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The weak form I am given is

$$\int_{U} \left(\left(u_{x} - \sin{(x)} u \right)\left(v_{x} - \sin{(x)} v\right)+ u_{y}v_{y}\right) dV = \int_{U} (fv) dV$$

EDIT: Based on the responses below, applying the lemma I get the following:

$$\int_{U} (u_{x}-\sin(x)u)v_{x} - (u_{x}-\sin(x)u)\sin(x)v +u_{y}v_{y} dV = \int_{U} (fv) dV$$

Integrating by parts yields

$$-\int_{U} (u_{x}-\sin(x)u)_{x}v dV - \int_{U} (u_{x}-\sin(x)u)\sin(x)v dV - \int_{U} u_{yy}v dV = \int_{U} (fv) dV$$

(Why do the boundary terms vanish?)

$$\int_{U} (u_{xx}-\sin(x)u_x-\cos(x)u)dV+\int_{U} (u_{x}\sin(x)-\sin^{2}(x)u)vdV+\int_{U} u_{yy}vdV=-\int_{U} (fv)dV$$

This will give us the result

$$u_{xx}-\sin(x)u_{x}-\cos(x)u+\sin(x)u_{x}-\sin^{2}(x)u+u_{yy}=-f$$ $$u_{xx}+u_{yy}-\cos(x)u-\sin^{2}(x)u=-f$$ $$\Delta u - (\cos(x)-\sin^{2}(x))u = -f$$

If I've made mistakes in my calculations I do apologize, it has been a very long day.

MrStormy83
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    You need to know information about $u$ and $v$ on the boundary of $U$ to apply integration by parts and not pick up boundary terms. In terms of the strong form, this corresponds to boundary conditions on the solution $u$, but in terms of the weak form, this corresponds to the definition of the spaces where $u$ and $v$ live in the case of homogeneous BCs. In the ase of nonhomogeneous BCs, one will get boundary terms in the weak form that must be translated into BCs for the strong form. – whpowell96 May 03 '23 at 22:03

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It's based on the fundamental lemma of calculus of variations:

Lemma. If $$ \int_U fv~dV = 0 $$ for all test functions (for a suitable class of test functions) $v$, then $f = 0$ a.e. on $U$.

Therefore what you want to do is to express all integrals in this above form: as integrating some function against a test function $v$. This means you need to manipulate your expressions so that at the end of everything, they depend on $v$ only through $v$ itself, rather than derived quantities like $v_x$ and so on.

So here for instance, one of the quantities we need to manipulate away is the term $$ \int_U (u_x - \sin(x)u)v_x~dV. $$ We would integrate by parts to convert the $v_x$-dependence into a $v$-dependence: $$ -\int_U (u_x - \sin(x)u)_x v~dV = -\int_U (u_{xx} - \sin(x)u_x - \cos(x)u)v~dV. $$ And so on and so forth, until you reach a form for which the fundamental lemma applies. I do wonder if the term $$ \int_U u_xu_y~dV $$ you included is a typo - normally you'd expect every term in the weak form of an equation to depend somehow on the variation $v$ or its derivatives.

Gyu Eun Lee
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