi}, 
      AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, 
      PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1}, 
      PlotRange->{Automatic,Automatic,{minZ,maxZ}}], 
      Row[{"time (sec)",Round@t,", N = ", 3 ", M = ",1}]],{t,0,80,.5}]; 
 
Export["anim_n_3_m_1.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] 
 
(*n=3,m=2*) 
r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], 
            Evaluate[u[r0,theta0,t,3,2]]},{r0,0,1},{theta0,0,2 Pi}, 
            AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, 
            PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, 
            BoxRatios->{1,1,1},PlotRange->{Automatic,Automatic,{minZ,maxZ}}], 
            Row[{"time (sec)",Round@t,", N = ", 3 ", M = ",2}]],{t,0,80,.5}]; 
 
Export["anim_n_3_m_2.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] 
 
(*n=3,m=3*) 
r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], 
           Evaluate[u[r0,theta0,t,3,3]]},{r0,0,1},{theta0,0,2 Pi}, 
           AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, 
           PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1}, 
           PlotRange->{Automatic,Automatic,{minZ,maxZ}}], 
           Row[{"time (sec)",Round@t,", N = ", 3 ", M = ",3}]],{t,0,80,.5}]; 
Export["anim_n_3_m_3.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]] 
 
(*n=3,m=4*) 
r=Table[Labeled[ParametricPlot3D[{r0 Cos[theta0],r0 Sin[theta0], 
           Evaluate[u[r0,theta0,t,3,4]]},{r0,0,1},{theta0,0,2 Pi}, 
           AxesLabel->{"r","theta","u"},BaseStyle->15,ImageMargins->5, 
           PerformanceGoal->"Quality",Mesh->10,MaxRecursion->1, BoxRatios->{1,1,1}, 
           PlotRange->{Automatic,Automatic,{minZ,maxZ}}], 
           Row[{"time (sec)",Round@t,", N = ", 3 ", M = ",4}]],{t,0,80,.5}]; 
 
Export["anim_n_3_m_4.gif",r,"DisplayDurations"->Table[.25,{Length[r]}]]

____________________________________________________________________________________

5.2.2.9 [412] $\theta$ dependency, fixed on edges, zero initial position, specific example

problem number 412

Added January 11, 2020

Math 322 UW exam problem. 2018.

Solve for $u(r,\theta ,t)$ with $0<r<a$ and $t>0$ and $-\pi <\theta <\pi$ $$\frac{{\partial }^{2}u}{\partial t^2}=c^2(\frac{{\partial }^{2}u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{{\partial }^{2}u}{\partial {\theta }^2})$$ With boundary conditions $$\begin{array}{rl}u(a,\theta ,t)& =0\\ |u(0,\theta ,t)|<\mathrm{\infty }\\ u(r,-\pi ,t)& =u(r,\pi ,t)\\ \frac{\partial u}{\partial \theta }(r,-\pi ,t)& =\frac{\partial u}{\partial \theta }(r,\pi ,t)\end{array}$$

With initial conditions $$\begin{array}{rl}u(r,\theta ,0)& =0\\ u_t(r,\theta ,0)& =\{\begin{array}{ccc}\frac{1}{\pi {ϵ}^2}& & \text{if }r\le ϵ\\ 0& & \text{otherwise}\end{array}\end{array}$$

Where $0<ϵ<1$

Using $a=1,c=1,ϵ=\frac{1}{2}$.

Figure 5.107: PDE specification

Mathematica ✓

ClearAll["Global`*"]; 
a=1; c=1; epsilon=1/2; 
f[r_,theta_]:=0; 
g[r_,theta_]:=Piecewise[{{1/(Pi*epsilon^2),r<epsilon},{0,True}}]; 
c=1; a=1; 
pde =  D[u[r, theta, t], {t, 2}] == c^2*Laplacian[u[r,theta,t],{r,theta},"Polar"]; 
ic  = {u[r, theta, 0] == f[r, theta], Derivative[0, 0, 1][u][r, theta, 0] == g[r,theta]}; 
bc  = {u[a, theta, t] == 0, u[r, -Pi, t] == u[r, Pi, t], Derivative[0, 1, 0][u][r, -Pi, t] == Derivative[0, 1, 0][u][r, Pi, t]}; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[r, theta, t], {r, theta, t}, Assumptions -> {0 < r < a,t > 0, -Pi < theta < Pi}], 60*10]];
 

$$\left\{\{u(r,\theta ,t)\to \begin{array}{cc}\{& \begin{array}{cc}\underset{K[3]=1}{\sum ^{\mathrm{\infty }}}\frac{2J_0\left(r{j}_{0,K[3]}\right){}_0{\tilde{F}}_1(;2;-\frac{1}{16}\left(j_{0,K[3]}\right){}^2)\sin \left(t{j}_{0,K[3]}\right)}{\pi J_1\left(j_{0,K[3]}\right){}^2{j}_{0,K[3]}}& (K[1]|K[3])\in \mathbb{Z}\wedge K[1]\ge 1\wedge K[3]\ge 1\\ \text{Indeterminate}& \text{True}\end{array}\end{array}\}\right\}$$

Maple ✗

restart; 
c:=1; 
a:=1; 
epsilon:=1/2; 
f:=(r,theta)->r*theta; 
g:=(r,theta)->piecewise(r<epsilon,1/(Pi*epsilon^2),true,0); 
pde := diff(u(r, theta, t), t$2) = c^2*VectorCalculus:-Laplacian(u(r,theta,t),'polar'[r,theta]); 
ic  := u(r, theta, 0) = f(r, theta) , (D[3](u))(r, theta, 0) = g(r,theta); 
bc  := u(a, theta, t) = 0, 
       u(r, -Pi, t) = u(r, Pi, t), 
       (D[2](u))(r, -Pi, t) = (D[2](u))(r, Pi, t); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic,bc], u(r, theta ,t),HINT = boundedseries(r=0))),output='realtime'));
 

sol=()

Hand solution

The basic solution for this type of PDE was already given in problem 5.2.2.6 on page 1557 as

$$\begin{array}{rl}u(r,\theta ,t)& =\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{A}_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\cos \left(n\theta \right)\\ & +\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{B}_{nm}\sin \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\cos \left(n\theta \right)\\ & +\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{C}_{nm}\cos \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\sin \left(n\theta \right)\\ & +\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{D}_{nm}\sin \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\sin \left(n\theta \right)\end{array}$$

$$A_{nm}=\frac{{\int }_{-\pi }^{\pi }\left({\int }_0^{a}f(r,\theta )J_n\left(\sqrt{{\lambda }_{nm}}r\right)rdr\right)\cos \left(n\theta \right)d\theta }{\pi {\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr}$$$$C_{nm}=\frac{{\int }_{-\pi }^{\pi }\left({\int }_0^{a}f(r,\theta )J_n\left(\sqrt{{\lambda }_{nm}}r\right)rdr\right)\sin \left(n\theta \right)d\theta }{\pi {\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr}$$$$B_{nm}=\frac{{\int }_{-\pi }^{\pi }{\int }_0^{a}g(r,\theta )J_n\left(\sqrt{{\lambda }_{nm}}r\right)\cos \left(n\theta \right)rd\theta dr}{c\pi \sqrt{{\lambda }_{nm}}{\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr}$$$$D_{nm}=\frac{{\int }_{-\pi }^{\pi }{\int }_0^{a}g(r,\theta )J_n\left(\sqrt{{\lambda }_{nm}}r\right)\sin \left(n\theta \right)rd\theta dr}{c\pi \sqrt{{\lambda }_{nm}}{\int }_0^a{J}_n^2\left(\sqrt{{\lambda }_{nm}}r\right)rdr}$$ With ${\lambda }_{nm}$ being the solutions for $0=J_n\left(\sqrt{{\lambda }_{nm}}a\right)$. For each $n$, we find ${\lambda }_{n,1},{\lambda }_{n,2},{\lambda }_{n,3},\cdots$, which are the zeros of the Bessel $J_n\left(x\right)$ function. So for each $n$, there are infinite number of zeros. This generates all eigenvalues ${\lambda }_{nm}$. Hence $\sqrt{{\lambda }_{nm}}a=BesselJZero(n,m)$, therefore $\sqrt{{\lambda }_{nm}}=\frac{a}{BesselJZero(n,m)}$.

In this problem $f(r,\theta )=0,a=1,c=1$, then $A_{nm}=0,C_{nm}=0$ and the solution simplifies to

$$u(r,\theta ,t)=\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{B}_{nm}\sin \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\cos \left(n\theta \right)+\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{D}_{nm}\sin \left(c\sqrt{{\lambda }_{nm}}t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\sin \left(n\theta \right)$$ Taking time derivative gives$$u_t(r,\theta ,t)=\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{B}_{nm}\cos \left(n\theta \right){\lambda }_{nm}\cos \left({\lambda }_{nm}t\right)J_n\left({\lambda }_{nm}r\right)+\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{D}_{nm}\sin \left(n\theta \right){\lambda }_{nm}\cos \left({\lambda }_{nm}t\right)J_n\left({\lambda }_{nm}r\right)$$ Applying the second initial condition at $t=0$ gives$$\begin{array}{}\text{(9)}& \sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{B}_{nm}\cos \left(n\theta \right){\lambda }_{nm}{J}_n\left({\lambda }_{nm}r\right)+\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{D}_{nm}\sin \left(n\theta \right){\lambda }_{nm}{J}_n\left({\lambda }_{nm}r\right)=\{\begin{array}{ccc}\frac{1}{\pi {ϵ}^2}& & \text{if }r\le ϵ\\ 0& & \text{otherwise}\end{array}\end{array}$$ Case $n=0$ (9) becomes$$\sum _{m=1}^{\mathrm{\infty }}{B}_{0m}{\lambda }_{0m}{J}_0\left({\lambda }_{0m}r\right)=\{\begin{array}{ccc}\frac{1}{\pi {ϵ}^2}& & \text{if }r\le ϵ\\ 0& & \text{otherwise}\end{array}$$ Applying orthogonality on $J_0\left({\lambda }_{0m}r\right)$ results in$$\begin{array}{rl}{B}_{0m}{\lambda }_{0m}{\int }_0^{1}r{J}_0^2\left({\lambda }_{0m}r\right)dr& =\frac{1}{\pi {ϵ}^2}{\int }_0^{ϵ}r{J}_0\left({\lambda }_{0m}r\right)dr\\ \text{(9A)}& B_{0m}& =\frac{1}{\pi {ϵ}^2{\lambda }_{0m}}\frac{{\int }_0^{ϵ}r{J}_0\left({\lambda }_{0m}r\right)dr}{{\int }_0^{1}r{J}_0^2\left({\lambda }_{0m}r\right)dr}\end{array}$$

Case $n>1$ Applying orthogonality on $\cos \left(n\theta \right),$ equation (9) becomes$$\begin{array}{rl}\sum _{m=1}^{\mathrm{\infty }}{B}_{nm}({\int }_{-\pi }^{\pi }{\cos }^2\left(n\theta \right)d\theta ){\lambda }_{nm}{J}_n\left({\lambda }_{nm}r\right)& =\{\begin{array}{ccc}\frac{1}{\pi {ϵ}^2}{\int }_{-\pi }^{\pi }\cos \left(n\theta \right)d\theta & & \text{if }{r}^2\le ϵ\\ 0& & \text{otherwise}\end{array}\\ \sum _{m=1}^{\mathrm{\infty }}\pi B_{nm}{\lambda }_{nm}{J}_n\left({\lambda }_{nm}r\right)& =\{\begin{array}{ccc}0& & \text{if }{r}^2\le ϵ\\ 0& & \text{otherwise}\end{array}\end{array}$$

Hence $B_{nm}=0$ for all $n>0$.

The same is now done to find ${\overline{D}}_{nm}$. Applying orthogonality on $\sin \left(n\theta \right),$ equation (9) becomes$$\begin{array}{rl}\sum _{m=1}^{\mathrm{\infty }}{D}_{nm}({\int }_{-\pi }^{\pi }{\sin }^2\left(n\theta \right)d\theta ){\lambda }_{nm}{J}_n\left({\lambda }_{nm}r\right)& =\{\begin{array}{ccc}\frac{1}{\pi {ϵ}^2}{\int }_{-\pi }^{\pi }\sin \left(n\theta \right)d\theta & & \text{if }{r}^2\le ϵ\\ 0& & \text{otherwise}\end{array}\\ \sum _{m=1}^{\mathrm{\infty }}{D}_{nm}({\int }_{-\pi }^{\pi }{\sin }^2\left(n\theta \right)d\theta ){\lambda }_{nm}{J}_n\left({\lambda }_{nm}r\right)& =\{\begin{array}{ccc}0& & \text{if }{r}^2\le ϵ\\ 0& & \text{otherwise}\end{array}\end{array}$$

Hence all $D_{nm}=0$ for all $n>0$.

Therefore the solution (8) reduces to only using $n=0,m=1,2,3,\cdots$. The solution can now be written as$$\begin{array}{}\text{(10)}& u(r,\theta ,t)=\sum _{m=1}^{\mathrm{\infty }}{B}_{0m}\sin \left({\lambda }_{0m}t\right)J_0\left({\lambda }_{0m}r\right)\end{array}$$ Where $B_{0m}=\frac{1}{\pi {ϵ}^2{\lambda }_{0m}}\frac{{\int }_0^{ϵ}r{J}_0\left({\lambda }_{0m}r\right)dr}{{\int }_0^{1}r{J}_0^2\left({\lambda }_{0m}r\right)dr}$ And ${\lambda }_{0m}$ are all the positive zeros of $J_0\left(z\right)$, $m=1,2,3,\cdots$.

$B_{0m}$ is now simplified more. Considering first the numerator of $B_{0m}$ which is ${\int }_0^{ϵ}r{J}_0\left({\lambda }_{0m}r\right)dr$. The hint given says that$$\frac{d}{dr}\left(r{J}_1\left(r\right)\right)=r{J}_0\left(r\right)$$ This is the same as saying$$\begin{array}{}\text{(10A)}& r{J}_1\left(r\right)=\int r{J}_0\left(r\right)dr\end{array}$$ However the integral in $B_{0m}$ is $\int r{J}_0\left({\lambda }_{0m}r\right)dr$ and not $\int r{J}_0\left(r\right)dr$. To transform it so that the hint can be used, let ${\lambda }_{0m}r=\overline{r}$, then $\frac{dr}{d\overline{r}}=\frac{1}{{\lambda }_{0m}}$ or $dr=\frac{d\overline{r}}{{\lambda }_{0m}}$. Now $\int r{J}_0\left({\lambda }_{0m}r\right)dr$ becomes $\int \frac{\overline{r}}{{\lambda }_{0m}}{J}_0\left(\overline{r}\right)\frac{d\overline{r}}{{\lambda }_{0m}}$ or $\frac{1}{{\lambda }_{0m}^2}\int \overline{r}{J}_0\left(\overline{r}\right)d\overline{r}$ and now the hint (10A) can be used on this integral giving$$\frac{1}{{\lambda }_{0m}^2}(\int \overline{r}{J}_0\left(\overline{r}\right)d\overline{r})=\frac{1}{{\lambda }_{0m}^2}\left(\overline{r}{J}_1\left(\overline{r}\right)\right)$$ Replacing $\overline{r}$ back by ${\lambda }_{0m}r$, gives the result needed$$\begin{array}{rl}\frac{1}{{\lambda }_{0m}^2}\left(\overline{r}{J}_1\left(\overline{r}\right)\right)& =\frac{1}{{\lambda }_{0m}^2}\left({\lambda }_{0m}r{J}_1\left({\lambda }_{0m}r\right)\right)\\ & =\frac{1}{{\lambda }_{0m}}r{J}_1\left({\lambda }_{0m}r\right)\end{array}$$

Now the limits are applied, using the fundamental theory of calculus$$\begin{array}{rl}{\int }_0^{ϵ}r{J}_0\left({\lambda }_{0m}r\right)dr& =\frac{1}{{\lambda }_{0m}}{\left[r{J}_1\left({\lambda }_{0m}r\right)\right]}_0^{ϵ}\\ \text{(10B)}& & =\frac{ϵ}{{\lambda }_{0m}}{J}_1\left({\lambda }_{0m}ϵ\right)\end{array}$$

This completes finding the numerator integral in $B_{0m}$. The denominator integral in $B_{0m}$ is ${\int }_0^{1}r{J}_0^2\left({\lambda }_{0m}r\right)dr$. This was found before which is $${\int }_0^{1}r{J}_0^2\left({\lambda }_{0m}r\right)dr=\frac{1}{2}{\left[J_0^{\prime }\left({\lambda }_{0m}\right)\right]}^2$$ But $J_0^{\prime }\left({\lambda }_{0m}\right)=-J_1\left({\lambda }_{0m}\right)$, hence the above becomes$$\begin{array}{}\text{(10C)}& {\int }_0^{1}r{J}_0^2\left({\lambda }_{0m}r\right)dr=\frac{1}{2}{J}_1^2\left({\lambda }_{0m}\right)\end{array}$$ Applying (10B) and (10C), $B_{0m}$ simplifies to the following expression$$\begin{array}{rl}{B}_{0m}& =\frac{1}{\pi {ϵ}^2{\lambda }_{0m}}\frac{\frac{ϵ}{{\lambda }_{0m}}{J}_1\left({\lambda }_{0m}ϵ\right)}{\frac{1}{2}{J}_1^2\left({\lambda }_{0m}\right)}\\ & =\frac{2}{\pi ϵ{\lambda }_{0m}^2}\frac{J_1\left({\lambda }_{0m}ϵ\right)}{J_1^2\left({\lambda }_{0m}\right)}\end{array}$$

Therefore the final solution becomes$$\begin{array}{rl}u(r,\theta ,t)& =\sum _{m=1}^{\mathrm{\infty }}{B}_{0m}\sin \left({\lambda }_{0m}t\right)J_0\left({\lambda }_{0m}r\right)\\ \text{(11)}& u(r,\theta ,t)& =\frac{2}{\pi ϵ}\sum _{m=1}^{\mathrm{\infty }}\frac{1}{{\lambda }_{0m}^2}\frac{J_1\left({\lambda }_{0m}ϵ\right)}{J_1^2\left({\lambda }_{0m}\right)}{J}_0\left({\lambda }_{0m}r\right)\sin \left({\lambda }_{0m}t\right)\end{array}$$

When $ϵ=\frac{1}{2}$, the above solution (11) becomes$$\begin{array}{}\text{(11A)}& u(r,\theta ,t)=\frac{4}{\pi }\sum _{m=1}^{\mathrm{\infty }}\frac{1}{{\lambda }_{0m}^2}\frac{J_1\left(\frac{1}{2}{\lambda }_{0m}\right)}{J_1^2\left({\lambda }_{0m}\right)}{J}_0\left({\lambda }_{0m}r\right)\sin \left({\lambda }_{0m}t\right)\end{array}$$ Here is animation for 5 seconds made in Mathematica

Mathematica Source code for all the above animations

(*By Nasser M. Abbasi. Animation of problem 4 solution*) 
 
ClearAll[t,r,m]; 
padIt2[v_,f_List]:=AccountingForm[v,f,NumberSigns->{"",""}, 
                          NumberPadding->{"0","0"},SignPadding->True]; 
nTerms=40; 
lam = Table[ BesselJZero[0,m],{m,1,nTerms}]//N; 
c   = Table[1/lam[[m]]^2 BesselJ[1,lam[[m]]/2]/BesselJ[1,lam[[m]] ]^2,{m,1,nTerms}]; 
mySol[r_,t_]:=4/Pi   Sum[c[[m]]BesselJ[0,lam[[m]] r] Sin[lam[[m]] t],{m,1,nTerms}]; 
 
frames=Table[ 
  Print["t=",t]; 
  Grid[{ 
   {Row[{"time = ",padIt2[t,{3,2}]}]}, 
   {ParametricPlot3D[{r Cos[theta],r Sin[theta],mySol[r,t]},{r,0,1},{theta,0,2 Pi}, 
    PlotRange->{Automatic,Automatic,{-0.6,0.6}}, 
    PerformanceGoal->"Speed",Boxed->True, 
    Axes->True,Mesh->20, 
    ViewPoint->{2.17,-2.4,1}, 
    ImageSize->400, 
    BoxRatios->{1, 1, 1}] 
  }}], 
    {t,0,5,0.05} 
]; 
 
Manipulate[ 
  frames[[i]], 
  {{i,1,"time"},1,Length@frames,1,Appearance->"Labeled"} 
] 
 
Export["anim.gif",frames,"DisplayDurations"->Table[.2,{Length@frames}]]

Here is the same animation made in Maple 2018

Maple source code for all the above animations

#by Nasser M. Abbasi, May 23,2018 
 
restart; 
currentdir("X:/data/public_html/my_notes/PDE_animations/problems/wave_disk_exam_problem_4"); 
nTerms := 20: 
lam    := evalf([BesselJZeros(0,1..nTerms)]): 
c      := seq(1/lam[n]^2*BesselJ(1,lam[n]/2)/BesselJ(1,lam[n])^2,n=1..nTerms): 
mySol  := proc(r,t) 
  local n; 
  4/Pi*sum(c[n]*BesselJ(0,lam[n]*r)*sin(lam[n]*t),n=1..nTerms); 
end proc: 
 
maxTime := 5: (*seconds*) 
delay   := 0.05: 
nFrames := round(maxTime/delay): 
 
frames  := [seq( plot3d([ r, theta, mySol(r,(i*delay)) ], 
                   r      = 0..1, 
                   theta  = 0..2*Pi, 
                   coords = cylindrical, 
                   axes   = none, 
                   title  = sprintf("%s %3.2f %s","time ",(i*delay),"seconds") 
                ), 
             i=0..nFrames-1) 
           ]: 
plots:-display(convert(frames,list),insequence=true);

Here is the same animation made in Matlab 2016a

Matlab source code for all the above animations

function nma_HW4_math_322 
%By Nasser M. Abbasi, May 23, 2018 
 
close all; 
 
GENERATE_GIF=true; %turn to false to not generate animated gif 
 
lam = zeros(80,1); %eigenvalues 
for i = 1:80 
  lam(i) = fzero(@(x)besselj(0,x),i); 
end; 
 
lam    = uniquetol(lam); %must use  uniquetol 
nTerms = 20; 
c      = zeros(nTerms,1); 
 
for i = 1:nTerms 
  c(i) = 1/lam(i)^2*besselj(1,lam(i)/2)/besselj(1,lam(i))^2; 
end 
 
  %------------- inner function -------- 
  function tot = mySol(r,t) 
    tot = 0; 
    for ii =1:nTerms 
        tot = tot + (c(ii)*besselj(0,lam(ii).*r).*sin(lam(ii)*t)); 
    end; 
    tot = 4/pi*tot; 
  end 
  %------------------- 
 
maxTime = 5; %seconds 
delay   = 0.05; 
nFrames = round(maxTime/delay); 
 
r   = 0:.05:1; 
phi = 0:pi/20:2*pi; 
[R,PHI] = meshgrid(r,phi); 
 
fig_handle   = figure(); 
set(fig_handle,'Name',.... 
   'Math 322, Final exam problem 4 animations by Nasser M. Abbasi'); 
 
for i=1:nFrames 
  Z = mySol(R,((i-1)*delay)); 
  surf(R.*cos(PHI), R.*sin(PHI), Z); 
  set(gca,'nextplot','replacechildren','visible','on'); 
  colormap cool ; 
  title(sprintf('time = %3.2f', (i-1)*delay)); 
  zlim([-0.6 0.6]); 
  drawnow; 
  pause(.01); 
  if GENERATE_GIF 
      frame = getframe(gcf); 
      im = frame2im(frame); 
      [imind,cm] = rgb2ind(im,256); 
      if i == 1 
          imwrite(imind,cm,'matlab_animations.gif','gif', ... 
                                     'DelayTime',0.1,'LoopCount',0); 
      else 
          imwrite(imind,cm,'matlab_animations.gif','gif',... 
                 'WriteMode','append','DelayTime',0.1,'LoopCount',0); 
      end 
  end 
end 
 
end

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5.2.2.10 [413] $\theta$ dependency, fixed on edges, zero initial position with internal source (Haberman 8.5.5. (b)

problem number 413

Added January 15, 2020

Problem 8.5.5. (b) Richard Haberman applied partial di fb00erential equations book, 5th edition

Solve wave PDE inside circular membrane for $u(r,\theta ,t)$ with $0<r<a$ and $t>0$ and $-\pi <\theta <\pi$ $$u_{tt}=c^2{\mathrm{\nabla }}^{2}u(r,\theta )+Q(r,\theta ,t)$$ With boundary conditions $$\begin{array}{rl}u(a,\theta ,t)& =0\\ |u(0,\theta ,t)|<\mathrm{\infty }\\ u(r,-\pi ,t)& =u(r,\pi ,t)\\ \frac{\partial u}{\partial \theta }(r,-\pi ,t)& =\frac{\partial u}{\partial \theta }(r,\pi ,t)\end{array}$$

With initial conditions $$\begin{array}{rl}u(r,\theta ,0)& =f(r,\theta )\\ u_t(r,\theta ,0)& =0\end{array}$$

Figure 5.108: PDE specification

Mathematica ✗

ClearAll["Global`*"]; 
pde =  D[u[r, theta, t], {t, 2}] == c^2*Laplacian[u[r,theta,t],{r,theta},"Polar"]+Q[r,theta,t]; 
ic  = {u[r, theta, 0] == f[r, theta], Derivative[0, 0, 1][u][r, theta, 0] == 0}; 
bc  = {u[a, theta, t] == 0, u[r, -Pi, t] == u[r, Pi, t], Derivative[0, 1, 0][u][r, -Pi, t] == Derivative[0, 1, 0][u][r, Pi, t]}; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, ic, bc}, u[r, theta, t], {r, theta, t}, Assumptions -> {0 < r < a,t > 0, -Pi < theta < Pi}], 60*10]];
 

Failed

Maple ✗

restart; 
pde := diff(u(r, theta, t), t$2) = c^2*VectorCalculus:-Laplacian(u(r,theta,t),'polar'[r,theta])+Q(r,theta,t); 
ic  := u(r, theta, 0) = f(r, theta) , (D[3](u))(r, theta, 0) = 0; 
bc  := u(a, theta, t) = 0, 
       u(r, -Pi, t) = u(r, Pi, t), 
       (D[2](u))(r, -Pi, t) = (D[2](u))(r, Pi, t); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, ic,bc], u(r, theta ,t),HINT = boundedseries(r=0)) assuming r>0,r<a),output='realtime'));
 

sol=()

Hand solution

The solution to the corresponding homogeneous wave PDE

$$\frac{{\partial }^{2}u}{\partial t^2}=c^2{\mathrm{\nabla }}^2$$ Is known to be $$u(r,\theta ,t)=\sum _{n=0}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{a}_n\left(t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\cos \left(n\theta \right)+\sum _{n=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}{a}_n\left(t\right)J_n\left(\sqrt{{\lambda }_{nm}}r\right)\sin \left(n\theta \right)$$ Where ${\lambda }_{nm}$ are found by solving roots of $J_n\left(\sqrt{{\lambda }_{nm}}a\right)=0$. To make things simpler, we will write $$u(r,\theta ,t)=\sum _i{a}_i\left(t\right){\mathrm{\Phi }}_i(r,\theta )$$ Where the above means the double sum of all eigenvalues ${\lambda }_i$. So ${\mathrm{\Phi }}_i(r,\theta )$ represents $J_n\left(\sqrt{{\lambda }_{nm}}r\right)\{\cos \left(n\theta \right),\sin \left(\theta \right)\}$ combined. So double sum is implied everywhere. Given this, we now expand the source term$$Q(r,\theta ,t)=\sum _i{q}_i\left(t\right){\mathrm{\Phi }}_i(r,\theta )$$ And the original PDE becomes$$\begin{array}{}\text{(1)}& \sum _i{a}_i^{\prime \prime }\left(t\right)\mathrm{\Phi }\left({\lambda }_i\right)=c^2\sum _i{a}_i\left(t\right){\mathrm{\nabla }}^2\left({\mathrm{\Phi }}_i(r,\theta )\right)+\sum _i{q}_i\left(t\right){\mathrm{\Phi }}_i(r,\theta )\end{array}$$ But $${\mathrm{\nabla }}^2\left({\mathrm{\Phi }}_i(r,\theta )\right)=-{\lambda }_i{\mathrm{\Phi }}_i(r,\theta )$$ Hence (1) becomes$$\begin{array}{rl}\sum _i{a}_i^{\prime \prime }\left(t\right){\mathrm{\Phi }}_i(r,\theta )+c^2{\lambda }_i{a}_i\left(t\right){\mathrm{\Phi }}_i(r,\theta )& =\sum _i{q}_i\left(t\right){\mathrm{\Phi }}_i(r,\theta )\\ \sum _i(a_i^{\prime \prime }\left(t\right)+c^2{\lambda }_i{a}_i\left(t\right)){\mathrm{\Phi }}_i(r,\theta )& =\sum _i{q}_i\left(t\right){\mathrm{\Phi }}_i(r,\theta )\end{array}$$

Applying orthogonality gives$$a_i^{\prime \prime }\left(t\right)+c^2{\lambda }_i{a}_i\left(t\right)=q_i\left(t\right)$$ Where $$q_i\left(t\right)=\frac{{\int }_0^a{\int }_{-\pi }^{\pi }Q(r,\theta ,t){\mathrm{\Phi }}_i(r,\theta )rdrd\theta }{{\int }_0^a{\int }_{-\pi }^{\pi }{\mathrm{\Phi }}_i^2(r,\theta )rdrd\theta }$$ The solution to the homogenous ODE is$$a_i^h\left(t\right)=A_i\cos \left(c\sqrt{{\lambda }_i}t\right)+B_i\sin \left(c\sqrt{{\lambda }_i}t\right)$$ And the particular solution is found if we know what $Q(r,\theta ,t)$ and hence $q_i\left(t\right)$. For now, lets call the particular solution as $a_i^p\left(t\right)$. Hence the solution for $a_i\left(t\right)$ is$$a_i\left(t\right)=A_i\cos \left(c\sqrt{{\lambda }_i}t\right)+B_i\sin \left(c\sqrt{{\lambda }_i}t\right)+a_i^p\left(t\right)$$ Plugging the above into the $u(r,\theta ,t)=\sum _i{a}_i\left(t\right){\mathrm{\Phi }}_i(r,\theta )$, gives$$\begin{array}{}\text{(2)}& u(r,\theta ,t)=\sum _i(A_i\cos \left(c\sqrt{{\lambda }_i}t\right)+B_i\sin \left(c\sqrt{{\lambda }_i}t\right)+a_i^p\left(t\right)){\mathrm{\Phi }}_i(r,\theta )\end{array}$$ We now find $A_i,B_i$ from initial conditions. At $t=0$$$f(r,\theta )=\sum _i(A_i+a_i^p\left(0\right)){\mathrm{\Phi }}_i(r,\theta )$$ Applying orthogonality$$\begin{array}{rl}{\int }_0^a{\int }_{-\pi }^{\pi }f(r,\theta ){\mathrm{\Phi }}_j(r,\theta )rdrd\theta & ={\int }_0^a{\int }_{-\pi }^{\pi }\sum _i(A_i+a_i^p\left(0\right)){\mathrm{\Phi }}_i(r,\theta ){\mathrm{\Phi }}_j(r,\theta )rdrd\theta \\ {\int }_0^a{\int }_{-\pi }^{\pi }f(r,\theta ){\mathrm{\Phi }}_j(r,\theta )rdrd\theta & =(A_j+a_j^p\left(0\right)){\int }_0^a{\int }_{-\pi }^{\pi }{\mathrm{\Phi }}_j^2(r,\theta )rdrd\theta \\ (A_i+a_i^p\left(0\right))& =\frac{{\int }_0^a{\int }_{-\pi }^{\pi }f(r,\theta ){\mathrm{\Phi }}_i(r,\theta )rdrd\theta }{{\int }_0^a{\int }_{-\pi }^{\pi }{\mathrm{\Phi }}_i^2(r,\theta )rdrd\theta }\end{array}$$

Taking time derivative of  (2)$$\frac{\partial u(r,\theta ,t)}{\partial t}=\sum _i(-A_{i}c\sqrt{{\lambda }_i}\sin \left(c\sqrt{{\lambda }_i}t\right)+c\sqrt{{\lambda }_i}{B}_i\cos \left(c\sqrt{{\lambda }_i}t\right)+\frac{d{a}_i^p\left(t\right)}{dt}){\mathrm{\Phi }}_i(r,\theta )$$ At $t=0$$$0=\sum _i(c\sqrt{{\lambda }_i}{B}_i+\frac{d{a}_i^p\left(0\right)}{dt}){\mathrm{\Phi }}_i(r,\theta )$$ Hence $B_i=0$. Therefore the final solution is$$u(r,\theta ,t)=\sum _i(A_i\cos \left(c\sqrt{{\lambda }_i}t\right)+a_i^p\left(t\right)){\mathrm{\Phi }}_i(r,\theta )$$ Where$$(A_i+a_i^p\left(0\right))=\frac{{\int }_0^a{\int }_{-\pi }^{\pi }f(r,\theta ){\mathrm{\Phi }}_i(r,\theta )rdrd\theta }{{\int }_0^a{\int }_{-\pi }^{\pi }{\mathrm{\Phi }}_i^2(r,\theta )rdrd\theta }$$ This complete the solution.