Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 5.2

1. Proof of Theorem 1. Let $K(x,y,t)$ be the Gaussian kernel (18).
   1. Show that $K(x,y,t)$ is $C^\infty$ and satisfies $(\partial_t-\Delta_x)K(x,y,t)=0$ for $x,y\in\mathbb R^n$ and $t>0$.
   2. Prove that for every $x\in\mathbb R^n$ and $t>0$, $\displaystyle\int_{\mathbb R^n}\!K(x,y,t)\,\mathrm dy=1$.
   3. For any $\delta>0$, show that

      $\displaystyle\lim_{t\to0^+}\int_{|x-y|>\delta}\!K(x,y,t)\,\mathrm dy=0$

      uniformly for $x\in\mathbb R^n$.
   4. Use (a)-(c) and $g\in C_B(\mathbb R^n)$ to show $u(x,t)\to g(x_0)$ as $(x,t)\to(x_0,0^+)$, proving the theorem.
Solution
1. We firstly recall that the heat kernel $K$ given by

   $\displaystyle K(x,y,t)=\frac1{(4\pi t)^{n/2}}\exp\left(-\frac{|x-y|^2}{4t}\right)\quad\text{for}~x,y\in\mathbb R^n~\text{and}~t>0$.

   It is clear that $K$ is $C^\infty(\mathbb R^n\times\mathbb R^n\times\mathbb R_+)$ because all functions composing $K$ are smooth. A direct computation gives

   $\begin{aligned} &\frac{\partial K}{\partial t}(x,y,t)=-\frac n2\frac1{(4\pi t)^{n/2}t}\exp\left(-\frac{|x-y|^2}{4t}\right)+\frac1{(4\pi t)^{n/2}}\exp\left(-\frac{|x-y|^2}{4t}\right)\cdot\frac{|x-y|^2}{4t^2},\\&\frac{\partial K}{\partial x_i}(x,y,t)=\frac1{(4\pi t)^{n/2}}\exp\left(-\frac{|x-y|^2}{4t}\right)\cdot\frac{-2(x_i-y_i)}{4t},\\&\frac{\partial^2K}{\partial x_i^2}(x,y,t)=\frac1{(4\pi t)^{n/2}}\exp\left(-\frac{|x-y|^2}{4t}\right)\cdot\frac{(x_i-y_i)^2}{4t^2}-\frac1{2(4\pi t)^{n/2}t}\exp\left(-\frac{|x-y|^2}{4t}\right).\end{aligned}$

   Hence it is easy to verify that

   $\begin{aligned}(\partial_t-\Delta_x)K(x,y,t)&=\frac{\partial K}{\partial t}-\sum_{i=1}^n\frac{\partial^2K}{\partial x_i^2}\\&=-\frac{n}2\frac1{(4\pi t)^{n/2}}\exp\left(-\frac{|x-y|^2}{4t}\right)+\frac1{(4\pi t)^{n/2}}\exp\left(-\frac{|x-y|^2}{4t}\right)\cdot\frac{|x-y|^2}{4t^2}\\&\quad-\frac1{(4\pi t)^{n/2}}\exp\left(-\frac{|x-y|^2}{4t}\right)\cdot\frac{|x-y|^2}{4t^2}-\frac n{2(4\pi t)^{n/2}t}\exp\left(-\frac{|x-y|^2}{4t}\right)\\&=0.\end{aligned}$

   The proof is complete.
2. For every $x\in\mathbb R^n$ and $t>0$, it is clear that

   $\begin{aligned}\int_{\mathbb R^n}\!K(x,y,t)\,\mathrm dy&=\frac1{(4\pi t)^{n/2}}\int_{\mathbb R^n}\exp\left(-\frac{|x-y|^2}{4t}\right)\,\mathrm dy=\frac1{(4\pi t)^{n/2}}\int_{\mathbb R^n}\exp\left(-\frac1{4t}\sum_{i=1}^n(x_i-y_i)^2\right)\,\mathrm dy\\&=\frac1{(4\pi t)^{n/2}}\prod_{i=1}^n\int_{\mathbb R}\exp\left(-\frac{(x_i-y_i)^2}{4t}\right)\,\mathrm dy_i=\frac1{(4\pi t)^{n/2}}\prod_{i=1}^n\cdot\sqrt{4\pi t}=1.\end{aligned}$

   Here we have used the fact that $\displaystyle\int_{\mathbb R}\!\exp\left(-k(x-y)^2\right)\,\mathrm dy=\sqrt{\frac\pi k}$ for $k>0$.
3. Using the polar coordinate, we have

   $\begin{aligned}\int_{|x-y|>\delta}\!K(x,y,t)\,\mathrm dy&=\frac1{(4\pi t)^{n/2}}\int_{\mathbb S^{n-1}}\int_\delta^\infty\!\exp\left(\frac{-r^2}{4t}\right)\cdot r^{n-1}\,\mathrm dr\,\mathrm d\omega_{\mathbb S^{n-1}}\\&=\frac{\mathcal H^{n-1}(\mathbb S^{n-1})}{(4\pi t)^{n/2}}\int_\delta^\infty\!\exp\left(\frac{-r^2}{4t}\right)\cdot r^{n-1}\,\mathrm dr,\end{aligned}$

   where $\mathcal H^{n-1}(\mathbb S^{n-1})$ denotes the $(n-1)$-dimensional Hausdorff measure of $\mathbb S^{n-1}$, i.e., the surface area of $\mathbb S^{n-1}$. Using the change of variable with $s=r/\sqrt{4t}$, the integral can be transformed into

   $\displaystyle\int_{|x-y|>\delta}\!K(x,y,t)\,\mathrm dy=\frac{\mathcal H^{n-1}(\mathbb S^{n-1})}{\pi^{n/2}}\int_{\delta/\sqrt{4t}}^\infty\!s^{n-1}\exp(-s^2)\,\mathrm ds$

   Since $\displaystyle\int_0^\infty\!s^{n-1}\exp(-s^2)\,\mathrm ds$ is convergent and $\displaystyle\lim_{t\to0^+}\frac{\delta}{\sqrt{4t}}=\infty$, we obtain

   $\displaystyle\lim_{t\to0^+}\int_{|x-y|>\delta}\!K(x,y,t)\,\mathrm dy=\frac{\mathcal H^{n-1}(\mathbb S^{n-1})}{\pi^{n/2}}\lim_{t\to0^+}\int_{\delta/\sqrt{4t}}^\infty\!s^{n-1}\exp(-s^2)\,\mathrm ds=0.$

   Note that this convergence rate is independent of $x$ so the convergnece is uniform in $x$.
4. Fix $x_0\in\mathbb R^n$ and $\epsilon>0$. Since $g$ is continuous in $\mathbb R^n$, there exists $\delta>0$ such that $|g(y)-g(x_0)|<\epsilon$ for $|y-x_0|<\delta$ and $y\in\mathbb R^n$. For $|x-x_0|<\delta/2$, we can use (b) to find that

   $\begin{aligned}|u(x,t)-g(x_0)|&=\left|\int_{\mathbb R^n}K(x,y,t)[g(y)-g(x_0)]\,\mathrm dy\right|\\&\leq\int_{B_\delta(x_0)}\!K(x,y,t)|g(y)-g(x_0)|\,\mathrm dy+\int_{\mathbb R^n-B_\delta(x_0)}\!K(x,y,t)|g(y)-g(x_0)|\,\mathrm dy.\end{aligned}$

   Clearly, we have

   $\displaystyle\int_{B_\delta(x_0)}\!K(x,y,t)|g(y)-g(x_0)|\,\mathrm dy\leq\epsilon\int_{B_\delta(x_0)}\!K(x,y,t)\,\mathrm dy\leq\epsilon\int_{\mathbb R^n}\!K(x,y,t)\,\mathrm dy=\epsilon.$

   Here we have used (b) again. On the other hand, if $|y-x_0|\geq\delta$, we observe that

   $\displaystyle|y-x_0|\leq|y-x|+|x-x_0|\leq|y-x|+\frac\delta2\leq|y-x|+\frac12|y-x_0|,$

   which implies $\displaystyle|y-x|\geq\frac12|y-x_0|$. Hence, by (c), we have

   $\displaystyle\int_{\mathbb R^n-B_\delta(x_0)}\!K(x,y,t)|g(y)-g(x_0)|\,\mathrm dy\leq2\sup_{y\in\mathbb R^n}|g(y)|\int_{\mathbb R^n-B_\delta(x_0)}\!K(x,y,t)\,\mathrm dy\to0\quad\text{as}~t\to0^+,$

   which means

   $\displaystyle\int_{\mathbb R^n-B_\delta(x_0)}\!K(x,y,t)|g(y)-g(x_0)|\,\mathrm dy\leq\epsilon\quad\text{for}~0<t\leq t_0,$

   where $t_0$ is sufficiently small. Therefore, $|u(x,t)-g(x_0)|<2\epsilon$ for $|x-x_0|<\delta/2$ and $0<t<t_0$ and the proof is complete.
   
   Remark. The original statement is that "$u(x,t)\to g(x)$ as $t\to0^+$", which cannot be regarded as the continuous extension of $u$ to $t\geq0$. A suitable continuous extension of $u$ has been shown.


2. Let $g(x)$ be bounded and continuous for $x\in\mathbb R^n$ and define $u$ by (19).
   1. Show $|u(x,t)|\leq\sup\{|g(y)|\,:\,y\in\mathbb R^n\}$.
   2. If, in addition, $\displaystyle\int_{\mathbb R^n}\!|g(y)|\,\mathrm dy<\infty$, show that $\displaystyle\lim_{t\to\infty}u(x,t)=0$ uniformly in $x\in\mathbb R^n$.
Solution
1. By part (b) of Exercise 1, we have

   $\begin{aligned}|u(x,t)|&=\left|\int_{\mathbb R^n}\!K(x,y,t)g(y)\,\mathrm dy\right|\leq\int_{\mathbb R^n}\!K(x,y,t)|g(y)|\,\mathrm dy\\&\leq\sup_{y\in\mathbb R^n}|g(y)|\int_{\mathbb R^n}\!K(x,y,t)\,\mathrm dy=\sup_{y\in\mathbb R^n}|g(y)|.\end{aligned}$

   Here we have used the fact that $K\geq0$ in $\mathbb R^n\times(0,\infty)$.
2. A direct estimate implies

   $\begin{aligned}|u(x,t)|=\left|\int_{\mathbb R^n}\!K(x,y,t)g(y)\,\mathrm dy\right|\leq\frac1{(4\pi t)^{n/2}}\int_{\mathbb R^n}\!\exp\left(-\frac{|x-y|^2}{4t}\right)|g(y)|\,\mathrm dy\leq\frac1{(4\pi t)^{n/2}}\int_{\mathbb R^n}|g(y)|\,\mathrm dy.\end{aligned}$

   Here we have used the fact that $K\geq0$ in $\mathbb R^n\times\mathbb R^n\times(0,\infty)$ and $\displaystyle\exp\left(-\frac{|x-y|^2}{4t}\right)\leq1$ for $x,y\in\mathbb R^n$ and $t\in(0,\infty)$. Since $\displaystyle\lim_{t\to\infty}\frac1{(4\pi t)^{n/2}}\int_{\mathbb R^n}\!|g(y)|\,\mathrm dy=0$ and is independent of $x\in\mathbb R^n$, we obtain $\displaystyle\lim_{t\to\infty}u(x,t)=0$ uniformly in $x\in\mathbb R^n$. The proof is complete.


3. Let $g(x)\in C^k(\mathbb R^n)$ with $D^\alpha g$ uniformly bounded on $\mathbb R^n$ for each $|\alpha|\leq k$. Show that $u$ defined by (19) satisfies $u\in C^{k,[k/2]}(\mathbb R^n\times[0,\infty))$.
Solution

We prove the statement by mathematical induction. Clearly, the statement holds true for $k=0$ due to Theorem 1 at page 149 in the textbook. Suppose the statement holds true for $k=m$, i.e., $u\in C^{m,[m/2]}(\mathbb R^n\times[0,\infty))$ when $g\in C^m(\mathbb R^n)$. Now we suppose that $g\in C^{m+1}(\mathbb R^n)$. Then by the inductive assumption, we have

$u\in C^{m,[m/2]}(\mathbb R^n\times[0,\infty))\cap C^{m+1,[(m+1)/2]}(\mathbb R^n\times(0,\infty))\subsetneq C^{m,[m/2]}(\mathbb R^n\times[0,\infty))\cap C^{\infty,\infty}(\mathbb R^n\times(0,\infty))$.

To show that $u\in C^{m+1,[(m+1)/2]}(\mathbb R^n\times[0,\infty))$, we need to prove $D_x^\alpha u$ ($|\alpha|=m+1$) and $\partial_t^{[(m+1)/2]}u$ can be continously extended to $\mathbb R^n\times\{0\}$. Since $u\in C^\infty(\mathbb R^n\times(0,\infty))$, it is clear that $D_x^\alpha u$ and $\partial_t^{[(m+1/2)]}$ satisfies the heat equation:

$\begin{aligned} &(D_x^\alpha u)_t=\Delta(D_x^\alpha u)\quad\text{in}~\mathbb R^n\times(0,\infty),\\&(\partial_t^{[(m+1)/2]}u)_t=\Delta(\partial_t^{[(m+1)/2]}u)\quad\text{in}~\mathbb R^n\times(0,\infty).\end{aligned}$

• For $D_x^\alpha u$, we find

  $\begin{aligned}D_x^\alpha u(x,t)&=D_x^\alpha\frac1{(4\pi t)^{n/2}}\int_{\mathbb R^n}\!\exp\left(-\frac{|x-y|^2}{4t}\right)g(y)\,\mathrm dy=\frac1{(4\pi t)^{n/2}}D_x^\alpha\int_{\mathbb R^n}\!\exp\left(-\frac{|y|^2}{4t}\right)g(x-y)\,\mathrm dy\\&=\frac1{(4\pi t)^{n/2}}\int_{\mathbb R^n}\!\exp\left(-\frac{|y|^2}{4t}\right)(D^\alpha g)(x-y)\,\mathrm dy=\frac1{(4\pi t)^{n/2}}\int_{\mathbb R^n}\!\exp\left(-\frac{|x-y|^2}{4t}\right)(D^\alpha g)(y)\,\mathrm dy.\end{aligned}$

  Since $D^\alpha g$ is bounded and continuous in $\mathbb R^n$, we can apply Theorem 1 for $D_x^\alpha u(x,t)$, which can be continuously extended to $\mathbb R^n\times\{0\}$.
• For $\partial_t^{[(m+1)/2]}u$, we can use $\partial_tK(x,y,t)=\Delta_xK(x,y,t)$ to find

  $\begin{aligned}\partial_t^{[(m+1)/2]}u(x,t)&=\partial_t^{[(m+1)/2]}\int_{\mathbb R^n}\!K(x,y,t)g(y)\,\mathrm dy=\int_{\mathbb R^n}\!(\Delta_x^{[(m+1)/2]}K)(x,y,t)g(y)\,\mathrm dy\\&=\int_{\mathbb R^n}\!(\Delta_y^{[(m+1)/2]}K)(x,y,t)g(y)\,\mathrm dy=\int_{\mathbb R^n}\!K(x,y,t)(\Delta^{[(m+1)/2]}g)(y)\,\mathrm dy.\end{aligned}$

  Since $\Delta^{[(m+1)/2]}g$ is bounded and continuous in $\mathbb R^n$, we can apply Theorem 1 for $\partial_t^{[(m+1)/2]}u(x,t)$, which can be continuously extended to $\mathbb R^n\times\{0\}$.

Since $D_x^\alpha u$ and $\partial_t^{[(m+1)/2]}u$ are continuous in $\mathbb R^n\times[0,\infty)$, we arrive at $u\in C^{m+1,[(m+1)/2]}(\mathbb R^n\times[0,\infty))$. By the mathematical induction, we complete the proof.

Remark. In fact, when $g\in C^k(\mathbb R^n)$, $D_x^\alpha\partial_t^su$ are continuous for multi-index $\alpha$ and nonnegative integer $s$ satisfying $|\alpha|+2s\leq k$.



4. Formally check that

   $\displaystyle u(x,t)=\sum_{k=0}^\infty\frac1{(2k)!}x^{2k}\frac{\mathrm d^k}{\mathrm dt^k}e^{-1/t^2}$

   satisfies $u_t=u_{xx}$ for $t>0$, $x\in\mathbb R$ with $u(x,0)=0$ for $x\in\mathbb R$. (Proving the convergence of the infinite series is not easy; see [John, 1] or [Widder].)
Solution
Here we suppose that the differentiation and summation can be intercahgned freely. A direct computation shows
$\begin{aligned}u_t(x,t)&=\frac\partial{\partial t}\sum_{k=0}^\infty\frac1{(2k)!}x^{2k}\frac{\mathrm d^{k+1}}{\mathrm dt^{k+1}}e^{-1/t^2}=\sum_{k=1}^\infty\frac1{(2k-2)!}x^{2k-2}\frac{\mathrm d^k}{\mathrm dt^k}e^{-1/t^2}\\&=\sum_{k=1}^\infty\frac1{(2k)!}\cdot(2k)(2k-1)x^{2k-2}\frac{\mathrm d^k}{\mathrm dt^k}e^{-1/t^2}=\sum_{k=1}^\infty\frac1{(2k)!}\left(\frac{\mathrm d^2}{\mathrm dx^2}x^{2k}\right)\frac{\mathrm d^k}{\mathrm dt^k}e^{-1/t^2}\\&=\sum_{k=0}^\infty\frac1{(2k)!}\left(\frac{\mathrm d^2}{\mathrm dx^2}x^{2k}\right)\frac{\mathrm d^k}{\mathrm dt^k}e^{-1/t^2}=\frac{\partial^2}{\partial x^2}\sum_{k=0}^\infty\frac1{(2k)!}x^{2k}\frac{\mathrm d^k}{\mathrm dt^k}e^{-1/t^2}=u_{xx}(x,t).\end{aligned}$
Next we prove that $\displaystyle\lim_{t\to0^+}u(x,t)=0$. For this purpose, it suffices to show $\displaystyle\lim_{t\to0^+}\frac{\mathrm d^k}{\mathrm dt^k}e^{-1/t^2}=0$ for all $k\in\mathbb N\cup\{0\}$, which implies
$\displaystyle\lim_{t\to0^+}u(x,t)=\lim_{t\to0^+}\sum_{k=0}^\infty\frac1{(2k)!}x^{2k}\frac{\mathrm d^k}{\mathrm dt^k}e^{-1/t^2}=\sum_{k=0}^\infty\frac1{(2k)!}x^{2k}\lim_{t\to0^+}\frac{\mathrm d^k}{\mathrm dt^k}e^{-1/t^2}=0=u(x,0)\quad\text{for}~x\in\mathbb R.$
Let $f(t)=e^{-1/t^2}$ for $t\neq0$ and $f(0)=0$. Then for $n\in\mathbb N$, we use the mathematical induction to prove that there exist $a_1^{(n)},\dots,a_{2n-1}^{(n)}$ such that
$f^{(n)}(t)=\begin{cases}\displaystyle\sum_{k=1}^{2n-1}\frac{a_k^{(n)}}{t^{n+k+1}}e^{-1/t^2}&\text{if}~t\neq0,\\0&\text{if}~t=0.\end{cases}$
When $n=1$, we have $\displaystyle f'(t)=\frac2{t^3}e^{-1/t^2}$ for $t\neq0$. For $t=0$, we observe that
$\displaystyle f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}h=\lim_{h\to0}\frac{e^{-1/h^2}}h=\lim_{h\to0}\frac{1/h}{e^{1/h^2}}=\lim_{h\to0}\frac{\displaystyle-\frac1{h^2}}{\displaystyle-\frac2{h^3}e^{1/h^2}}=\lim_{h\to0}\frac h{2e^{1/h^2}}=0.$
Thus, we get
$f'(t)=\begin{cases}\displaystyle\frac2{t^3}e^{-1/t^2}&\text{if}~t\neq0,\\0&\text{if}~t=0.\end{cases}$
Here $a_1^{(1)}=2$. Now we suppose that the statement holds true for $n=m$, i.e., there exists $a_1^{(m)},\dots,a_{2m-1}^{(m)}$ such that
$f^{(m)}(t)=\begin{cases}\displaystyle\sum_{k=1}^{2m-1}\frac{a_k^{(m)}}{x^{m+k+1}}e^{-1/x^2}&\text{if}~t\neq0,\\0&\text{if}~t=0.\end{cases}$
For $x\neq0$, we have
$\begin{aligned}f^{(m+1)}(t)&=\sum_{k=1}^{2m-1}\frac{-a_k^{(m)}(m+k+1)}{t^{m+k+2}}e^{-1/t^2}+\sum_{k=1}^{2m-1}\frac{-2a_k^{(m)}}{t^{m+k+4}}e^{-1/t^2}\\&=\sum_{k=1}^{2m-1}\frac{-a_k^{(m)}(m+k+1)}{t^{m+k+2}}e^{-1/t^2}+\sum_{k=3}^{2m+1}\frac{-2a_{k-2}^{(m)}}{t^{m+k+2}}e^{-1/t^2}\\&=\sum_{k=1}^{2m+1}\frac{a_k^{(m+1)}}{t^{m+k+2}}e^{-1/t^2},\end{aligned}$
where $a_1^{(m+1)}=-(m+2)a_1^{(m)}$, $a_2^{(m+1)}=-(m+3)a_2^{(m)}$, $a_{2m}^{(m+1)}=-2a_{2m-2}^{(m)}$, $a_{2m+1}^{(m+1)}=-2a_{2m-1}^{(m)}$, and $a_k^{(m+1)}=-(m+k+1)a_k^{(m)}-2a_{k-2}^{(m)}$ for $3\leq k\leq2m-1$. For $x=0$, we have
$\displaystyle f^{(m+1)}(0)=\lim_{h\to0}\frac{f^{(m)}(h)-f^{(m)}(0)}h=\lim_{h\to0}\sum_{k=1}^{2m-1}\frac{a_k^{(m)}}{h^{m+k+2}}e^{-1/h^2}=\sum_{k=1}^{2m-1}\lim_{h\to0}\frac{a_k^{(m)}}{h^{m+k+2}}e^{-1/h^2}=0.$
Here we have used the same technique to calculuate the limit. Therefore, by mathematical induction, we complete the proof of statement, which tells us $f^{(n)}(0)=0$, i.e., $\displaystyle\lim_{t\to0}\frac{\mathrm d^k}{\mathrm dt^k}e^{-1/t^2}=0$ for all $k\in\mathbb N$.


5. 1. Prove the following weak maximum principle for the heat equation in $U=\mathbb R^n\times(0,T)$: Let $u$ be bounded and continuous on $\bar U=\mathbb R^n\times[0,T]$ with $u_t,u_{x_ix_j}\in C(U)$ and $u_t-\Delta u\leq0$ in $U$. Then

      $\displaystyle M\equiv\sup_{(x,t)\in U}u(x,t)=\sup_{x\in\mathbb R^n}u(x,0)\equiv m$.

   2. Use the maximum principle of (a) to show that the solution (19) is the unique solution that is bounded in $\mathbb R^n\times[0,T]$.
Solution
1. Clearly, $M\geq m$ follows from the fact that $(\mathbb R^n\times\{0\})\subsetneq\bar U$. It suffices to show that $M\leq m$. For this purpose, we fix $x_0\in\mathbb R^n$ and consider the function

   $v(x,t)=u(x,t)-\epsilon(3nt+|x-x_0|^2)\quad\text{for}~(x,t)\in\bar U~\text{and}~\epsilon>0$.

   It is easy to verify that

   $v_t(x,t)-\Delta v(x,t)=u_t(x,t)-\Delta u(x,t)-\epsilon n<0\quad\text{in}~U.$

   Since $u$ is bounded, $\displaystyle\lim_{|x|\to\infty}v(x,t)=-\infty$ for any $t\in[0,T]$, which implies there exists $R>0$ (independent of $t$) such that $\displaystyle v(x,t)\leq-1+\inf_{x\in\mathbb R^n}u(x,0)$ for $|x-x_0|\geq R$ and $t\in\mathbb R$. On the other hand, $\displaystyle v(x_0,t)=u(x_0,0)\geq\inf_{x\in\mathbb R^n}u(x,0)>v(x,t)$ for $|x-x_0|\geq R$ and $t\in[0,T]$. Hence for any $\tau\in(0,T)$, we can apply the weak maximum principle for heat equation on the bounded domain:

   $\displaystyle\sup_{\mathbb R^n\times(0,\tau)}v(x,t)=\sup_{B_R(x_0)\times(0,\tau)}v(x,t)=\max_{\bar B_R(x_0)\times[0,\tau]}v(x,t)=\max_{(\bar B_R(x_0)\times\{0\})\cup(\partial B_R(x_0)\times[0,\tau])}v(x,t)$.

   Let $(x^*,t^*)\in(\bar B_R(x_0)\times\{0\})\cup(\partial B_R(x_0)\times[0,\tau])$ be the maximum point of $v$. Suppose by contradiction that $(x^*,t^*)\in\partial B_R(x_0)\times(0,\tau]$. Then $v_t(x^*,t^*)\geq0$ and $\Delta v(x^*,t^*)\leq0$, which implies $v_t(x^*,t^*)-\Delta v(x^*,t^*)\geq0$. This leads a contradiction. Due to the arbitrariness of $\tau\in(0,T)$, we have

   $\begin{aligned}u(x_0,t)-3\epsilon nT&\leq u(x_0,t)-3\epsilon nt=v(x_0,t)\leq\sup_{(x,t)\in\mathbb R^n\times[0,T]}v(x,t)=\max_{x\in\bar B_R(x_0)}v(x,0)\\&\leq\sup_{x\in\mathbb R^n}v(x,0)=\sup_{x\in\mathbb R^n}[u(x,0)-\epsilon|x-x_0|^2]\leq\sup_{x\in\mathbb R^n}u(x,0)=m.\end{aligned}$

   Letting $\epsilon\to0^+$, we obtain $u(x_0,t)\leq m$ for any $(x_0,t)\in\mathbb R^n\times[0,T]$. Therefore, we obtain $M\leq m$ and complete the proof.
2. Suppose $\tilde u$ and $u$ are bounded solution of the heat equation $u_t(x,t)-\Delta u(x,t)=0$ in $U$ with initial condition $u(x,0)=g(x)$ for $x\in\mathbb R^n$, where $g\in C_B(\mathbb R^n)$. Let $w(x,t)=\tilde u(x,t)-u(x,t)$ for $(x,t)\in\mathbb R^n\times\mathbb R_+$. Then $w$ is also bounded and satisfies

   $\left\{\begin{aligned} &w_t(x,t)-\Delta w(x,t)=0\quad\text{for}~(x,t)\in U,\\&w(x,0)=0\quad\text{for}~x\in\mathbb R^n.\end{aligned}\right.$

   By the result of part (a) for $w$ and $-w$, we have

   $\begin{aligned}0=-\sup_{x\in\mathbb R^n}(-w)(x,0)=-\sup_{(x,t)\in U}(-w)(x,t)=\inf_{(x,t)\in U}w(x,t)\leq\sup_{(x,t)\in U}w(x,t)=\sup_{x\in\mathbb R^n}w(x,0)=0,\end{aligned}$

   which shows $w\equiv0$, and hence $\tilde u\equiv u$ in $\bar U$. Therefore, we complete the proof of the uniqueness of bounded solution of heat equation with initial condition.


6. For $|x|\neq0$, show that $\tilde K$ (given in (22)) is smooth ($C^\infty$) in $t\in\mathbb R$. Conclude that $\tilde K$ is smooth for $(x,t)\neq(0,0)$.
Solution

Recall that

$\tilde K(x,t)=\begin{cases}\displaystyle\frac1{(4\pi t)^{n/2}}e^{-|x|^2/4t}&\text{for}~t>0,\\0&\text{for}~t\leq0.\end{cases}$

To show $\tilde K$ is smooth in $t$ for $|x|\neq0$, it suffices to prove $\partial_t^m\tilde K(x,0)=0$ for $|x|\neq0$ and $m\in\mathbb N$. The major observation is based on the calculation in the solution of Exercise 4, which shows that

$\displaystyle\frac{\partial^m\tilde K}{\partial t^m}(x,t)=\begin{cases}\displaystyle\sum_{i=0}^m\left(\frac{\mathrm d^i}{\mathrm dt^i}(4\pi t)^{-n/2}\right)\left(\frac{\partial^{m-i}}{\partial t^{m-i}}e^{-|x|^2/4t}\right)&\text{for}~t>0,\\0&\text{for}~t<0.\end{cases}$

Clearly $\displaystyle\lim_{t\to0}\frac{\partial^m\tilde K}{\partial t^m}(x,t)=0$ and hence $\tilde K$ is smooth in $t$. On the other hand, for $t\neq0$, $\tilde K$ is obviously smooth for all $x\in\mathbb R^n$. Therefore, $\tilde K$ is smooth for $(x,t)\neq(0,0)$.

The original statement does not follow English grammar so I correct it directly.



7. Heat conduction in a semi-infinite rod with initial temperature $g(x)$ leads to the equations

   $\left\{\begin{aligned} &u_t=u_{xx}\quad\text{for}~x>0~\text{and}~t>0,\\&u(x,0)=g(x)\quad\text{for}~x>0.\end{aligned}\right.$

   Assume that $g$ is continuous and bounded for $x\geq0$.
   1. If $g(0)=0$ and the rod has its end maintained at zero temperature, then we must include the boundary condition $u(0,t)=0$ for $t>0$. Find a formula for the solution $u(x,t)$.
   2. If the rod has its end insulated so that there is no heat flow at $x=0$, then we must include the boundary condition $u_x(0,t)=0$ for $t>0$. Find a formula for the solution $u(x,t)$. Do you need to require $g'(0)=0$?
Solution
1. The problem can be formulated as

   $\left\{\begin{aligned} &u_t=u_{xx}\quad\text{for}~x>0~\text{and}~t>0,\\&u(x,0)=g(x)\quad\text{for}~x>0,\\&u(0,t)=0\quad\text{for}~t>0.\end{aligned}\right.$

   Here $g$ is continuous and bouneded for $x\geq0$ with $g(0)=0$. To solve this problem, we use the reflection method to extend $g$ by

   $\tilde g(x)=\begin{cases}g(x)&\text{if}~x\geq0,\\-g(-x)&\text{if}~x<0.\end{cases}$

   Note that $\tilde g$ is continuous and bounded in $\mathbb R^n$ since $g(0)=0$. Now we pay attention on the following pure initial value problem

   $\left\{\begin{aligned} &\tilde u_t=\tilde u_{xx}\quad\text{for}~x\in\mathbb R~\text{and}~t>0,\\&\tilde u(x,0)=\tilde g(x)\quad\text{for}~x\in\mathbb R.\end{aligned}\right.$

   By Theorem 1 at page 149 in the textbook, we get

   $\begin{aligned}\tilde u(x,t)&=\frac1{\sqrt{4\pi t}}\int_{\mathbb R}\!\exp\left(-\frac{(x-y)^2}{4t}\right)\tilde g(y)\,\mathrm dy\\&=\frac1{\sqrt{4\pi t}}\left[\int_0^\infty\!\exp\left(-\frac{(x-y)^2}{4t}\right)\tilde g(y)\,\mathrm dy+\int_{-\infty}^0\!\exp\left(-\frac{(x-y)^2}{4t}\right)\tilde g(y)\,\mathrm dy\right]\\&=\frac1{\sqrt{4\pi t}}\left[\int_0^\infty\!\exp\left(-\frac{(x-y)^2}{4t}\right)g(y)\,\mathrm dy-\int_{-\infty}^0\!\exp\left(-\frac{(x-y)^2}{4t}\right)g(-y)\,\mathrm dy\right]\\&=\frac1{\sqrt{4\pi t}}\left[\int_0^\infty\!\exp\left(-\frac{(x-y)^2}{4t}\right)g(y)\,\mathrm dy-\int_0^\infty\!\exp\left(-\frac{(x+y)^2}{4t}\right)g(y)\,\mathrm dy\right]\\&=\frac1{\sqrt{4\pi t}}\int_0^\infty\!\left[\exp\left(-\frac{(x-y)^2}{4t}\right)-\exp\left(-\frac{(x+y)^2}{4t}\right)\right]g(y)\,\mathrm dy\\&=\frac1{\sqrt{\pi t}}\int_0^\infty\!\exp\left(-\frac{x^2+y^2}{4t}\right)\sinh\left(\frac{xy}{2t}\right)g(y)\,\mathrm dy.\end{aligned}$

   It is clear that $\tilde u(0,t)=0$ because $\sinh(0)=0$. Therefore, the solution $u$ can be obtained from the restriction of $\tilde u$ over $[0,\infty)\times[0,\infty)$, i.e., $u=\tilde u\Big|_{[0,\infty)\times[0,\infty)}$.
2. The problem can be formulated as

   $\left\{\begin{aligned} &u_t=u_{xx}\quad\text{for}~x>0~\text{and}~t>0,\\&u(x,0)=g(x)\quad\text{for}~x>0,\\&u_x(0,t)=0\quad\text{for}~t>0.\end{aligned}\right.$

   To solve this problem, we use the reflection method to extend $g$ by

   $\tilde g(x)=\begin{cases}g(x)&\text{if}~x\geq0,\\g(-x)&\text{if}~x<0.\end{cases}$

   Clearly, $\tilde g$ is continuous and bounded in $\mathbb R$. Now we pay attention on the following pure initial value problem

   $\left\{\begin{aligned} &\tilde u_t=\tilde u_{xx}\quad\text{for}~x\in\mathbb R~\text{and}~t>0,\\&\tilde u(x,0)=\tilde g(x)\quad\text{for}~x\in\mathbb R.\end{aligned}\right.$

   By Theorem 1 at page 149 in the textbook, we get

   $\begin{aligned}\tilde u(x,t)&=\frac1{\sqrt{4\pi t}}\int_{\mathbb R}\!\exp\left(-\frac{(x-y)^2}{4t}\right)\tilde g(y)\,\mathrm dy\\&=\frac1{\sqrt{4\pi t}}\left[\int_0^\infty\!\exp\left(-\frac{(x-y)^2}{4t}\right)\tilde g(y)\,\mathrm dy+\int_{-\infty}^0\!\exp\left(-\frac{(x-y)^2}{4t}\right)\tilde g(y)\,\mathrm dy\right]\\&=\frac1{\sqrt{4\pi t}}\left[\int_0^\infty\!\exp\left(-\frac{(x-y)^2}{4t}\right)g(y)\,\mathrm dy+\int_{-\infty}^0\!\exp\left(-\frac{(x-y)^2}{4t}\right)g(-y)\,\mathrm dy\right]\\&=\frac1{\sqrt{4\pi t}}\left[\int_0^\infty\!\exp\left(-\frac{(x-y)^2}{4t}\right)g(t)\,\mathrm dy+\int_0^\infty\!\exp\left(-\frac{(x+y)^2}{4t}\right)g(y)\,\mathrm dy\right]\\&=\frac1{\sqrt{4\pi t}}\int_0^\infty\!\left[\exp\left(-\frac{(x-y)^2}{4t}\right)+\exp\left(-\frac{(x+y)^2}{4t}\right)\right]g(y)\,\mathrm dy\\&=\frac1{\sqrt{\pi t}}\int_0^\infty\!\exp\left(-\frac{x^2+y^2}{4t}\right)\cosh\left(\frac{xy}{2t}\right)g(y)\,\mathrm dy.\end{aligned}$

   To verify that the boundary condition holds, we notice that

   $\begin{aligned}\tilde u_x(x,t)=\frac1{\sqrt{\pi t}}\int_0^\infty\!\left[\exp\left(-\frac{x^2+y^2}{4t}\right)\cdot-\frac x{2t}\cosh\left(\frac{xy}{2t}\right)+\exp\left(-\frac{x^2+y^2}{4t}\right)\cdot\sinh\left(\frac{xy}{2t}\right)\cdot\frac y{2t}\right]g(y)\,\mathrm dy,\end{aligned}$

   which implies

   $\displaystyle\tilde u_x(0,t)=\frac1{\sqrt{\pi t}}\int_0^\infty\!\left[\exp\left(-\frac{y^2}{4t}\right)\cdot0+\exp\left(-\frac{y^2}{4t}\right)\cdot0\cdot\frac y{2t}\right]g(y)\,\mathrm dy=0.$

   Therefore, the solution $u$ can be obtained from the restriction of $\tilde u$ over $[0,\infty)\times[0,\infty)$, i.e., $u=\tilde u\Big|_{[0,\infty)\times[0,\infty)}$. Clearly, we do not need to require $g'(0)=0$.


8. If $u\in C^{2,1}(\mathbb R^n\times[0,\infty))$ and $f\in C(\mathbb R^{n+1})$, show that (29) is equivalent to (27).
Solution
Suppose that $u\in C^{2,1}(\mathbb R^n\times[0,\infty))$ and $f\in C(\mathbb R^{n+1})$ satisfies (29), i.e.,
$\displaystyle\int_0^\infty\!\int_{\mathbb R^n}\!(uv_t+u\Delta v+fv)\,\mathrm dx\,\mathrm dt=0\quad\text{for}~v\in C_c^\infty(\mathbb R^{n+1}).$
Using the intergration by parts, we find
$\begin{aligned}0&=\int_0^\infty\!\int_{\mathbb R^n}\!(uv_t+u\Delta v+fv)\,\mathrm dx\,\mathrm dt\\&=\int_{\mathbb R^n}\left[u(x,t)v(x,t)\Big|_{t=0}^{t=\infty}-\int_0^\infty\!u_tv\,\mathrm dt\right]\,\mathrm dx+\int_0^\infty\!\int_{\mathbb R^n}\!(\Delta u)v\,\mathrm dx\,\mathrm dt+\int_0^\infty\!\int_{\mathbb R^n}\!fv\,\mathrm dx\,\mathrm dt\\&=-\int_{\mathbb R^n}u(x,0)v(x,0)\,\mathrm dx+\int_0^\infty\!\int_{\mathbb R^n}\!(-u_t+\Delta u+f)v\,\mathrm dx\,\mathrm dt.\end{aligned}$
Due to arbitrariness of $v\in C_c^\infty(\mathbb R^{n+1})$, we get $-u_t+\Delta u+f\equiv0$ in $\mathbb R^n\times(0,\infty)$ and $u(x,0)=0$ on $\mathbb R^n$, which gives (27).
Conversely, we suppose that $u\in C^{2,1}(\mathbb R^n\times[0,\infty))$ and $f\in C(\mathbb R^{n+1})$ satisfies (27), i.e.,
$\left\{\begin{aligned} &u_t=\Delta u+f(x,t)\quad\text{for}~(x,t)\in\mathbb R^n\times(0,\infty),\\&u(x,0)=0\quad\text{for}~x\in\mathbb R^n.\end{aligned}\right.$
Multiplying it by $v\in C_c^\infty(\mathbb R^{n+1})$ and then integrating it over $\mathbb R^n\times[0,\infty)$, we get
$\displaystyle\int_0^\infty\!\int_{\mathbb R^n}\!(-u_t+\Delta u+f(x,t))v\,\mathrm dx\,\mathrm dt=0$.
Using the integration by parts, we have
$\displaystyle\int_0^\infty\!\int_{\mathbb R^n}(uv_t+u\Delta v+fv)\,\mathrm dx\,\mathrm dt=0$,
which means (29). Here we have used the condition that $v$ has compact support in $\mathbb R^{n+1}$.
Therefore these calculations end the proof.


9. 1. Use (30) to prove Theorem 3.
   2. Verify the following elementary fact: If $0<\alpha<1$ and $\beta\geq0$, then there exists a constant $M=M(\alpha,\beta)>0$ so that $z^\beta e^{-z}\leq Me^{-\alpha z}$ for all $z\geq0$.
   3. Use (b) to verify the following estimates:

      $\begin{aligned} &\partial_t\tilde K(x,t)\leq\frac{M_1}t\tilde K(x,2t),\\&\partial_{x_i}\tilde K(x,t)\leq\frac{M_2}{\sqrt t}\tilde K(x,2t),\\&\partial_{x_ix_j}\tilde K(x,t)\leq\frac{M_3}t\tilde K(x,2t).\end{aligned}$

   4. Use (30) and (c) to confirm the Remark following Theorem 3.
Solution
1. In order to prove Theorem 3, we need to check that function $u$ defined by

   $\displaystyle u(x,t)=\int_0^t\!\int_{\mathbb R^n}\!\tilde K(x-y,t-s)f(y,s)\,\mathrm dy\,\mathrm ds$

   satisfies nonhomogeneous heat equation with homogeneous initial condition:

   $\left\{\begin{aligned} &u_t=\Delta u+f(x,t)\quad\text{for}~x\in\mathbb R^n~\text{and}~t>0,\\&u(x,0)=0\quad\text{for}~x\in\mathbb R^n.\end{aligned}\right.$

   Here $f\in C^{2,1}(\mathbb R^n\times[0,T])$ is bounded for every $T>0$ and $\tilde K$ is defined in (22) (also see Exercise 6). In addition, we set

   $\displaystyle M=\max\left\{\sup_{(x,t)\in\mathbb R^n}|f_t(x,t)|,\sup_{(x,t)\in\mathbb R^n}|\nabla f(x,t)|,\sup_{(x,t)\in\mathbb R^n}|D_x^2f(x,t)|\right\}$.

   Firstly we change variable to write

   $\displaystyle u(x,t)=\int_0^t\!\int_{\mathbb R^n}\!\tilde K(y,s)f(x-y,t-s)\,\mathrm dy\,\mathrm ds$.

   Since $f\in C^{2,1}(\mathbb R^n\times[0,T])$ is bounded for every $T>0$ and $\tilde K=\tilde K(y,s)$ is smooth near $s=t>0$, we get

   $\begin{aligned} &u_t(x,t)=\int_0^t\!\int_{\mathbb R^n}\tilde K(y,s)f_t(x-y,t-s)\,\mathrm dy\,\mathrm ds+\int_{\mathbb R^n}\!\tilde K(y,t)f(x-y,0)\,\mathrm dy,\\&u_{x_ix_j}(x,t)=\int_0^t\!\int_{\mathbb R^n}\!\tilde K(y,s)f_{x_ix_j}(x-y,t-s)\,\mathrm dy\,\mathrm ds.\end{aligned}$

   Clearly, $u_t$ and $u_{x_ix_j}$ are continuous so $u\in C^{2,1}(\mathbb R^n\times(0,\infty))$. Then we find

   $\begin{aligned}u_t(x,t)-\Delta u(x,t)&=\int_0^t\!\int_{\mathbb R^n}\!\tilde K(y,s)\left[\partial_t-\Delta_x\right]f(x-y,t-s)\,\mathrm dy\,\mathrm ds+\int_{\mathbb R^n}\!\tilde K(y,t)f(x-y,0)\,\mathrm dy\\&=\left(\int_0^\epsilon+\int_\epsilon^t\right)\!\int_{\mathbb R^n}\!\tilde K(y,s)[-\partial_s-\Delta_y]f(x-y,t-s)\,\mathrm dy\,\mathrm ds+\int_{\mathbb R^n}\!\tilde K(y,t)f(x-y,0)\,\mathrm dy\end{aligned}$

   For $0\leq t\leq T$, we note that

   $\displaystyle\left|\int_0^\epsilon\!\int_{\mathbb R^n}\!\tilde K(y,s)[-\partial_s-\Delta_y]f(x-y,t-s)\,\mathrm dy\,\mathrm ds\right|\leq2M\int_0^\epsilon\!\int_{\mathbb R^n}\!\tilde K(y,s)\,\mathrm dy\,\mathrm ds=2M\epsilon.$

   Using the integrating by parts, we find

   $\begin{aligned}\int_\epsilon^t\!\int_{\mathbb R^n}\!\tilde K(y,s)[-\partial_s-\Delta_y]f(x-y,t-s)\,\mathrm dy\,\mathrm ds&=\int_\epsilon^t\!\int_{\mathbb R^n}\![\partial_s-\Delta_y]\tilde K(y,s)f(x-y,t-s)\,\mathrm dy\,\mathrm ds\\&\quad+\int_{\mathbb R^n}\!\tilde K(y,\epsilon)f(x-y,t-\epsilon)\,\mathrm dy-\int_{\mathbb R^n}\!\tilde K(y,t)f(x-y,0)\,\mathrm dy,\end{aligned}$

   which implies

   $\begin{aligned} &\quad\int_\epsilon^t\!\int_{\mathbb R^n}\!\tilde K(y,s)[-\partial_s-\Delta_y]f(x-y,t-s)\,\mathrm dy\,\mathrm ds+\int_{\mathbb R^n}\!\tilde K(y,t)f(x-y,0)\,\mathrm dy\\&=\int_{\mathbb R^n}\!\tilde K(y,\epsilon)f(x-y,t-\epsilon)\,\mathrm dy.\end{aligned}$

   Here we have used the fact that $\tilde K_s(y,s)=\Delta_y\tilde K(y,s)$ for $y\in\mathbb R^n$ and $\epsilon\leq s\leq t$. Thus, by Proposition we obtain

   $\displaystyle u_t(x,t)-\Delta u(x,t)=\lim_{\epsilon\to0^+}\int_{\mathbb R^n}\!\tilde K(y,\epsilon)f(x-y,t-\epsilon)\,\mathrm dy=f(x,t).$

   Here we have used the fact that (21), i.e., $\displaystyle\lim_{t\to0^+}\tilde K(x,t)=\delta_0(x)=\delta(x)$ as distribution on $\mathbb R^n$. Finally, we note that $|u(x,t)|\leq tM\to0$ as $t\to0^+$, which gives $u(x,0)=0$. The proof is complete.
2. Consider the function $g:[0,\infty)\to\mathbb R$ defined by

   $g(z)=z^\beta e^{(\alpha-1)z}\quad\text{for}~z\geq0$.

   It is easy to see

   $\displaystyle g'(z)=\beta z^{\beta-1}e^{(\alpha-1)z}+z^\beta e^{(\alpha-1)z}(\alpha-1)=\frac{g(z)}z(\beta+(\alpha-1)z)$.

   By solving $g'(z)=0$, we get $z=\beta/(1-\alpha)$. Since $g(z)\geq0$, we have $g'(z)>0$ on $(0,(1-\alpha)^{-1}\beta)$ and $g'(z)<0$ on $((1-\alpha)^{-1}\beta,\infty)$, which implies $g$ attina its maximum value at $z=(1-\alpha)^{-1}\beta$. Therefore, we find

   $\displaystyle g(z)\leq g\left(\frac\beta{1-\alpha}\right):=M(\alpha,\beta)=M$,

   which gives $z^\beta e^{-z}\leq Me^{-\alpha z}$ for all $z\geq0$.
3. Some direct computations give

   $\begin{aligned}\partial_t\tilde K(x,t)&=\frac\partial{\partial t}\left[(4\pi t)^{-n/2}\exp\left(-\frac{|x|^2}{4t}\right)\right]=\left(-\frac n2\right)(4\pi t)^{-(n+2)/2}\exp\left(-\frac{|x|^2}{4t}\right)+(4\pi t)^{-n/2}\exp\left(-\frac{|x|^2}{4t}\right)\cdot\frac{|x|^2}{4t^2}\\&=\left(-\frac n{8\pi t}+\frac{|x|^2}{4t^2}\right)(4\pi t)^{-n/2}\exp\left(-\frac{|x|^2}{4t}\right)\leq\frac{2^{n/2}}t(4\pi(2t))^{-n/2}\cdot\frac{|x|^2}{4t}\exp\left(-\frac{|x|^2}{4t}\right)\\&\leq\frac{2^{n/2}}t(4\pi(2t))^{-n/2}\cdot M\left(\frac12,1\right)\exp\left(-\frac{|x|^2}{8t}\right)=\frac{2^{n/2}M(1/2,1)}t\tilde K(x,2t):=\frac{M_1}{t}\tilde K(x,2t),\end{aligned}$

   where $M_1=2^{n/2}M(1/2,1)$. On the other hand, we have

   $\begin{aligned}\partial_{x_i}\tilde K(x,t)&=\frac\partial{\partial x_i}(4\pi t)^{-n/2}\exp\left(-\frac{|x|^2}{4t}\right)=(4\pi t)^{-n/2}\exp\left(-\frac{|x|^2}{4t}\right)\cdot-\frac{x_i}{2t}\\&\leq\frac{2^{n/2}}{\sqrt t}(4\pi(2t))^{-n/2}\frac{|x|}{2\sqrt t}\exp\left(-\frac{|x|^2}{4t}\right)\frac{2^{n/2}}{\sqrt t}(8\pi t)^{-n/2}\cdot M\left(\frac12,\frac12\right)\exp\left(-\frac{|x|^2}{8t}\right)\\&=\frac{2^{n/2}M(1/2,1/2)}{\sqrt t}\tilde K(x,2t):=\frac{M_2}{\sqrt t}\tilde K(x,2t),\end{aligned}$

   where $M_2=2^{n/2}M(1/2,1/2)$. Moreover, we have

   $\begin{aligned}\partial_{x_ix_j}\tilde K(x,t)&=\frac\partial{\partial x_j}\left[(4\pi t)^{-n/2}\exp\left(-\frac{|x|^2}{4t}\right)\cdot-\frac{x_i}{2t}\right]=(4\pi t)^{-n/2}\exp\left(-\frac{|x|^2}{4t}\right)\left[\frac{x_ix_j}{4t^2}-\frac{\delta_{ij}}{2t}\right]\\&\leq\frac{2^{n/2}}t\cdot(4\pi(2t))^{-n/2}\cdot\frac{|x|^2}{4t}\exp\left(-\frac{|x|^2}{4t}\right)\leq\frac{2^{n/2}}t(8\pi t)^{-n/2}\cdot M\left(\frac12,1\right)\exp\left(-\frac{|x|^2}{8t}\right)\\&=\frac{2^{n/2}M(1/2,1)}t\cdot \tilde K(x,2t):=\frac{M_3}t\tilde K(x,2t),\end{aligned}$

   where $M_3=2^{n/2}M(1/2,1)$. The proof is complete.
4. I am not sure why we need to use (c). In my opinion, we need to prove the equality $u_t=\Delta u+f(x,t)$ so we cannot use too weak inequality unless we can drop those terms. I think there is a formal derivation as follows.

   $\begin{aligned} u_t(x,t)&=\int_{\mathbb R^n}\!(\Delta_x\tilde K)(x-y,0)f(y,t)\,\mathrm dy\,\mathrm ds+\int_0^t\!\int_{\mathbb R^n}\!\tilde K_t(x-y,t-s)f(y,s)\,\mathrm dy\,\mathrm ds\\&=f(x,t)+\int_0^t\!\int_{\mathbb R^n}\!(\Delta_x\tilde K)(x-y,t-s)f(y,s)\,\mathrm dy\,\mathrm ds\\&=\Delta u(x,t)+f(x,t)\quad\text{in}~\mathbb R^n\times(0,T).\end{aligned}$

   Since $f$ is continuous and bounded in $\mathbb R^n\times[0,T]$, we have $\displaystyle|u(x,t)|\leq t\sup_{(y,s)\in\mathbb R^n\times[0,T]}|f(y,s)|$, which implies $\displaystyle\lim_{t\to0^+}u(x,t)=0$ for any $x\in\mathbb R^n$. Therefore, we complete the proof.


10. If $f(x,t)$ is continuous and bounded on $\mathbb R^{n+1}$ and vanishes for $t$ sufficiently negative [i.e., $f(x,t)=0$ for $t\leq\tau<0$], show that the convolution $u=\tilde K\ast f$ defines a continuous weak solution of the nonhomogeneous heat equation $u_t=\Delta u+f(x,t)$ in $\mathbb R^{n+1}$.
Solution
Since $f$ is defined in $\mathbb R^{n+1}$ and vanishes for $t$ sufficiently negative, the convolution $u$ can be denoted as
$\begin{aligned}u(x,t)&=\int_0^t\!\int_{\mathbb R^n}\!\tilde K(x-y,t-s)f(y,s)\,\mathrm dy\,\mathrm ds\\&=\int_{-\infty}^\infty\!\int_{\mathbb R^n}\!\tilde K(x-y,t-s)f(y,s)\,\mathrm dy\,\mathrm ds\\&=\int_{\mathbb R^{n+1}}\!\tilde K(x-y,t-s)f(y,s)\,\mathrm dy\,\mathrm ds.\end{aligned}$
Here we have used the fact that $\tilde K(x-y,t-s)=0$ for $s\geq t$. Using the integration by parts, we can verify that
$\displaystyle\int_{\mathbb R^{n+1}}\!\int_{\mathbb R^{n+1}}\![\tilde K(x-y,t-s)f(y,s)(\phi_t(x,t)+\Delta\phi(x,t))]\,\mathrm dy\,\mathrm ds\,\mathrm dx\,\mathrm dt+\int_{\mathbb R^{n+1}}\!f(x,t)\phi(x,t)\,\mathrm dx\,\mathrm dt=0$
for any $\phi\in C_c^\infty(\mathbb R^{n+1})$. Hecne we need to show that $u$ is continuous in $\mathbb R^{n+1}$. Clearly, $u(x,t)=0$ for $t<0$. Let us consider the function $\displaystyle v(x,t,s)=\int_{\mathbb R^n}\!K(x,y,t-s)f(y,s)\,\mathrm dy$, which is continuous in $(x,s)\in\mathbb R^n\times[0,t]$ and by Theorem 1 with the intial data $g(y)=f(y,s)$. Therefore, the solution $\displaystyle u=\int_0^t\!v(x,t,s)\,\mathrm ds$ is also continuous in $(x,t)\in\mathbb R^{n+1}$. The proof is complete.


11. Find a formula for the solution of the initial value problem

    $\left\{\begin{aligned} &u_t=\Delta u-u\quad\text{for}~x\in\mathbb R^n~\text{and}~t>0,\\&u(x,0)=g(x)\quad\text{for}~x\in\mathbb R^n,\end{aligned}\right.$

    where $g$ is continuous and bouneded. Is the solution bounded? Is it the only bounded solution?
Solution
Let $v(x,t)=e^tu(x,t)$ for $x\in\mathbb R^n$ and $t>0$. Then $v$ satisfies
$\left\{\begin{aligned} &v_t=\Delta v\quad\text{for}~x\in\mathbb R^n~\text{and}~t>0,\\&v(x,0)=g(x)\quad\text{for}~x\in\mathbb R^n.\end{aligned}\right.$
Since $g$ is continuous and bouneded in $\mathbb R^n$, we can apply Theorem 1 to get
$\displaystyle v(x,t)=\int_{\mathbb R^n}\!K(x,y,t)g(y)\,\mathrm dy=\frac1{(4\pi t)^{n/2}}\int_{\mathbb R^n}\exp\left(-\frac{|x-y|^2}{4t}\right)g(y)\,\mathrm dy$,
which gives
$\displaystyle u(x,t)=\frac{e^{-t}}{(4\pi t)^{n/2}}\int_{\mathbb R^n}\!\exp\left(-\frac{|x-y|^2}{4t}\right)g(y)\,\mathrm dy.$
Since $v$ is bounded and $e^{-t}\leq1$, this solution $u$ is also bounded in $\mathbb R^n\times[0,\infty)$. To show the uniqueness of bounded solution, we suppose that $\tilde u$ is also a bounded solution of
$\left\{\begin{aligned} &\tilde u_t=\Delta\tilde u-\tilde u\quad\text{for}~x\in\mathbb R^n~\text{and}~t>0,\\&\tilde u(x,0)=g(x)\quad\text{for}~x\in\mathbb R^n.\end{aligned}\right.$
Let $w=e^{-t}(\tilde u-u)$ in $\mathbb R^n\times[0,\infty)$. Then $w$ is also bounded in $\mathbb R^n\times[0,\infty)$ and satisfies
$\left\{\begin{aligned} &w_t=\Delta w\quad\text{for}~x\in\mathbb R^n~\text{and}~t>0,\\&w(x,0)=0\quad\text{for}~x\in\mathbb R^n.\end{aligned}\right.$
By the result of part (b) of Exercise 5, we have $w\equiv0$ in $\mathbb R^n\times[0,T]$ for any $T>0$, which means $w\equiv0$ and hence $\tilde u\equiv u$ in $\mathbb R^n\times[0,\infty)$.