Partial Differential Equations III/IV

[6pt] Exercise Sheet 1: Solutions

1. Examples of PDEs.

   • The Schrödinger equation:

     $$i\mathrm{\hslash }{\mathrm{\Psi }}_t=-\frac{{\mathrm{\hslash }}^2}{2m}\mathrm{\Delta }\mathrm{\Psi }+V\mathrm{\Psi }.$$

     This is a linear second-order PDE.

   • The $p$–Laplace equation:

     $$\mathrm{div}({|\nabla u|}^{p-2}\nabla u)=0.$$

     If $p=2$ this is just Laplace’s equation and therefore a linear second-order PDE. If $p\in (2,\mathrm{\infty })$, then this is a quasilinear second-order PDE. One way of seeing this is to expand the divergence on the left-hand side. By using the vector identity

     $$\mathrm{div}(\phi 𝒇)=\nabla \phi \cdot 𝒇+\phi \mathrm{div}𝒇$$

     we obtain

     $$\mathrm{div}({|\nabla u|}^{p-2}\nabla u)=\nabla ({|\nabla u|}^{p-2})\cdot \nabla u+{|\nabla u|}^{p-2}\mathrm{div}\nabla u.$$

     Recall that $\nabla |𝒙|=\frac{𝒙}{|𝒙|}$. By the Chain Rule, the $i$–th component of $\nabla ({|\nabla u|}^{p-2})$ is

     $$\frac{\partial }{\partial x_i}{|\nabla u|}^{p-2}=(p-2){|\nabla u|}^{p-3}\sum _{j=1}^n\frac{u_{x_j}}{|\nabla u|}\frac{\partial u_{x_j}}{\partial x_i}=(p-2){|\nabla u|}^{p-4}\sum _{j=1}^n{u}_{x_i{x}_j}{u}_{x_j}.$$

     In vector notation:

     $$\nabla ({|\nabla u|}^{p-2})=(p-2){|\nabla u|}^{p-4}{D}^{2}u\nabla u$$

     where $D^{2}u$ is the matrix of second partial derivatives of $u$, which has components ${[D^{2}u]}_{ij}=u_{x_i{x}_j}$. Therefore the $p$–Laplace equation is

     $$(p-2){|\nabla u|}^{p-4}\nabla u\cdot D^{2}u\nabla u+{|\nabla u|}^{p-2}\mathrm{\Delta }u=0,$$

     which is clearly quasilinear and second-order.

   • The Cahn-Hilliard equation:

     $$u_t=k\mathrm{\Delta }(u^3-u-{\epsilon }^2\mathrm{\Delta }u).$$

     This is a semilinear fourth-order PDE. If you can’t see this immediately, then expand the right-hand side of the PDE to obtain

     $$u_t=k\mathrm{\Delta }(u^3)-k\mathrm{\Delta }u-k{\epsilon }^2{\mathrm{\Delta }}^{2}u$$

     where ${\mathrm{\Delta }}^{2}u=\mathrm{\Delta }(\mathrm{\Delta }u)$ is the bilaplacian operator. You should already be able to see that this is a semilinear fourth-order PDE but, if not, then expand the right-hand side further to obtain

     $$u_t=6ku{|\nabla u|}^2+3k{u}^2\mathrm{\Delta }u-k\mathrm{\Delta }u-k{\epsilon }^2\left(\sum _{i=1}^n\sum _{j=1}^n\frac{{\partial }^{4}u}{\partial x_i^2{x}_j^2}\right)$$

     where we have used that $\nabla (u^3)=3u^2\nabla u$ and so

     $$\mathrm{\Delta }(u^3)=\mathrm{div}\nabla (u^3)=\mathrm{div}(3u^2\nabla u)=\nabla (3u^2)\cdot \nabla u+3u^2\mathrm{div}\nabla u=6u{|\nabla u|}^2+3u^2\mathrm{\Delta }u.$$

   • The nonlinear wave equation:

     $$\rho {𝒓}_{tt}=\frac{\partial }{\partial x}\left(N(|{𝒓}_x|)\frac{{𝒓}_x}{|{𝒓}_x|}\right)+𝒇.$$

     Expanding the right-hand side gives

     $\rho {𝒓}_{tt}$ $=N^{\prime }(|{𝒓}_x|){\displaystyle \frac{\partial |{𝒓}_x|}{\partial x}}{\displaystyle \frac{{𝒓}_x}{|{𝒓}_x|}}+N(|{𝒓}_x|){\displaystyle \frac{{𝒓}_{xx}}{|{𝒓}_x|}}-N(|{𝒓}_x|){\displaystyle \frac{{𝒓}_x}{{|{𝒓}_x|}^2}}{\displaystyle \frac{\partial |{𝒓}_x|}{\partial x}}+𝒇$

     $=N^{\prime }(|{𝒓}_x|)\left({\displaystyle \frac{{𝒓}_x}{|{𝒓}_x|}}\cdot {𝒓}_{xx}\right){\displaystyle \frac{{𝒓}_x}{|{𝒓}_x|}}+N(|{𝒓}_x|){\displaystyle \frac{{𝒓}_{xx}}{|{𝒓}_x|}}-N(|{𝒓}_x|){\displaystyle \frac{{𝒓}_x}{{|{𝒓}_x|}^2}}\left({\displaystyle \frac{{𝒓}_x}{|{𝒓}_x|}}\cdot {𝒓}_{xx}\right)+𝒇.$

     The highest derivatives ${𝒓}_{tt}$ and ${𝒓}_{xx}$ appear linearly and their coefficients only depend on lower order derivatives. Therefore the nonlinear wave equation is a quasilinear second-order system of PDEs.

2. Characterisation of PDEs.

   • General form of a second-order scalar PDE in two independent variables:

     $$F(x,y,u,u_x,u_y,u_{xx},u_{x,y},u_{yy})=0$$

     or in vector notation

     $$F(𝒙,u,\nabla u,D^{2}u)=0,$$

     where $F:\mathrm{\Omega }\times ℝ\times {ℝ}^2\times {ℝ}^4\to ℝ$ is a given function, $\mathrm{\Omega }\subseteq {ℝ}^2$ is a given domain and $u:\mathrm{\Omega }\to ℝ$ stands for the unknown function.

   • General form of a linear, second-order scalar PDE in two independent variables:

     $$a_{11}(x,y)u_{xx}+2a_{12}(x,y)u_{xy}+a_{22}(x,y)u_{yy}+b_1(x,y)u_x+b_2(x,y)u_y+c(x,y)u+d(x,y)=0.$$

     The factor of $2$ is just so that we can write this equation more compactly in vector notation as

     $$A(x,y):D^{2}u+𝒃(x,y)\cdot \nabla u+c(x,y)u+d(x,y)=0$$

     where $A:\mathrm{\Omega }\to {ℝ}^{2\times 2}$ (a symmetric matrix valued function), $𝒃:\mathrm{\Omega }\to {ℝ}^2$ and $c,d:\mathrm{\Omega }\to ℝ$ are given, $\mathrm{\Omega }\subseteq {ℝ}^2$ is a given domain and $u:\mathrm{\Omega }\to ℝ$ stands for the unknown function.

   • General form of a semilinear, second-order scalar PDE in two independent variables:

     $$a_{11}(x,y)u_{xx}+2a_{12}(x,y)u_{xy}+a_{22}(x,y)u_{yy}+b(x,y,u,u_x,u_y)=0$$

     or in vector notation

     $$A(x,y):D^{2}u+b(x,y,u,\nabla u)=0$$

     where $A:\mathrm{\Omega }\to {ℝ}^{2\times 2}$ (a symmetric matrix valued function) and $b:\mathrm{\Omega }\times ℝ\times {ℝ}^2\to ℝ$ are given, $\mathrm{\Omega }\subseteq {ℝ}^2$ is a given domain and $u:\mathrm{\Omega }\to ℝ$ stands for the unknown function.

   • General form of a quasilinear, second-order scalar PDE in two independent variables:

     $$a_{11}(x,y,u,u_x,u_y)u_{xx}+2a_{12}(x,y,u,u_x,u_y)u_{xy}+a_{22}(x,y,u,u_x,u_y)u_{yy}+b(x,y,u,u_x,u_y)=0$$

     or in vector notation

     $$A(x,y,u,\nabla u):D^{2}u+b(x,y,u,\nabla u)=0$$

     where $A:\mathrm{\Omega }\times ℝ\times {ℝ}^2\to {ℝ}^{2\times 2}$ (a symmetric matrix valued function) and $b:\mathrm{\Omega }\times ℝ\times {ℝ}^2\to ℝ$ are given, $\mathrm{\Omega }\subseteq {ℝ}^2$ is a given domain and $u:\mathrm{\Omega }\to ℝ$ stands for the unknown function.

3. The transport equation: Derivation of the travelling wave solution.

   • Observe that

     $$\nabla u\cdot \left(\begin{array}{c}\hfill c\hfill \\ \hfill 1\hfill \end{array}\right)=\left(\begin{array}{c}\hfill u_x\hfill \\ \hfill u_t\hfill \end{array}\right)\cdot \left(\begin{array}{c}\hfill c\hfill \\ \hfill 1\hfill \end{array}\right)=c{u}_x+u_t.$$

     Therefore

     $$u_t+c{u}_x=0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}\nabla u\cdot \left(\begin{array}{c}\hfill c\hfill \\ \hfill 1\hfill \end{array}\right)=0.$$

   • It follows that $u$ is constant in the direction ${(c,1)}^{\mathrm{T}}$. In particular, for each $x_0\in ℝ$, $u$ is constant on the line that passes through $(x,t)=(x_0,0)$ and has direction ${(c,1)}^{\mathrm{T}}$. Points on this line have the form

     $$\left(\begin{array}{c}\hfill x\hfill \\ \hfill t\hfill \end{array}\right)=\left(\begin{array}{c}\hfill x_0\hfill \\ \hfill 0\hfill \end{array}\right)+\lambda \left(\begin{array}{c}\hfill c\hfill \\ \hfill 1\hfill \end{array}\right)=\left(\begin{array}{c}\hfill x_0+c\lambda \hfill \\ \hfill \lambda \hfill \end{array}\right),\lambda \in ℝ.$$

     Since $u$ is constant on this line, $u(x_0+c\lambda ,\lambda )$ is independent of $\lambda$. In particular, we obtain the same value if we take $\lambda =t$ and $\lambda =0$:

     $$u(x_0+ct,t)=u(x_0+c\cdot 0,0)=u(x_0,0)=g(x_0)$$

     by the initial condition. Therefore

     $$u(x_0+ct,t)=g(x_0).$$

     Making the change of variables $x=x_0+ct$ gives

     $$u(x,t)=g(x-ct)$$

     as required.

4. The transport equation on ${ℝ}^n$. The function

   $$u(𝒙,t)=g(𝒙-𝒄t)$$

   satisfies the transport equation

   $u_t+𝒄\cdot \nabla u=0$ $\text{for }(𝒙,t)\in {ℝ}^n\times (0,\mathrm{\infty }),$

   $u(𝒙,0)=g(𝒙)$ $\text{for }x\in ℝ.$

5. The transport equation with boundary conditions.

   • We want to solve the initial-boundary value problem

     $u_t+4u_x=0$ $\text{for }(x,t)\in (0,\mathrm{\infty })\times (0,\mathrm{\infty }),$

     $u(x,0)=0$ $\text{for }x\in (0,\mathrm{\infty }),$

     $u(0,t)=t^2{e}^{-t}$ $\text{for }t\in [0,\mathrm{\infty }).$

     Method 1: Use physical intuition to guess the solution. We know that the transport equation has the property of transporting information with velocity $c$. In this case $c=4$ and so information is transported to the right with speed $4$. Initially $u$ is zero everywhere. Fix a point $x>0$. The boundary data $u(0,t)=t^2{e}^{-t}$ is transported to the right with speed $4$ and so will take $x/4$ amount of time to reach the point $x$ from the boundary point $0$ (recall that time = distance/speed). Therefore $u(x,t)=0$ up until time $t=x/4$.

     From this point in time onwards, the value of $u$ will be determined by the boundary data. The value of $u$ at point $x$ at time $t$ will be the same as the value of $u$ at the boundary point $0$ at time $t-x/4$, i.e., $x/4$ amount of time in the past. Therefore $u(x,t)=u(0,t-x/4)={(t-x/4)}^2\exp (-(t-x/4))$. We have argued that

     We can rewrite this as

     $u(x,t)$

     $=f(x-4t)$

     where

     Note that $f$ is continuously differentiable since the left and right limits of $f$ and $f^{\prime }$ at 0 agree: $f(0^{+})=f(0^{-})=0$, $f^{\prime }(0^{+})=f^{\prime }(0^{-})=0$. (This is not the case for the original problem in Shearer and Levy (2015).) Therefore $u$ is also continuously differentiable. It is easy to check that $u$ satisfies the PDE.

     Method 2: Seek a travelling wave solution of the form

     $$u(x,t)=f(x-4t)$$

     and use the boundary and initial conditions to find $f$. Clearly $u$ satisfies $u_t+4u_x=0$. The initial condition $u(x,0)=0$ for $x>0$ implies that

     $$f(x)=0\text{ for }x>0.$$

     The boundary condition $u(0,t)=t^2{e}^{-t}$ for $t\ge 0$ implies that

     $$f(-4t)=t^2{e}^{-t}\text{ for }t\ge 0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}f(x)={\left(\frac{x}{4}\right)}^2{e}^{\frac{x}{4}}\text{ for }x\le 0.$$

     Therefore

     as we found using Method 1.

   • Now we seek a solution of the form $u(x,t)=f(x+4t)$ of

     $u_t-4u_x=0$ $\text{for }(x,t)\in (0,\mathrm{\infty })\times (0,\mathrm{\infty }),$

     $u(x,0)=0$ $\text{for }x\in (0,\mathrm{\infty }),$

     $u(0,t)=t^2{e}^{-t}$ $\text{for }t\in [0,\mathrm{\infty }).$

     Clearly $u$ satisfies $u_t-4u_x=0$. The initial condition $u(x,0)=0$ for $x>0$ implies that

     $$f(x)=0\text{ for }x>0.$$

     The boundary condition $u(0,t)=t^2{e}^{-t}$ for $t\ge 0$ implies that

     $$f(4t)=t^2{e}^{-t}\text{ for }t\ge 0\mathit{\hspace{1em}}⟺\mathit{\hspace{1em}}f(x)={\left(\frac{x}{4}\right)}^2{e}^{-\frac{x}{4}}\text{ for }x\ge 0.$$

     We have arrived at the contradiction $f(x)=0$ and $f(x)={(x/4)}^2\exp (-x/4)$ for $x\ge 0$. This also has a physical explanation: In this case, since $c=-4$, information is transported to the left with speed $4$, i.e., towards the boundary. The boundary $x=0$ is an outflow boundary, whereas in part (i) the boundary $x=0$ was an inflow boundary. This means that the initial data $u=0$ is transported to the boundary. But this is not compatible with the boundary condition $u(0,t)=t^2{e}^{-t}\ne 0$ for $t>0$.

6. The heat equation on the real line. Define $u:ℝ\times (0,\mathrm{\infty })\to ℝ$ by

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

   • By bringing the partial derivatives inside the integral and using the product rule and chain rule, we compute

     $u_t$ $=-{\displaystyle \frac{1}{2t}}{\displaystyle \frac{1}{\sqrt{4\pi kt}}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-\frac{{(x-y)}^2}{4kt}}g(y)𝑑y+{\displaystyle \frac{1}{\sqrt{4\pi kt}}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}\frac{{(x-y)}^2}{4k{t}^2}{e}^{-\frac{{(x-y)}^2}{4kt}}g(y)𝑑y,$

     $u_x$ $=-{\displaystyle \frac{1}{\sqrt{4\pi kt}}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}\frac{2(x-y)}{4kt}{e}^{-\frac{{(x-y)}^2}{4kt}}g(y)𝑑y,$

     $u_{xx}$ $=-{\displaystyle \frac{1}{\sqrt{4\pi kt}}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}\frac{1}{2kt}{e}^{-\frac{{(x-y)}^2}{4kt}}g(y)𝑑y+{\displaystyle \frac{1}{\sqrt{4\pi kt}}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{\left(\frac{2(x-y)}{4kt}\right)}^2{e}^{-\frac{{(x-y)}^2}{4kt}}g(y)𝑑y$

     $=-{\displaystyle \frac{1}{2kt}}{\displaystyle \frac{1}{\sqrt{4\pi kt}}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-\frac{{(x-y)}^2}{4kt}}g(y)𝑑y+{\displaystyle \frac{1}{\sqrt{4\pi kt}}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}\frac{{(x-y)}^2}{4k^2{t}^2}{e}^{-\frac{{(x-y)}^2}{4kt}}g(y)𝑑y$

     $={\displaystyle \frac{1}{k}}{u}_t.$

     Therefore $u$ satisfies $u_t=k{u}_{xx}$ for $x\in ℝ$, $t>0$. To make this proof completely rigorous, you should prove that the partial derivatives can be brought inside the integral, but this goes beyond the scope of this course.

   • Substituting $g(x)=u_0$ into the integral formula for $u$ gives

     $u(x,t)$ $={\displaystyle \frac{1}{\sqrt{4\pi kt}}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-\frac{{(x-y)}^2}{4kt}}{u}_0𝑑y$

     $={\displaystyle \frac{u_0}{\sqrt{4\pi kt}}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-z^2}\sqrt{4kt}𝑑z$ $\text{(by substituting }z=(y-x)/\sqrt{4kt}\text{)}$

     $={\displaystyle \frac{u_0}{\sqrt{\pi }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-z^2}𝑑z$

     $=u_0$

     since the integral on the right-hand side is the Gaussian integral. Therefore $u(x,t)=u_0$.

     This answer should not come as a surprise since clearly $u=u_0$ satisfies $u_t=k{u}_{xx}$ and $u(x,0)=u_0$. Also, thinking in physical terms, a metal bar at constant initial temperature $g(x)=u_0$ remains at this temperature if there are no external heat sources or sinks; every point is already at the same temperature and so there is no diffusion of heat.

     The fact that the integral of the Gaussian over $ℝ$ is $\sqrt{\pi }$ can be proved using polar coordinates as follows:

     ${\left({\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-z^2}𝑑z\right)}^2$ $=\left({\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-x^2}𝑑x\right)\left({\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-y^2}𝑑y\right)$

     $={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-(x^2+y^2)}𝑑x𝑑y$

     $={\displaystyle {\int }_{{ℝ}^2}}{e}^{-{|𝒙|}^2}𝑑𝒙$

     $={\displaystyle {\int }_0^{2\pi }}{\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-r^2}r𝑑r𝑑\theta$ (using polar coordinates)

     $={2\pi \left(-\frac{1}{2}\right)e^{-r^2}|}_0^{\mathrm{\infty }}$

     $=\pi .$

   • We estimate

     $|u(x,t)|$ $=\left|{\displaystyle \frac{1}{\sqrt{4\pi kt}}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-\frac{{(x-y)}^2}{4kt}}g(y)𝑑y\right|$

     $\le {\displaystyle \frac{1}{\sqrt{4\pi kt}}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-\frac{{(x-y)}^2}{4kt}}|g(y)|𝑑y$

     $\le {\displaystyle \frac{1}{\sqrt{4\pi kt}}}\underset{y\in ℝ}{sup}{e}^{-\frac{{(x-y)}^2}{4kt}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}|g(y)|𝑑y$ $(\text{since }g\in L^1(ℝ))$

     $={\displaystyle \frac{1}{\sqrt{4\pi kt}}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}|g(y)|𝑑y$ $\left(\text{since }\underset{y\in ℝ}{sup}{e}^{-\frac{{(x-y)}^2}{4kt}}=\underset{z\in ℝ}{sup}{e}^{-z^2}=1\right)$

     $={\displaystyle \frac{1}{\sqrt{4\pi kt}}}{\parallel g\parallel }_{L^1(ℝ)}.$

     We have shown that

     $$|u(x,t)|\le \frac{1}{\sqrt{4\pi kt}}{\parallel g\parallel }_{L^1(ℝ)}\to 0\mathit{\hspace{1em}}\text{as }t\to \mathrm{\infty }.$$

     Therefore, for each $x\in ℝ$, $u(x,t)\to 0$ as $t\to \mathrm{\infty }$, as desired. This does not contradict part (ii) since $g\notin L^1(ℝ)$ if $g(x)=u_0$ and $u_0$ is a nonzero constant.

   • Interchanging the order of integration (which is allowed by Tonelli’s Theorem) gives

     ${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}u(x,t)𝑑x$ $={\displaystyle \frac{1}{\sqrt{4\pi kt}}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}\left({\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-\frac{{(x-y)}^2}{4kt}}g(y)𝑑y\right)𝑑x$

     $={\displaystyle \frac{1}{\sqrt{4\pi kt}}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}\left({\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-\frac{{(x-y)}^2}{4kt}}g(y)𝑑x\right)𝑑y$

     $={\displaystyle \frac{1}{\sqrt{4\pi kt}}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}\left({\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-\frac{{(x-y)}^2}{4kt}}𝑑x\right)g(y)𝑑y$

     $={\displaystyle \frac{1}{\sqrt{4\pi kt}}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}\left({\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-\frac{z^2}{4kt}}𝑑z\right)g(y)𝑑y$ $(z=x-y)$

     $={\displaystyle \frac{1}{\sqrt{4\pi kt}}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}\sqrt{4\pi kt}g(y)𝑑y$

     $={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}g(y)𝑑y$

     as required. We used the following formula for the integral of the Gaussian:

     $${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-a{z}^2}𝑑z=\sqrt{\frac{\pi }{a}}.$$

7. Revision of vector calculus and integration by parts in many variables.

   • By the definition of the divergence operator and the product rule

     $\mathrm{div}(\phi 𝒇)$ $={\displaystyle \sum _{i=1}^n}{\displaystyle \frac{\partial }{\partial x_i}}{(\phi 𝒇)}_i$

     $={\displaystyle \sum _{i=1}^n}{\displaystyle \frac{\partial }{\partial x_i}}(\phi f_i)$

     $={\displaystyle \sum _{i=1}^n}\left({\displaystyle \frac{\partial \phi }{\partial x_i}}{f}_i+\phi {\displaystyle \frac{\partial f_i}{\partial x_i}}\right)$

     $=\nabla \phi \cdot 𝒇+\phi \mathrm{div}𝒇.$

   • Integrate the identity $\nabla \phi \cdot 𝒇+\phi \mathrm{div}𝒇=\mathrm{div}(\phi 𝒇)$ over $\mathrm{\Omega }$ to obtain

     $${\int }_{\mathrm{\Omega }}\left(\nabla \phi \cdot 𝒇+\phi \mathrm{div}𝒇\right)𝑑𝒙={\int }_{\mathrm{\Omega }}\mathrm{div}(\phi 𝒇)𝑑𝒙={\int }_{\partial \mathrm{\Omega }}\phi 𝒇\cdot 𝒏𝑑S$$

     by the Divergence Theorem. Rearranging this equation gives the desired result:

     $${\int }_{\mathrm{\Omega }}\phi \mathrm{div}𝒇𝑑𝒙={\int }_{\partial \mathrm{\Omega }}\phi 𝒇\cdot 𝒏𝑑S-{\int }_{\mathrm{\Omega }}\nabla \phi \cdot 𝒇d𝒙.$$ (1)

   • By definition

     $$\mathrm{div}\nabla u=\sum _{i=1}^n\frac{\partial }{\partial x_i}{(\nabla u)}_i=\sum _{i=1}^n\frac{\partial }{\partial x_i}\frac{\partial u}{\partial x_i}=\sum _{i=1}^n\frac{{\partial }^{2}u}{\partial x_i^2}=\mathrm{\Delta }u.$$

   • Substituting $𝒇=\nabla u$ into equation (1) and using that div$\nabla u=\mathrm{\Delta }u$ gives

     $${\int }_{\mathrm{\Omega }}\phi \mathrm{\Delta }u𝑑𝒙={\int }_{\partial \mathrm{\Omega }}\phi \nabla u\cdot 𝒏dS-{\int }_{\mathrm{\Omega }}\nabla u\cdot \nabla \phi d𝒙$$

     as required.

8. Poisson’s equation with Neumann boundary conditions. Consider Poisson’s equation with Neumann boundary conditions:

   $-\mathrm{\Delta }u=f$ $\text{in }\mathrm{\Omega },$ (2)

   $\nabla u\cdot 𝒏=g$ $\text{on }\partial \mathrm{\Omega }.$ (3)

   • Suppose that there exists a function $u$ satisfying (2), (3). Then we can integrate (2) to obtain

     $-{\displaystyle {\int }_{\mathrm{\Omega }}}\mathrm{\Delta }u𝑑𝒙={\displaystyle {\int }_{\mathrm{\Omega }}}f𝑑𝒙$ $⟺\mathit{\hspace{1em}}-{\displaystyle {\int }_{\mathrm{\Omega }}}\text{div}\nabla ud𝒙={\displaystyle {\int }_{\mathrm{\Omega }}}f𝑑𝒙$ $\text{(since }\mathrm{\Delta }u=\text{div}\nabla u\text{)}$

     $⟺\mathit{\hspace{1em}}-{\displaystyle {\int }_{\partial \mathrm{\Omega }}}\nabla u\cdot 𝒏dS={\displaystyle {\int }_{\mathrm{\Omega }}}f𝑑𝒙$ (by the Divergence Theorem)

     $⟺\mathit{\hspace{1em}}-{\displaystyle {\int }_{\partial \mathrm{\Omega }}}g𝑑S={\displaystyle {\int }_{\mathrm{\Omega }}}f𝑑𝒙$ $\text{(by equation (}\text{<a href="\#S0.E3" title="(3) ‣ item 8." class="ltx\_ref"><span class="ltx\_text ltx\_ref\_tag">3</span></a>}\text{))}.$

     Rearranging gives

     $${\int }_{\mathrm{\Omega }}f𝑑𝒙+{\int }_{\partial \mathrm{\Omega }}g𝑑S=0$$ (4)

     as required.

   • If $u$ is a solution of (2), (3), then so is $u+c$ for any constant $c\in ℝ$. Therefore if the PDE has a solution, it has infinitely many. On the other hand, if the data $f$ and $g$ are such that the necessary condition (4) is violated (take for instance $f\equiv 1$ and $g\equiv 0$), then the problem does not have any solutions.

9. Implicit form of Burger’s equation. Clearly

   $$u(x,t)=u_0(x-u(x,t)t)$$ (5)

   satisfies the initial condition $u(x,0)=u_0(x)$ for all $x\in ℝ$. We need to show that $u_t+u{u}_x=0$ for all $(x,t)\in ℝ\times (0,t_{\mathrm{c}})$. Differentiating equation (5) yields

   $u_t$ $=-u_0^{\prime }(u_{t}t+u),$ (6)

   $u_x$ $=u_0^{\prime }(1-u_{x}t),$ (7)

   where we have omitted the argument $x-u(x,t)t$ of $u_0$. Therefore

   $$u_t+u{u}_x=-u_0^{\prime }(u_{t}t+u)+u{u}_0^{\prime }(1-u_{x}t)=-u_0^{\prime }t(u_t+u{u}_x).$$

   Rearranging gives

   $$(u_t+u{u}_x)(1+u_0^{\prime }t)=0.$$ (8)

   Either $u_0^{\prime }(x-u(x,t)t)\ge 0$, in which case $1+u_0^{\prime }(x-u(x,t)t)t\ge 1$ (since $t>0$). Or , in which case

   $$1+u_0^{\prime }(x-u(x,t)t)t>1+u_0^{\prime }(x-u(x,t)t)t_{\mathrm{c}}\ge 1+u_0^{\prime }(x-u(x,t)t)\frac{-1}{u_0^{\prime }(x-u(x,t)t)}=0$$

   by definition of $t_{\mathrm{c}}$. In either case, $1+u_0^{\prime }t>0$, and hence equation (8) implies that $u_t+u{u}_x=0$, as required.