#### 4.2.2 Cylinderical coordinates

4.2.2.1  Haberman 7.9.1 (a)
4.2.2.2  Haberman 7.9.1 (b)
4.2.2.3  Haberman 7.9.1 (c)
4.2.2.4  Haberman 7.9.1 (d)
4.2.2.5  Haberman 7.9.1 (e)
4.2.2.6  Haberman 7.9.2 (a)
4.2.2.7  Haberman 7.9.2 (b)
4.2.2.8  Haberman 7.9.2 (c)
4.2.2.9  Haberman 7.9.2 (d)

##### 4.2.2.1  Haberman 7.9.1 (a)

problem number 318

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

Failed

Maple

sol=()

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##### 4.2.2.2  Haberman 7.9.1 (b)

problem number 319

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

Failed

Maple

sol=()

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##### 4.2.2.3  Haberman 7.9.1 (c)

problem number 320

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

Failed

Maple

sol=()

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##### 4.2.2.4  Haberman 7.9.1 (d)

problem number 321

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

Failed

Maple

sol=()

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##### 4.2.2.5  Haberman 7.9.1 (e)

problem number 322

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

Failed

Maple

sol=()

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##### 4.2.2.6  Haberman 7.9.2 (a)

problem number 323

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

Failed

Maple

sol=()

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##### 4.2.2.7  Haberman 7.9.2 (b)

problem number 324

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

Failed

Maple

sol=()

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##### 4.2.2.8  Haberman 7.9.2 (c)

problem number 325

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

Failed

Maple

sol=()

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##### 4.2.2.9  Haberman 7.9.2 (d)

problem number 326

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

Failed

Maple

sol=()