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For given $f \in L^2(\Omega)$ Poisson's equation reads $$- \Delta u=f \quad \text{on }\Omega.$$ So the variational problem becomes: For given $f \in H^{-1}(\Omega)$ find $u \in H_0^1(\Omega)$ such that $$\int_{\Omega} \nabla u \cdot \nabla \varphi \, \mathrm dx=\int_{\Omega}f\varphi \, \mathrm dx.$$ for all $\varphi \in H_0^1(\Omega).$

Why don't we keep $f \in L^2(\Omega)$?

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  • $\begingroup$ The notation $\int_\Omega f \, \varphi \, \mathrm{d}x$ does not make sense for $f \in H^{-1}(\Omega)$. $\endgroup$
    – gerw
    Jun 12, 2019 at 11:25

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For the poisson equation the natural space of the weak formulation is $H^{-1}$. However it is wrong or at least bad style to write it as an integral. It is advised to use the bracket notation $(f,\phi)_{H^{-1}}$.

Notice that for all functions $f\in L^2$ there exists an associated element $f^*\in H^{-1}$ given by $$ (f^*,\phi)_{H^{-1}}=\int f \phi dx $$

To this end it is common to simply use $f$ instead of $f^*$. But it is not possible the other way round: not every element of $f\in H^{-1}$ can be represented via an integral.

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  • $\begingroup$ Thanks a lot - it's the Riesz representation theorem right? $\endgroup$
    – Tesla
    Jun 12, 2019 at 12:10
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    $\begingroup$ The standard Riesz theorem applies only for Hilbert spaces so in this case for $L^2$ and it dual. But we get the chain $H^{-1}\subset (L^2)^* \simeq L^2\subset H^1$ $\endgroup$
    – maxmilgram
    Jun 12, 2019 at 12:21

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