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?

  • 1
    $\begingroup$ 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? $\endgroup$ Nov 13, 2015 at 17:17

1 Answer 1


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

  • $\begingroup$ Why is $\|u_m\|_{H_0^1}^2=\|Du_m\|_{L^2}^2$? $\endgroup$
    – Mathman
    Apr 29, 2022 at 18:49
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    $\begingroup$ @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 $\|u\|_{H_0^1}^2:=\sum_{\|\alpha\|=1}\|D^\alpha u\|_{L^2}^2=\sum_{i=1}^n\|u_{x_i}\|_{L^2}^2=\sum_{i=1}^n\int u_{x_i}^2=\int\sum_{i=1}^n u_{x_i}^2=\int|Du|^2=\||Du|\|_{L^2}^2=\|Du\|^2_{L^2}$ $\endgroup$
    – Pedro
    May 2, 2022 at 20:05

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