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In the finite element method, at a certain point we arrive at the following Galerkian problem where it is desired to find the solution $u_h \space \in V_h$ that solves the following equation:

$$ a(u_h,v_h)=L(v_h) \space \space \space \forall v_h \in V_h $$

where $a$ and $L$ are, respectively, a bilinear and linear operators. I cannot understand why is normally stated that it is enough to test against a set of basis functions $\Phi_i \in V_h$ (which are linearly combined to form $u_h$)and not against all functions $v\in V_h$

Thank you very much in advance and I hope you may help me understanding this issue.

Kind regards

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The idea of FEM is to find a finite set of equations after getting the Galkerin form. Once we have reached $$a(u_h,v_h)=L(v_h),$$ we want to reduce it to a system of equations of the form $Au=L$. Now an easy way to do this is if we choose $v_h=\{\phi_i\}$ because we have finite dimension and hence finite $\{\phi_i\}$, and the biggest thing is that we know these elements.

Now, the question is why only $\{\phi_i\}$ and not any other $n$ elements of $V_h$. Well if we choose $\phi_i$ then we can easily compute our matrix $A=\{a_{ij}\}$ as then we have $a_{ij}=a(\phi_i,\phi_j)$ and we already know these elements.

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