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Of course, here L is the differential operator given by:
:<math>
\mathrm{L}f=f_{xx}+f_{yy}
</math>
L is known as the Laplace operator. We are now looking for a ''u'' in V so that L''u''=''g''.
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for all ''v'' in V. From now on, we will use ∫<sub>T</sub> for the double integral ∫<sub>0</sub><sup>1</sup>∫<sub>0</sub><sup>1</sup>. One can see, via [[integration by parts]], and noting that because of the periodic boundary condition the first right hand side ''fg'' term
:<math>
\int_0^1 \left[ v_x u_x \right]_0^1 dy +\int_0^1 \left[ v_y u_y \right]_0^1 dx
</math>
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vanishes, that
:<math>
\int_0^1 \int_0^1 v\left(u_{xx}+u_{yy}\right)dxdy = -\int_0^1\int_0^1 \left(v_x u_x + v_y u_y \right) dxdy.
</math>
Thus we find that the
:ψ(''u'',''v''):=-∫<sub>T</sub>(''u<sub>x</sub>v<sub>x</sub>''+''u<sub>y</sub>v<sub>y</sub>'') = ∫<sub>T</sub>''g·v'' = φ<sub>''v''</sub>(''g'')
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