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4.2.2 Cylinderical coordinates

4.2.2.1 [317] Haberman 7.9.1 (a)
4.2.2.2 [318] Haberman 7.9.1 (b)
4.2.2.3 [319] Haberman 7.9.1 (c)
4.2.2.4 [320] Haberman 7.9.1 (d)
4.2.2.5 [321] Haberman 7.9.1 (e)
4.2.2.6 [322] Haberman 7.9.2 (a)
4.2.2.7 [323] Haberman 7.9.2 (b)
4.2.2.8 [324] Haberman 7.9.2 (c)
4.2.2.9 [325] Haberman 7.9.2 (d)

4.2.2.1 [317] Haberman 7.9.1 (a)

problem number 317

Added May 25, 2019.

Problem 7.9.1 (a) from Richard Haberman Applied Partial Differential Equations, 4th edition.

Solve Laplace PDE inside circular cylinder subject to boundary conditions \(u(r,\theta ,0)=f(r,\theta )\), \(u(r,\theta ,H)=0\), \(u(a,\theta ,z)=0\).

\begin {align*} u_{rr} + \frac {1}{r} u_r + \frac {1}{r^2} u_{\theta \theta } + u_{zz} = 0 \end {align*}

Mathematica ✗ 

ClearAll["Global`*"]; 
lap = Laplacian[u[r, theta, z], {r, theta, z}, "Cylindrical"]; 
bc  = {u[r, theta, 0] == f[r, theta], u[r, theta, H] == 0, u[a, theta, z] == 0}; 
sol = AbsoluteTiming[TimeConstrained[DSolve[{lap == 0, bc}, u[r, theta, z], {r, theta, z}, Assumptions -> {a > 0, r < a, H > 0}], 60*10]];

Failed

Maple ✗ 

restart; 
pde :=VectorCalculus:-Laplacian(u(r,theta,z),'cylindrical'[r,theta,z])=0; 
bc  := u(r, theta, 0) = f(r, theta), u(r, theta, H) = 0, u(a, theta, z) = 0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(r,theta,z)) assuming a>0,r<a,H>0),output='realtime'));

sol=()

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4.2.2.2 [318] Haberman 7.9.1 (b)

problem number 318

Added May 25, 2019.

Problem 7.9.1 (b) from Richard Haberman Applied Partial Differential Equations, 4th edition.

Solve Laplace PDE inside circular cylinder subject to boundary conditions \(u(r,\theta ,0)=f(r) \sin (7\theta )\), \(u(r,\theta ,H)=0\), \(u(a,\theta ,z)=0\).

\begin {align*} u_{rr} + \frac {1}{r} u_r + \frac {1}{r^2} u_{\theta \theta } + u_{zz} = 0 \end {align*}

Mathematica ✗ 

ClearAll["Global`*"]; 
lap = Laplacian[u[r, theta, z], {r, theta, z}, "Cylindrical"]; 
bc  = {u[r, theta, 0] == f[r]*Sin[7*theta], u[r, theta, H] == 0, u[a, theta, z] == 0}; 
sol = AbsoluteTiming[TimeConstrained[DSolve[{lap == 0, bc}, u[r, theta, z], {r, theta, z}, Assumptions -> {a > 0, r < a, H > 0}], 60*10]];

Failed

Maple ✗ 

restart; 
pde :=VectorCalculus:-Laplacian(u(r,theta,z),'cylindrical'[r,theta,z])=0; 
bc  := u(r, theta, 0) = f(r)*sin(7*theta), u(r, theta, H) = 0, u(a, theta, z) = 0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(r,theta,z)) assuming a>0,r<a,H>0),output='realtime'));

sol=()

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4.2.2.3 [319] Haberman 7.9.1 (c)

problem number 319

Added May 25, 2019.

Problem 7.9.1 (c) from Richard Haberman Applied Partial Differential Equations, 4th edition.

Solve Laplace PDE inside circular cylinder subject to boundary conditions \(u(r,\theta ,0)=0\), \(u(r,\theta ,H)=f(r) \cos (3 \theta )\), \(u_r(a,\theta ,z)=0\).

\begin {align*} u_{rr} + \frac {1}{r} u_r + \frac {1}{r^2} u_{\theta \theta } + u_{zz} = 0 \end {align*}

Mathematica ✗ 

ClearAll["Global`*"]; 
lap = Laplacian[u[r, theta, z], {r, theta, z}, "Cylindrical"]; 
bc  = {u[r, theta, 0] == 0, u[r, theta, H] == f[r]*Cos[3*theta], Derivative[1,0,0][u][a, theta, z] == 0}; 
sol = AbsoluteTiming[TimeConstrained[DSolve[{lap == 0, bc}, u[r, theta, z], {r, theta, z}, Assumptions -> {a > 0, r < a, H > 0}], 60*10]];

Failed

Maple ✗ 

restart; 
pde :=VectorCalculus:-Laplacian(u(r,theta,z),'cylindrical'[r,theta,z])=0; 
bc  := u(r, theta, 0) = 0, u(r, theta, H) = f(r)*cos(3*theta), eval(diff(u(r,theta,z),r),r=a)=0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(r,theta,z)) assuming a>0,r<a,H>0),output='realtime'));

sol=()

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4.2.2.4 [320] Haberman 7.9.1 (d)

problem number 320

Added May 25, 2019.

Problem 7.9.1 (d) from Richard Haberman Applied Partial Differential Equations, 4th edition.

Solve Laplace PDE inside circular cylinder subject to boundary conditions \(u_z(r,\theta ,0)=f(r) \sin (3 \theta )\), \(u_z(r,\theta ,H)=0\), \(u_r(a,\theta ,z)=0\).

\begin {align*} u_{rr} + \frac {1}{r} u_r + \frac {1}{r^2} u_{\theta \theta } + u_{zz} = 0 \end {align*}

Mathematica ✗ 

ClearAll["Global`*"]; 
lap = Laplacian[u[r, theta, z], {r, theta, z}, "Cylindrical"]; 
bc  = {Derivative[0,0,1][u][r, theta, 0] == f[r]*Sin[3*theta], Derivative[0,0,1][u][r, theta, H] == 0, Derivative[1,0,0][u][a, theta, z] == 0}; 
sol = AbsoluteTiming[TimeConstrained[DSolve[{lap == 0, bc}, u[r, theta, z], {r, theta, z}, Assumptions -> {a > 0, r < a, H > 0}], 60*10]];

Failed

Maple ✗ 

restart; 
pde :=VectorCalculus:-Laplacian(u(r,theta,z),'cylindrical'[r,theta,z])=0; 
bc:=eval(diff(u(r,theta,z),z),z=0)=f(r)*sin(3*theta), eval(diff(u(r,theta,z),z),z=H)= 0, eval(diff(u(r,theta,z),r),r=a)=0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(r,theta,z)) assuming a>0,r<a,H>0),output='realtime'));

sol=()

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4.2.2.5 [321] Haberman 7.9.1 (e)

problem number 321

Added May 25, 2019.

Problem 7.9.1 (e) from Richard Haberman Applied Partial Differential Equations, 4th edition.

Solve Laplace PDE inside circular cylinder subject to boundary conditions \(u_z(r,\theta ,0)=f(r,\theta )\), \(u_z(r,\theta ,H)=0\), \(u_r(a,\theta ,z)=0\).

\begin {align*} u_{rr} + \frac {1}{r} u_r + \frac {1}{r^2} u_{\theta \theta } + u_{zz} = 0 \end {align*}

Mathematica ✗ 

ClearAll["Global`*"]; 
lap = Laplacian[u[r, theta, z], {r, theta, z}, "Cylindrical"]; 
bc  = {Derivative[0,0,1][u][r, theta, 0] == f[r,theta], Derivative[0,0,1][u][r, theta, H] == 0, Derivative[1,0,0][u][a, theta, z] == 0}; 
sol = AbsoluteTiming[TimeConstrained[DSolve[{lap == 0, bc}, u[r, theta, z], {r, theta, z}, Assumptions -> {a > 0, r < a, H > 0}], 60*10]];

Failed

Maple ✗ 

restart; 
pde :=VectorCalculus:-Laplacian(u(r,theta,z),'cylindrical'[r,theta,z])=0; 
bc:=eval(diff(u(r,theta,z),z),z=0)=f(r,theta), eval(diff(u(r,theta,z),z),z=H)= 0, eval(diff(u(r,theta,z),r),r=a)=0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(r,theta,z)) assuming a>0,r<a,H>0),output='realtime'));

sol=()

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4.2.2.6 [322] Haberman 7.9.2 (a)

problem number 322

Added May 25, 2019.

Problem 7.9.2 (a) from Richard Haberman Applied Partial Differential Equations, 4th edition.

Solve Laplace PDE inside semicircular cylinder subject to boundary conditions \(u(r,\theta ,0)=0\), \(u(r,\theta ,H)=f(r,\theta )\), \(u(r,0,z)=0\), \(u(r,\pi ,z)=0\), \(u(a,\theta ,z)=0\).

\begin {align*} u_{rr} + \frac {1}{r} u_r + \frac {1}{r^2} u_{\theta \theta } + u_{zz} = 0 \end {align*}

Mathematica ✗ 

ClearAll["Global`*"]; 
lap = Laplacian[u[r, theta, z], {r, theta, z}, "Cylindrical"]; 
bc  = {u[r, theta, 0] == 0, u[r, theta, H] == f[r,theta], u[r, 0, z] == 0,u[r,Pi,z]==0,u[a,theta,z]==0}; 
sol = AbsoluteTiming[TimeConstrained[DSolve[{lap == 0, bc}, u[r, theta, z], {r, theta, z}, Assumptions -> {a > 0, r < a, H > 0,theta>0,theta<Pi}], 60*10]];

Failed

Maple ✗ 

restart; 
pde :=VectorCalculus:-Laplacian(u(r,theta,z),'cylindrical'[r,theta,z])=0; 
bc  := u(r,theta,0)=0, u(r,theta,H)= f(r,theta), u(r,0,z)=0, u(r,Pi,z)=0,u(a,theta,z)=0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(r,theta,z)) assuming a>0,r<a,H>0,theta>0,theta<Pi),output='realtime'));

sol=()

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4.2.2.7 [323] Haberman 7.9.2 (b)

problem number 323

Added May 25, 2019.

Problem 7.9.2 (b) from Richard Haberman Applied Partial Differential Equations, 4th edition.

Solve Laplace PDE inside semicircular cylinder subject to boundary conditions \(u(r,\theta ,0)=0\), \(u_z(r,\theta ,H)=0\), \(u(r,0,z)=0\), \(u(r,\pi ,z)=0\), \(u(a,\theta ,z)=g(\theta ,z)\).

\begin {align*} u_{rr} + \frac {1}{r} u_r + \frac {1}{r^2} u_{\theta \theta } + u_{zz} = 0 \end {align*}

Mathematica ✗ 

ClearAll["Global`*"]; 
lap = Laplacian[u[r, theta, z], {r, theta, z}, "Cylindrical"]; 
bc  = {u[r, theta, 0] == 0, Derivative[0,0,1][u][r, theta, H] == 0, u[r, 0, z] == 0,u[r,Pi,z]==0,u[a,theta,z]==g[theta,z]}; 
sol = AbsoluteTiming[TimeConstrained[DSolve[{lap == 0, bc}, u[r, theta, z], {r, theta, z}, Assumptions -> {a > 0, r < a, H > 0,theta>0,theta<Pi}], 60*10]];

Failed

Maple ✗ 

restart; 
pde :=VectorCalculus:-Laplacian(u(r,theta,z),'cylindrical'[r,theta,z])=0; 
bc  := u(r,theta,0)=0, eval(diff(u(r,theta,z),z),z=H)=0, u(r,0,z)=0, u(r,Pi,z)=0,u(a,theta,z)=g(theta,z); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(r,theta,z)) assuming a>0,r<a,H>0,theta>0,theta<Pi),output='realtime'));

sol=()

____________________________________________________________________________________

4.2.2.8 [324] Haberman 7.9.2 (c)

problem number 324

Added May 25, 2019.

Problem 7.9.2 (c) from Richard Haberman Applied Partial Differential Equations, 4th edition.

Solve Laplace PDE inside semicircular cylinder subject to boundary conditions \(u_z(r,\theta ,0)=0\), \(u_z(r,\theta ,H)=0\), \(u_\theta (r,0,z)=0\), \(u_\theta (r,\pi ,z)=0\), \(u_r(a,\theta ,z)=g(\theta ,z)\).

\begin {align*} u_{rr} + \frac {1}{r} u_r + \frac {1}{r^2} u_{\theta \theta } + u_{zz} = 0 \end {align*}

Mathematica ✗ 

ClearAll["Global`*"]; 
lap = Laplacian[u[r, theta, z], {r, theta, z}, "Cylindrical"]; 
bc  = {Derivative[0,0,1][u][r, theta, 0] == 0, Derivative[0,0,1][u][r, theta, H] == 0, Derivative[0,1,0][u][r, 0, z] == 0,Derivative[1,0,0][u][r,Pi,z]==0,Derivative[1,0,0][u][a,theta,z]==g[theta,z]}; 
sol = AbsoluteTiming[TimeConstrained[DSolve[{lap == 0, bc}, u[r, theta, z], {r, theta, z}, Assumptions -> {a > 0, r < a, H > 0,theta>0,theta<Pi}], 60*10]];

Failed

Maple ✗ 

restart; 
pde :=VectorCalculus:-Laplacian(u(r,theta,z),'cylindrical'[r,theta,z])=0; 
bc:=eval(diff(u(r,theta,z),z),z=0)=0, eval(diff(u(r,theta,z),z),z=H)=0, eval(diff(u(r,theta,z),theta),theta=0)=0, eval(diff(u(r,theta,z),theta),theta=Pi)=0,eval(diff(u(r,theta,z),r),r=a)=g(theta,z); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(r,theta,z)) assuming a>0,r<a,H>0,theta>0,theta<Pi),output='realtime'));

sol=()

____________________________________________________________________________________

4.2.2.9 [325] Haberman 7.9.2 (d)

problem number 325

Added May 26, 2019.

Problem 7.9.2 (d) from Richard Haberman Applied Partial Differential Equations, 4th edition.

Solve Laplace PDE inside semicircular cylinder subject to boundary conditions \(u(r,\theta ,0)=0\), \(u(r,0,z)=0\), \(u(a,\theta ,z)=0\), \(u(r,\theta ,H)=0\), \(u_\theta (r,\pi ,z)=f(r,z)\).

\begin {align*} u_{rr} + \frac {1}{r} u_r + \frac {1}{r^2} u_{\theta \theta } + u_{zz} = 0 \end {align*}

Mathematica ✗ 

ClearAll["Global`*"]; 
lap = Laplacian[u[r, theta, z], {r, theta, z}, "Cylindrical"]; 
bc  = {u[r, theta, 0] == 0, u[r, 0, z] == 0, u[a, theta, z] == 0,u[r,theta,H]==0,Derivative[0,1,0][u][r,Pi,z]==f[r,z]}; 
sol = AbsoluteTiming[TimeConstrained[DSolve[{lap == 0, bc}, u[r, theta, z], {r, theta, z}, Assumptions -> {a > 0, r < a, H > 0,theta>0,theta<Pi}], 60*10]];

Failed

Maple ✗ 

restart; 
pde :=VectorCalculus:-Laplacian(u(r,theta,z),'cylindrical'[r,theta,z])=0; 
bc  := u(r,theta,0)=0,u(r,0,z)=0, u(a,theta,z)=0, u(r,theta,H)=0,eval(diff(u(r,theta,z),theta),theta=Pi)=f(r,z); 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(r,theta,z)) assuming a>0,r<a,H>0,theta>0,theta<Pi),output='realtime'));

sol=()

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