Solution. As in the heat equation on a disc, we perform separation of variables. The \(R\) and \(\Theta \) are identical to the heat equation case from Chapter 5; the \(T\) equation becomes

\[ T''=\lambda T. \]

As in Chapters 5 and 12 we obtain

\[ \lambda _{km}=-\left (\frac {z_{km}}{b}\right )^2, \]

where \(z_{km}\) is the \(m\)-th zero of the Bessel function of order \(k\), and \(\Theta _k\) is given by

\[ \frac {1}{\sqrt {2\pi }},\qquad \frac {1}{\sqrt {\pi }}\sin (k\theta ),\qquad \frac {1}{\sqrt {\pi }}\cos (k\theta ), \]

and

\[ R_{km}(r)=\frac {\sqrt {2}}{bJ_{k+1}(z_{km})}J_k\left (\frac {z_{km}}{b}r\right ). \]

Writing

\[ \omega _{km}=\frac {z_{km}}{b}, \]

we have that the general solution of \(T''=\lambda _{km} T\) is

\[ C_{km}\sin (\omega _{km}t)+D_{km}\cos (\omega _{km}t). \]

We therefore obtain the separated solutions

\[ u_{km}(t,\theta ,r)=\left [C_{km}\sin (\omega _{km}t)+D_{km}\cos (\omega _{km}t)\right ]\Theta _k(\theta )R_{km}(r). \]

The general solution then is (for \(k=0\) we obtain one separated solution of the above form, for \(k=1\) we obtain two since there are two choices for \(\Theta _k\))

\begin{multline*} u=\sum _{m=1}^\infty \left [C_{0m}\sin \left (\omega _{0m}t\right )+D_{0m}\cos \left (\omega _{0m}t\right )\right ]\frac {1}{\sqrt {2\pi }}R_{0m}(r) \\ + \sum _{k=1}^\infty \sum _{m=1}^\infty \left [C_{km1}\sin \left (\omega _{km}t\right )+D_{km1}\cos \left (\omega _{km}t\right )\right ] \frac {1}{\sqrt {\pi }}\cos (k\theta )R_{km}(r) \\ + \left [C_{km2}\sin \left (\omega _{km}t\right )+D_{km2}\cos \left (\omega _{km}t\right )\right ] \frac {1}{\sqrt {\pi }}\sin (k\theta )R_{km}(r). \end{multline*} The initial conditions then become

\[ u^0(x,y)=\sum _{m=1}^\infty D_{0m}\frac {1}{\sqrt {2\pi }}R_{0m}(r) +\sum _{k=1}^\infty \sum _{m=1}^\infty D_{km1} \frac {1}{\sqrt {\pi }}\cos (k\theta )R_{km}(r) + D_{km2} \frac {1}{\sqrt {\pi }}\sin (k\theta )R_{km}(r), \]

and

\begin{multline*} u^1(x,y)=\sum _{m=1}^\infty C_{0m}\omega _{0m}\frac {1}{\sqrt {2\pi }}R_{0m}(r) \\ +\sum _{k=1}^\infty \sum _{m=1}^\infty C_{km1}\omega _{km} \frac {1}{\sqrt {\pi }}\cos (k\theta )R_{km}(r) + C_{km2}\omega _{km} \frac {1}{\sqrt {\pi }}\sin (k\theta )R_{km}(r). \end{multline*} Using orthonormality of \(\Theta _k R_{km}\) gives (using a mixture of Cartesian and polar coordinates)

\begin{gather*} D_{0m}=\int _{B_b(0)} u^0(x,y)\frac {1}{\sqrt {2\pi }}R_{0m}(r)\,dxdy,\\ D_{km1}=\int _{B_b(0)} u^0(x,y)\frac {1}{\sqrt {\pi }}\cos (k\theta )R_{km}(r)\,dxdy,\\ D_{km2}=\int _{B_b(0)} u^0(x,y)\frac {1}{\sqrt {\pi }}\sin (k\theta )R_{km}(r)\,dxdy,\\ C_{0m}=\frac {1}{\omega _{0m}}\int _{B_b(0)} u^1(x,y)\frac {1}{\sqrt {2\pi }}R_{0m}(r)\,dxdy,\\ C_{km1}=\frac {1}{\omega _{km}}\int _{B_b(0)} u^1(x,y)\frac {1}{\sqrt {\pi }}\cos (k\theta )R_{km}(r)\,dxdy,\\ C_{km2}=\frac {1}{\omega _{km}}\int _{B_b(0)} u^1(x,y)\frac {1}{\sqrt {\pi }}\sin (k\theta )R_{km}(r)\,dxdy. \end{gather*} □

Solution. From Problem B.1 we obtain a solution \(T(t)R(r)\Theta (\theta )Z(z)\) where \(\Theta \) and \(R\) satisfy the same equations as for the disc as treated in Chapters 5 and 12. We therefore have that \(\Theta _k\) is given by

\[ \frac {1}{\sqrt {2\pi }},\qquad \frac {1}{\sqrt {\pi }}\sin (k\theta ),\qquad \frac {1}{\sqrt {\pi }}\cos (k\theta ), \]

and

\[ R_{km}(r)=\frac {\sqrt {2}}{bJ_{k+1}(z_{km})}J_k\left (\frac {z_{km}}{b}r\right ), \]

where \(z_{km}\) is the \(m\)-th zero of the Bessel function of order \(k\). We further have

\[ \lambda ^R_{km}=-\left (\frac {z_{km}}{b}\right )^2. \]

The equation for \(Z\) from Problem B.1 is the same as that in Chapter 3 and therefore we obtain

\[ \lambda ^Z_n=-\left (\frac {n\pi }{L}\right )^2,\qquad Z_n(z)=\sqrt {\frac {2}{L}}\sin \left (\frac {n\pi z}{L}\right ). \]

From Problem B.1 the equation for \(T\) then is

\[ T'=\lambda _{kmn}T,\qquad \lambda _{kmn}=\lambda ^R_{km}+\lambda ^Z_n=-\left (\frac {z_{km}}{b}\right )^2-\left (\frac {n\pi }{L}\right )^2. \]

The general solution of this is

\[ T_{kmn}(t)=C_{kmn}\e ^{\lambda _{kmn}t}. \]

We therefore obtain

\begin{multline*} u=\sum _{n=1}^\infty \sum _{m=1}^\infty C_{0mn}\e ^{\lambda _{0mn}t}R_{km}(r)Z_n(z)\frac {1}{\sqrt {2\pi }} \\ +\sum _{k=1}^\infty \sum _{n=1}^\infty \sum _{m=1}^\infty C_{kmn1}\e ^{\lambda _{kmn}t}R_{km}(r)Z_n(z)\frac {1}{\sqrt {\pi }}\sin (k\theta ) +C_{kmn2}\e ^{\lambda _{kmn}t}R_{km}(r)Z_n(z)\frac {1}{\sqrt {\pi }}\cos (k\theta ). \end{multline*} The initial condition gives

\begin{multline*} u^0=\sum _{n=1}^\infty \sum _{m=1}^\infty C_{0mn}R_{0m}(r)Z_n(z)\frac {1}{\sqrt {2\pi }} \\ +\sum _{k=1}^\infty \sum _{n=1}^\infty \sum _{m=1}^\infty C_{kmn1}R_{km}(r)Z_n(z)\frac {1}{\sqrt {\pi }}\sin (k\theta ) +C_{kmn2}R_{km}(r)Z_n(z)\frac {1}{\sqrt {\pi }}\cos (k\theta ). \end{multline*} Orthonormality of \(R\Theta Z\) then gives

\begin{gather*} C_{0mn}=\int _{B_b(0)\times [0,L]} u^0(x,y,z)\frac {1}{\sqrt {2\pi }}R_{0m}(r)Z_n(z)\,dxdydz, \\ C_{kmn1}=\int _{B_b(0)\times [0,L]} u^0(x,y,z)R_{km}(r)Z_n(z)\frac {1}{\sqrt {\pi }}\sin (k\theta )\,dxdydz, \\ C_{kmn2}=\int _{B_b(0)\times [0,L]} u^0(x,y,z)R_{km}(r)Z_n(z)\frac {1}{\sqrt {\pi }}\cos (k\theta )\,dxdydz. \end{gather*} □

Solution. From Problem B.2 we obtain a solution \(T(t)R(r)\Theta (\theta )\Phi (\phi )\) where \(\lambda ^\Theta =-k^2\) for \(k\in \mN _0\) and \(\Theta _k\) is given by

\[ \frac {1}{\sqrt {2\pi }},\qquad \frac {1}{\sqrt {\pi }}\sin (k\theta ),\qquad \frac {1}{\sqrt {\pi }}\cos (k\theta ). \]

The \(\Phi \)-equation then is

\[ \Phi ''+\cot (\phi )\Phi '-k^2\csc ^2(\phi )\Phi =\lambda ^\Phi \Phi , \]

which is the associated Legendre equation with index \(k\) from Problem Sheet E after the change of variables \(x=\cos (\phi )\) from Problem C.1. Therefore

\[ \lambda ^\Phi =-(n+1)n,\qquad n\in \mN _0, \]

and \(\Phi _{kn}(\phi )\) is a (scaled) associated Legendre function \(P^k_n(\cos (\phi ))\). The \(R\)-equation then is

\[ R''+\frac {2}{r}R'-\frac {(n+1)n}{r^2}R=\lambda R,\qquad R(b)=0, \]

which is the parametric form of the spherical Bessel equation with \(\sigma =n\) and \(\lambda =-\eta ^2\). Denote the solution of the spherical Bessel equation obtained in Problem E.10 by \(j_n\). From Problem E.12 we then obtain that the boundary condition gives

\[ j_n(\eta b)=0. \]

This means that \(\eta b\) should be a zero of \(j_n\). Denote the \(m\)-th zero of \(j_n\) by \(z_{nm}\) (where \(m\in \mN \)). Then

\[ \eta =\frac {z_{nm}}{b}, \]

so that

\[ \lambda _{nm}=-\left (\frac {z_{nm}}{b}\right )^2, \]

and (up to normalization) \(R_{nm}\) equals

\[ j_n\left (\frac {z_{nm}}{b} r\right ). \]

Therefore we have the separated solutions

\[ u_{knm}(t,r,\theta ,\phi )=\e ^{\lambda _{nm}t}j_n\left (\frac {z_{nm}}{b} r\right )\Theta _k(\theta )P^k_n(\cos (\phi )). \]

From these separated solutions (which we would have to normalize) we then obtain the solution of the heat equation in a ball which satisfies the given initial condition similarly as in the notes for the disc case. □