This is not homework. I'm going over my lecture notes to study for an exam.

For an Abstract Elliptic Problem such as the problem with V a Hilbert Space

$$\begin{cases} \text{Find } u \in V \text{ such that} \\ a(u,v) = L(v) && \forall v\in V \end{cases}$$

where $a$ is a bounded, coercive bilinear form. $L \in V^*$. We know that by Lax-Milgram that $u$ is the unique solution

We search for approximations to $u$, $u_m \in V_m \subset V$ where $V_m$ is a finite dimensional subset of $V$.

In the Galerkin method we seek the best approximation $u_m$ of $u$ with respect to the bilinear form $a$ described above.

$$\begin{cases} \text{Find } u_m \in V_m \text{ such that} \\ a(u_m,v_m) = L(v_m) && \forall v_m\in V_m \end{cases}$$

The Galerkin orthogonality states that:

$$a(u-u_m, v_m) = 0$$

which implies that orthogonality of the error with respect to $a$. The usual proof is given as follows:

$$a(u-u_m, v_m) = a(u,v_m) - a(u_m,v_m) = L(v_m) - L(v_m) = 0$$

What I don't understand is why the two $L$s in the above are the same. Since from Riesz Representation Theorem, $L$ is the unique functional in the dual of $V$ or $V_m$, how can they possibly be equal otherwise $u=u_m$?

Edit: From comments below, the two functionals (L) are not technically operating in the same spaces. On the other hand, they both represent some inner product ($L^2$ or other) and they both produce the same result, hence equality in result without necessarily equivalance.

  • 2
    $\begingroup$ $L$ is given. The problem is to find $u$ or $u_m$ satisfying these variational equations. $\endgroup$ – daw Feb 19 at 15:45
  • $\begingroup$ but if $L$ is given, since $a$ is also given how can $u_m$ be different to $u$ since the mapping to the dual space is an isometry? $\endgroup$ – Not a chance Feb 19 at 16:15
  • $\begingroup$ because $u$ and $u_m$ solve different problems $\endgroup$ – daw Feb 19 at 20:37
  • $\begingroup$ OK, but if the functional analysis theory holds that there is a unique $L$ such that $a(u,v)=L(v) \forall v$ then $a(u_m,\cdot)\neq a(u,\cdot)$ unless $u_m=u$ $\endgroup$ – Not a chance Feb 20 at 7:43
  • $\begingroup$ The second problem is in a subspace unlike the original one $\endgroup$ – VorKir Feb 20 at 16:00

$L$ is a given linear and continuous function $L:V\to\mathbb R$.

As you write in the comments, there exists a unique functional $M\in V^\star$ such that $ a(u,v) = M(v) \forall v\in V$. (I am giving this object a name different from $L$ to avoid confusion).

Also, there exists a unique functional $M_m\in V_m^\star$ such that $ a(u_m,v_m) = M_m(v_m) \forall v_m\in V_m$.

In general, $M_m\neq M$.

However, the proof uses the functional $L$ and not $M,M_m$, so $L(v_m)-L(v_m)=0$ in the proof is always true.

  • $\begingroup$ thanks for completing this $\endgroup$ – Not a chance May 15 at 9:30

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