Chapter 13 The heat equation on the whole line

Consider the heat equation \(\partial _tu=\partial _{xx}u\) on the spatial interval \((-\infty ,\infty )\) (rather than a bounded interval considered in earlier chapters). Similarly to the bounded spatial interval case, we look for solutions which for each \(t\) belong to \(L^2(\mR )\).

Separation of variables as before gives

\[ T'=\lambda T,\qquad X''=\lambda X. \]

For \(\lambda >0\) the equation for \(X\) can be solved to obtain \(A\e ^{\sqrt {\lambda }x}+B\e ^{-\sqrt {\lambda }x}\). This is unbounded and does not belong to \(L^2(\mR )\) unless \(A=B=0\). For \(\lambda =0\) we obtain \(A+Bx\), which again is unbounded (unless \(B=0\)) and does not belong to \(L^2(\mR )\) (unless \(A=B=0\)). For \(\lambda <0\) and writing \(\lambda =-\omega ^2\) we obtain the two linearly independent solutions \(\e ^{i\omega x}\) and \(\e ^{-i\omega x}\) which do not belong to \(L^2(\mR )\), but at least are both bounded. This is the same form that we obtained in the bounded interval case (where we used \(\sin (\omega x)\) and \(\cos (\omega x)\) rather than complex exponentials).

We further consider only the case \(\lambda \leq 0\) (where we have at least bounded solutions) and consider the solutions \(\e ^{i\omega x}\) for \(\omega \in \mR \) (which for \(\omega =0\) reduces to \(1\) and because now \(\omega \in \mR \) includes both solutions for \(\lambda <0\)). We now do not have boundary conditions to whittle down the number of solutions further, so (here the factor \(\frac {1}{\sqrt {2\pi }}\) is for normalization as the factor \(\sqrt {\frac {2}{L}}\) was in the bounded interval case)

\[ X(x;\omega )=\frac {1}{\sqrt {2\pi }}\e ^{i\omega x},\qquad \omega \in \mR . \]

Here the continuous variable \(\omega \) plays the role that the discrete variable \(k\) played in the bounded interval case. We then have

\[ T'=-\omega ^2 T, \]

which gives

\[ T(t;\omega )=C(\omega ) \e ^{-\omega ^2 t}. \]

We therefore obtain the solutions

\[ u(t,x;\omega )=\frac {1}{\sqrt {2\pi }} C(\omega ) \e ^{-\omega ^2 t}\e ^{i\omega x}. \]

We combine these by integrating over the continuous variable \(\omega \) (rather than summing over the discrete variable \(k\) as we did in the bounded interval case):

\[ u(t,x)=\frac {1}{\sqrt {2\pi }} \int _\mR C(\omega ) \e ^{-\omega ^2 t}\e ^{i\omega x}\,d\omega . \]

This gives for the initial condition

\[ u^0(x)=u(0,x)=\frac {1}{\sqrt {2\pi }} \int _\mR C(\omega ) \e ^{i\omega x}\,d\omega . \]

As in the bounded interval case, we multiply by \(X(x;-\eta )\) and integrate over \(x\) to obtain

\[ \frac {1}{\sqrt {2\pi }}\int _\mR u^0(x) \e ^{i\eta x}\,dx =\frac {1}{2\pi }\int _\mR \int _\mR C(\omega ) \e ^{i\omega x}\e ^{-i\eta x}\,d\omega \,dx. \]

It turns out that (suitably interpreted) the integral on the right equals \(C(\eta )\) so that we obtain

\[ C(\eta )=\frac {1}{\sqrt {2\pi }}\int _\mR u^0(x) \e ^{-i\eta x}. \]

We therefore obtain the solution

\[ u(t,x)=\frac {1}{\sqrt {2\pi }} \int _\mR C(\omega ) \e ^{-\omega ^2 t}\e ^{i\omega x}\,d\omega ,\qquad C(\omega )=\frac {1}{\sqrt {2\pi }}\int _\mR u^0(x) \e ^{-i\omega x}\,dx. \]

It turns out (we will see this in Example 15.1) that this can be re-written as

\[ u(t,x)=\frac {1}{\sqrt {4\pi t}}\int _\mR \e ^{-(x-y)^2/(4t)}u^0(y)\,dy. \]