Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 2.3

1. 1. For complex-valued functions $u$ and $v$ on $\Omega$, let $\displaystyle\langle u,v\rangle=\int_\Omega\!u\bar v\,\mathrm dx$, where $\bar v$ denotes the complex conjugate of $v$. If the coefficients $a_\alpha(x)$ in (32a) are complex-valued functions, define the adjoint $L^*$ so that $\langle Lu,v\rangle=\langle u,L^*v\rangle$ for all $u\in C^m(\Omega)$, $v\in C_0^m(\Omega)$.
   2. For (complex) vector-valued functions $\vec u(x)=(u_1(x),\dots,u_N(x))$ on $\Omega$, let $\displaystyle\langle\vec u,\vec v\rangle=\int_\Omega\!(u_1\bar v_1+\cdots+u_N\bar v_N)\,\mathrm dx$. If the coefficients $a_\alpha(x)$ in (32a) are $N\times N$ matrix-valued functions, then (32a) defines a system of $m$th-order operators. Define the adjoint $L^*$ so that $\langle L\vec u,\vec v\rangle=\langle\vec u,L^*\vec v\rangle$ for all $\vec u\in C^m(\Omega,\mathbb R^N)$, $\vec v\in C_0^m(\Omega,\mathbb R^N)$.
Solution
1. From (32a), we find that

   $\begin{aligned}\langle Lu,v\rangle&=\int_\Omega\!(Lu)(\bar v)\,\mathrm dx=\int_\Omega\!\left(\sum_{|\alpha|\leq m}a_\alpha(x)D^\alpha u\right)\bar v\,\mathrm dx\\&=\sum_{|\alpha|\leq m}\int_\Omega\!(D^\alpha u)(a_\alpha(x)\bar v)\,\mathrm dx=\sum_{|\alpha|\leq m}(-1)^{|\alpha|}\int_\Omega\!(u)(D^\alpha(a_\alpha(x)\bar v))\,\mathrm dx\\&=\int_\Omega\!(u)\left[\sum_{|\alpha|\leq m}(-1)^{|\alpha|}D^\alpha(a_\alpha(x)\bar v)\right]\,\mathrm dx:=\langle u,L^*v\rangle,\end{aligned}$

   where the adjoint operator $L^*$ is given by

   $\displaystyle L^*v:=\overline{\sum_{|\alpha|\leq m}(-1)^{|\alpha|}D^{\alpha}(a_\alpha(x)\bar v)}=\sum_{|\alpha|\leq m}(-1)^{|\alpha|}D^\alpha(\overline{a_\alpha(x)}v)\quad\text{for}~v\in C_0^m(\Omega).$

2. Suppose $a_\alpha(x)=[a_\alpha^{ij}(x)]_{1\leq i,j\leq N}$ is a $N\times N$ matrix-valued function. Then we have

   $\begin{aligned}\langle L\vec u,\vec v\rangle&=\int_\Omega\!\left[\left(\sum_{j=1}^N\sum_{|\alpha|\leq m}a_\alpha^{1j}(x)D^\alpha u_j\right)\bar v_1+\cdots+\left(\sum_{j=1}^N\sum_{|\alpha|\leq m}a_\alpha^{Nj}(x)D^\alpha u_j\right)\bar v_N\right]\,\mathrm dx\\&=\int_\Omega\!\left(\sum_{j=1}^N\sum_{|\alpha|\leq m}a_\alpha^{1j}(x)D^\alpha u_j\right)\bar v_1\,\mathrm dx+\cdots+\int_\Omega\!\left(\sum_{j=1}^N\sum_{|\alpha|\leq m}a_\alpha^{Nj}(x)D^\alpha u_j\right)\bar v_N\,\mathrm dx\\&=\sum_{j=1}^N\sum_{|\alpha|\leq m}\left[\int_\Omega\!(D^\alpha u_j)(a_\alpha^{1j}(x)\bar v_1)\,\mathrm dx+\cdots+\int_\Omega\!(D^\alpha u_j)(a_\alpha^{Nj}(x)\bar v_N)\,\mathrm dx\right]\\&=\sum_{j=1}^N\sum_{|\alpha|\leq m}(-1)^{|\alpha|}\left[\int_\Omega\!(u_j)D^\alpha(a_\alpha^{1j}(x)\bar v_1)\,\mathrm dx+\cdots+\int_\Omega\!(u_j)D^\alpha(a_\alpha^{Nj}(x)\bar v_N)\,\mathrm dx\right]\\&=\sum_{j=1}^N\int_\Omega\!(u_j)\sum_{|\alpha|\leq m}(-1)^{|\alpha|}D^\alpha(a_\alpha^{1j}(x)\bar v_1+\cdots+a_\alpha^{Nj}(x)\bar v_N)\,\mathrm dx=\langle\vec u,L^*\vec v\rangle,\end{aligned}$

   where the adjoint operator $L^*$ is given by

   $\displaystyle L^*\vec v=\overline{\sum_{|\alpha|\leq m}(-1)^{|\alpha|}D^\alpha(a_\alpha^{\mathsf T}(x)\bar v)}=\sum_{|\alpha|\leq m}(-1)^{|\alpha|}D^\alpha(\overline{a_\alpha^{\mathsf T}(x)}v)\quad\text{for}~v\in C_0^m(\Omega,\mathbb R^N).$

   Here $A^{\mathsf T}$ denotes the tranpose of the matrix $A$.


2. Let $Lu=u_{\mu\eta}$ as in Example 1. Derive the transmission condition (44) along the characteristic $\mu=0$.
Solution

Suppose $u$ is a weak solution of $Lu=0$, i.e., $\int_\Omega\!u_{\mu\eta}v\,\mathrm d\mu\,\mathrm d\eta=\int_\Omega(Lu)v\,\mathrm d\mu\,\mathrm d\eta=0$ for $v\in C_0^2(\Omega)$. Recall that the operator $L$ is self-adjoint, i.e., $L=L'$. From (40), we have

$\begin{aligned}0&=\int_\Omega\!u(L'v)\,\mathrm d\mu\,\mathrm d\eta=\int_{\Omega^+}\!u^+(L'v)\,\mathrm d\mu\,\mathrm d\eta+\int_{\Omega^-}\!u^-(L'v)\,\mathrm d\mu\,\mathrm d\eta\\&=\int_{\Omega^+}\!u^+v_{\mu\eta}\,\mathrm d\mu\,\mathrm d\eta+\int_{\Omega^-}\!u^-v_{\mu\eta}\,\mathrm d\mu\,\mathrm d\eta\\&=\int_0^\infty\!\int_{-\infty}^{\infty}\!u^+v_{\mu\eta}\,\mathrm d\eta\,\mathrm d\mu+\int_{-\infty}^0\!\int_{-\infty}^\infty\!u^-v_{\mu\eta}\,\mathrm d\eta\,\mathrm d\mu\\&=-\int_0^\infty\!\int_{-\infty}^\infty\!u_\eta^+v_\mu\,\mathrm d\eta\,\mathrm d\mu-\int_{-\infty}^0\!\int_{-\infty}^\infty\!u_\eta^-v_\mu\,\mathrm d\eta\,\mathrm d\mu\\&=-\int_{-\infty}^\infty\!\int_0^\infty\!u_\eta^+v_\mu\,\mathrm d\mu\,\mathrm d\eta-\int_{-\infty}^\infty\!\int_{-\infty}^0\!u_\eta^-v_\mu\,\mathrm d\mu\,\mathrm d\eta\\&=-\int_{-\infty}^\infty\!\left[u_\eta^+v\Big|_{\mu=0}^{\mu=\infty}-\int_0^\infty\!u_{\mu\eta}^+v\,\mathrm d\mu\right]\,\mathrm d\eta-\int_{-\infty}^\infty\left[u_\eta^-v\Big|_{\mu=-\infty}^{\mu=0}-\int_{-\infty}^0\!u_{\mu\eta}^-v\,\mathrm d\mu\right]\,\mathrm d\eta\\&=\int_{-\infty}^\infty\!u_\eta^+(0,\eta)v(0,\eta)\,\mathrm d\eta-\int_{-\infty}^\infty\!u_\eta^-(0,\eta)v(0,\eta)\,\mathrm d\eta=\int_{-\infty}^\infty\![u_\eta^+(0,\eta)-u_\eta^-(0,\eta)]v(0,\eta)\,\mathrm d\eta.\end{aligned}$

Thus, the transmission condition is

$\displaystyle\int_{-\infty}^\infty\![u_\eta^+(0,\eta)-u_\eta^-(0,\eta)]v(0,\eta)\,\mathrm d\eta=0\quad\text{for}~v\in C_0^2(\Omega)$.



3. Consider the first-order equation $u_t+cu_x=0$.
   1. If $f\in C(\mathbb R)$, show that $u(x,t)=f(x-ct)$ is a weak solution.
   2. Can you find any discontinuous weak solutions?
   3. Is there a transmission condition for a weak solution with jump discontinuity along the characteristic $x=ct$?
Solution
1. To show $u(x,t)=f(x-ct)$ is a weak solution, it suffices to verify that $\displaystyle\int_{\mathbb R^2}\!f(x-ct)(L'v)(x,t)\,\mathrm dx\,\mathrm dt=0$ for $v\in C_0^1(\mathbb R^2)$, where $L'=-\partial_t-c\partial_x$ is the adjoint operator $L=\partial_t+c\partial_x$. Let $s=x-ct$, $t=t$ and $\tilde v(s,t)=v(s+ct,t)$ for $s,t\in\mathbb R^2$. Clearly, $\tilde v\in C_0^1(\mathbb R^2)$ and satisfies $\tilde v_t(s,t)=v_t(s+ct,t)+cv_x(s+ct,t)$. Thus, the improper double integral becomes

   $\begin{aligned}\int_{\mathbb R^2}f(x-ct)(L'v)(x,t)\,\mathrm dx\,\mathrm dt&=-\int_{-\infty}^\infty\!\int_{-\infty}^\infty\!f(x-ct)(v_t(x,t)+cv_x(x,t))\,\mathrm dx\,\mathrm dt\\&=-\int_{-\infty}^\infty\!\int_{-\infty}^\infty\!f(s)\tilde v_t(s,t)\,\mathrm ds\,\mathrm dt=-\int_{-\infty}^\infty\!f(s)\int_{-\infty}^\infty\tilde v_t(s,t)\,\mathrm dt\,\mathrm ds\\&=-\int_{-\infty}^\infty\!f(s)\cdot0\,\mathrm ds=0\quad\text{for}~v\in C_0^1(\mathbb R^2).\end{aligned}$

   Here the Jacobian is $1$. Therefore, $u(x,t)=f(x-ct)$ is a weak solution for $u_t+cu_x=0$.
2. When $f$ is discontinuous but Lebesgue integrable, $u(x,t)=f(x-ct)$ is still a weak solution because the computation in part (a) also holds true.
3. Suppose $u$ is a weak solution with jump discontinuity along the characteristic $x=ct$. Let $\tilde u(s,t)=u(s+ct,t)$. Then $\tilde u$ is a weak solution of $\tilde L\tilde u:=\tilde u_t(s,t)=0$ with jump discontinuity along the line $s=0$, where $\tilde L=\partial_t$. Clearly, the adjoint operator $\tilde L'$ of $\tilde L$ is $-\partial_t$. Let $\mathbb R^2_\pm=\{(s,t)\in\mathbb R^2:\,\pm s>0,\,t\in\mathbb R\}$ and $\tilde u^\pm=\tilde u\Big|_{\mathbb R^2_\pm}$. Then for any test function $v\in C_0^1(\mathbb R^2)$, we have

   $\begin{aligned}0&=\int_{\mathbb R^2}\tilde u(s,t)\tilde L'v(s,t)\,\mathrm ds\,\mathrm dt=\int_{\mathbb R^2}\tilde u(s,t)\cdot(-\partial_tv(s,t))\,\mathrm ds\,\mathrm dt\\&=-\int_0^\infty\!\int_{-\infty}^\infty\!\tilde u^+(s,t)v_t(s,t)\,\mathrm dt\,\mathrm ds-\int_{-\infty}^0\!\int_{-\infty}^\infty\!\tilde u^-(s,t)v_t(s,t)\,\mathrm dt\,\mathrm ds\\&=\int_0^\infty\!\left[-\tilde u^+(s,t)v(s,t)\Big|_{t=-\infty}^{t=\infty}+\int_{-\infty}^\infty\!\tilde u_t^+(s,t)v(s,t)\,\mathrm dt\right]\,\mathrm ds\\&\quad+\int_{-\infty}^0\!\left[-\tilde u^-(s,t)v(s,t)\Big|_{t=-\infty}^{t=\infty}+\int_{-\infty}^\infty\!\tilde u_t^-(s,t)v(s,t)\,\mathrm dt\right]\,\mathrm ds\\&=\int_0^\infty\!\int_{-\infty}^\infty\!\tilde u_t^+(s,t)v(s,t)\,\mathrm dt\,\mathrm ds+\int_{-\infty}^0\!\int_{-\infty}^\infty\!\tilde u_t^-(s,t)v(s,t)\,\mathrm dt\,\mathrm ds=0.\end{aligned}$

   Thus, there is no transmission condition for a weak solution with jump discontinuity along the characteristic $x=ct$.


4. If $f\in L_{\text{loc}}^1(\Omega)$, define $\displaystyle\langle F_f,v\rangle\equiv\int_\Omega\!f(x)v(x)\,\mathrm dx$. Show that $F_f$ is a distribution in $\Omega$.
Solution
Let $a,b\in\mathbb R$ and $v,w\in\mathcal D(\Omega)$. It is easy to verify that
$\begin{aligned}F_f(av+bw)&:=\langle F_f,av+bw\rangle=\int_\Omega\!f(x)[av(x)+bw(x)]\,\mathrm dx\\&=a\int_\Omega\!f(x)v(x)\,\mathrm dx+b\int_\Omega\!f(x)w(x)\,\mathrm dx=a\langle F_f,v\rangle+b\langle F_f,w\rangle=aF_f(v)+bF_f(w),\end{aligned}$
which shows $F_f$ is a linear mapping on $\mathcal D(\Omega)$. Moreover, let $\{v_j\}_{j=1}^\infty$ be a sequence of functions in $\mathcal D(\Omega)$ with $\text{supp}(v_j)\subseteq K$ for some compact set $K\subsetneq\Omega$ (independent of $j$) and $D^\alpha v_j\to0$ uniformly in $K$. Then for any given $\varepsilon>0$, there exists a $N\in\mathbb N$ such that $|v_j(x)|<\varepsilon\left(\int_K\!|f(x)|\,\mathrm dx\right)^{-1}$ for $j>N$ and $x\in K$. For $j>N$, we have
$\begin{aligned}|F_f(v_j)-0|&=|\langle F_f,v_j\rangle|=\left|\int_\Omega\!f(x)v_j(x)\,\mathrm dx\right|=\left|\int_K\!f(x)v_j(x)\,\mathrm dx\right|\\&\leq\int_K\!|f(x)||v_j(x)|\,\mathrm dx\leq\int_K\!|f(x)|\cdot\varepsilon\left(\int_K\!|f(x)|\,\mathrm dx\right)^{-1}\,\mathrm dx=\varepsilon.\end{aligned}$
This implies $\displaystyle\lim_{j\to\infty}F_f(v_j)=0$. Therefore, $F_f$ is continuous on $\mathcal D(\Omega)$ and $F_f$ is a distribution in $\Omega$.


5. 1. If $\xi_1,\dots,\xi_k$ are points in $\Omega$ and $a_1,\dots,a_k$ are real numbers, show that $F\equiv a_1\delta_{\xi_1}+\cdots+a_k\delta_{\xi_k}$ is a distribution in $\Omega$.
   2. For $\Omega=\mathbb R^n$, find an infinite sequence $\{\xi_k\}_{k=1}^\infty$ for which $f\equiv a_1\delta_{\xi_1}+\cdots$ is a distribution. (This gives an example of a distribution that cannot be realized as a distributional derivative of an integrable function.)
Solution
1. Let $a,b\in\mathbb R$ and $v,w\in\mathcal D(\Omega)$. It is easy to verify that

   $\begin{aligned}F(av+bw)&=\langle F,av+bw\rangle=\left\langle\sum_{i=1}^ka_i\delta_{\xi_i},av+bw\right\rangle=\int_\Omega\!\sum_{i=1}^ka_i\delta_{\xi_i}(x)[av(x)+bw(x)]\,\mathrm dx\\&=\sum_{i=1}^ka_i\int_\Omega\!\delta_{\xi_i}(x)[av(x)+bw(x)]\,\mathrm dx=\sum_{i=1}^ka_i[av(\xi_i)+bw(\xi_i)]\\&=a\sum_{i=1}^ka_iv(\xi_i)+b\sum_{i=1}^ka_iw(\xi_i)=a\sum_{i=1}^ka_i\int_\Omega\!\delta_{\xi_i}(x)v(x)\,\mathrm dx+b\sum_{i=1}^ka_i\int_\Omega\!\delta_{\xi_i}(x)w(x)\,\mathrm dx\\&=a\int_\Omega\!\sum_{i=1}^ka_i\delta_{\xi_i}(x)[v(x)]\,\mathrm dx+b\int_\Omega\!\sum_{i=1}^ka_i\delta_{\xi_i}(x)[w(x)]\,\mathrm dx=a\left\langle\sum_{i=1}^ka_i\delta_{\xi_i},v\right\rangle+ba\left\langle\sum_{i=1}^ka_i\delta_{\xi_i},w\right\rangle\\&=a\langle F,v\rangle+b\langle F,w\rangle=aF(v)+bF(w),\end{aligned}$

   which shows $F$ is a linear mapping on $\mathcal D(\Omega)$. Moreover, let $\{v_j\}_{j=1}^\infty$ be a sequence of functions in $\mathcal D(\Omega)$ with $\text{supp}(v_j)\subseteq K$ for some compact set $K\subsetneq\Omega$ (independent of $j$) and $D^\alpha v_j\to0$ uniformly in $K$. Since $K\cap\{\xi_1,\dots,\xi_k\}$ is finite, we may write it as $\{\xi_{j_1},\dots,\xi_{j_r}\}$ for $r\in\mathbb N$. Set $A=\max\limits_{1\leq i\leq r}|a_{j_i}|$. Then for any $\varepsilon>0$, there exists $N\in\mathbb N$ such that $|v_j(x)|\leq\varepsilon/(rA)$ for $j>N$ and $x\in K$. For $j>N$, we have

   $\begin{aligned}|F(v_j)-0|&=|\langle F,v_j\rangle|=\left|\int_\Omega\!F(x)v_j(x)\,\mathrm dx\right|=\left|\int_\Omega\!\sum_{i=1}^ka_i\delta_{\xi_i}(x)v_j(x)\,\mathrm dx\right|\\&\leq\sum_{i=1}^r|a_{j_i}|\left|\int_K\!\delta_{\xi_{j_i}}(x)v_j(x)\,\mathrm dx\right|=\sum_{i=1}^r|a_{j_i}||v_j(\xi_{j_i})|\leq\sum_{i=1}^rA\cdot\frac{\varepsilon}{rA}=\varepsilon.\end{aligned}$

   This implies $\displaystyle\lim_{j\to\infty}F(v_j)=0$. Therefore, $F$ is continuous on $\mathcal D(\Omega)$ and $F$ is a distribution in $\Omega$.
2. Choose $\xi_j=je_1:=(j,0,\dots,0)$ for $j\in\mathbb N$, and then we prove $F=\sum_{i=1}^{\infty}a_i\delta_{\xi_i}$ is a distribution in $\mathbb R^n$. Let $a,b\in\mathbb R$ and $v,w\in\mathcal D(\mathbb R^n)$. Then there exists a compact subset $K\subsetneq\mathbb R^n$ such that $\text{supp}(v)\subseteq K$ and $\text{supp}(w)\subseteq K$. Since $K$ is bounded, $K\cap\{\xi_j\}_{j=1}^\infty$ is finite and $K\cap\{\xi_j\}_{j=1}^\infty=\{\xi_{j_1},\dots,\xi_{j_r}\}$ for $r\in\mathbb N$. Then one may check that

   $\begin{aligned}F(av+bw)&=\langle F,av+bw\rangle=\left\langle\sum_{j=1}^\infty a_j\delta_{\xi_j},av+bw\right\rangle=\int_{\mathbb R^n}\!\sum_{j=1}^\infty\!a_j\delta_{\xi_j}(x)[av(x)+bw(x)]\,\mathrm dx\\&=\int_K\!\sum_{i=1}^ra_{j_i}\delta_{\xi_{j_i}}(x)[av(x)+bw(x)]\,\mathrm dx=\sum_{i=1}^ra_{j_i}\int_K\!\delta_{\xi_{j_i}}(x)[av(x)+bw(x)]\,\mathrm dx\\&=\sum_{i=1}^ra_{j_i}[av(\xi_{j_i})+bw(\xi_{j_i})]=a\sum_{i=1}^ra_{j_i}v(\xi_{j_i})+b\sum_{i=1}^ra_{j_i}w(\xi_{j_i})\\&=a\sum_{i=1}^ra_{j_i}\int_K\!\delta_{\xi_{j_i}}(x)v(x)\,\mathrm dx+b\sum_{i=1}^ra_{j_i}\int_K\!\delta_{\xi_{j_i}}(x)w(x)\,\mathrm dx\\&=a\int_K\!\sum_{i=1}^ra_{j_i}\delta_{\xi_{j_i}}(x)v(x)\,\mathrm dx+b\int_K\!\sum_{i=1}^ra_{j_i}\delta_{\xi_{j_i}}(x)w(x)\,\mathrm dx\\&=a\int_{\mathbb R^n}\!\sum_{j=1}^\infty a_j\delta_{\xi_j}(x)v(x)\,\mathrm dx+b\int_{\mathbb R^n}\!\sum_{j=1}^\infty a_j\delta_{\xi_j}(x)w(x)\,\mathrm dx\\&=a\left\langle\sum_{j=1}^\infty a_j\delta_{\xi_j},v\right\rangle+b\left\langle\sum_{j=1}^\infty a_j\delta_{\xi_j},w\right\rangle\\&=a\langle F,v\rangle+b\langle F,w\rangle=aF(v)+bF(w),\end{aligned}$

   which shows $F$ is a linear mapping on $\mathcal D(\mathbb R^n)$. Moreover, let $\{v_k\}_{k=1}^\infty$ be a sequence of functions in $\mathcal D(\mathbb R^n)$ with $\text{supp}(v_k)\subseteq K$ for some compact set $K\subsetneq\mathbb R^n$ (independent of $j$) and $D^\alpha v_k\to0$ uniformly in $K$. Since $K$ is bounded, $K\cap\{\xi_j\}_{j=1}^\infty$ is finite and $K\cap\{\xi_j\}_{j=1}^\infty=\{\xi_{j_i}\}_{i=1}^r$ for $r\in\mathbb N$. Set $A=\max\limits_{1\leq i\leq r}|a_{j_i}|$. Then for any $\varepsilon>0$, there exists $N\in\mathbb N$ such that $|v_k(x)|<\varepsilon/(rA)$ for $k>N$ and $x\in K$. For $j>N$, we have

   $\begin{aligned}|F(v_k)-0|&=|\langle F,v_k\rangle|=\left|\int_{\mathbb R^n}F(x)v_k(x)\,\mathrm dx\right|=\left|\int_{\mathbb R^n}\!\sum_{j=1}^\infty a_j\delta_{\xi_j}(x)v_k(x)\,\mathrm dx\right|\\&=\left|\int_K\!\sum_{i=1}^ra_{j_i}\delta_{\xi_{j_i}}(x)v_k(x)\,\mathrm dx\right|=\left|\sum_{i=1}^ra_{j_i}v_k(\xi_{j_i})\right|\leq\sum_{i=1}^r|a_{j_i}||v_k(\xi_{j_i})|\leq\sum_{i=1}^rA\cdot\frac{\varepsilon}{rA}=\varepsilon.\end{aligned}$

   This implies $\displaystyle\lim_{k\to\infty}F(v_k)=0$. Therefore, $F$ is continuous on $\mathcal D(\mathbb R^n)$ and $F$ is a distribution in $\mathbb R^n$.


6. 1. Prove the Lemma on p. 67 when $f\in C(\mathbb R^n)$ has a compact support.
   2. Find an example to show that the result fails to hold if $g\in L_{\text{loc}}^1(\mathbb R^n)$ is replaced by $g\in\mathcal D'(\mathbb R^n)$.
Solution
1. Suppose that $K=\text{supp}(f)$ is a compact subset of $\mathbb R^n$. Let $x\in\mathbb R^n$ and $y\in\bar B_1(x)$ arbitrarily and define $h(z)=f(x-z)-f(y-z)$ for $z\in\mathbb R^n$. Clearly, $h$ is continuous function compact support, and

   $\text{supp}(h)\subseteq(x-K)\cup(\bar B_1(x)-K):=\tilde K$,

   where

   $x-K=\{x-z\in\mathbb R^n\,:\,z\in K\}$ and $\bar B_1(x)-K=\{y-z\in\mathbb R^n\,:y\in\bar B_1(x),\,z\in K\}$.

   On the other hand, $f$ is uniformly continuous in $\mathbb R^n$. Then for any $\varepsilon>0$, there exists $\delta=\delta(\varepsilon)>0$ such that $\displaystyle|f(x-z)-f(y-z)|<\varepsilon\left(\int_{\tilde K}\!|g(z)|\,\mathrm dz\right)^{-1}$ whenever $|x-y|<\delta$ and $z\in\mathbb R^n$. Thus, we find that

   $\begin{aligned}|(f\ast g)(x)-(f\ast g)(y)|&=\left|\int_{\mathbb R^n}\![f(x-z)-f(y-z)]g(z)\,\mathrm dz\right|\\&=\left|\int_{\tilde K}[f(x-z)-f(y-z)]g(z)\,\mathrm dz\right|\\&\leq\int_{\tilde K}|f(x-z)-f(y-z)||g(z)|\,\mathrm dz<\varepsilon.\end{aligned}$

   This shows $\displaystyle\lim_{y\to x}(f\ast g)(y)=(f\ast g)(x)$, which means $f\ast g\in C(\mathbb R^n)$.
2. For $n=1$, let $g=\delta'\in\mathcal D'(\mathbb R)$ and $f(x)=|x|$ for $x\in\mathbb R$. Then one may check that

   $\begin{aligned}(f\ast g)(x)&=\int_{\mathbb R}\!f(x-y)g(y)\,\mathrm dy=\int_{\mathbb R}\!|x-y|\delta'(y)\,\mathrm dy\\&=-\int_{\mathbb R}\!\frac{\mathrm d}{\mathrm dy}|x-y|\delta(y)\,\mathrm dy=\int_{\mathbb R}\text{sgn}(x-y)\delta(y)\,\mathrm dy=\text{sgn}(x),\end{aligned}$

   which is not continuous on $\mathbb R$.


7. Let $\Gamma$ be a hypersurface (such as a sphere) in $\mathbb R^n$, let $a(z)$ be a continuous function of $z\in\Gamma$, and let $\mathrm dz$ denote the surface measure on $\Gamma$.
   1. Show that $\displaystyle\langle F,v\rangle=\int_\Gamma\!a(z)v(z)\,\mathrm dz$ is a distribution in $\mathbb R^n$. (It is natural to denote this distribution by $F=a\delta_\Gamma$.)
   2. Suppose we can choose a unit normal $\nu$ along $\Gamma$. Formulate a definition for the "conormal distribution" $a\partial_\nu\delta_\Gamma$ in $\mathbb R^n$.
Solution
1. Note that $\Gamma$ is a hypersurface in $\mathbb R^n$ so $\Gamma$ is a closed set in $\mathbb R^n$. For $c_1,c_2\in\mathbb R$ and $v_1,v_2\in\mathcal D(\mathbb R^n)$, we have

   $\begin{aligned}\langle F,c_1v_1+c_2v_2\rangle&=\int_{\Gamma}\!a(z)[c_1v_1(z)+c_2v_2(z)]\,\mathrm dz\\&=c_1\int_{\Gamma}\!a(z)v_1(z)\,\mathrm dz+c_2\int_{\Gamma}\!a(z)v_2(z)\,\mathrm dz\\&=c_1\langle F,v_1\rangle+c_2\langle F,v_2\rangle,\end{aligned}$

   which shows $F$ is a linear mapping on $\mathcal D(\mathbb R^n)$. Here we have used the fact that $v_1,v_2\in C^\infty(\Gamma)\subsetneq\mathcal D(\mathbb R^n)$. Moreoever, let $\{v_j\}_{j=1}^\infty$ be a sequence of functions in $\mathcal D(\mathbb R^n)$ with $\text{supp}(v_j)\subseteq K$ for some compact set $K\subsetneq\mathbb R^n$ (independent of $j$) and $D^\alpha v_j\to0$ uniformly in $K$. Since $a$ is continuous on the compact set $\Gamma\cap K$, the maximum of $|a|$ over $\Gamma\cap K$ is finite and can be denoted by $A:=\max\limits_{z\in\Gamma\cap K}a(z)$. For any $\varepsilon>0$, there exists $N\in\mathbb N$ such that

   $\displaystyle|v_j(x)|\leq\varepsilon\left[A\int_{\Gamma\cap K}\!\,\mathrm dz\right]^{-1}$ for $j> N$ and $x\in K$.

   For $j>N$, we have

   $\begin{aligned}|\langle F,v_j\rangle-0|&=\left|\int_{\Gamma}\!a(z)v_j(z)\,\mathrm dz\right|=\left|\int_{\Gamma\cap K}\!a(z)v_j(z)\,\mathrm dz\right|\\&\leq\int_{\Gamma\cap K}\!|a(z)||v_j(z)|\,\mathrm dz\leq\int_{\Gamma\cap K}\!A\cdot\varepsilon\left[A\int_{\Gamma\cap K}\!\,\mathrm dz\right]^{-1}\,\mathrm dz=\varepsilon.\end{aligned}$

   This implies $\displaystyle\lim_{j\to\infty}F(v_j)=0$. Therefore, $F$ is continuous on $\mathcal D(\mathbb R^n)$ and $F$ is a distribution in $\mathbb R^n$.
2. Suppose $a\in C^1(\Gamma)$. Then we define the conformal distribution $a\partial_\nu\delta_\Gamma$ by

   $\displaystyle\langle a\partial_\nu\delta_\Gamma,v\rangle=-\int_\Gamma\nabla(a(z)v(z))\cdot\nu(z)\,\mathrm dz\quad\text{for}~v\in\mathcal D(\mathbb R^n)$.

   For $c_1,c_2\in\mathbb R$ and $v_1,v_2\in\mathcal D(\mathbb R^n)$, we have

   $\begin{aligned}\langle a\partial_\nu\delta_\Gamma,c_1v_1+c_2v_2\rangle&=-\int_\Gamma\!\nabla(a(z)[c_1v_1(z)+c_2v_2(z)])\cdot\nu(z)\,\mathrm dz\\&=-c_1\int_\Gamma\!\nabla(a(z)v_1(z))\cdot\nu(z)\,\mathrm dz-c_2\int_\Gamma\!\nabla(a(z)v_2(z))\cdot\nu(z)\,\mathrm dz\\&=c_1\langle a\partial_\nu\delta_\Gamma,v_1\rangle+c_2\langle a\partial_\nu\delta_\Gamma,v_2\rangle,\end{aligned}$

   which shows $\langle a\partial_\nu\delta_\Gamma,\cdot\rangle$ is a linear mapping on $\mathcal D(\mathbb R^n)$. Here we have used the fact that $v_1,v_2\in C^\infty(\Gamma)\subsetneq\mathcal D(\mathbb R^n)$. Moreover, let $\{v_j\}_{j=1}^\infty$ be a sequence of functions in $\mathcal D(\mathbb R^n)$ with $\text{supp}(v_j)\subseteq K$ for some compact set $K\subsetneq\mathbb R^n$ (independent of $j$) and $D^\alpha v_j\to0$ uniformly in $K$. Since $a$ is continuously differentiable on the compact set $\Gamma\cap K$, $\|a\|_{C^1(\Gamma\cap K)}:=\max\limits_{z\in\Gamma\cap K}|a(z)|+\max\limits_{z\in\Gamma\cap K}|\nabla a(z)|$ is finite. For any $\varepsilon>0$, there exists $N\in\mathbb N$ such that

   $\displaystyle v_j(x)\leq\frac\varepsilon2\left[\|a\|_{C^1(\Gamma\cap K)}\int_{\Gamma\cap K}\mathrm dz\right]^{-1},\quad|\nabla v_j(x)|\leq\frac\varepsilon2\left[\|a\|_{C^1(\Gamma\cap K)}\int_{\Gamma\cap K}\mathrm dz\right]^{-1}\quad\text{for}~j>N~\text{and}~x\in K$.

   For $j>N$, we have

   $\begin{aligned}|\langle a\partial_\nu\delta_\Gamma,v_j\rangle|&=\left|\int_\Gamma\!\nabla(a(z)v_j(z))\cdot\nu(z)\,\mathrm dz\right|=\left|\int_\Gamma\![v_j(z)\nabla a(z)+a(z)\nabla v_j(z)]\cdot\nu(z)\,\mathrm dz\right|\\&=\left|\int_{\Gamma\cap K}\![v_j(z)\nabla a(z)+a(z)\nabla v_j(z)]\cdot\nu(z)\,\mathrm dz\right|\leq\int_{\Gamma\cap K}(|v_j(z)||\nabla a(z)|+|a(z)||\nabla v_j(z)|)\,\mathrm dz\\&\leq\int_{\Gamma\cap K}\|a\|_{C^1(\Gamma\cap K)}(|v_j(z)|+|\nabla v_j(z)|)\,\mathrm dz\\&\leq\|a\|_{C^1(\Gamma\cap K)}\int_{\Gamma\cap K}\varepsilon\left[\|a\|_{C^1(\Gamma\cap K)}\int_{\Gamma\cap K}\mathrm dz\right]^{-1}\,\mathrm dz=\varepsilon.\end{aligned}$

   This implies $\displaystyle\lim_{j\to\infty}\langle a\partial_\nu\delta_\Gamma,v_j\rangle=0$. Therefore, $\langle a\partial_\nu\delta_\Gamma,\cdot\rangle$ is continuous on $\mathcal D(\mathbb R^n)$ and $F$ is a distribution in $\mathbb R^n$.


8. Let

   $f_n(x)=\begin{cases}\displaystyle\frac n2&\displaystyle\text{for}~-\frac1n<x<\frac1n;\\0&\displaystyle\text{for}~|x|\geq\frac1n.\end{cases}$

   Show that $f_n(x)\to\delta(x)$ as distributions on $\mathbb R$.
Solution
For $v\in C_0^\infty(\mathbb R)$, we find that
$\displaystyle\int_{\mathbb R}\!f_n(x)v(x)\,\mathrm dx=\frac n2\int_{-1/n}^{1/n}\!v(x)\,\mathrm dx$.
Since $v$ is continuous on $\mathbb R$, for any $\varepsilon>0$, there exists $\delta>0$ such that $|v(x)-v(0)|<\varepsilon$ whenever $|x|<\delta$. Then for $n>1/\delta$, we have
$\displaystyle\left|\int_{\mathbb R}\!f_n(x)v(x)\,\mathrm dx-v(0)\right|=\frac n2\left|\int_{-1/n}^{1/n}\![v(x)-v(0)]\,\mathrm dx\right|\leq\frac n2\int_{-1/n}^{1/n}|v(x)-v(0)|\,\mathrm dx<\frac n2\int_{-1/n}^{1/n}\!\varepsilon\,\mathrm dx=\varepsilon$.
This shows $\displaystyle\lim_{n\to\infty}\int_{\mathbb R}f_n(x)v(x)\,\mathrm dx=v(0)=\int_{\mathbb R}\!\delta(x)v(x)\,\mathrm dx$, i.e., $f_n(x)\to\delta(x)$ as distributions on $\mathbb R$.


9. If $f_n(x)$ and $f(x)$ are integrable functions such that for any compact set $K\subset\Omega$ we have $\displaystyle\int_K\!|f_n(x)-f(x)|\,\mathrm dx\to0$ as $n\to\infty$, then $f_n\to f$ as distributions.
Solution
Let $v\in C_0^\infty(\mathbb R)$ and $K=\text{supp}(v)$. Then we observe that
$\begin{aligned}|\langle f_n,v\rangle-\langle f,v\rangle|&=|\langle f_n-f,v\rangle|=\left|\int_\Omega\!(f_n(x)-f(x))v(x)\,\mathrm dx\right|\\&=\left|\int_K\!(f_n(x)-f(x))v(x)\,\mathrm dx\right|\leq\int_K\!|f_n(x)-f(x)||v(x)|\,\mathrm dx\\&\leq\|v\|_{C^0(K)}\int_K\!|f_n(x)-f(x)|\,\mathrm dx\to0\quad\text{as}~n\to\infty.\end{aligned}$
This means $\displaystyle\lim_{n\to\infty}\langle f_n,v\rangle=\langle f,v\rangle$ for $v\in\mathcal D(\Omega)$, i.e., $f_n\to f$ as distributions on $\Omega$.


10. Let $a\in\mathbb R$, $a\neq0$.
    1. Find a fundamental solution for $L=\mathrm d/\mathrm dx-a$ on $\mathbb R$ (i.e., solve $\mathrm dF/\mathrm dx-aF=\delta$).
    2. Show that a fundamental solution for $L=\mathrm d^2/\mathrm dx^2-a^2=(\mathrm d/\mathrm dx+a)(\mathrm d/\mathrm dx-a)$ on $\mathbb R$ is given by

       $F(x)=\begin{cases}a^{-1}\sinh ax&\text{if}~x>0;\\0&\text{if}~x<0.\end{cases}$

Solution
1. Multiplying equation by $e^{-ax}$, it gives

   $\displaystyle\frac{\mathrm d}{\mathrm dx}(e^{-ax}F)=e^{-ax}\frac{\mathrm dF}{\mathrm dx}-ae^{-ax}F=e^{-ax}\delta(x)=\delta(x)$,

   which implies

   $e^{-ax}F(x)=H(x)=\begin{cases}1&\text{if}~x>0;\\0&\text{if}~x<0.\end{cases}$

   Hence $F(x)=e^{ax}H(x)=\begin{cases}e^{ax}&\text{if}~x>0;\\0&\text{if}~x<0.\end{cases}$
   
   Note that the adjoint operator $L'$ of $L$ is $-\mathrm d/\mathrm dx-a$. It is easy to verify that

   $\begin{aligned}\langle LF,v\rangle&=\langle F,L'v\rangle=\int_{\mathbb R}F(x)\left(-\frac{\mathrm dv}{\mathrm dx}-av\right)\,\mathrm dx\\&=-\int_0^\infty\!e^{ax}\frac{\mathrm dv}{\mathrm dx}\,\mathrm dx-a\int_0^\infty\!e^{ax}v(x)\,\mathrm dx\\&=-e^{ax}v(x)\Big|_0^\infty+a\int_0^\infty\!e^{ax}v(x)\,\mathrm dx-a\int_0^\infty\!e^{ax}v(x)\,\mathrm dx\\&=v(0)=\langle\delta,v\rangle,\end{aligned}$

   which shows $LF=\delta$.
2. Note that $L$ is self-adjoint, i.e., $L=L'$. Then for $v\in\mathcal D(\mathbb R)$, it is easy to verify that

   $\begin{aligned}\langle LF,v\rangle&=\langle F,L'v\rangle=\int_{\mathbb R}F(x)\left(\frac{\mathrm d^2v}{\mathrm dx^2}-a^2v(x)\right)\,\mathrm dx\\&=\int_0^\infty\!a^{-1}\sinh(ax)\left(\frac{\mathrm d^2v}{\mathrm dx^2}-a^2v(x)\right)\,\mathrm dx\\&=\left.a^{-1}\sinh(ax)\frac{\mathrm dv}{\mathrm dx}\right|_0^\infty-\int_0^\infty\!\cosh(ax)\frac{\mathrm dv}{\mathrm dx}\,\mathrm dx-\int_0^\infty\!a\sinh(ax)v(x)\,\mathrm dx\\&=-\cosh(ax)v(x)\Big|_0^\infty+\int_0^\infty\!a\sinh(ax)v(x)\,\mathrm dx-\int_0^\infty\!a\sinh(ax)v(x)\,\mathrm dx\\&=v(0)=\langle\delta,v\rangle,\end{aligned}$

   which implies $LF=\delta$.


11. Regularity of $u$ defined by (60):
    1. If $f(x)\in L^1(\mathbb R)$ has a compact support, show that (60) defines a continuous weak solution of $u''=f(x)$.
    2. If, in addition to (a), $f(x)$ is a bounded function on $\mathbb R$, show $u\in C^1(\mathbb R)$, and $\displaystyle u'(x)=\frac12\left(\int_{-\infty}^x\!f(y)\,\mathrm dy-\int_x^\infty\!f(y)\,\mathrm dy\right)$.
    3. If, in addition to (a), $f(x)$ is continuous, show that $u$ is in fact a classical solution: $u\in C^2(\mathbb R)$ with $u''=f$.
Solution
1. Since $f\in L^1(\mathbb R)$ has a compact support, $f\in L_{\text{loc}}^1(\mathbb R)$. On the other hand, $\displaystyle g(x)=\frac12|x|\in C(\mathbb R)$. Thus, by Lemma at Page 67, $\displaystyle u(x)=(g\ast f)(x)=\frac12\int_{\mathbb R}\!|x-y|f(y)\,\mathrm dy$ is continuous on $\mathbb R$.
2. From (60), we have

   $\displaystyle u(x)=\frac12\int_{-\infty}^x\!(x-y)f(y)\,\mathrm dy+\frac12\int_x^\infty\!(y-x)f(y)\,\mathrm dy$.

   A direct computation gives

   $\begin{aligned}u(x+h)-u(x)&=\frac12\int_{-\infty}^{x+h}\!(x+h-y)f(y)\,\mathrm dy-\frac12\int_{-\infty}^x\!(x-y)f(y)\,\mathrm dy\\&\quad+\frac12\int_{x+h}^\infty\!(y-x-h)f(y)\,\mathrm dy-\frac12\int_x^\infty\!(y-x)f(y)\,\mathrm dy\\&=\frac h2\int_{-\infty}^{x+h}\!f(y)\,\mathrm dy+\frac12\int_x^{x+h}\!(x-y)f(y)\,\mathrm dy-\frac h2\int_{x+h}^\infty\!f(y)\,\mathrm dy-\frac12\int_x^{x+h}\!(y-x)f(y)\,\mathrm dy.\end{aligned}$

   Since $f$ is bounded, there exists $M>0$ such that $|f(y)|\leq M$ for $y\in\mathbb R$. Then it is easy to observe that

   $\displaystyle\left|\int_x^{x+h}\!\pm(x-y)f(y)\,\mathrm dy\right|\leq\left|\int_x^{x+h}\!(y-x)M\,\mathrm dy\right|=\frac{Mh^2}2$,

   which implies $\displaystyle\lim_{h\to0}\frac1h\int_x^{x+h}\!\pm(x-y)f(y)\,\mathrm dy=0$ by the squeeze theorem. Therefore, by the definition, we find

   $\begin{aligned}u'(x)&=\lim_{h\to0}\frac{u(x+h)-u(x)}h\\&=\frac12\int_{-\infty}^{x+h}\!f(y)\,\mathrm dy-\frac12\int_{x+h}^\infty\!f(y)\,\mathrm dy+\frac1{2h}\int_x^{x+h}(x-y)f(y)\,\mathrm dy-\frac1{2h}\int_x^{x+h}\!(y-x)f(y)\,\mathrm dy\\&=\frac12\int_{-\infty}^x\!f(y)\,\mathrm dy-\frac12\int_x^\infty\!f(y)\,\mathrm dy.\end{aligned}$

   Here we have used the fact that functions $\displaystyle\int_x^\infty\!f(y)\,\mathrm dy$ and $\displaystyle\int_{-\infty}^x\!f(y)\,\mathrm dy$ are continuous in $x$ because $f\in L^1(\mathbb R)$ is bounded with compact support. This also shows $u'\in C(\mathbb R)$, and hence $u\in C^1(\mathbb R)$.
3. When continuous function $f\in L^1(\mathbb R)$ has a compact support, $f$ is bounded on $\mathbb R$. Thus, by (b), $u\in C^1(\mathbb R)$. Moreover, by the fundamental theorem of calculus, functions $\displaystyle\int_{-\infty}^x\!f(y)\,\mathrm dy$ and $\displaystyle\int_x^\infty\!f(y)\,\mathrm dy$ are differentiable in $x$, and then $u'$ is also differentiable. By the fundamental theorem of calculus, we arrive at $\displaystyle u''(x)=\frac12f(x)-\frac12\cdot(-f(x))=f(x)$ for $x\in\mathbb R$. Since $f\in C(\mathbb R)$, $u''\in C(\mathbb R)$ and then $u\in C^2(\mathbb R)$.


12. 1. Use the fundamental solution (59) to solve the initial $u''=f(x)$ for $x>0$ with $u(0)=u_0$ and $u'(0)=u_0'$, where $f\in C^\infty([0,\infty))$ and $f=O(|x|^{-2-\epsilon})$ as $|x|\to\infty$.
    2. Do the same for the boundary value problem $u''=f(x)$ for $0<x<\ell$ with $u(0)=0=u(\ell)$.
Solution

1. Extend $\tilde f$ by $\tilde f(x)=f(x)$ if $x>0$; $0$ if $x\leq0$. Then we consider the equation $\tilde u''(x)=\tilde f(x)$ for $x\in\mathbb R$. Plugging (59) into (56), we obtain the particular solution

   $\begin{aligned}u_p(x)&=(F\ast f)(x)=\int_{\mathbb R}F(x-y)\tilde f(y)\,\mathrm dy\\&=\frac12\int_{\mathbb R}|x-y|\tilde f(y)\,\mathrm dy=\frac12\int_0^\infty\!|x-y|f(y)\,\mathrm dy\quad\text{for}~x\in\mathbb R.\end{aligned}$

   To fulfill the initial condition, we need to find a function $u_h$ satisfying

   $\begin{aligned} &u_h''(x)=0,\\&u_h(0)=u_0-u_p(0)=u_0-\frac12\int_0^\infty\!yf(y)\,\mathrm dy,\\&u_h'(0)=u_0'-u_p'(0)=u_0'+\frac12\int_0^\infty\!f(y)\,\mathrm dy.\end{aligned}$

   Here we have used the result of Exercise 11(b). It is easy to obtain

   $\displaystyle u_h(x)=u_h(0)+u_h'(0)x=u_0-\frac12\int_0^\infty\!yf(y)\,\mathrm dy+\left(u_0'+\frac12\int_0^\infty\!f(y)\,\mathrm dy\right)x\quad\text{for}~x\in\mathbb R$.

   Thus, we find

   $\begin{aligned}\tilde u(x)&=u_h(x)+u_p(x)\\&=u_0-\frac12\int_0^\infty\!yf(y)\,\mathrm dy+\left(u_0'+\frac12\int_0^\infty\!f(y)\,\mathrm dy\right)x+\frac12\int_0^\infty\!|x-y|f(y)\,\mathrm dy\quad\text{for}~x\in\mathbb R.\end{aligned}$

   In particular, the solution is $u(x)=\tilde u\Big|_{x\geq0}(x)$ for $x\geq0$.
   
   It is remark that the solution $u$ can be represented as

   $\begin{aligned} u(x)&=u(0)+\int_0^x\!u'(t)\,\mathrm dt=u_0+\int_0^x\!\left[u'(0)+\int_0^t\!u''(s)\,\mathrm ds\right]\,\mathrm dt\\&=u_0+u_0'x+\int_0^x\!\int_0^t\!f(s)\,\mathrm ds\,\mathrm dt=u_0+u_0'x+\int_0^x\!\int_s^x\!f(s)\,\mathrm dt\,\mathrm ds\\&=u_0+u_0'x+\int_0^x\!(x-s)f(s)\,\mathrm ds\quad\text{for}~x\geq0.\end{aligned}$.

2. Extend $\tilde f$ by $\tilde f(x)=f(x)$ if $0<x<\ell$; $0$ if $x\leq0$ or $x\geq\ell$. Then we consider the equation $\tilde u''(x)=\tilde f(x)$ for $x\in\mathbb R$. Plugging (59) into (56), we obtain the particular solution

   $\begin{aligned}u_p(x)&=(F\ast f)(x)=\int_{\mathbb R}\!F(x-y)\tilde f(y)\\&=\frac12\int_{\mathbb R}\!|x-y|\tilde f(y)\,\mathrm dy=\frac12\int_0^\ell\!|x-y|f(y)\,\mathrm dy\quad\text{for}~x\in\mathbb R.\end{aligned}$

   To fulfill the boundary condition, we need to find a function $u_h$ satisfying

   $\begin{aligned} &u_h''(x)=0,\\&u_h(0)=-u_p(0)=-\frac12\int_0^\ell\!yf(y)\,\mathrm dy,\\&u_h(\ell)=-u_p(\ell)=-\frac12\int_0^\ell\!(\ell-y)f(y)\,\mathrm dy.\end{aligned}$

   It is easy to obtain

   $\begin{aligned}u_h(x)&=-\left(\frac1{2\ell}\int_0^\ell\!(\ell-y)f(y)\,\mathrm dy\right)x-\left(\frac1{2\ell}\int_0^\ell\!yf(y)\,\mathrm dy\right)(\ell-x)\\&=\frac{x}{\ell}\int_0^\ell\!(2y-\ell)f(y)\,\mathrm dy-\frac12\int_0^\ell\!yf(y)\,\mathrm dy\quad\text{for}~x\in\mathbb R.\end{aligned}$

   Thus, we find

   $\begin{aligned}\tilde u(x)&=u_h(x)+u_p(x)\\&=\frac{x}{\ell}\int_0^\ell\!(2y-\ell)f(y)\,\mathrm dy-\frac12\int_0^\ell\!yf(y)\,\mathrm dy+\frac12\int_0^\ell\!|x-y|f(y)\,\mathrm dy\quad\text{for}~x\in\mathbb R.\end{aligned}$

   In particular, the solution is $u(x)=\tilde u\Big|_{0\leq x\leq\ell}(x)$ for $0\leq x\leq\ell$.
   
   Note that $u(\ell)=0=u(0)$, which implies $\displaystyle\int_0^\ell\!u'(t)\,\mathrm dt=0$, and hence

   $\displaystyle0=\int_0^\ell\!\left[u'(0)+\int_0^t\!u''(s)\,\mathrm ds\right]\,\mathrm dt=\ell u'(0)+\int_0^\ell\int_0^t\!f(s)\,\mathrm ds\,\mathrm dt.$

   Thus, $\displaystyle u'(0)=-\frac1\ell\int_0^\ell\!\int_0^t\!f(s)\,\mathrm ds\,\mathrm dt$. Therefore, the solution can be represented as

   $\begin{aligned}u(x)&=u(0)+\int_0^x\!u'(t)\,\mathrm dt=\int_0^x\!\left[u'(0)+\int_0^t\!u''(s)\,\mathrm ds\right]\,\mathrm dt\\&=-\left(\frac1\ell\int_0^\ell\!\int_0^t\!f(s)\,\mathrm ds\,\mathrm dt\right)x+\int_0^x\!\int_0^t\!f(s)\,\mathrm ds\,\mathrm dt\\&=-\frac x\ell\int_0^\ell\!(\ell-s)f(s)\,\mathrm ds+\int_0^x\!(x-s)f(s)\,\mathrm ds\quad\text{for}~0\leq x\leq\ell.\end{aligned}$



13. 1. Using $\delta(\mu,\eta)=\delta(\mu)\delta(\eta)$, show that each of the following functions is a fundamental solution of $L=\partial^2/\partial\mu\partial\eta$:

       $\begin{aligned} &F_1(\mu,\eta)=H(\mu)H(\eta),\quad F_2(\mu,\eta)=-H(\mu)H(-\eta),\\&F_3(\mu,\eta)=-H(-\mu)H(\eta),\quad F_4(\mu,\eta)=H(-\mu)H(-\eta).\end{aligned}$

    2. Use part (a) with the change of variables $\mu=x+ct$, $\eta=x-ct$ to obtain four distinct fundamental solutions for the one-dimensional wave operator, each having support in one of the wedges determined by the lines $x=\pm ct$.
Solution

1. We firstly note that $L$ is self-adjoint, i.e., $L=L'$. For $v\in\mathcal D(\mathbb R^2)$, it is easy to verify that

   $\begin{aligned}\langle LF_1,v\rangle&=\langle F_1,L'v\rangle=\int_{\mathbb R^2}H(\mu)H(\eta)v_{\mu\eta}(\mu,\eta)\,\mathrm d\mu\,\mathrm d\eta=\int_0^\infty\!\int_0^\infty\!v_{\mu\eta}(\mu,\eta)\,\mathrm d\mu\,\mathrm d\eta\\&=\int_0^\infty\!v_\eta(\mu,\eta)\Big|_{\mu=0}^{\mu=\infty}\,\mathrm d\eta=-\int_0^\infty\!v_\eta(0,\eta)\,\mathrm d\eta=-v(0,\eta)\Big|_{\eta=0}^\infty=v(0,0)=\langle\delta,v\rangle,\\\langle LF_2,v\rangle&=\langle F_2,L'v\rangle=-\int_{\mathbb R^2}\!H(\mu)H(-\eta)v_{\mu\eta}(\mu,\eta)\,\mathrm d\mu\,\mathrm d\eta=-\int_{-\infty}^0\!\int_0^\infty\!v_{\mu\eta}(\mu,\eta)\,\mathrm d\mu\,\mathrm d\eta\\&=-\int_{-\infty}^0\!v_{\eta}(\mu,\eta)\Big|_{\mu=0}^{\mu=\infty}\,\mathrm d\eta=\int_{-\infty}^0\!v_\eta(0,\eta)\,\mathrm d\eta=v(0,\eta)\Big|_{\eta=-\infty}^{\eta=0}=v(0,0)=\langle\delta,v\rangle,\\\langle LF_3,v\rangle&=\langle F_3,L'v\rangle=-\int_{\mathbb R^2}H(-\mu)H(\eta)v_{\mu\eta}(\mu,\eta)\,\mathrm d\mu\,\mathrm d\eta=-\int_0^\infty\!\int_{-\infty}^0\!v_{\mu\eta}(\mu,\eta)\,\mathrm d\mu\,\mathrm d\eta\\&=-\int_0^\infty\!v_{\eta}(\mu,\eta)\Big|_{\mu=-\infty}^{\mu=0}\,\mathrm d\eta=-\int_0^\infty\!v_\eta(0,\eta)\,\mathrm d\eta=-v(0,\eta)\Big|_{\eta=0}^{\eta=\infty}=v(0,0)=\langle\delta,v\rangle\\\langle LF_4,v\rangle&=\langle F_4,L'v\rangle=\int_{\mathbb R^2}H(-\mu)H(-\eta)v_{\mu\eta}(\mu,\eta)\,\mathrm d\mu\,\mathrm d\eta==\int_{-\infty}^0\!\int_{-\infty}^0\!v_{\mu\eta}(\mu,\eta)\,\mathrm d\mu\,\mathrm d\eta\\&=\int_{-\infty}^0\!v_\eta(\mu,\eta)\Big|_{\mu=-\infty}^{\mu=0}\,\mathrm d\eta=\int_{-\infty}^0\!v_\eta(0,\eta)\,\mathrm d\eta=v(0,\eta)\Big|_{\eta=-\infty}^{\eta=0}=v(0,0)=\langle\delta,v\rangle.\end{aligned}$

   This shows $F_i$ is a fundamental solution of $LF_i=\delta$ for $i=1,2,3,4$.
2. By (a) with the change of variables $\mu=x+ct$ and $\eta=x-ct$, the fundamental solutions for $-4c^2\tilde u_{\mu\eta}(\mu,\eta)=u_{tt}(x,t)-c^2u_{xx}(x,t)=\delta$ are

   $\begin{aligned} &U_1(x,t)=-(2c)^{-1}H(x+ct)H(x-ct),\quad U_2(x,t)=(2c)^{-1}H(x+ct)H(-x+ct),\\&U_3(x,t)=(2c)^{-1}H(-x-ct)H(x-ct),\quad U_4(x,t)=-(2c)^{-1}H(-x-ct)H(-x+ct).\end{aligned}$



14. If $K(x,t)$ is a distribution in $\mathbb R^n$ depending on the parameter $t$, then we say $K$ depends continuously on $t$ if, for every $v\in C_0^\infty(\mathbb R^n)$, $\displaystyle f(t)\equiv\langle K(\cdot,t),v\rangle=\int\!K(x,t)v(x)\,\mathrm dx$ defines a continuous function of $t$. Use this idea to define $K$ is continuously differentiable in $t$.
Solution
$K$ is continuously differentiable in $t$ if for every $v\in C_0^\infty(\mathbb R^n)$, $\displaystyle f(t)=\int_{\mathbb R^n}\!K(x,t)v(x)\,\mathrm dx$ defines a continuously differentiable function of $t$.


15. According to Section 1.2, a weak solution of $u_y+(G(u))_x=0$ in a rectangle $\Omega=(a,b)\times (c,d)$ satisfies

    $\displaystyle\frac{\mathrm d}{\mathrm dy}\int_{x_1}^{x_2}\!u(x,y)\,\mathrm dx+G(u(x,y))\Big|_{x=x_1}^{x=x_2}=0$

    for every $[x_1,x_2]\subset(a,b)$ and $c<y<d$. Show that this implies

    $\displaystyle\int_\Omega\!uv_y+G(u)v_x\,\mathrm dx\,\mathrm dy=0\quad\text{for every}~v\in C_0^1(\Omega)$

    (i.e., the notion of weak solution of Section 1.2 is consistent with that of Section 2.3).
Solution

Let $x_i=a+(b-a)i/m$ for $i=0,1,\dots,m$ and $y_j=c+(d-c)j/n$ for $j=1,\dots,n$. Set $\Omega_{ij}=[x_{i-1},x_i]\times[y_{j-1},y_j]$ for $1\leq i\leq n$ and $1\leq j\leq m$. Then $\{\Omega_{ij}\}_{1\leq i\leq n,1\leq j\leq m}$ forms a partition of $\Omega$. Put $\Delta x=(b-a)/m$ and $\Delta y=(d-c)/n$. Then using the summation by parts, we have

$\begin{aligned} &\quad\sum_{i=1}^m\sum_{j=1}^n\left[u(x_i,y_j)\frac{v(x_i,y_j)-v(x_i,y_{j-1})}{\Delta y}+G(u(x_i,y_j))\frac{v(x_i,y_j)-v(x_{i-1},y_j)}{\Delta x}\right]\Delta y\Delta x\\&=\sum_{i=1}^m\left[\frac{u(x_i,y_n)v(x_i,y_n)-u(x_i,y_1)v(x_i,y_0)}{\Delta y}-\sum_{j=2}^nv(x_i,y_{j-1})\frac{u(x_i,y_j)-u(x_i,y_{j-1})}{\Delta y}\right]\Delta y\Delta x\\&\quad+\sum_{j=1}^n\left[\frac{G(u(x_m,y_j))v(x_m,y_j)-G(u(x_1,y_j))v(x_0,y_j)}{\Delta x}-\sum_{i=2}^mv(x_{i-1},y_j)\frac{G(u(x_i,y_j))-G(u(x_{i-1},y_j))}{\Delta x}\right]\Delta y\Delta x\\&=-\sum_{i=1}^m\sum_{j=2}^n\left[v(x_i,y_{j-1})\frac{u(x_i,y_j)-u(x_i,y_{j-1})}{\Delta y}\right]\Delta y\Delta x-\sum_{i=2}^m\sum_{j=1}^n\left[v(x_{i-1},y_j)\frac{G(u(x_i,y_j))-G(u(x_{i-1},y_j))}{\Delta x}\right]\Delta y\Delta x\\&=-\sum_{i=1}^{m-1}\sum_{j=1}^{n-1}\left\{v(x_i,y_j)\left[\frac{u(x_i,y_{j+1})-u(x_i,y_j)}{\Delta y}+\frac{G(u(x_{i+1},y_j))-G(u(x_i,y_j))}{\Delta x}\right]\right\}\Delta y\Delta x\end{aligned}$

Here we have used the fact that $v\in C_0^\infty(\Omega)$ so $v(x_0,y_j)=v(x_m,v_j)=v(x_i,y_0)=v(x_i,y_n)=0$ for $1\leq i\leq m$ and $1\leq j\leq n$. Note that

$\displaystyle G(u(x_{i+1},y_j))-G(u(x_i,y_j))=-\int_{x_i}^{x_{i+1}}u_y(x,y_j)\,\mathrm dx\quad\text{for}~1\leq i\leq m-1,~1\leq j\leq n-1$.

Then the double sum becomes

$\begin{aligned} &\quad-\sum_{i=1}^{m-1}\sum_{j=1}^{n-1}\left\{v(x_i,y_j)\left[\frac{u(x_i,y_{j+1})-u(x_i,y_j)}{\Delta y}-\frac1{\Delta x}\int_{x_i}^{x_{i+1}}\!u_y(x,y_j)\,\mathrm dx\right]\right\}\Delta y\Delta x\\&\to\int_a^b\!\int_c^d\!v(x,y)\left[u_y(x,y)-u_y(x,y)\right]\,\mathrm dx\,\mathrm dy=0\quad\text{as}~n,m\to\infty.\end{aligned}$

This shows $\displaystyle\int_\Omega\!uv_y+G(u)v_x\,\mathrm dx\,\mathrm dy=0$ for $v\in C_0^1(\Omega)$.

Warning: This problem is quite difficult so I cannot promise this solution is correct. Please provide some references or theorem that I may use to make this solution better.



16. The $m$th-order operator (32a) is elliptic at $x$ if its principal symbol (32c) has no nonzero characteristics [i.e., $\sigma_L(x,\xi)\neq0$ for all $\xi\in\mathbb R^n\backslash\{0\}$]. In this case, show that $m$ must be an even integer.
Solution
Suppose by contradiction that $m$ is an odd integer. Then $\displaystyle\sigma_L(x,\xi)=\sum_{|\alpha|=m}a_\alpha(x)\xi^\alpha$ is a polynomial of odd degree. Since $a_\alpha(x)$ is not all zero, we may assume $a_{e_1}(x)=a_{(m,0,\dots,0)}(x)\neq0$ by rearranging the indices. Thus, we have
$\displaystyle\sigma_L(x,\xi)=a_{me_1}(x)\xi_1^m+\sum_{|\alpha|=m,\,\alpha\neq me_1}a_\alpha(x)\xi^\alpha$.
Since $m$ is odd and $a_\alpha(x)$ are real numbers, by the fundamental theorem of algebra, there exists $\xi_1=\xi_1(\xi_2,\dots,\xi_m,x)$ for any $(\xi_2,\dots,\xi_m)\neq0$. This means the principal symbol has nonzero characteristics, which leads a contradiction. Hence $m$ must be an even integer.