Chapter 15 Solving PDEs using the Fourier transform
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Example 15.1. We reconsider the calculations in Chapter 13 in light of the Fourier transform from Chapter 14.
We consider the heat equation \(\partial _tu=\partial _{xx}u\) on the spatial interval \(\mR \). We Fourier transform in \(x\) (really: Fourier–Plancherel transform as we consider the solution space \(L^2(\mR )\)) and use Remark 14.9 to obtain
\[ \partial _t \hat {u}(t,\omega )=-\omega ^2\hat {u}(t,\omega ). \]
The initial value transforms to
\[ \hat {u}(0,\omega )=\hat {u}_0(\omega ). \]
We therefore want to solve the ODE initial value problem for \(t\mapsto \hat {u}(t,\omega )\) (where \(\omega \in \mR \) is a parameter):
\[ \partial _t \hat {u}(t,\omega )=-\omega ^2\hat {u}(t,\omega ),\qquad \hat {u}(0,\omega )=\hat {u}_0(\omega ). \]
This gives
\(\seteqnumber{0}{15.}{0}\)\begin{equation} \label {eq:uhatheat} \hat {u}(t,\omega )=\e ^{-\omega ^2t}\hat {u}_0(\omega ). \end{equation}
From Remark 14.11, we have
\[ \mathcal {F}\left (x\mapsto \frac {1}{\sqrt {2t}}\e ^{-x^2/(4t)}\right )= \omega \mapsto \e ^{-\omega ^2t}, \]
so that (15.1) is
\[ \mathcal {F}(u)=\mathcal {F}\left (\frac {1}{\sqrt {2t}}\e ^{-x^2/(4t)}\right )\mathcal {F}(u_0). \]
Using the convolution theorem (Theorem 14.14) we then obtain
\[ \mathcal {F}(u)=\frac {1}{\sqrt {2\pi }}\mathcal {F}\left (\frac {1}{\sqrt {2t}}\e ^{-x^2/(4t)}\ast u_0\right ), \]
so that
\[ u(t,x)=\frac {1}{\sqrt {4\pi t}}\e ^{-x^2/(4t)}\ast u_0, \]
which we can more explicitly write as
\[ u(t,x)=\frac {1}{\sqrt {4\pi t}}\int _{-\infty }^\infty \e ^{-(x-y)^2/(4t)}u_0(y)\,dy. \]
The function \((x,t)\mapsto \frac {1}{\sqrt {4\pi t}}\e ^{-x^2/(4t)}\) is called the heat kernel.
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Remark 15.2. The calculation in Chapter 13 more or less coincides with the above calculation (but the above calculation is a bit more streamlined). The solution
\[ u(t,x)=\frac {1}{\sqrt {2\pi }} \int _\mR C(\omega ) \e ^{-\omega ^2 t}\e ^{i\omega x}\,d\omega ,\qquad C(\omega )=\frac {1}{\sqrt {2\pi }}\int _\mR u^0(x) \e ^{-i\omega x}\,dx, \]
as given in Chapter 13 is
\[ C=\mathcal {F}(u^0),\qquad u=\mathcal {F}^{-1}\left (\mathcal {F}(u^0)\mathcal {F}\left (x\mapsto \frac {1}{\sqrt {2t}}\e ^{-x^2/(4t)}\right )\right ). \]
Applying the convolution theorem gives that this is
\[ u=\mathcal {F}^{-1}\left (\mathcal {F}\left (\frac {1}{\sqrt {2\pi }}u^0\ast \frac {1}{\sqrt {2t}}\e ^{-x^2/(4t)}\right )\right )=\frac {1}{\sqrt {2\pi }}u^0\ast \frac {1}{\sqrt {2t}}\e ^{-x^2/(4t)}, \]
which coincides with the solution given in this chapter.
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Proof. We have (using the substitution \(y=x-a\))
\(\seteqnumber{0}{15.}{1}\)\begin{multline*} \mathcal {F}\left (x\mapsto u(x-a)\right ) =\frac {1}{\sqrt {2\pi }}\int _{-\infty }^\infty \e ^{-i\omega x}u(x-a)\,dx =\frac {1}{\sqrt {2\pi }}\int _{-\infty }^\infty \e ^{-i\omega (y+a)}u(y)\,dy \\ =\e ^{-i\omega a}~\frac {1}{\sqrt {2\pi }}\int _{-\infty }^\infty \e ^{-i\omega y}u(y)\,dy =\e ^{-ia\omega }\hat {u}(\omega ). \end{multline*} □
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Remark 15.4. The term time-shift formula is also used for Lemma 15.3.
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Example 15.5. Let \(c>0\) and \(u_0:\mR \to \mR \) be given. We solve the transport equation
\[ \partial _tu=c\,\partial _xu,\qquad u(0,x)=u_0(x), \]
where \(x\in \mR \) and \(t>0\), using the Fourier transform.
Taking the Fourier transform in \(x\) of the differential equation and using the relation between differentiation and Fourier transformation (Remark 14.9) gives
\[ \partial _t \hat {u}(t,\omega )=ci\omega \hat {u}(t,\omega ),\qquad \hat {u}(0,\omega )=\hat {u}_0(\omega ). \]
We therefore want to solve this ODE initial value problem for \(t\mapsto \hat {u}(t,\omega )\) (where \(\omega \in \mR \) is a parameter). Solving this ODE initial value problem gives
\[ \hat {u}(t,\omega )=\e ^{ci\omega t}\hat {u}_0(\omega ). \]
By the space-shift formula (Lemma 15.3) we recognize the right-hand side as the Fourier transform of \(x\mapsto u_0(x+ct)\). Therefore
\[ u(t,x)=u_0(x+ct). \]