By definition $\{ {w_k}\} _{k = 1}^\infty \ $ is an orthorgonal basis in $H_0^1(U)$ and an orthornomal basis in $L^2(U)$.

suppose $f \in {L^2}(U)$ and assume that ${u_m} = \sum\limits_{k = 1}^m {d_m^k{w_k}}$ solves

$$\int\limits_U {D{u_m} \cdot D{w_k}dx = } \int\limits_U {f \cdot {w_k}dx} $$ for $k=1,2,\cdots\ m$.

Show that a subsequence of $\{ {u_m}\} _{m = 1}^\infty $converges weakly in $H_0^1$ to the weak solution of

$$\left\{\begin{align*} &- \Delta u = f&&\text{in }U\\ &u = 0&&\text{on }\partial U \end{align*} \right.$$

Actually I know that we can according to the usual energy estimation to obtain the subsequence, and integrate by parts,while I still don't konw how to pass to limit to in the integrating equation.Wish to somebody to give the details, Thanks a lot!

  • 1
    $\begingroup$ I think weak convergence in $H_0^1$ exactly means that the integrals involving $u_m$ will converge.... By the way, what are $w_k$ and $d_m^k$? $\endgroup$ – user53153 Jan 4 '13 at 4:57
  • $\begingroup$ As stated it is false. You really must give all details, including what $d^k_m$ and $w_k$ are. Counterexample to the current version: let $w_k \equiv 1$ (assuming $U$ is bounded open) and $f \equiv 0$. Then $d^k_m = \delta_{km}$ certainly leads to $u$ that satisfy the integral equation. The only weak solution to Laplace's equation with Dirichlet boundary conditions is $u\equiv 0$. But all our $u_m \equiv 1$. $\endgroup$ – Willie Wong Jan 4 '13 at 15:18
  • $\begingroup$ My best guess is that you want $w_k$ to be some basis functions for $L^2(U)$ and $d^k_m$ are coefficients of approximate solutions? Your formulation reminds me a bit of finite element methods. $\endgroup$ – Willie Wong Jan 4 '13 at 15:24
  • $\begingroup$ @FengxiaoHe Please add the missing information to the question by clicking edit under it. It should not be left in the comment thread only. While doing so, you should clarify what kind of basis the functions $w_k$ form (orthogonal, orthonormal,...?) $\endgroup$ – user53153 Jan 5 '13 at 3:58

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