From weak to strong formulation: ∫Ω∇u∇v+∫Ω1u⋅v=∫Ω1v+∫Ωf⋅v

Stonelord

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From weak to strong formulation: ∫Ω∇u∇v+∫Ω1u⋅v=∫Ω1v+∫Ωf⋅v

Hello there,

part of an excercise I've been asked to do is about converting a weak form of a differential equation into its strong form,being the solution \(\displaystyle u\) defined in \(\displaystyle \Omega\, =\, [0,10]\, \times\,[0,10],\, \Omega_1\,=\, [0,10]\, \times\,\{0\}\),

\(\displaystyle \displaystyle{\int_{\Omega}\,\nabla\, u\, \nabla\, v\, +\, \int_{\Omega_1}\,u \,\cdot\, v\, = \,\int_{\Omega_1}\, v\, +\, \int_{\Omega}\, f\, \cdot \,v}\)

\(\displaystyle u(x)\,=\, g,\, x \,\in\, \partial\Omega \,\setminus \,\Omega_1\)

\(\displaystyle g\,\equiv\, {0},\, f\, \equiv\,{1}\)

The problem statement did not specify the space function to which the solution \(\displaystyle u\) as well as the test function \(\displaystyle v\) belong, so I guess we can suppose \(\displaystyle v \) has the same boundary conditions \(\displaystyle u\) has. I guess the way the weak form was obtained was the usual; multiplying the test funcion in both sides of the equation and using the integration by parts formula on the left one applying the boundary conditions when needed. Yet I don't get how to do it the reverse way.

Thank you in advance.
 
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