Partial Differential Equations III/IV

[6pt] Exercise Sheet 6: Solutions

1. The Fourier transform: The heat equation with source term.
   Remark: In this HTML solution we will use the notation ${ℱ}^{-1}\left(g\right)$ instead of $\check{g}$.

   • By the Fundamental Theorem of Calculus

     $\dot{x}(t)$ $=\lambda G{e}^{\lambda t}+{e^{\lambda (t-s)}F(s)|}_{s=t}+{\displaystyle {\int }_0^t}\lambda e^{\lambda (t-s)}F(s)𝑑s$

     $=\lambda G{e}^{\lambda t}+F(t)+\lambda {\displaystyle {\int }_0^t}{e}^{\lambda (t-s)}F(s)𝑑s$

     $=\lambda x(t)+F(t)$

     as claimed.

   • Taking the Fourier transform of $u_t=k{u}_{xx}+f$ with respect to the $x$ variable gives

     $$\widehat{u_t}=k\widehat{u_{xx}}+\widehat{f}\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}{\widehat{u}}_t(\xi ,t)=k{(i\xi )}^2\widehat{u}(\xi ,t)+\widehat{f}(\xi ,t)=-k{\xi }^2\widehat{u}(\xi ,t)+\widehat{f}(\xi ,t).$$

     Taking the Fourier transform of the initial condition $u(x,0)=g(x)$ gives

     $$\widehat{u}(\xi ,0)=\widehat{g}(\xi ).$$

     We have reduced the PDE to a one-parameter family of uncoupled ODEs, indexed by $\xi$:

     $${\widehat{u}}_t=-k{\xi }^2\widehat{u}+\widehat{f},\widehat{u}(\xi ,0)=\widehat{g}(\xi ).$$

     Applying part (i) with $x=\widehat{u}$, $\lambda =-k{\xi }^2$, $F=\widehat{f}$, $G=\widehat{g}$ gives

     $$\widehat{u}(\xi ,t)=\widehat{g}(\xi )e^{-k{\xi }^{2}t}+{\int }_0^t{e}^{-k{\xi }^2(t-s)}\widehat{f}(\xi ,s)𝑑s.$$

     Therefore

     $$u(x,t)={ℱ}^{-1}\left(\widehat{g}(\xi )e^{-k{\xi }^{2}t}\right)+{\int }_0^t{ℱ}^{-1}\left(e^{-k{\xi }^2(t-s)}\widehat{f}(\xi ,s)\right)𝑑s.$$ (1)

     Recall that

     $$\widehat{e^{-a{x}^2}}(\xi )=\frac{1}{\sqrt{2a}}{e}^{-\frac{{\xi }^2}{4a}}.$$

     Therefore

     $$e^{-k{\xi }^{2}t}=\sqrt{2a}\widehat{e^{-a{x}^2}}(\xi )\mathit{\hspace{1em}}\text{for}\mathit{\hspace{1em}}a=\frac{1}{4kt}.$$

     Since the product of Fourier transforms is the Fourier transform of a convolution, we obtain

     $\widehat{g}(\xi )e^{-k{\xi }^{2}t}$ $=\sqrt{2a}\widehat{g}(\xi )\widehat{e^{-a{x}^2}}(\xi )$

     $=\sqrt{2a}{\displaystyle \frac{1}{\sqrt{2\pi }}}\widehat{g*e^{-a{x}^2}}(\xi )$

     $=\sqrt{{\displaystyle \frac{a}{\pi }}}\widehat{g*e^{-a{x}^2}}(\xi )$

     $={\displaystyle \frac{1}{\sqrt{4\pi kt}}}\widehat{g*e^{-\frac{x^2}{4kt}}}(\xi )$

     since $a=\frac{1}{4kt}$. Therefore

     $${ℱ}^{-1}\left(\widehat{g}(\xi )e^{-k{\xi }^{2}t}\right)=\frac{1}{\sqrt{4\pi kt}}g*e^{-\frac{x^2}{4kt}}=\mathrm{\Phi }(\cdot ,t)*g={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathrm{\Phi }(x-y,t)g(y)𝑑y$$ (2)

     where $\mathrm{\Phi }$ is the fundamental solution of the heat equation in $ℝ$. Similarly

     $${ℱ}^{-1}\left(e^{-k{\xi }^2(t-s)}\widehat{f}(\xi ,s)\right)=\frac{1}{\sqrt{4\pi k(t-s)}}{e}^{-\frac{x^2}{4k(t-s)}}*f(\cdot ,s)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathrm{\Phi }(x-y,t-s)f(y,s)𝑑y.$$ (3)

     Combining equations (1), (2) and (3) yields

     $$u(x,t)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathrm{\Phi }(x-y,t)g(y)𝑑y+{\int }_0^t{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathrm{\Phi }(x-y,t-s)f(y,s)𝑑y𝑑s$$

     as required.

2. The Fourier transform: The transport equation.

   • By definition

     $\widehat{{\tau }_{a}v}(\xi )$ $={\displaystyle \frac{1}{\sqrt{2\pi }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{\tau }_{a}v(x)e^{-i\xi x}𝑑x$

     $={\displaystyle \frac{1}{\sqrt{2\pi }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}v(x-a)e^{-i\xi x}𝑑x$

     $={\displaystyle \frac{1}{\sqrt{2\pi }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}v(y)e^{-i\xi (y+a)}𝑑y$ $(y=x-a)$

     $=e^{-i\xi a}{\displaystyle \frac{1}{\sqrt{2\pi }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}v(y)e^{-i\xi y}𝑑y$

     $=e^{-i\xi a}\widehat{v}(\xi )$

     as required.

   • Taking the Fourier transform of the transport equation $u_t+c{u}_x=0$ gives

     $$\widehat{u_t}+c\widehat{u_x}=0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}{\widehat{u}}_t(\xi ,t)+ci\xi \widehat{u}(\xi ,t)=0$$

     and taking the Fourier transform of the initial condition $u(x,0)=g(x)$ gives

     $$\widehat{u}(\xi ,0)=\widehat{g}(\xi ).$$

     We have reduced the PDE to a one-parameter family of uncoupled ODEs, indexed by $\xi$:

     $${\widehat{u}}_t=-ci\xi \widehat{u},\widehat{u}(\xi ,0)=\widehat{g}(\xi ).$$

     Recall that the ODE $\dot{x}=\lambda x$ has solution $x(t)=x(0)e^{\lambda t}$. Applying this with $x=\widehat{u}$, $\lambda =-ci\xi$ yields

     $\widehat{u}(\xi ,t)$ $=\widehat{u}(\xi ,0)e^{-ci\xi t}$

     $=\widehat{g}(\xi )e^{-ci\xi t}$

     $=\widehat{{\tau }_{a}g}(\xi )$

     where $a=ct$, by part (i). By taking the inverse Fourier transform we obtain

     $$u(x,t)={\tau }_{a}g(x)=g(x-a)=g(x-ct)$$

     as desired.

3. The Fourier transform: Schrödinger’s equation.

   • Taking the Fourier transform of $i{u}_t=-u_{xx}$ with respect to the $x$ variable gives

     $$i\widehat{u_t}=-\widehat{u_{xx}}\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}i{\widehat{u}}_t(\xi ,t)=-{(i\xi )}^2\widehat{u}(\xi ,t)={\xi }^2\widehat{u}(\xi ,t).$$

     By multiplying by $-i$ we can rewrite this as ${\widehat{u}}_t=-i{\xi }^2\widehat{u}$. Taking the Fourier transform of the initial condition $u(x,0)=g(x)$ gives

     $$\widehat{u}(\xi ,0)=\widehat{g}(\xi ).$$

     We have reduced the PDE to a one-parameter family of uncoupled ODEs, indexed by $\xi$:

     $${\widehat{u}}_t=-i{\xi }^2\widehat{u},\widehat{u}(\xi ,0)=\widehat{g}(\xi ).$$

     Recall that the ODE $\dot{x}=\lambda x$ has solution $x(t)=x(0)e^{\lambda t}$. Applying this with $x=\widehat{u}$, $\lambda =-i{\xi }^2$ gives

     $$\widehat{u}(\xi ,t)=\widehat{u}(\xi ,0)e^{-i{\xi }^{2}t}=\widehat{g}(\xi )e^{-i{\xi }^{2}t}.$$ (4)

     To obtain $u$ we need to compute the following inverse Fourier transform:

     $${ℱ}^{-1}\left(\widehat{g}(\xi )e^{-i{\xi }^{2}t}\right).$$

     The trick is to recognise that $\widehat{g}(\xi )e^{-i{\xi }^{2}t}$ is the product of Fourier transforms, which follows from the fact that the Fourier transform of a Gaussian is a Gaussian. Recall that

     $$\widehat{e^{-a{x}^2}}(\xi )=\frac{1}{\sqrt{2a}}{e}^{-\frac{{\xi }^2}{4a}}.$$

     Therefore

     $$e^{-i{\xi }^{2}t}=\sqrt{2a}\widehat{e^{-a{x}^2}}(\xi )\mathit{\hspace{1em}}\text{for}\mathit{\hspace{1em}}a=\frac{1}{4it}.$$ (5)

     Since the product of Fourier transforms is the Fourier transform of a convolution, we obtain

     $\widehat{g}(\xi )e^{-k{\xi }^{2}t}$ $=\sqrt{2a}\widehat{g}(\xi )\widehat{e^{-a{x}^2}}(\xi )$ (by equation (5))

     $=\sqrt{2a}{\displaystyle \frac{1}{\sqrt{2\pi }}}\widehat{g*e^{-a{x}^2}}(\xi )$

     $=\sqrt{{\displaystyle \frac{a}{\pi }}}\widehat{g*e^{-a{x}^2}}(\xi ).$

     Combining this with equation (4) and taking the inverse Fourier transform gives

     $$\widehat{u}(\xi ,t)=\sqrt{\frac{a}{\pi }}\widehat{g*e^{-a{x}^2}}(\xi )\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}u(x,t)=\sqrt{\frac{a}{\pi }}g*e^{-a{x}^2}.$$

     Since $a=\frac{1}{4it}$ and the convolution is commutative we arrive at

     $$u(x,t)=\frac{1}{\sqrt{4\pi it}}g*e^{-\frac{x^2}{4it}}=\frac{1}{\sqrt{4\pi it}}{e}^{\frac{i{x}^2}{4t}}*g=\frac{1}{\sqrt{4\pi it}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{\frac{i{(x-y)}^2}{4t}}g(y)𝑑y$$

     as required.

   • In part (i) we showed that

     $$\widehat{u}(\xi ,t)=\widehat{g}(\xi )e^{-i{\xi }^{2}t}.$$

     Since the Fourier transform preserves the $L^2$–norm,

     ${\parallel u(\cdot ,t)\parallel }_{L^2(ℝ)}^2$ $={\parallel \widehat{u}(\cdot ,t)\parallel }_{L^2(ℝ)}^2$

     $={\parallel \widehat{g}(\xi )e^{-i{\xi }^{2}t}\parallel }_{L^2(ℝ)}^2$

     $={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{\left|\widehat{g}(\xi )e^{-i{\xi }^{2}t}\right|}^2𝑑\xi$

     $={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{\left|\widehat{g}(\xi )\right|}^2𝑑\xi$

     $={\parallel g\parallel }_{L^2(ℝ)}^2$

     as required.

4. The Fourier transform: The wave equation. Use the Fourier transform to derive the solution

   $$u(x,t)=\frac{1}{2}[g(x-ct)+g(x+ct)]$$

   of the wave equation

   $u_{tt}=c^2{u}_{xx}$ $\text{for }(x,t)\in ℝ\times (0,\mathrm{\infty }),$

   $u(x,0)=g(x)$ $\text{for }x\in ℝ,$

   $u_t(x,0)=0$ $\text{for }x\in ℝ,$

   where the constant $c>0$ is the wave speed. This is known as D’Alembert’s solution.
   Hint: Use Q2(i) and the fact that $\cos (c\xi t)=[\exp (ic\xi t)+\exp (-ic\xi t)]/2$.

   Taking the Fourier transform of $u_{tt}=c^2{u}_{xx}$ with respect to the $x$ variable gives

   $$\widehat{u_{tt}}=\widehat{c^2{u}_{xx}}\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}{\widehat{u}}_{tt}(\xi ,t)=c^2{(i\xi )}^2\widehat{u}(\xi ,t)=-c^2{\xi }^2\widehat{u}(\xi ,t).$$

   Taking the Fourier transform of the initial condition $u(x,0)=g(x)$ gives

   $$\widehat{u}(\xi ,0)=\widehat{g}(\xi ).$$

   Taking the Fourier transform of the initial condition $u_t(x,0)=0$ gives

   $${\widehat{u}}_t(\xi ,0)=0.$$

   We have reduced the PDE to a one-parameter family of uncoupled ODEs, indexed by $\xi$:

   $${\widehat{u}}_{tt}=-c^2{\xi }^2\widehat{u},\widehat{u}(\xi ,0)=\widehat{g}(\xi ),{\widehat{u}}_t(\xi ,0)=0.$$

   Recall that the ODE $\ddot{x}=-{\lambda }^{2}x$ has solution of the form $x(t)=A\cos (\lambda t)+B\sin (\lambda t)$. Applying this with $x=\widehat{u}$, $\lambda =c\xi$ gives

   $$\widehat{u}(\xi ,t)=A\cos (c\xi t)+B\sin (c\xi t).$$

   The initial conditions $\widehat{u}(\xi ,0)=\widehat{g}(\xi )$, ${\widehat{u}}_t(\xi ,0)=0$ imply that $A=\widehat{g}(\xi )$ and $B=0$. Therefore

   $\widehat{u}(\xi ,t)$ $=\widehat{g}(\xi )\cos (c\xi t)$

   $=\widehat{g}(\xi )\left[{\displaystyle \frac{\exp (ic\xi t)+\exp (-ic\xi t)}{2}}\right]$

   $={\displaystyle \frac{1}{2}}\exp (ic\xi t)\widehat{g}(\xi )+{\displaystyle \frac{1}{2}}\exp (-ic\xi t)\widehat{g}(\xi )$

   $={\displaystyle \frac{1}{2}}\widehat{{\tau }_{-a}g}(\xi )+{\displaystyle \frac{1}{2}}\widehat{{\tau }_{a}g}(\xi )$

   where $a=ct$, by Q2(i). Taking the inverse Fourier transform gives

   $u(x,t)$ $={\displaystyle \frac{1}{2}}{\tau }_{-a}g(x)+{\displaystyle \frac{1}{2}}{\tau }_{a}g(x)$

   $={\displaystyle \frac{1}{2}}g(x+a)+{\displaystyle \frac{1}{2}}g(x-a)$

   $={\displaystyle \frac{1}{2}}[g(x+ct)+g(x-ct)]$

   as required.

5. The Fourier transform of a derivative. By definition

   $\widehat{u^{\prime }}(\xi )$ $={\displaystyle \frac{1}{\sqrt{2\pi }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{u}^{\prime }(x)e^{-i\xi x}𝑑x$

   $=-{\displaystyle \frac{1}{\sqrt{2\pi }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}u(x){\displaystyle \frac{d}{dx}}{e}^{-i\xi x}𝑑x$ (integration by parts)

   $=-{\displaystyle \frac{1}{\sqrt{2\pi }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}u(x)(-i\xi )e^{-i\xi x}𝑑x$

   $=i\xi {\displaystyle \frac{1}{\sqrt{2\pi }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}u(x)e^{-i\xi x}𝑑x$

   $=i\xi \widehat{u}(\xi )$

   as required.

6. The Fourier transform of a convolution. By definition

   $\widehat{u*v}(\xi )$ $={\displaystyle \frac{1}{\sqrt{2\pi }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}(u*v)(x)e^{-i\xi x}𝑑x$

   $={\displaystyle \frac{1}{\sqrt{2\pi }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}\left({\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}u(z)v(x-z)𝑑z\right)e^{-i\xi x}𝑑x$

   $={\displaystyle \frac{1}{\sqrt{2\pi }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}\left({\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}u(z)v(x-z)𝑑z\right)e^{-i\xi z}{e}^{-i\xi (x-z)}𝑑x$

   $={\displaystyle \frac{1}{\sqrt{2\pi }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}\left({\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}v(x-z)e^{-i\xi (x-z)}𝑑x\right)u(z)e^{-i\xi z}𝑑z$

   $={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}\left({\displaystyle \frac{1}{\sqrt{2\pi }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}v(\tilde{x})e^{-i\xi \tilde{x}}𝑑\tilde{x}\right)u(z)e^{-i\xi z}𝑑z$ $(\tilde{x}=x-z)$

   $={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}\widehat{v}(\xi )u(z)e^{-i\xi z}𝑑z$

   $=\sqrt{2\pi }\widehat{v}(\xi ){\displaystyle \frac{1}{\sqrt{2\pi }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}u(z)e^{-i\xi z}𝑑z$

   $=\sqrt{2\pi }\widehat{v}(\xi )\widehat{u}(\xi )$

   as desired.

7. Proof of the Sobolev embedding using the Fourier transform.

   • Since the Fourier transform preserves the $L^2$–norm

     ${\parallel u\parallel }_{H^1(ℝ)}^2$ $={\parallel u\parallel }_{L^2(ℝ)}^2+{\parallel u^{\prime }\parallel }_{L^2(ℝ)}^2$

     $={\parallel \widehat{u}\parallel }_{L^2(ℝ)}^2+{\parallel \widehat{u^{\prime }}\parallel }_{L^2(ℝ)}^2$

     $={\parallel \widehat{u}\parallel }_{L^2(ℝ)}^2+{\parallel i\xi \widehat{u}\parallel }_{L^2(ℝ)}^2$

     $={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{|\widehat{u}(\xi )|}^2𝑑\xi +{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{|i\xi \widehat{u}(\xi )|}^2𝑑\xi$

     $={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}(1+{|\xi |}^2){|\widehat{u}(\xi )|}^2𝑑\xi .$

     as required.

   • Following the hint

     ${\parallel \widehat{u}\parallel }_{L^1(ℝ)}$ $={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}|\widehat{u}(\xi )|𝑑\xi$

     $={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{\displaystyle \frac{1}{{(1+{|\xi |}^2)}^{1/2}}}{(1+{|\xi |}^2)}^{1/2}|\widehat{u}(\xi )|𝑑\xi$

     $\le {\left({\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{\displaystyle \frac{1}{(1+{|\xi |}^2)}}𝑑\xi \right)}^{1/2}{\left({\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}(1+{|\xi |}^2){|\widehat{u}(\xi )|}^2𝑑\xi \right)}^{1/2}$ (Cauchy-Schwarz)

     $={\left({\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{\displaystyle \frac{1}{(1+{|\xi |}^2)}}𝑑\xi \right)}^{1/2}{\parallel u\parallel }_{H^1(ℝ)}$

     $=C{\parallel u\parallel }_{H^1(ℝ)}$

     where

     $$C={\left({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{(1+{|\xi |}^2)}𝑑\xi \right)}^{1/2}.$$

     If we can show that $C$ is finite, then we’ve completed the proof. This is a simple calculus exercise; one way is as follows:

     $C^2$ $=2{\displaystyle {\int }_0^{\mathrm{\infty }}}{\displaystyle \frac{1}{(1+{\xi }^2)}}𝑑\xi$

     $={\displaystyle {\int }_0^1}{\displaystyle \frac{1}{(1+{\xi }^2)}}𝑑\xi +{\displaystyle {\int }_1^{\mathrm{\infty }}}{\displaystyle \frac{1}{(1+{\xi }^2)}}𝑑\xi$

     $\le {\displaystyle {\int }_0^1}{\displaystyle \frac{1}{(1+0)}}𝑑\xi +{\displaystyle {\int }_1^{\mathrm{\infty }}}{\displaystyle \frac{1}{{\xi }^2}}𝑑\xi$

     $=1+1$

     as required.

   • By the Fourier Inversion Theorem

     $|u(x)|$ $=\left|{\displaystyle \frac{1}{\sqrt{2\pi }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}\widehat{u}(\xi )e^{i\xi x}𝑑\xi \right|$

     $\le {\displaystyle \frac{1}{\sqrt{2\pi }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}|\widehat{u}(\xi )||e^{i\xi x}|𝑑\xi$

     $={\displaystyle \frac{1}{\sqrt{2\pi }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}|\widehat{u}(\xi )|𝑑\xi$

     $={\displaystyle \frac{1}{\sqrt{2\pi }}}{\parallel \widehat{u}\parallel }_{L^1(ℝ)}$

     $\le {\displaystyle \frac{C}{\sqrt{2\pi }}}{\parallel u\parallel }_{H^1(ℝ)}$

     by part (ii). Since this holds for all $x\in ℝ$,

     $${\parallel u\parallel }_{L^{\mathrm{\infty }}(ℝ)}\le \frac{C}{\sqrt{2\pi }}{\parallel u\parallel }_{H^1(ℝ)}$$

     as required.

8. Fundamental Solution of the Heat Equation. We compute

   ${\mathrm{\Phi }}_t(𝒙,t)$ $={\displaystyle \frac{1}{{(4\pi k)}^{\frac{n}{2}}}}{e}^{-\frac{{|𝒙|}^2}{4kt}}\left(-{\displaystyle \frac{n}{2}}{t}^{-\frac{n}{2}-1}+{\displaystyle \frac{{|𝒙|}^2}{4k{t}^2}}\right)$

   ${\displaystyle \frac{\partial \mathrm{\Phi }}{\partial x_i}}(𝒙,t)$ $={\displaystyle \frac{1}{{(4\pi kt)}^{\frac{n}{2}}}}{e}^{-\frac{{|𝒙|}^2}{4kt}}\left(-{\displaystyle \frac{2x_i}{4kt}}\right),$

   ${\displaystyle \frac{{\partial }^2\mathrm{\Phi }}{\partial x_i^2}}(𝒙,t)$ $={\displaystyle \frac{1}{{(4\pi kt)}^{\frac{n}{2}}}}{e}^{-\frac{{|𝒙|}^2}{4kt}}\left(-{\displaystyle \frac{2}{4kt}}+{\left({\displaystyle \frac{2x_i}{4kt}}\right)}^2\right)={\displaystyle \frac{1}{{(4\pi kt)}^{\frac{n}{2}}}}{e}^{-\frac{{|𝒙|}^2}{4kt}}\left(-{\displaystyle \frac{1}{2kt}}+{\displaystyle \frac{x_i^2}{4k^2{t}^2}}\right),$

   $\mathrm{\Delta }\mathrm{\Phi }(𝒙,t)$ $={\displaystyle \sum _{i=1}^n}{\displaystyle \frac{{\partial }^2\mathrm{\Phi }}{\partial x_i^2}}(𝒙,t)={\displaystyle \frac{1}{{(4\pi kt)}^{\frac{n}{2}}}}{e}^{-\frac{{|𝒙|}^2}{4kt}}\left(-{\displaystyle \frac{n}{2kt}}+{\displaystyle \frac{{|𝒙|}^2}{4k^2{t}^2}}\right).$

   Therefore $\mathrm{\Phi }$ satisfies ${\mathrm{\Phi }}_t(𝒙,t)=k\mathrm{\Delta }\mathrm{\Phi }(𝒙,t)$ for all $𝒙\in {ℝ}^n$, $t>0$, as required.

9. Finite speed of propagation for a degenerate diffusion equation.

   • Observe that

     $$\frac{1}{2}{\left(\frac{3}{k\pi t}\right)}^{\frac{1}{3}}-\frac{1}{6k}\frac{x^2}{t}=0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}|x|=3^{\frac{2}{3}}{k}^{\frac{1}{3}}{\pi }^{-\frac{1}{6}}{t}^{\frac{1}{3}}$$

     and so

     Clearly $\mathrm{\Phi }$ satisfies the degenerate diffusion equation for $|x|>3^{\frac{2}{3}}{k}^{\frac{1}{3}}{\pi }^{-\frac{1}{6}}{t}^{\frac{1}{3}}$. For ,

     ${\mathrm{\Phi }}_t(x,t)$ $={\displaystyle \frac{1}{6}}{\left({\displaystyle \frac{3}{k\pi t}}\right)}^{-\frac{2}{3}}\left(-{\displaystyle \frac{3}{k\pi t^2}}\right)+{\displaystyle \frac{x^2}{6k{t}^2}}=-{\displaystyle \frac{1}{6t}}{\left({\displaystyle \frac{3}{k\pi t}}\right)}^{\frac{1}{3}}+{\displaystyle \frac{x^2}{6k{t}^2}},$

     ${\mathrm{\Phi }}_x(x,t)$ $=-{\displaystyle \frac{x}{3kt}},$

     $(a(\mathrm{\Phi }){\mathrm{\Phi }}_x)(x,t)$ $=-{\displaystyle \frac{x}{6t}}{\left({\displaystyle \frac{3}{k\pi t}}\right)}^{\frac{1}{3}}+{\displaystyle \frac{x^3}{18k{t}^2}},$

     ${(a(\mathrm{\Phi }){\mathrm{\Phi }}_x)}_x(x,t)$ $=-{\displaystyle \frac{1}{6t}}{\left({\displaystyle \frac{3}{k\pi t}}\right)}^{\frac{1}{3}}+{\displaystyle \frac{x^2}{6k{t}^2}}.$

     Therefore $\mathrm{\Phi }$ satisfies the degenerate diffusion equation for all $t>0$ and all $x\in ℝ$ with $|x|\ne 3^{\frac{2}{3}}{k}^{\frac{1}{3}}{\pi }^{-\frac{1}{6}}{t}^{\frac{1}{3}}$.

   • For fixed $t>0$, $\mathrm{\Phi }(x,t)$ is the maximum of a concave quadratic function and the zero function. The support of the map $x\mapsto \mathrm{\Phi }(x,t)$ is the compact interval

     $$[-3^{\frac{2}{3}}{k}^{\frac{1}{3}}{\pi }^{-\frac{1}{6}}{t}^{\frac{1}{3}},3^{\frac{2}{3}}{k}^{\frac{1}{3}}{\pi }^{-\frac{1}{6}}{t}^{\frac{1}{3}}].$$

10. The mathematical equation that caused the banks to crash. Let $t(x,\tau )$ satisfy

    $$\tau =T-t(x,\tau ).$$

    Differentiating this expression with respect to $\tau$ and $x$ gives

    $$t_{\tau }=-1,t_x=0.$$

    Let $S(x,\tau )$ satisfy

    $$x=\ln \left(\frac{S(x,\tau )}{K}\right)+\left(r-\frac{1}{2}{\sigma }^2\right)\tau .$$ (6)

    Differentiating this expressions with respect to $\tau$ gives

    $0={\displaystyle \frac{K}{S}}{\displaystyle \frac{S_{\tau }}{K}}+r-{\displaystyle \frac{1}{2}}{\sigma }^2\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}{S}_{\tau }=\left({\displaystyle \frac{1}{2}}{\sigma }^2-r\right)S.$

    Differentiating equation (6) with respect to $x$ gives

    $$1=\frac{K}{S}\frac{S_x}{K}\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}{S}_x=S.$$

    Therefore

    $u_{\tau }$ $=C{e}^{r\tau }\left[rV+V_S{S}_{\tau }+V_t{t}_{\tau }\right]=C{e}^{r\tau }\left[rV+\left({\displaystyle \frac{1}{2}}{\sigma }^2-r\right)S{V}_s-V_t\right],$

    $u_x$ $=C{e}^{r\tau }\left[V_S{S}_x+V_t{t}_x\right]=C{e}^{r\tau }S{V}_S,$

    $u_{xx}$ $=C{e}^{r\tau }\left[S_x{V}_S+S{V}_{SS}{S}_x\right]=C{e}^{r\tau }\left[S{V}_S+S^2{V}_{SS}\right].$

    Therefore

    $u_{\tau }-{\displaystyle \frac{1}{2}}{\sigma }^2{u}_{xx}$ $=C{e}^{r\tau }\left[rV+\left({\displaystyle \frac{1}{2}}{\sigma }^2-r\right)S{V}_s-V_t-{\displaystyle \frac{1}{2}}{\sigma }^2(S{V}_S+S^2{V}_{SS})\right]$

    $=-C{e}^{r\tau }\left[V_t+{\displaystyle \frac{1}{2}}{\sigma }^2{S}^2{V}_{SS}+rS{V}_S-rV\right]$

    $=0$

    since $V$ satisfies the Black-Scholes PDE. This completes the proof.

11. The energy method: Uniqueness for the heat equation in a time dependent domain. Let $u$ and $v$ be solutions and let $w=u-v$. Then $w$ satisfies

    $w_t-k{w}_{xx}=0$ $\text{for }(x,t)\in U,$

    $w(a(t),t)=0$ $\text{for }t\in [0,T],$

    $w(b(t),t)=0$ $\text{for }t\in [0,T],$

    $w(x,0)=0$ $\text{for }x\in (a(0),b(0)).$

    Multiply the equation $w_t=k{w}_{xx}$ by $w$ and integrate over $(a(t),b(t))$ to obtain

    $${\int }_{a(t)}^{b(t)}w{w}_t𝑑x=k{\int }_{a(t)}^{b(t)}w{w}_{xx}𝑑x.$$ (7)

    Recall the Fundamental Theorem of Calculus:

    $$\frac{d}{dt}{\int }_{a(t)}^{b(t)}f(x,t)𝑑x={\int }_{a(t)}^{b(t)}{f}_t(x,t)𝑑x+\dot{b}(t)f(b(t),t)-\dot{a}(t)f(a(t),t).$$

    We can use this to rewrite the left-hand side of equation (7) as follows:

    ${\displaystyle {\int }_{a(t)}^{b(t)}}w{w}_t𝑑x$ $={\displaystyle \frac{d}{dt}}{\displaystyle \frac{1}{2}}{\displaystyle {\int }_{a(t)}^{b(t)}}{w}^2(x,t)𝑑x-{\displaystyle \frac{1}{2}}\dot{b}(t)w^2(b(t),t)+{\displaystyle \frac{1}{2}}\dot{a}(t)w^2(a(t),t)$

    $={\displaystyle \frac{d}{dt}}{\displaystyle \frac{1}{2}}{\displaystyle {\int }_{a(t)}^{b(t)}}{w}^2(x,t)𝑑x$ (8)

    since $w(a(t),t)=w(b(t),t)=0$. We rewrite the right-hand side of equation (7) using integration by parts:

    $$k{\int }_{a(t)}^{b(t)}w{w}_{xx}𝑑x={kw{w}_x|}_{a(t)}^{b(t)}-k{\int }_{a(t)}^{b(t)}{w}_x^2𝑑x=-k{\int }_{a(t)}^{b(t)}{w}_x^2𝑑x,$$ (9)

    again using the fact that $w(a(t),t)=w(b(t),t)=0$. Substituting (8) and (9) into (7) gives

    $$\frac{d}{dt}\frac{1}{2}{\int }_{a(t)}^{b(t)}{w}^2(x,t)𝑑x=-k{\int }_{a(t)}^{b(t)}{w}_x^2𝑑x\le 0.$$

    Let

    $$E(t)=\frac{1}{2}{\int }_{a(t)}^{b(t)}{w}^2(x,t)𝑑x.$$

    We have shown that $\dot{E}(t)\le 0$. Hence

    $$0\le E(t)\le E(0)=0$$

    since $w=0$ for $t=0$. Consequently $E(t)=0$ for all $t$. Therefore $w=0$ in $U$ and so $u=v$, as required.

12. The energy method: Uniqueness for a 4th-order heat equation. Let $u$ and $v$ be solutions and let $w=u-v$. Then $w$ satisfies

    $w_t+k{w}_{xxxx}=0$ $\text{for }(x,t)\in (a,b)\times (0,T],$ (10)

    $w(a,t)=w(b,t)=0$ $\text{for }t\in [0,T],$ (11)

    $w_x(a,t)=w_x(b,t)=0$ $\text{for }t\in [0,T],$ (12)

    $w(x,0)=0$ $\text{for }x\in (a,b).$ (13)

    Multiplying the equation $w_t=-k{w}_{xxxx}$ by $w$ and integrating over $(a,b)$ gives

    ${\displaystyle {\int }_a^b}\underset{\frac{1}{2}\frac{\partial }{\partial t}{w}^2}{\underbrace{w{w}_t}}𝑑x$ $=-k{\displaystyle {\int }_a^b}w{w}_{xxxx}𝑑x$

    $=-{kw{w}_{xxx}|}_a^b+k{\displaystyle {\int }_a^b}{w}_x{w}_{xxx}𝑑x$

    $=k{\displaystyle {\int }_a^b}{w}_x{w}_{xxx}𝑑x$ (by (11))

    $={k{w}_x{w}_{xx}|}_a^b-k{\displaystyle {\int }_a^b}{w}_{xx}{w}_{xx}𝑑x$

    $=-k{\displaystyle {\int }_a^b}{w}_{xx}^2𝑑x$ (by (12)).

    Therefore

    $$\frac{d}{dt}\frac{1}{2}{\int }_a^b{w}^2𝑑x=-k{\int }_a^b{w}_{xx}^2𝑑x\le 0$$

    and so

    $$0\le {\int }_a^b{w}^2(x,t)𝑑x\le {\int }_a^b{w}^2(x,0)𝑑x=0$$

    by (13). We conclude that $w=0$ and hence $u=v$, as required.

    We consider $u_t+k{u}_{xxxx}=0$ to be the 4th-order version of the heat equation $u_t-k{u}_{xx}=0$ since it has the same energy-decay property:

    $$\frac{d}{dt}{\parallel u\parallel }_{L^2([a,b])}^2\le 0$$

    (provided that $u$ and $u_x$ vanish at $x=a$ and $x=b$). The equation $u_t-k{u}_{xxxx}=0$ looks more similar to the heat equation $u_t-k{u}_{xx}=0$ (because of the minus sign), but its $L^2$–energy grows with time:

    $$\frac{d}{dt}{\parallel u\parallel }_{L^2([a,b])}^2\ge 0.$$

13. Asymptotic behaviour of the heat equation with time independent data. Let $w(𝒙,t)=u(𝒙,t)-v(𝒙)$. We need to prove that ${lim}_{t\to \mathrm{\infty }}{\parallel w\parallel }_{L^2(\mathrm{\Omega })}=0$. By subtracting the PDEs for $u$ and $v$ we find that $w$ satisfies

    $w_t(𝒙,t)-k\mathrm{\Delta }w(𝒙,t)=0$ $\text{for }(𝒙,t)\in \mathrm{\Omega }\times (0,\mathrm{\infty }),$

    $w(𝒙,t)=0$ $\text{for }(𝒙,t)\in \partial \mathrm{\Omega }\times [0,\mathrm{\infty }),$

    $w(𝒙,0)=u_0(𝒙)-v(𝒙)$ $\text{for }𝒙\in \mathrm{\Omega }.$

    Multiplying the equation $w_t=k\mathrm{\Delta }w$ by $w$ and integrating by parts over $\mathrm{\Omega }$ gives

    ${\displaystyle {\int }_{\mathrm{\Omega }}}w{w}_t𝑑𝒙=k{\displaystyle {\int }_{\mathrm{\Omega }}}w\mathrm{\Delta }w𝑑𝒙$ $⟺\mathit{\hspace{1em}}{\displaystyle \frac{d}{dt}}{\displaystyle \frac{1}{2}}{\displaystyle {\int }_{\mathrm{\Omega }}}{w}^2𝑑𝒙=k{\displaystyle {\int }_{\partial \mathrm{\Omega }}}w\nabla w\cdot 𝒏dS-k{\displaystyle {\int }_{\mathrm{\Omega }}}\nabla w\cdot \nabla wd𝒙$

    $⟺\mathit{\hspace{1em}}{\displaystyle \frac{d}{dt}}{\displaystyle \frac{1}{2}}{\displaystyle {\int }_{\mathrm{\Omega }}}{w}^2𝑑𝒙=-k{\displaystyle {\int }_{\mathrm{\Omega }}}{|\nabla w|}^2𝑑𝒙$ (14)

    since $w=0$ on $\partial \mathrm{\Omega }$. By the Poincaré inequality, there exists a constant $C_p>0$ such that

    $${\int }_{\mathrm{\Omega }}{|w|}^2𝑑𝒙\le C_p{\int }_{\mathrm{\Omega }}{|\nabla w|}^2𝑑𝒙.$$

    Multiplying this by $-k/C_p$ gives

    $$-\frac{k}{C_p}{\int }_{\mathrm{\Omega }}{|w|}^2𝑑𝒙\ge -k{\int }_{\mathrm{\Omega }}{|\nabla w|}^2𝑑𝒙.$$ (15)

    Combining equations (14), (15) yields

    $$\frac{d}{dt}\frac{1}{2}{\int }_{\mathrm{\Omega }}{w}^2𝑑𝒙\le -\frac{k}{C_p}{\int }_{\mathrm{\Omega }}{|w|}^2𝑑𝒙.$$

    Define

    $$E(t)={\int }_{\mathrm{\Omega }}{w}^2(𝒙,t)𝑑𝒙={\parallel w\parallel }_{L^2(\mathrm{\Omega })}^2$$

    and $\lambda =\frac{2k}{C_p}$. We have shown that

    $$\dot{E}\le -\lambda E.$$

    By the Gr̈onwall inequality,

    $$E(t)\le e^{-\lambda t}E(0).$$

    Since $\lambda >0$, we conclude that $E(t)\to 0$ as $t\to \mathrm{\infty }$. Therefore $w\to 0$ in $L^2(\mathrm{\Omega })$ as $t\to \mathrm{\infty }$, as required.

14. Asymptotic behaviour of the heat equation with time independent data in the $L^{\mathrm{\infty }}$–norm.

    • By linearity, $w$ satisfies

      $w_t(x,t)-k{w}_{xx}(x,t)=0$ $\text{for }(x,t)\in (a,b)\times (0,\mathrm{\infty }),$

      $w(x,0)=u_0(x)-v(x)$ $\text{for }x\in (a,b),$

      $w(a,t)=w(b,t)=0$ $\text{for }t\in [0,\mathrm{\infty }).$

      Multiplying the PDE by $w$ and integrating over $[a,b]$ gives

      $${\int }_a^{b}w{w}_t𝑑x=k{\int }_a^{b}w{w}_{xx}𝑑x\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}\frac{d}{dt}\frac{1}{2}{\int }_a^b{w}^2𝑑x=\underset{=0}{\underbrace{{kw{w}_x|}_a^b}}-k{\int }_a^b{w}_x^2𝑑x$$

      where we have used the Chain Rule and integration by parts. Note that the boundary terms vanish when we perform integration by parts since $w(a,t)=w(b,t)=0$. Therefore

      $$\frac{d}{dt}{\int }_a^b{w}^2(x,t)𝑑x=-2k{\int }_a^b{w}_x^2(x,t)𝑑x$$

      as required.

    • We can write the result of part (i) in terms of $L^2$–norms as

      $$\frac{d}{dt}{\parallel w\parallel }_{L^2([a,b])}^2=-2k{\parallel w_x\parallel }_{L^2([a,b])}^2.$$ (16)

      By the Poincaré inequality, there exists a constant $C>0$ such that

      $${\parallel w\parallel }_{L^2([a,b])}^2\le C{\parallel w_x\parallel }_{L^2([a,b])}^2.$$ (17)

      Combining equations (16), (17) gives

      $$\frac{d}{dt}{\parallel w\parallel }_{L^2([a,b])}^2=-2k{\parallel w_x\parallel }_{L^2([a,b])}^2\le -\frac{2k}{C}{\parallel w\parallel }_{L^2([a,b])}^2.$$ (18)

      Define

      $$E(t)={\parallel w\parallel }_{L^2([a,b])}^2.$$

      We can rewrite equation (18) as

      $$\dot{E}\le -\lambda E$$

      with $\lambda =2k/C>0$. By the Grönwall inequality,

      $$E(t)\le E(0)e^{-\lambda t}\to 0\mathit{\hspace{1em}}\text{as }t\to \mathrm{\infty }.$$

      Therefore, by definition of $E$, $w\to 0$ in $L^2([a,b])$ as $t\to \mathrm{\infty }$, as required.

    • By differentiating the PDE for $w$ with respect to $t$ we obtain

      $w_{tt}(x,t)-k{w}_{txx}(x,t)=0$ $\text{for }(x,t)\in (a,b)\times (0,\mathrm{\infty }),$

      $w_t(a,t)=w_t(b,t)=0$ $\text{for }t\in [0,\mathrm{\infty }).$

      In particular, $w_t$ satisfies the heat equation with Dirichlet boundary conditions, just like $w$. Therefore the argument we applied in parts (i) and (ii) to $w$ can also be applied to $w_t$, which yields $w_t\to 0$ in $L^2([a,b])$ as $t\to \mathrm{\infty }$.

    • We have

      ${\parallel w_x\parallel }_{L^2([a,b])}^2$ $={\displaystyle {\int }_a^b}{w}_x^2(x,t)𝑑x$

      $=-{\displaystyle \frac{1}{2k}}{\displaystyle \frac{d}{dt}}{\displaystyle {\int }_a^b}{w}^2(x,t)𝑑x$ (by part (i))

      $=-{\displaystyle \frac{1}{k}}{\displaystyle {\int }_a^b}w(x,t)w_t(x,t)𝑑x$

      $\le {\displaystyle \frac{1}{k}}{\left({\displaystyle {\int }_a^b}{w}^2(x,t)𝑑x\right)}^{1/2}{\left({\displaystyle {\int }_a^b}{w}_t^2(x,t)𝑑x\right)}^{1/2}$ (Cauchy-Schwarz)

      $={\displaystyle \frac{1}{k}}{\parallel w\parallel }_{L^2([a,b])}{\parallel w_t\parallel }_{L^2([a,b])}\to 0\mathit{\hspace{1em}}\text{as }t\to \mathrm{\infty }$

      by parts (ii) and (iii). Therefore $w_x\to 0$ in $L^2([a,b])$ as $t\to \mathrm{\infty }$, as required. Note that we don’t really need $w_t\to 0$ in $L^2([a,b])$ as $t\to \mathrm{\infty }$, we just need ${\parallel w_t\parallel }_{L^2([a,b])}$ to be uniformly bounded in $t$.

    • This final result follows from the Sobolev inequality: There exists a constant $C>0$ such that

      $${\parallel w\parallel }_{L^{\mathrm{\infty }}([a,b])}\le C{\parallel w\parallel }_{H^1([a,b])}=C{\left({\parallel w\parallel }_{L^2([a,b])}+{\parallel w_x\parallel }_{L^2([a,b])}\right)}^{1/2}\to 0\mathit{\hspace{1em}}\text{as }t\to \mathrm{\infty }$$

      by parts (ii) and (iv).

15. Applications of the maximum principle: Uniqueness and bounds on solutions.

    • Let ${\mathrm{\Gamma }}_T=[a,b]\times \{0\}\cup \{a,b\}\times [0,T]$ be the parabolic boundary of ${\mathrm{\Omega }}_T$. Let $u,v\in C_1^2({\mathrm{\Omega }}_T)\cap C({\overline{\mathrm{\Omega }}}_T)$ satisfy

      $u_t-u_{xx}=1$ $\text{in }{\mathrm{\Omega }}_T,$

      $u=0$ $\text{in }{\mathrm{\Gamma }}_T.$

      Then $w=u-v\in C_1^2({\mathrm{\Omega }}_T)\cap C({\overline{\mathrm{\Omega }}}_T)$ satisfies

      $w_t-w_{xx}=0$ $\text{in }{\mathrm{\Omega }}_T,$

      $w=0$ $\text{in }{\mathrm{\Gamma }}_T.$

      By the weak maximum principle

      $$\underset{{\overline{\mathrm{\Omega }}}_T}{\max }w=\underset{{\mathrm{\Gamma }}_T}{\max }w=0,\underset{{\overline{\mathrm{\Omega }}}_T}{\min }w=\underset{{\mathrm{\Gamma }}_T}{\min }w=0.$$

      Therefore $w=0$ and $u=v$, as required.

    • Since $u_t-u_{xx}=1>0$, the weak maximum principle gives

      $$\underset{{\overline{\mathrm{\Omega }}}_T}{\min }u=\underset{{\mathrm{\Gamma }}_T}{\min }u=0.$$

      This is the desired lower bound on $u$. We still need to prove the upper bound. Let $v(x,t)=t$. Then $v_t-v_{xx}=1$ and $w=u-v$ satisfies

      $w_t-w_{xx}=0$ $\text{in }{\mathrm{\Omega }}_T,$

      $w=-t$ $\text{in }{\mathrm{\Gamma }}_T.$

      By the weak maximum principle

      $$\underset{{\overline{\mathrm{\Omega }}}_T}{\max }w=\underset{{\mathrm{\Gamma }}_T}{\max }w=\underset{{\mathrm{\Gamma }}_T}{\max }(-t)=0.$$

      Therefore $w\le 0$ in ${\mathrm{\Omega }}_T$ and hence $u\le v=t$ in ${\mathrm{\Omega }}_T$, which is the desired upper bound.

16. Application of the maximum principle: Comparison Principle. Define $v=u_1-u_2$. Then $v$ satisfies

    ${\displaystyle \frac{\partial v}{\partial t}}(𝒙,t)-k\mathrm{\Delta }v(𝒙,t)=f_1(𝒙)-f_2(𝒙)$ $\text{for }(𝒙,t)\in \mathrm{\Omega }\times (0,T],$

    $v(𝒙,t)=g_1(𝒙)-g_2(𝒙)$ $\text{for }(𝒙,t)\in \partial \mathrm{\Omega }\times [0,T],$

    $v(𝒙,0)=u_1^0(𝒙)-u_2^0(𝒙)$ $\text{for }𝒙\in \mathrm{\Omega }.$

    Since $f_1\le f_2$, then

    $$v_t-k\mathrm{\Delta }v=f_1-f_2\le 0\mathit{\hspace{1em}}\text{in }{\mathrm{\Omega }}_T.$$

    Therefore the weak maximum principle implies that

    $$\underset{\overline{{\mathrm{\Omega }}_T}}{\max }v=\underset{{\mathrm{\Gamma }}_T}{\max }v.$$

    For $(𝒙,t)\in {\mathrm{\Gamma }}_T$,

    But

    $$g_1-g_2\le 0,u_1^0-u_2^0\le 0.$$

    Therefore $v\le 0$ on ${\mathrm{\Gamma }}_T$ and hence

    $$\underset{\overline{{\mathrm{\Omega }}_T}}{\max }v=\underset{{\mathrm{\Gamma }}_T}{\max }v\le 0.$$

    Hence $v\le 0$ in ${\mathrm{\Omega }}_T$ and so $u_1\le u_2$ in ${\mathrm{\Omega }}_T$, as required.

17. Eigenfunctions of the Laplacian and an application to the heat equation. Formally (not worrying about interchanging limits and infinite sums),

    $0$ $=v_t-k\mathrm{\Delta }v$

    $={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\dot{c}}_n(t)u_n(𝒙)-k{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{c}_n(t)\mathrm{\Delta }{u}_n(𝒙)$

    $={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\dot{c}}_n(t)u_n(𝒙)+k{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{c}_n(t){\lambda }_n{u}_n(𝒙)$

    $={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}({\dot{c}}_n(t)+k{\lambda }_n{c}_n(t))u_n(𝒙).$

    Since ${\{u_n\}}_{n\in \mathbb{N}}$ forms an orthogonal basis, it follows that

    $${\dot{c}}_n(t)+k{\lambda }_n{c}_n(t)=0$$

    for all $n$. We also have

    $$v(𝒙,0)=g(𝒙)\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}\sum _{n=1}^{\mathrm{\infty }}{c}_n(0)u_n(𝒙)=\sum _{n=1}^{\mathrm{\infty }}{g}_n{u}_n(𝒙)\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}\sum _{n=1}^{\mathrm{\infty }}(c_n(0)-g_n)u_n(𝒙)=0.$$

    Again, since ${\{u_n\}}_{n\in \mathbb{N}}$ forms an orthogonal basis, it follows that

    $$c_n(0)=g_n$$

    for all $n$. We have reduced the PDE for $v$ to a one-parameter family of uncoupled ODEs, indexed by $n$:

    $${\dot{c}}_n(t)=-k{\lambda }_n{c}_n(t),c_n(0)=g_n.$$

    These ODEs have solutions

    $$c_n(t)=g_n{e}^{-k{\lambda }_{n}t}.$$

    Therefore

    $$v(𝒙,t)=\sum _{n=1}^{\mathrm{\infty }}{g}_n{e}^{-k{\lambda }_{n}t}{u}_n(𝒙)$$

    as required.