Chapter B Problem Sheet 2 (Lectures 4-5)

The relation between cylindrical and Cartesian coordinates is

\[ x=r\cos (\theta ),\qquad y=r\sin (\theta ),\qquad z=z. \]

The Jacobian determinant for cylindrical coordinates is \(r\).

The Laplacian in cylindrical coordinates is

\[ \Delta u=\partial _{rr}u+\frac {1}{r}\partial _ru+\frac {1}{r^2}\partial _{\theta \theta }u+\partial _{zz}u. \]

A cylinder in cylindrical coordinates is described by \(r\in [0,b]\), \(\theta \in [0,2\pi ]\) (where \(0\) and \(2\pi \) should really be identified) and \(z\in [0,L]\).

The relation between spherical and Cartesian coordinates is

\[ x=r\cos (\theta )\sin (\phi ),\qquad y=r\sin (\theta )\sin (\phi ),\qquad z=r\cos (\phi ). \]

The Laplacian in spherical coordinates is

\[ \Delta u=\partial _{rr}u+\frac {2}{r}\partial _ru+\frac {1}{r^2}\left (\partial _{\phi \phi }u +\cot (\phi )\partial _\phi u+\csc ^2(\phi )\partial _{\theta \theta }u\right ). \]

Here

\[ \cot (\phi )=\frac {1}{\tan (\phi )}=\frac {\cos (\phi )}{\sin (\phi )},\qquad \csc (\phi )=\frac {1}{\sin (\phi )}. \]

Here we have named variables so that for \(\phi =\frac {\pi }{2}\) and \(u\) independent of \(\phi \) we reduce to polar coordinates \((r,\theta )\); many books have \(\theta \) and \(\phi \) the other way around than we do.

The Jacobian determinant for spherical coordinates is \(r^2\sin (\phi )\).

A ball in spherical coordinates is described by \(r\in [0,b]\), \(\theta \in [0,2\pi ]\) (where \(0\) and \(2\pi \) should really be identified) and \(\phi \in [0,\pi ]\).

B.1 Heat equation in a cylinder 1

Show that separation of variables for the heat equation with Dirichlet boundary conditions in a cylinder leads to

\begin{gather*} \Theta ''=\lambda ^\Theta \Theta ,\qquad \Theta (0)=\Theta (2\pi ),\quad \Theta '(0)=\Theta '(2\pi ),\\ Z''=\lambda ^Z Z,\qquad Z(0)=Z(L)=0,\\ R''+\frac {1}{r}R'+\frac {\lambda ^\Theta }{r^2}R=\lambda ^R R,\qquad R(b)=0,\\ T'=(\lambda ^R+\lambda ^Z)T. \end{gather*}

  • Solution. The heat equation is

    \[ \partial _tu=\Delta u. \]

    The boundary conditions are

    \[ u(t,b,\theta ,z)=u(t,r,\theta ,0)=u(t,r,\theta ,L)=0. \]

    The Ansatz is that

    \[ u(t,r,\theta ,z)=T(t)R(r)\Theta (\theta )Z(z). \]

    The boundary conditions give

    \[ T(t)R(b)\Theta (\theta )Z(z)=0,\quad T(t)R(r)\Theta (\theta )Z(0)=0,\quad T(t)R(r)\Theta (\theta )Z(L)=0, \]

    and for non-trivial functions these are satisfied if and only if \(R(b)=0\), \(Z(0)=0\) and \(Z(L)=0\). That \(\theta =0\) and \(\theta =2\pi \) should really be identified leads to the boundary conditions

    \[ \Theta (0)=\Theta (2\pi ),\quad \Theta '(0)=\Theta '(2\pi ). \]

    Substituting the Ansatz in the heat equation gives

    \[ T'R\Theta Z=T\Delta R\Theta Z, \]

    which leads to

    \[ \frac {T'}{T}=\frac {\Delta R\Theta Z}{R\Theta Z}, \]

    which leads to

    \[ T'=\lambda T,\qquad \Delta R\Theta Z=\lambda R\Theta Z. \]

    Using the cylindrical coordinate expression for the Laplacian, the latter equation is

    \[ R''\Theta Z+\frac {1}{r}R'+\Theta Z+\frac {1}{r^2}R\Theta '' Z+R\Theta Z''=\lambda R\Theta Z, \]

    and when divided by \(R\Theta Z\) it is

    \begin{equation} \label {eq:cyl:Laplac}\tag {$\dagger $} \frac {R''}{R}+\frac {1}{r}\frac {R'}{R}+\frac {1}{r^2}\frac {\Theta ''}{\Theta }+\frac {Z''}{Z}=\lambda . \end{equation}

    We can re-arrange (\(\dagger \)) as

    \[ \frac {Z''}{Z}= \lambda -\left [\frac {R''}{R}+\frac {1}{r}\frac {R'}{R}+\frac {1}{r^2}\frac {\Theta ''}{\Theta }\right ]. \]

    The right-hand side depends only on \((r,\theta )\) and the left-hand side depends only on \(z\). Therefore, both sides must by constant. Denoting this constant \(\lambda ^Z\) we obtain

    \[ Z''=\lambda ^Z Z,\qquad \frac {R''}{R}+\frac {1}{r}\frac {R'}{R}+\frac {1}{r^2}\frac {\Theta ''}{\Theta }=\lambda -\lambda ^Z. \]

    We rewrite the second equation as

    \[ \frac {\Theta ''}{\Theta }=r^2\left [\lambda -\lambda ^Z-\left (\frac {R''}{R}+\frac {1}{r}\frac {R'}{R}\right )\right ]. \]

    The left-hand side depends only on \(\theta \) and the right-hand side only on \(r\). Therefore, both sides must be constant. Denoting this constant \(\lambda ^\Theta \) we obtain

    \[ \Theta ''=\lambda ^\Theta \Theta ,\qquad \frac {R''}{R}+\frac {1}{r}\frac {R'}{R}+\frac {\lambda ^\Theta }{r^2}=\lambda -\lambda ^Z. \]

    Defining \(\lambda ^R=\lambda -\lambda ^Z\) (so that \(\lambda =\lambda ^R+\lambda ^Z\)), the latter equation is

    \[ R''+\frac {1}{r}R'+\frac {\lambda ^\Theta }{r^2}R=\lambda ^R R. \]

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B.2 Heat equation in a ball 1

Show that separation of variables for the heat equation with Dirichlet boundary conditions in a ball leads to

\begin{gather*} \Theta ''=\lambda ^\Theta \Theta ,\qquad \Theta (0)=\Theta (2\pi ),\quad \Theta '(0)=\Theta '(2\pi ),\\ \Phi ''+\cot (\phi )\Phi '+\lambda ^\Theta \csc ^2(\phi )\Phi =\lambda ^\Phi \Phi ,\\ R''+\frac {2}{r}R'+\frac {\lambda ^\Phi }{r^2}R=\lambda R,\qquad R(b)=0,\\ T'=\lambda T. \end{gather*}

  • Solution. The boundary condition is

    \[ u(t,b,\theta ,\phi )=0. \]

    The Ansatz is that

    \[ u(t,r,\theta ,\phi )=T(t)R(r)\Theta (\theta )\Phi (\phi ). \]

    The boundary condition gives

    \[ T(t)R(b)\Theta (\theta )\Phi (\phi )=0,\quad \]

    and for non-trivial functions this are satisfied if and only if \(R(b)=0\). That \(\theta =0\) and \(\theta =2\pi \) should really be identified leads to the boundary conditions

    \[ \Theta (0)=\Theta (2\pi ),\quad \Theta '(0)=\Theta '(2\pi ). \]

    As before, separation of variables leads to

    \[ T'=\lambda T,\qquad \Delta R\Theta \Phi =\lambda R\Theta \Phi . \]

    Using the spherical coordinate expression for the Laplacian, the latter equation is

    \[ R''\Theta \Phi +\frac {2}{r}R'\Theta \Phi +\frac {1}{r^2}\left (R\Theta \Phi ''+\cot (\phi )R\Theta \Phi '+\csc ^2(\phi )R\Theta ''\Phi \right )=\lambda R\Theta \Phi , \]

    which after dividing by \(R\Theta \Phi \) gives

    \begin{equation} \label {eq:sp:Laplac}\tag {$\diamond $} \frac {R''}{R}+\frac {2}{r}\frac {R'}{R}+\frac {1}{r^2}\left (\frac {\Phi ''}{\Phi }+\cot (\phi )\frac {\Phi '}{\Phi }+\csc ^2(\phi )\frac {\Theta ''}{\Theta }\right )=\lambda . \end{equation}

    Re-arranging (\(\diamond \)) gives

    \[ \frac {\Phi ''}{\Phi }+\cot (\phi )\frac {\Phi '}{\Phi }+\csc ^2(\phi )\frac {\Theta ''}{\Theta } = r^2\left [\lambda -\left (\frac {R''}{R}+\frac {2}{r}\frac {R'}{R}\right )\right ]. \]

    Since the left-hand side depends only on \((\theta ,\phi )\) and the right-hand side depends only on \(r\), both sides must be constant. Denoting this constant \(\lambda ^\Phi \) we obtain

    \[ R''+\frac {2}{r}R'+\frac {\lambda ^\Phi }{r^2}R=\lambda R,\qquad \frac {\Phi ''}{\Phi }+\cot (\phi )\frac {\Phi '}{\Phi }+\csc ^2(\phi )\frac {\Theta ''}{\Theta }=\lambda ^\Phi . \]

    Re-arranging the latter equation gives

    \[ \frac {\Theta ''}{\Theta }=\frac {1}{\csc ^2(\phi )}\left [\lambda ^\Phi -\left (\frac {\Phi ''}{\Phi }+\cot (\phi )\frac {\Phi '}{\Phi }\right )\right ]. \]

    Since the left-hand side depends only on \(\theta \) and the right-hand side depends only on \(\phi \), both sides must be constant. Denoting this constant \(\lambda ^\Theta \) we obtain

    \[ \Theta ''=\lambda ^\Theta \Theta ,\qquad \Phi ''+\cot (\phi )\Phi '+\lambda ^\Theta \csc ^2(\phi )\Phi =\lambda ^\Phi \Phi . \]

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