Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 6.2

1. Assume that $u\in H_0^{1,2}(\Omega)$ and $f\in L^2(\Omega)$, and (13) holds for all $v\in C_0^1(\Omega)$. Show that (13) holds for all $v\in H_0^{1,2}(\Omega)$.
Solution
For $v\in H_0^{1,2}(\Omega)$, there exists a sequence of functions $\{v_n\}_{n=1}^\infty\subseteq C_0^1(\Omega)$ such that $\displaystyle\lim_{n\to\infty}\|v_n-v\|_{1,2}=0$. Since $v_n\in C_0^1$ for $n\in\mathbb N$, then by (13) and (14), we have
$\displaystyle-(u,v_n)_1=-\int_\Omega\!\nabla u\cdot\nabla v_n\,\mathrm dx=\int_\Omega\!fv_n\,\mathrm dx$.
Note that $|(u,v_n-v)_1|\leq\|\nabla u\|_2\|\nabla(v_n-v)\|_2\leq|\nabla u|_2\|v_n-v\|_{1,2}\to0$ as $n\to\infty$ and
$\displaystyle\left|\int_\Omega\!f(v_n-v)\,\mathrm dx\right|\leq\|f\|_2\|v_n-v\|_2\leq\|f\|_2\|v_n-v\|_{1,2}\to0\quad\text{as}~n\to\infty$.
Therefore, we have
$\begin{aligned}-\int_\Omega\!\nabla u\cdot\nabla v\,\mathrm dx&=-(u,v)_1=-(u,v_n+(v-v_n))_1=\displaystyle\lim_{n\to\infty}\left[-(u,v_n)_1+(u,v_n-v)_1\right]\\&=\lim_{n\to\infty}\int_\Omega\!fv_n\,\mathrm dx=\int_\Omega\!fv\,\mathrm dx+\lim_{n\to\infty}\int_\Omega\!f(v_n-v)\,\mathrm dx=\int_\Omega\!fv\,\mathrm dx.\end{aligned}$
This shows (13) holds for $v\in H_0^{1,2}(\Omega)$.


2. Show that the Poincaré inequality generalizes to $p\neq2$; that is, if $\Omega$ is a bounded domain and $1\leq p<\infty$, then

   $\|u\|_p\leq C\|\nabla u\|_p,\quad\text{where}~C=C(\Omega,p)$

   for all $u\in C_0^1(\Omega)$, and by completion for all $u\in H_0^{1,p}(\Omega)$.
Solution
Since $\Omega$ is a bounded domain, there exists $L>0$ such that $\Omega\subseteq Q\equiv\{x=(x_1,\dots,x_n)\in\mathbb R^n\,:\,|x_i|\leq L,~i=1,\dots,n\}$, a cube centerd with $(0,\dots,0)$ and with side length of $2L$. Write $Q=[-L,L]\times Q'$, where $Q'=\{x'=(x_2,\dots,x_n)\in\mathbb R^{n-1}\,:\,|x_i|\leq L,\,i=2,\dots,n\}$. Clearly, $u\in C_0^1(\Omega)$ implies $u\in C_0^1(Q)$. Then we can use the integration by parts in the $x_1$-direction and the Cauchy-Schwarz inequality to get
$\begin{aligned}\|u\|_p^p&=\int_\Omega\!|u|^p\,\mathrm dx=\int_Q\!|u|^p\,\mathrm dx=\int_{Q'}\!\int_{-L}^L\!1\cdot|u|^p\,\mathrm dx_1\,\mathrm dx'\\&=\int_{Q'}\!\left[x_1|u|^p\Big|_{x_1=-L}^{x_1=L}-\int_{-L}^L\!x_1\cdot pu|u|^{p-2}\frac{\partial u}{\partial x_1}\,\mathrm dx_1\right]\mathrm dx'\\&=-p\int_{Q'}\!\int_{-L}^L\!x_1u|u|^{p-2}\frac{\partial u}{\partial x_1}\,\mathrm dx_1\,\mathrm dx'\leq pL\int_Q\!|u|^{p-1}|\nabla u|\,\mathrm dx\\&\leq pL\||u|^{p-1}\|_{p'}\|\nabla u\|_p=pL\left(\int_\Omega\!|u|^{(p-1)p'}\,\mathrm dx\right)^{1/p'}\|\nabla u\|_p\\&=pL\left[\left(\int_\Omega\!|u|^p\,\mathrm dx\right)^{1/p}\right]^{p-1}\|\nabla u\|_p=pL\|u\|_p^{p-1}\|\nabla u\|_p,\end{aligned}$
gives $\|u\|_p\leq pL\|\nabla u\|_p$. Hence we choose $C=pL$, which depends on $\Omega$ and $p$.


3. Suppose $\Omega$ is unbounded but lies between two parallel hyperplanes $P_1$ and $P_2$.
   1. Show that the Poincaré inequality holds for $\Omega$.
   2. Show that there is a weak soluiton of the Dirichlet problem $\Delta u=f$ in $\Omega$ and $u=0$ on $\partial\Omega$, where $f\in L^2(\Omega)$.
Solution
1. Suppose that $P_1$ and $P_2$ can be writen as $P_i=\{x\in\mathbb R^n\,:\,\vec n\cdot x=d_i\}$ for $i=1,2$, where $\vec n$ is the normal vector of $P_i$, and $d_i$'s are distinct real numbers. There exists a rotation operator $R:\mathbb R^n\to\mathbb R^n$ such that

   $\tilde P_i:=R(P_i)\equiv\{y=Rx\in\mathbb R^n\,:\,x\in P_i\}=\{y=(y_1,\dots,y_n)\in\mathbb R^n\,:\,y_n=\tilde d_i\}$.

   Without loss of generality, we assume $\tilde d_1<\tilde d_2$. Clearly, $R(\Omega)$ is also between $\tilde P_1$ and $\tilde P_2$, i.e., $\tilde\Omega:=R(\Omega)\equiv\{y\in Rx\,:\,x\in\Omega\}\subseteq S\equiv\{y=(y',y_n)\in\mathbb R^n\,:\,\tilde d_1\leq y_n\leq\tilde d_2\}$. Now for $u\in C_0^1(\Omega)$, we define $\tilde u:C_0^1(\tilde\Omega)\to\mathbb R$ by $\tilde u(y)=u(R^{-1}y)=u(x)$ for $y\in\tilde\Omega$. Recall that the Jacobian of rotation is $1$. Then using the change of variables, we have

   $\begin{aligned}\|u\|_p^p&=\int_\Omega\!|u|^p\,\mathrm dx=\int_{\tilde\Omega}\!|\tilde u|^p\,\mathrm dy=\int_{\mathbb R^{n-1}}\!\int_{\tilde d_1}^{\tilde d_2}\!|\tilde u|^p\,\mathrm dy_n\,\mathrm dy'\\&=\int_{\mathbb R^{n-1}}\left[y_n|\tilde u(y)|^p\Big|_{y_n=\tilde d_1}^{y_n=\tilde d_2}-p\int_{\tilde d_1}^{\tilde d_2}y_n\tilde u|\tilde u|^{p-2}\frac{\partial\tilde u}{\partial y_n}\,\mathrm dy_n\right]\mathrm dy'\\&=-p\int_{\mathbb R^{n-1}}\!\int_{\tilde d_1}^{\tilde d_2}\!y_n\tilde u|\tilde u|^{p-2}\frac{\partial\tilde u}{\partial y_n}\,\mathrm dy_n\,\mathrm dy'\leq p(\tilde d_2-\tilde d_1)\int_{\tilde\Omega}\!|\tilde u|^{p-1}|\nabla_y\tilde u|\,\mathrm dy\\&=p(\tilde d_2-\tilde d_1)\int_\Omega\!|u|^{p-1}|\nabla_xu|\,\mathrm dx=p(\tilde d_2-\tilde d_1)\|u\|_p^{p-1}\|\nabla u\|_p,\end{aligned}$

   which means $\|u\|_p\leq C(\Omega,p)\|\nabla u\|_p$ with $C(\Omega,p)=p(\tilde d_2-\tilde d_1)$. The proof is complete.
2. Let $(H_0^{1,2}(\Omega),|\cdot|_{1,2})$ be the Hilbert space. Define $F:H_0^{1,2}(\Omega)\to\mathbb R$ by

   $\displaystyle F(v)=-\int_\Omega\!fv\,\mathrm dx$ for $v\in H_0^{1,2}(\Omega)$.

   Then $F$ is a bounded linear functional because

   $|F(v)|=\left|\int_\Omega\!fv\,\mathrm dx\right|\leq\|f\|_2\|v\|_2\leq\|f\|_2\sqrt{2(\tilde d_2-\tilde d_1)}\|v\|_{1,2}$,

   where we have used the Cauchy-Schwarz inequality and then the Poincaré inequality. By the Riesz representation theorem, there exists an unique element $u\in H_0^{1,2}(\Omega)$ such that $F(v)=(v,u)_1$, which implies

   $\displaystyle-\int_\Omega\!\nabla v\cdot\nabla u\,\mathrm dx=-(v,u)_1=-F(v)=\int_\Omega\!fv\,\mathrm dx\quad\text{for all}~v\in H_0^{1,2}(\Omega)$.

   Therefore, $u$ is the weak solution of the Dirichlet problem.


4. For a bounded domain $\Omega\subset\mathbb R^n$, let

   $\displaystyle\lambda_1=\inf_{u\in C_0^1(\Omega)}\frac{\int_\Omega\!|\nabla u|^2\,\mathrm dx}{\int_\Omega\!|u|^2\,\mathrm dx}$.

   As we see in Section 7.2, $\lambda_1$ is indeed the first eigenvalue of the Laplacian as defined in Section 4.4.
   1. Prove that $\lambda_1>0$.
   2. Prove that, for every $f\in L^2(\Omega)$ and constant $c<\lambda_1$, the Dirichlet problem $\Delta u+cu=f$ in $\Omega$ in $\Omega$ with $u=0$ on $\partial\Omega$ has a weak solution $u\in H_0^{1,2}(\Omega)$.
Solution
1. Since $\Omega\subseteq\mathbb R^n$ is a bounded domain, then by Theorem 1 (The Poincaré Inequality), there eixsts a constant $C=C(\Omega)$ such that $\displaystyle\int_\Omega\!|u|^2\,\mathrm dx\leq C(\Omega)\int_\Omega\!|\nabla u|^2\,\mathrm dx$ for $u\in C_0^1(\Omega)$, which means

   $\displaystyle\frac{\int_\Omega\!|\nabla u|^2\,\mathrm dx}{\int_\Omega\!|u|^2\,\mathrm dx}\geq\frac1{C(\Omega)}\quad\text{for all}~u\in C_0^1(\Omega)$.

   Taking the infimum over $C_0^1(\Omega)$, we get $\lambda_1\geq1/C(\Omega)>0$. The proof is complete.
2. It suffices to show that there exists $u\in H_0^{1,2}(\Omega)$ satisfying

   $\displaystyle-\int_\Omega\!\nabla u\cdot\nabla v\,\mathrm dx+\int_\Omega\!cuv\,\mathrm dx=\int_\Omega\!fv\,\mathrm dx\quad\text{for all}~v\in H_0^{1,2}(\Omega)$.

   Define the bilinear form $B:H_0^{1,2}(\Omega)\times H_0^{1,2}(\Omega)\to\mathbb R$ by

   $\displaystyle B(u,v)=\int_\Omega\!(\nabla u\cdot\nabla v-cuv)\,\mathrm dx\quad\text{for}~u,v\in H_0^{1,2}(\Omega).$

   It is clear that

   $\displaystyle|B(u,v)|\leq\int_\Omega\!(|\nabla u||\nabla v|+|c||u||v|)\,\mathrm dx\leq\max\{1,|c|\}(|u|_{1,2}|v|_{1,2}+\|u\|_2\|v\|_2)\leq\max\{1,|c|\}\|u\|_{1,2}\|v\|_{1,2}$,

   which means $B$ is bounded. On the other hand, $B$ is positive on $H_0^{1,2}(\Omega)$ because

   $\begin{aligned}B(u,u)&=\int_\Omega\!(|\nabla u|^2-cu^2)\,\mathrm dx\geq\begin{cases}\displaystyle\int_\Omega|\nabla u|^2\,\mathrm dx&\text{if}~c\leq0;\\\displaystyle\left(1-\frac c{\lambda_1}\right)\int_\Omega|\nabla u|^2\,\mathrm dx&\text{if}~0<c<\lambda_1\end{cases}\\&\geq\begin{cases}(1+C(\Omega))^{-1}\|u\|_{1,2}&\text{if}~c\leq0;\\\displaystyle(1+C(\Omega))^{-1}\left(1-\frac{c}{\lambda_1}\right)\|u\|_{1,2}&\text{if}~0<c<\lambda_1.\end{cases}\end{aligned}$

   Hence by Theorem 3 at page 179 in the textbook, there exists a unique weak solution $u\in H_0^{1,2}(\Omega)$ of the Dirichlet problem $\Delta u+cu=f$ in $\Omega$ with $u=0$ on $\partial\Omega$.


5. Consider the Dirichlet problem for the Laplace equation as studied in Chapter 4: $\Delta u=0$ in $\Omega$ with $u=g$ on $\partial\Omega$, where $\Omega$ is a bounded domain. Suppose that $g\in C(\partial\Omega)$ satisfies $g=G\Big|_{\partial\Omega}$ for some $G\in C^2(\bar\Omega)$. How can you use the results in this section to find a weak solution for this problem?
Solution
Define $\tilde u=u-G$ in $\bar\Omega$. Then $\tilde u$ satisfies $\Delta\tilde u=f$ in $\Omega$ with $\tilde u=0$ on $\partial\Omega$, where $f=\Delta G$. Since $G\in C^2(\bar\Omega)$, we have $f\in C(\bar\Omega)\subseteq L^2(\Omega)$ and hence we can obtain the weak solution $\tilde u\in H_0^{1,2}(\Omega)$. Thus, the weak solution $u$ can be obtained from $u=\tilde u+G\in H^{1,2}(\Omega)$.


6. For $n=2$, compare the definition (21) of uniform ellipticity with the definition of ellipticity given in Section 2.2, and explain why they agree.
Solution
Theses two definition, i.e., (21b) of Section 6.2 and (10) of Section 2.2, are agree because they both are equivalent to the matrix $\begin{bmatrix}a_{11}(x)&a_{12}(x)\\a_{21}(x)&a_{22}(x)\end{bmatrix}$ is positive definite.


7. Use (29) to show that the space $Y$ defined by the condition (28) is closed.
Solution
Let $\{y_n\}_{n=1}^\infty\subseteq Y$ be a sequence converging to $y^*\in X$. To show the closeness of $Y$, we need to prove $y^*\in Y$. Clearly, $\{y_n\}_{n=1}^\infty$ is Cauchy. By the definition of $Y$ (cf. (28)), there exists a corresponding sequence $\{w_{y_n}\}_{n=1}^\infty\subseteq X$ such that $B(w_{y_n},v)=\langle y_n,v\rangle$ for all $v\in X$ and $n\in\mathbb N$, which implies $B(w_{y_n}-w_{y_m},v)=\langle y_n-y_m,v\rangle$ for all $n,m\in\mathbb N$ and $v\in X$. Then by the positivity of $B$ or by (29), we have
$\displaystyle\|w_{y_n}-w_{y_m}\|\leq\frac1\epsilon\|y_n-y_m\|\quad\text{for}~n,m\in\mathbb N$.
This shows that $\{w_{y_n}\}_{n=1}^\infty\subseteq X$ is also Cauchy. Since $X$ is a Hilbert space, $X$ is complete and hence $\{w_{y_n}\}_{n=1}^\infty$ is convergent. We denote the limit of $\{w_{y_n}\}_{n=1}^\infty$ by $w^*\in X$. Hence it suffices to show $B(w^*,v)=\langle y^*,v\rangle$ for all $v\in X$. Note that by the boundedness of $B$, we have
$|B(w^*-w_{y_n},v)|\leq C\|w^*-w_{y_n}\|\|v\|\to0\quad\text{as}~n\to\infty$.
Thus, we find
$\begin{aligned}B(w^*,v)&=\lim_{n\to\infty}[B(w_{y_n},v)+B(w^*-w_{y_n},v)]=\lim_{n\to\infty}\langle y_n,v\rangle\\&=\lim_{n\to\infty}[\langle y^*,v\rangle+\langle y_n-y^*,v\rangle]=\langle y^*,v\rangle.\end{aligned}$
Therefore, $y^*\in Y$ and $Y$ is closed subspace of $X$, which completes the proof.


8. Assume that the operator in (27) satisfies (21b), $|b_k(x)|\leq\eta$ for $x\in\Omega$ in (27), and $c(x)\equiv0$. Find a smallness condition on $\eta>0$ (in terms of $\epsilon$ and the Poincaré constant) that guarantees that the bilinear form (30) is positive, so that the Lax-Milgram theorem applies to (26).
Solution

From (30) with $c(x)\equiv0$, we can use (21b) and $|b_k(x)|\leq\eta$ for $x\in\Omega$ to get

$\begin{aligned}|B_L(u,u)|&=\left|\int_\Omega\!\left[\sum_{i,j=1}^na_{ij}(x)\frac{\partial u}{\partial x_i}\frac{\partial u}{\partial x_j}-\sum_{k=1}^nb_k(x)u\frac{\partial u}{\partial x_k}\right]\,\mathrm dx\right|\\&\geq\int_\Omega\!(\epsilon|\nabla u|^2-n\eta|u||\nabla u|)\,\mathrm dx\geq\epsilon(1+C)^{-1}\|u\|_{1,2}^2-\frac{n\eta}2\|u\|_{1,2}^2\\&=\left(\frac\epsilon{1+C}-\frac{n\eta}2\right)\|u\|_{1,2}^2,\end{aligned}$

where $C$ is the Poincaré constant. Thus, we find a sufficient condition $\displaystyle\eta<\frac{2\epsilon}{n(1+C)}$ such that $B_L$ is positive. Here we have used the facts that

$\displaystyle\int_\Omega\!|u||\nabla u|\,\mathrm dx\leq\int_\Omega\!\frac{|u|^2+|\nabla u|^2}2\,\mathrm dx=\frac{\|u\|_{1,2}^2}2$

and

$\displaystyle\|u\|_{1,2}^2=\int_\Omega\!|u|^2\,\mathrm dx+\int_\Omega\!|\nabla u|^2\,\mathrm dx\leq(1+C)\int_\Omega\!|\nabla u|^2\,\mathrm dx$.