Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 5.3

1. Suppose $u$ is a distribution solution of the nonhomogeneous heat equation $u_t=\Delta u+f(x,t)$ in $U\subset\mathbb R^{n+1}$, where $f\in C^\infty(U)$. Show that $u\in C^\infty(U)$.
Solution

Pick $\xi=(x_0,t_0)\in U$ and $\epsilon>0$ so small that the ball $B_{4\epsilon}(\xi)\subseteq U$. Choose $\phi\in C_0^\infty(B_{4\epsilon}(\xi))$ with $\phi\equiv1$ on $B_{3\epsilon}(\xi)$. Let $w(x,t)=\phi(x,t)u(x,t)$ and $v(x,t)=w_t(x,t)-\Delta w(x,t)-f(x,t)$ in $U$, which are both distributions in $U$ with support in $B_{4\epsilon}(\xi)$. Clearly, $v=0$ on $B_{3\epsilon}(\xi)$. In particular, the convolution

$\displaystyle(\tilde K\ast v)(x,t)=\int_{\mathbb R^{n+1}}\!\tilde K(x-y,t-s)v(y,s)\,\mathrm dy\,\mathrm ds=\int_{\mathbb R^{n+1}}\!\tilde K(y,s)v(x-y,t-s)\,\mathrm dy\,\mathrm ds$

is well-defined as a distribution. But

$\begin{aligned}v(x-y,t-s)&=w_t(x-y,t-s)-\Delta w(x-y,t-s)-f(x-y,t-s)\\&=(\partial_t-\Delta_x)w(x-y,t-s)-f(x-y,t-s)\\&=(-\partial_s-\Delta_y)w(x-y,t-s)-f(x-y,t-s),\end{aligned}$

so for $g\in C_0^\infty(U)$, we find

$\begin{aligned}\langle\tilde K\ast v,g\rangle&=\int_U\!\int_{\mathbb R^{n+1}}\!\tilde K(y,s)[(-\partial_s-\Delta_y)w(x-y,t-s)-f(x-y,t-s)]g(x,t)\,\mathrm dy\,\mathrm ds\,\mathrm dx\,\mathrm dt\\&=\int_U\!\int_{\mathbb R^{n+1}}(\partial_s-\Delta_y)\tilde K(y,s)w(x-y,t-s)g(x,t)\,\mathrm dy\,\mathrm ds\,\mathrm dx\,\mathrm dt\\&\quad+\int_U\!\int_{\mathbb R^{n+1}}\!\tilde K(y,s)f(x-y,t-s)g(x,t)\,\mathrm dy\,\mathrm ds\,\mathrm dx\,\mathrm dt\\&=\int_U\!\int_{\mathbb R^{n+1}}\![\delta(y,s)w(x-y,t-s)+\tilde K(y,s)f(x-y,t-s)]g(x,t)\,\mathrm dy\,\mathrm ds\,\mathrm dx\,\mathrm dt\\&=\int_U\![w(x,t)+(\tilde K\ast f)(x,t)]g(x,t)\,\mathrm dx\,\mathrm dt=\langle w+\tilde K\ast f,g\rangle.\end{aligned}$

In other words, $\tilde K\ast(v-f)=w$ as distributions.

Now we choose $\psi\in C^\infty(\mathbb R^{n+1})$ with $\psi\equiv0$ on $B_\epsilon(0)$ and $\psi\equiv1$ on $\mathbb R^{n+1}-B_{2\epsilon}(0)$. Then $\psi$ vanishes in a neighborhood of the singularity of $\tilde K$, so $\psi\tilde K$ is smooth function, and hence so is $(\psi\tilde K)\ast(v-f)$. But for $(x,t)\in B_\epsilon(\xi)$ and $(y,s)\in\text{supp}(v)\subset B_{4\epsilon}(\xi)\cap(\mathbb R^{n+1}-B_{3\epsilon}(\xi))$, we have $|(x,t)-(y,s)|\geq2\epsilon$, so $\psi(x-y,t-s)=1$. In other words, $(\psi\tilde K\ast(v-f))(x,t)=(\tilde K\ast(v-f))(x,t)=w(x,t)$ for $(x,t)\in B_\epsilon(\xi)$, which shows that $w$ is smooth on $B_\epsilon(\xi)$. But $w(x,t)=\phi(x,t)u(x,t)$, so $u$ is also smooth on $B_\epsilon(\xi)$. Since $\xi\in U$ was arbitrary, we conclude that $u$ is smooth on $U$.



2. Consider the heat equation $u_t=\Delta u$ in $(x,t)\in\mathbb R^n\times(0,\infty)$ with the initial condition $u(x,0)=\delta(x)$. Since $\displaystyle\int\!u(x,t)\,\mathrm dx$ represents the total heat content in $\mathbb R^n$ at time $t$, we encounter the condition $\displaystyle\int\!u(x,t)\,\mathrm dx=1$ for all $t\geq0$.
   1. Show that the choice $\alpha=-n/2$, which was made to simplify (41) to (42), is a consequence of the scale invariance of $\displaystyle\int\!u(x,t)\,\mathrm dx$ for $t>0$.
   2. Show that (43b) is a consequence of $\displaystyle\int\!u(x,t)\,\mathrm dx<\infty$.
   3. Use $\displaystyle\int\!u(x,t)\,\mathrm dx=1$ to evaluate the constant $C$ in (44), and conclude that $u(x,t)$ coincides with the heat kernel.
Solution
1. Suppose it is known that $u(x,t)=t^\alpha w(x/\sqrt t)$ (see (39)). Then we use the change of variable to find

   $\displaystyle C=\int_{\mathbb R^n}\!u(x,t)\,\mathrm dx=t^\alpha\int_{\mathbb R^n}\!w(x/\sqrt t)\,\mathrm dx=t^\alpha\int_{\mathbb R^n}\!w(y)\cdot\sqrt t^n\,\mathrm dy=t^{\alpha+n/2}\int_{\mathbb R^n}\!w(y)\,\mathrm dy$.

   Due to the $t$-independence, we have $\alpha=-n/2$. The proof is complete.
2. From (43a), we have

   $\displaystyle w'+\frac r2w=Ar^{1-n}.$

   Multiplying it by $e^{r^2/4}$, it yields

   $\displaystyle(e^{r^2/4}w)'=Ar^{1-n}e^{r^2/4},$

   which gives

   $\displaystyle w(r)=w(0)e^{-r^2/4}+Ae^{-r^2/4}\int_0^r\!s^{1-n}e^{s^2/4}\,\mathrm ds\quad\text{for}~r\in(0,\infty)$.

   From part (a), the condition $\displaystyle\int_{\mathbb R^n}\!u(x,t)\,\mathrm dx<\infty$ implies $\displaystyle\int_0^\infty\!r^{n-1}w(r)\,\mathrm dr<\infty$. This means

   $\displaystyle A\int_0^\infty\!r^{n-1}e^{-r^2/4}\int_0^r\!s^{1-n}e^{s^2/4}\,\mathrm ds\,\mathrm dr<\infty.$

   Here we have used the fact that $\displaystyle\int_0^\infty\!w(0)r^{n-1}e^{-r^2/4}\,\mathrm dr$ is finite. However, we apply L'Hôpital's rule to observe that

   $\begin{aligned}\lim_{r\to\infty}\left[r\cdot r^{n-1}e^{-r^2/4}\int_0^r\!s^{1-n}e^{s^2/4}\,\mathrm ds\right]&=\lim_{r\to\infty}\frac{\displaystyle\int_0^r\!s^{1-n}e^{s^2/4}\,\mathrm ds}{r^{-n}e^{r^2/4}}\\&=\lim_{r\to\infty}\frac{r^{1-n}e^{r^2/4}}{-nr^{-n-1}e^{r^2/4}+r^{-n}e^{r^2/4}\cdot r/2}=\lim_{r\to\infty}\frac2{-2nr^{-2}+1}=2.\end{aligned}$

   This means the function $\displaystyle r^{n-1}e^{-r^2/4}\int_0^r\!s^{-1n}e^{s^2/4}\,\mathrm ds\sim\frac1r$ as $r$ sufficiently large, and the integral $\displaystyle\int_0^\infty\!r^{n-1}e^{-r^2/4}\int_0^r\!s^{-1n}e^{s^2/4}\,\mathrm ds\,\mathrm dr$ must be divergent. Therefore, $A$ must be zero.
3. By (44) and $\displaystyle\int_{\mathbb R^n}\!u(x,t)\,\mathrm dx=1$, we have

   $\begin{aligned}1&=\frac{C}{t^{n/2}}\int_{\mathbb R^n}\!\exp\left(-\frac{|x|^2}{4t}\right)\,\mathrm dt=\frac{C}{t^{n/2}}\prod_{i=1}^n\int_{\mathbb R}\exp\left(-\frac{x_i^2}{4t}\right)\,\mathrm dx_i\\&=\frac{C}{t^{n/2}}\prod_{i=1}^n\cdot\sqrt{4\pi t}=\frac{C}{t^{n/2}}\cdot(4\pi t)^{n/2}=(4\pi)^{n/2}C.\end{aligned}$

   Therefore, we get $C=(4\pi)^{-n/2}$ and hence

   $\displaystyle u(x,t)=\frac1{(4\pi t)^{n/2}}\exp\left(-\frac{|x|^2}{4t}\right)$,

   which conincides with the heat kernel.


3. Apply the similarity method to the linear transport equation $u_t+au_x=0$ to obtain the special solutions $u(x,t)=c(x-at)^\alpha$.
Solution

Let $\tilde u=\epsilon^\alpha u$, $\tilde x=\epsilon^\beta x$ and $\tilde t=\epsilon^\gamma t$, where $\alpha$, $\beta$ and $\gamma$ are constants, i.e.,

$\tilde u(\tilde x,\tilde t)=\epsilon^\alpha u(\epsilon^{-\beta}\tilde x,\epsilon^{-\gamma}\tilde t)$.

To scale out the $t$-dependence means that we should take $\epsilon^\gamma=t^{-1}$ so that $\tilde t=1$; that is if we introduce $w(\tilde x)=\tilde u(\tilde x,1)$ and $z=\tilde x$, then $u(x,t)=t^{\alpha/\gamma}w(z)$ and $z=t^{-\beta/\gamma}x$ defines the similarity transformation to the similarity variables $w$ and $z$. For convenience, we take $\gamma=1$ and hence

$u(x,t)=t^{\alpha}w(z),\quad z=t^{-\beta}x$.

The linear transport equation becomes

$\begin{aligned}0&=u_t+au_x=\alpha t^{\alpha-1}w(z)+t^\alpha w'(z)\cdot-\beta t^{-\beta-1}x+at^\alpha w'(z)\cdot t^{-\beta}\\&=\alpha t^{\alpha-1}w(z)+at^{\alpha-\beta}w'(z)-\beta t^{\alpha-\beta-1}xw'(z)\\&=\alpha t^{\alpha-1}w(z)-\beta t^{\alpha-1}zw'(z)+at^{\alpha-\beta}w'(z)\\&=t^{\alpha-1}(\alpha w(z)-\beta zw'(z))+at^{\alpha-\beta}w'(z).\end{aligned}$

If we choose $\beta=1$, then

$u(x,t)=t^\alpha w(z),\quad z=x/t$.

Here $w$ satisfies

$\alpha w(z)+(a-z)w'(z)=0$,

which can be solved by $w(z)=c|z-a|^\alpha$. Therefore, we obtain

$\displaystyle u(x,t)=t^\alpha\cdot c|z-a|^\alpha=t^\alpha\cdot c\left|\frac xt-a\right|^\alpha=c|x-at|^\alpha.$



4. Consider the inviscid Burgers equation $u_t+uu_x=0$.
   1. Use the similarity method to reduce this equation to the ODE $ww'-(\alpha+1)zw'+\alpha w=0$.
   2. Take $\alpha=-1$ and obtain the rarefaction wave solutions discussed in Section 1.2.
Solution

1. Let $\tilde u=\epsilon^\alpha u$, $\tilde x=\epsilon^\beta x$ and $\tilde t=\epsilon^\gamma t$, where $\alpha$, $\beta$ and $\gamma$ are constants, i.e.,

   $\tilde u(\tilde x,\tilde t)=\epsilon^\alpha u(\epsilon^{-\beta}\tilde x,\epsilon^{-\gamma}\tilde t)$.

   To scale out the $t$-dependence means that we should take $\epsilon^\gamma=t^{-1}$ so that $\tilde t=1$; that is if we introduce $w(\tilde x)=\tilde u(\tilde x,1)$ and $z=\tilde x$, then $u(x,t)=t^{\alpha/\gamma}w(z)$ and $z=t^{-\beta/\gamma}x$ defines the similarity transformation to the similarity variables $w$ and $z$. For convenience, we take $\gamma=1$ and hence

   $u(x,t)=t^\alpha w(z),\quad z=t^{-\beta}x$.

   The inviscid Burgers equation becomes

   $\begin{aligned}0&=u_t+uu_x=\alpha t^{\alpha-1}w(z)+t^\alpha w'(z)\cdot-\beta t^{-\beta-1}x+t^\alpha w(z)\cdot t^\alpha w'(z)\cdot t^{-\beta}\\&=\alpha t^{\alpha-1}w(z)-\beta t^{\alpha-1}zw'(z)+t^{2\alpha-\beta}w(z)w'(z)\end{aligned}$

   If we choose $\beta=\alpha+1$, then

   $u(x,t)=t^\alpha w(z),\quad z=t^{-\alpha-1}x$.

   Here $w$ satisfies

   $w(z)w'(z)-(\alpha+1)zw'(z)+\alpha w(z)=0$,

   which is desired.
2. Taking $\alpha=-1$, we get $w'(z)-1=0$, which gives $w(z)=z+C$. Therefore, we obtain

   $\displaystyle u(x,t)=t^{-1}(z+C)=t^{-1}(x+C)=\frac{x+C}t.$



5. 1. For $n=1$ and $\alpha=0$, express the solution of (41) in terms of the error function

      $\displaystyle\text{erf}(y)=\frac2{\sqrt\pi}\int_0^y\!e^{-s^2}\,\mathrm ds$.

   2. Use (a) to solve the heat equation on a half-line: $u_t=u_{xx}$ for $x,t>0$ with $u(x,0)=1$ for $x>0$, and $u(0,t)=0$ for $t>0$.
Solution

1. For $n=1$ and $\alpha=0$, equation (41) becomes

   $\displaystyle w''+\frac12rw'=0.$

   Multiplying it by $e^{r^2/4}$ and then integrating over $[0,t]$, we get$(e^{r^2/4}w)'=0$ and

   $e^{r^2/4}w'(r)=w'(0)$.

   It is clear that

   $\displaystyle w(r)=w(0)+w'(0)\int_0^r\!e^{-t^2/4}\,\mathrm dt=w(0)+2w'(0)\int_0^{r/2}\!e^{-s^2}\,\mathrm ds=w(0)+w'(0)\sqrt\pi\text{erf}(r/2).$

2. By (a), we get the solution $u$ in the following form:

   $\displaystyle u(x,t)=w(z)=w\left(\frac x{\sqrt t}\right)=w(0)+w'(0)\sqrt\pi\text{erf}\left(\frac{x}{\sqrt{4t}}\right)\quad\text{for}~x,t>0$.

   Clearly, $\text{erf}(0)=0$ and $\displaystyle\lim_{y\to\infty}\text{erf}(y)=1$. By the boundary condition $u(0,t)=0$, we get $w(0)=0$. Using the initial data $u(x,0)=1$ for $x>0$, we get $w'(0)=1/\sqrt\pi$. Therefore, we obtain the solution $u$ given by

   $\displaystyle u(x,t)=\text{erf}\left(\frac x{\sqrt{4t}}\right)\quad\text{for}~x,t>0$.



6. Apply the similarity method to the porous medium equation $u_t=\Delta(u^\gamma)$ in $\mathbb R^n$, where $\gamma>1$, to obtain Barenblatt's solution:

   $\displaystyle u(x,t)=t^\alpha\left(C-\frac{(\gamma-1)\beta|x|^2}{2\gamma t^{2\beta}}\right)^{1/(\gamma-1)},$

   where $C$ is a positive constant, $\beta=(n(\gamma-1)+2)^{-1}$, and $\alpha=-n\beta$.

   Remark. Here $u(x,t)$ is defined and positive for $(\gamma-1)\beta|x|^2<C2\gamma t^{2\beta}$, but may be extended by zero to a weak solution on $\mathbb R^n\times(0,\infty)$. In this way, $u$ exhibits finite propagation speed, unlike the solution of the heat equation $\gamma=1$.

Solution
Let $\tilde u=\epsilon^\alpha u$, $\tilde x=\epsilon^\beta x$ and $\tilde t=\epsilon^\delta t$, where $\alpha$, $\beta$ and $\delta$ are constants, i.e.,
$\tilde u(\tilde x,\tilde t)=\epsilon^\alpha u(\epsilon^{-\beta}\tilde x,\epsilon^{-\delta}\tilde t)$.
To scale out the $t$-dependence means that we should take $\epsilon^\delta=t^{-1}$ so that $\tilde=1$; that if we introduce $w(\tilde x)=\tilde u(\tilde x,1)$ and $z=\tilde x$, then $u(x,t)=t^{\alpha/\delta}w(z)$ and $z=t^{-\beta/\delta}x$ defines the similarity transformation to the similarity variables $w$ and $z$. For convenience, we take $\delta=1$ and hence
$u(x,t)=t^\alpha w(z),\quad z=t^{-\beta}x$.
The porous medium equation becomes
$\begin{aligned}0&=u_t-\Delta(u^\gamma)=\alpha t^{\alpha-1}w(z)+t^\alpha\nabla w(z)\cdot-\beta t^{-\beta-1}x-t^{\alpha\gamma}\Delta_x(w^\gamma(z))\\&=\alpha t^{\alpha-1}w(z)-\beta t^{\alpha-1}z\cdot\nabla w(z)-t^{\alpha\gamma}\nabla_x\cdot(t^{-\beta}\gamma w^{\gamma-1}(z)\nabla w(z))\\&=\alpha t^{\alpha-1}w(z)-\beta t^{\alpha-1}z\cdot\nabla w(z)-\gamma t^{\alpha\gamma-\beta}[t^{-\beta}(\gamma-1)w^{\gamma-2}(z)|\nabla w(z)|^2+t^{-\beta}w^{\gamma-1}(z)\Delta w(z)]\\&=\alpha t^{\alpha-1}w(z)-\beta t^{\alpha-1}z\cdot\nabla w(z)-\gamma t^{\alpha\gamma-2\beta}[(\gamma-1)w^{\gamma-2}(z)|\nabla w(z)|^2+w^{\gamma-1}(z)\Delta w(z)].\end{aligned}$
In order to obtain a time-independent partial differential equation for $w$, the powers of $t$ must agree, which gives $\alpha-1=\alpha\gamma-2\beta$, and hence $\beta=(\alpha(\gamma-1)+1)/2$. Thus, we find that
$u(x,t)=t^\alpha w(z),\quad z=t^{-(\alpha(\gamma-1)+1)/2}x$.
Here $w$ satisfies
$\displaystyle\alpha w(x)-\beta z\cdot\nabla w(z)-\gamma[(\gamma-1)w^{\gamma-2}(z)|\nabla w(z)|^2+w^{\gamma-1}(z)\Delta w(z)]=0$.
Now we assume that $w$ is radial, i.e., $\tilde w(r)=w(z)$ for $|z|=r$. Then $\tilde w$ satisfies
$\displaystyle\alpha\tilde w(r)-\beta\cdot r\tilde w'(r)-\gamma\left[(\gamma-1)\tilde w^{\gamma-2}(r)\tilde w'^2(r)+w^{\gamma-1}(z)\left(\tilde w''(r)+\frac{n-1}r\tilde w'(r)\right)\right]=0$,
which implies
$\displaystyle\alpha\tilde w(r)-\beta\cdot r\tilde w'(r)-\left[(w^\gamma(r))''+\frac{n-1}r(\tilde w^\gamma(r))'\right]=0.$
If we choose $\alpha=-n\beta$, then we observe that $r^{n-1}(\alpha\tilde w-\beta r\tilde w'(r))=-\beta(r^n\tilde w)'$. Multiplying the differential equation by $r^{n-1}$, we obtain
$(r^{n-1}(\tilde w^\gamma)')'+\beta(r^n\tilde w)'=0$,
which implies
$r^{n-1}(\tilde w^\gamma)'+\beta r^n\tilde w=A$.
For the sake of simplicity, we assume that $w$ and $w'$ decay to zero as $r$ tends to infinity, and hence $A=0$. Thus, we get
$\displaystyle\tilde w^{\gamma-2}\tilde w'=-\frac{\beta}{\gamma}r$,
which gives us
$\displaystyle\tilde w^{\gamma-1}(r)=C-\frac{(\gamma-1)\beta}{2\gamma}r^2$.
Therefore, we arrive at
$\begin{aligned} &w(z)=\left(C-\frac{(\gamma-1)\beta}{2\gamma}|z|^2\right)^{1/(\gamma-1)},\\&u(x,t)=t^\alpha w(t^{-\beta}x)=t^\alpha\left(C-\frac{(\gamma-1)\beta}{2\gamma t^{2\beta}}|x|^2\right)^{1/(\gamma-1)},\end{aligned}$
where $\alpha=-n(n(\gamma-1)+2)^{-1}$ and $\beta=(n(\gamma-1)+2)^{-1}$.