Chapter 5 The heat equation on a disc

Consider the heat equation on a disc in polar coordinates from Section 4.1

\begin{equation} \label {eq:heat:polar} \partial _t u=\partial _{rr}u+\frac {1}{r}\partial _ru+\frac {1}{r^2}\partial _{\theta \theta }u,\qquad u(t,b,\theta )=0,\qquad u(0,r,\theta )=u^0(r,\theta ). \end{equation}

We again look for a solution which is separated, i.e. of the form

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

Substituting this in \(\partial _tu=\Delta u\) gives

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

which leads to

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

which similarly as before leads to

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

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

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

After dividing by \(R\Theta \) and multiplying by \(r^2\), we can re-write this as

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

Since the left-hand side depends only on \(\theta \) and the right-hand side depends only on \(r\), there must exist a constant \(\lambda ^\Theta \) such that

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

This gives

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

We therefore do not have full separation of variables: the constant \(\lambda ^\Theta \) from the \(\Theta \) equation appears in the \(R\) equation. In practice, this simply means that we have to solve the \(\Theta \) equation before solving the \(R\) equation. This is in contrast to the rectangle case from Chapter 3 where we could solve the \(X\) and \(Y\) equations completely separately (and therefore in either order). Recapitulating:

\[ T'=\lambda T,\qquad \Theta ''=\lambda ^\Theta \Theta ,\qquad R''+\frac {1}{r}R'+\frac {\lambda ^\Theta }{r^2}R=\lambda R. \]

We now consider boundary conditions. In polar coordinates \(\theta =0\) and \(\theta =2\pi \) are really the same, so we shouldn’t really have considered the interval \((0,2\pi )\) for \(\theta \) (we have to instead identify the two end-points of the interval as the same). However, we can still consider \(\theta \in (0,2\pi )\) if we impose

\begin{equation} \label {eq:periodicbc} \Theta (0)=\Theta (2\pi ),\qquad \Theta '(0)=\Theta '(2\pi ), \end{equation}

which means we don’t quite identify the points \(\theta =0\) and \(\theta =2\pi \), but we do only consider functions which agree (together with their first derivatives, and by the differential equation then with all their derivatives) at these two values for \(\theta \). The boundary conditions (5.2) are called periodic boundary conditions. We therefore aim to solve the ODE boundary value problem

\begin{equation} \label {eq:periodic} \Theta ''=\lambda ^\Theta \Theta ,\qquad \Theta (0)=\Theta (2\pi ),\qquad \Theta '(0)=\Theta '(2\pi ). \end{equation}

Similarly as in Chapter 1, we first want to show that we must have \(\lambda ^\Theta \leq 0\). The easiest way to do this is the following argument. Let \(\Theta \) be a nonzero solution of (5.3). Then (integrating by parts for the second equality and using the periodic boundary conditions to obtain the third equality)

\[ \lambda ^\Theta \int _0^{2\pi }\Theta (\theta )^2\,d\theta = \int _0^{2\pi }\Theta ''(\theta )\Theta (\theta )\,d\theta = \left [\Theta '(\theta )\Theta (\theta )\right ]_0^{2\pi }-\int _0^{2\pi }\Theta '(\theta )^2\,d\theta = -\int _0^{2\pi }\Theta '(\theta )^2\,d\theta . \]

We therefore have

\[ \lambda ^\Theta =-\dfrac {\int _0^{2\pi }\Theta '(\theta )^2\,d\theta }{\int _0^{2\pi }\Theta (\theta )^2\,d\theta } \leq 0. \]

We can therefore write \(\lambda ^\Theta =-\omega ^2\) for \(\omega \geq 0\) and (5.3) can then be written as

\[ \Theta ''=-\omega ^2 \Theta ,\qquad \Theta (0)=\Theta (2\pi ),\qquad \Theta '(0)=\Theta '(2\pi ). \]

For \(\omega >0\) the general solution of the differential equation (ignoring the boundary conditions for now) is

\[ \Theta (\theta )=A\cos (\omega \theta )+B\sin (\omega \theta ). \]

The easiest way to deal with the boundary conditions is to observe that they imply that the above sine and cosine must have period \(2\pi \) and therefore we must have \(\omega =n\), where \(n\in \mN _0\). We then obtain the solutions \(1\) (corresponding to \(n=0\)), \(\cos (n\theta )\) and \(\sin (n\theta )\). We normalize these so that \(\int _0^{2\pi } \Theta (\theta )^2\,d\theta =1\) and obtain

\begin{gather*} \frac {1}{\sqrt {2\pi }},\qquad \lambda ^\Theta _0=0, \\ \frac {1}{\sqrt {\pi }}\cos (k\theta ),\qquad \frac {1}{\sqrt {\pi }}\sin (k\theta ),\qquad \lambda ^\Theta _k=-k^2,\qquad k\in \mN , \end{gather*} as the normalized (basis of) solutions of (5.3). Note that for \(k\in \mN \) we obtain a two-dimensional space of solutions of (5.3) (and for \(k=0\) a one-dimensional space of solutions).

We now return to the equation for \(R\) which now knowing what \(\lambda ^\Theta \) is becomes

\begin{equation} \label {eq:BesselR} R''+\frac {1}{r}R'-\frac {\sigma ^2}{r^2}R=\lambda R, \end{equation}

where \(\sigma \in \mN _0\). We can however more generally consider \(\sigma \geq 0\) a real number without added difficulty. The boundary condition we obtain from (5.1) is

\begin{equation} \label {eq:BesselRbc} R(b)=0. \end{equation}

It should be clear from the discussion in Section 1.1 that this is an insufficient number of boundary conditions: for any \(\lambda \in \mC \) we obtain a two-dimensional space of solutions for (5.4) and the boundary condition (5.5) will restrict this to a one-dimensional space of solutions. However, that still leaves us with no restriction on \(\lambda \). We therefore need to somehow find additional boundary conditions for the \(R\) equation (in addition to the obvious one (5.5)). Furthermore, (5.4) is a second order linear differential equation with non-constant coefficients, for which in earlier modules no method of solution was provided; therefore we need to discuss a solution method for such equations (we will do that in Chapters 10 and 11).