# Galerkin method for Poisson's equation

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?

• Let us start by the things you've already tried. What did you try to do? What worked, what did not? Where do you get stuck? Commented Nov 13, 2015 at 17:17

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 $(4)$.$\;\square$

• Why is $\|u_m\|_{H_0^1}^2=\|Du_m\|_{L^2}^2$? Commented Apr 29, 2022 at 18:49
• @IvanMathman According to Evans notation (see Appendix A), $\|Du\|_{L^2}^2=\||Du|\|_{L^2}^2$, where $|Du|=|(u_{x_1},...,u_{x_n})|=(\sum_{i=1}^n u_{x_i}^2)^{1/2}$. Therefore, the usual equivalent norm in $H_0^1$ is given by $\|u\|_{H_0^1}^2:=\sum_{\|\alpha\|=1}\|D^\alpha u\|_{L^2}^2=\sum_{i=1}^n\|u_{x_i}\|_{L^2}^2=\sum_{i=1}^n\int u_{x_i}^2=\int\sum_{i=1}^n u_{x_i}^2=\int|Du|^2=\||Du|\|_{L^2}^2=\|Du\|^2_{L^2}$ Commented May 2, 2022 at 20:05

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 $$v\in H_0^1$$: $$\int f v dx=\int_U Du Dv dx$$