This is problem 3 from chapter 7 of Evans book:
Suppose $f\in L^2(U)$ and assume that $u_m=\sum_{k=1}^md_m^kw_k$ solves $$\int_UDu_m\cdot Dw_k=\int_Uf\cdot w_kdx$$ for $k=1,...,m$. Show that a subsequence of $\{u_m\}_{m=1}^\infty$ converges weakly in $H_0^1(U)$ to the weak solution $u$ of $-\Delta u=f$ in $U$ and a zero Dirichlet condition.
How do I solve this?
How do I solve this?
Follow the ideas presented in theorems 1-3 of section 7.1.2. Here is a detailed answer:
We want to prove that there exists $u\in H_0^1(U)$, weak limit of a subsequence of $\{u_m\}$, satisfying $$\int_UDu\cdot Dv\ dx=\int_Uf\cdot v\ dx,\qquad\forall\ v\in H_0^1(U)\tag{$*$}$$ because this is the definition of weak solution for the problem. We know that $$\int_UDu_m\cdot Dw_k\ dx=\int_Uf\cdot w_k\ dx,\tag{1}$$ where $u_m=\sum_{k=1}^md_m^kw_k$. Here, $d_m^k\in\mathbb{R}$ and $\{w_k\}$ is an orthogonal basis of $H_0^1(U)$ and an orthonormal basis of $L^2(U)$.
Multiplying $(1)$ by $d^k_m$ and summing from $k = 1$ to $k=m$ we get \begin{align} \|u_m\|_{H_0^1}^2&=\|Du_m\|_{L^2}^2=\int_U |Du_m|^2\ dx=\int_UDu_m\cdot Du_m\ dx=\int_Uf\cdot u_m\ dx\\ &\leq \|f\|_{L^2}\|u_m\|_{L^2}\leq \|f\|_{L^2}\|u_m\|_{H^1}\leq C\|f\|_{L^2}\|u_m\|_{H_0^1}\leq C_\varepsilon\|f\|_{L^2}^2+\varepsilon\|u_m\|_{H_0^1}^2 \end{align} for all $m\in\mathbb{N}$. Taking $\varepsilon$ small enough, we get a constante $C$ such that
$$\|u_m\|_{H_0^1}^2\leq C\|f\|_{L^2}^2,\qquad\forall\ m\in\mathbb{N}.$$
So, $\{u_m\}$ is bounded in $H_0^1(U)$. Since $H_0^1(U)$ is reflexive, there exists a subsequence (which will not be relabeled) such that $$u_m\rightharpoonup u\quad \text{in}\quad H_0^1(U)$$ which implies (why?) $$\int_U Du_m\cdot Dg\ dx\to\int_UDu\cdot Dg\ dx,\qquad\forall\ g\in H_0^1(U).\tag{2}$$
Fix a positive integer $N$ and choose a function $g$ having the form $$g=\sum_{k=1}^N d^kw_k.\tag{3}$$ Multiplying $(1)$ by $d^k$ and summing from $k = 1$ to $k=N$ we get $$\int_UDu_m\cdot Dg\ dx=\int_Uf\cdot g\ dx,\qquad\forall\ m\in\mathbb{N}.$$
Taking the limit with respect to $m$, it follows from $(2)$ that
$$\int_U Du\cdot Dg\ dx=\int_Uf\cdot g\ dx.\tag{4}$$
As functions of the form $(3)$ are dense in $H_0^1(U)$ (why?), we obtain $(*)$ from $(4)$.$\;\square$
Due to Theorem 1 in 6.5.1, there is an orthonormal basis of $L^2$ consisting of smooth functions $w_k\in H_0^1$ which satisfy $-\nabla^2 w=\lambda_k w_k$ where $\lambda_k\rightarrow \infty$ and $\lambda_1>0$. We will first try to prove boundedness in $L^2(U)$ of the $u_m$ sequence. It is clear by orthonormality that: $$\int_U u_m^2 dx=\sum_{k=1}^m (d_m^k)^2$$ On the other hand we have $w_k \in H_0^1 \cap C^\infty(\overline{U})$, so it is zero in the boundary (by the Trace Theorem) and we may perform integration by parts and Parseval to conclude that: $$\langle f,w_k \rangle=\int_U \nabla u_m\cdot \nabla w_kdx=\int_U -u_m\nabla^2 w_k dx=\lambda_k\int_U u_m w_kdx=d_{m}^k\lambda_k$$ $$\therefore \:\:\sum_{k=1}^m (d_m^k \lambda_k)^2\leq \sum_k |\langle f,w_k\rangle|^2= \Vert f\Vert_{L^2(U)}$$ Finally, we have that there is $k_o$ such that if $k\geq k_o$, then $\lambda_k\geq 1$. This of course means that: $$\Vert u_m\Vert_{L^2(U)}^2=\sum_{k=1}^m (d_m^k)^2\leq \sum_{k=1}^{k_o}\left(\frac{\langle f,w_k\rangle}{\lambda_k}\right)^2+\sum_{k_o}^m (d_m^k \lambda_k)^2\leq $$ $$\frac{1}{\lambda_1^2}\sum_{k=1}^{k_o}|\langle f,w_k\rangle|^2+\sum_{k=k_o}^m |\langle f,w_k\rangle|^2\leq \frac{1+\lambda_1^2}{\lambda_1^2} \Vert f\Vert_{L^2(U)} $$ Furthermore, by linearity of our identity, we have that: $$\int_U \nabla u_m\cdot \nabla u_mdx=\int_U f u_m dx\leq \Vert f\Vert_{L^2(U)}\Vert u_m\Vert_{L^2(U)}\leq \frac{\sqrt{1+\lambda_1^2}}{\lambda_1}\Vert f\Vert_{L^2(U)}^{3/2} $$ All of this allows us to conclude that there is an uniform bound $\Vert u_m\Vert_{H^1}\leq C$ and since we are in a Hilbert (hence reflexive space), there exists a subsequence such that $u_{m_j}\rightharpoonup u$. Taking the weak limit: $$\int_U f w_k dx=\lim_j\int_U Du_{m_j}\cdot Dw_k dx=\int_U Du \cdot Dw_k$$ Because $w_k$ is dense in $H^1$ (actually it is an orthogonal basis as proven in Theorem 2 of section 6.5.1), we have that this is enough to conclude that for every $v\in H_0^1$: $$\int f v dx=\int_U Du Dv dx$$
