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Write $q_i=a^{ij}{D}_{j}u$. If $u$ is a classical solution, multiply $-{\partial }_i{q}_i=f$ by a smooth function $v$ and apply the divergence theorem. The Neumann boundary condition $q_i{N}_i=0$ removes the boundary term and givesThe density of smooth functions in a Sobolev space and boundedness of the coefficients extend this identity to every $v\in H^1(U)$, so $u$ is a weak solution.

Conversely, take $v$ to be a test function compactly supported in $U$. The weak formulation saysThe fundamental lemma of the calculus of variations gives the equation in $U$. Under the regularity implicit in the stated notion of a classical solution, it holds pointwise. Applying integration by parts again with arbitrary $v\in H^1(U)$ leaveswhere $T$ is the trace operator. Traces of smooth functions can be chosen arbitrarily on the boundary, so the boundary fundamental lemma of the calculus of variations gives $(a^{ij}{\partial }_{j}u)N_i=0$. Thus $u$ is a classical solution.

Solved by gpt-5.6-sol high.

Write points of $U$ as $(x,y)$ with $x>0$. For $h>0$, translate $u$ into the domain byContinuity of translations in $L^p$ applied to $u,D_{x}u,D_{y}u$ givesas $h\downarrow 0$. Choose a standard mollifier ${\rho }_{\epsilon }$ supported in a ball of radius $\epsilon <h/2$. For $(x,y)\in U$, the convolutiononly samples points with first coordinate greater than $-h$, so it is well-defined and smooth throughout $U$. The approximation-to-the-identity theorem, applied also to each weak derivative, allows $\epsilon (h)<h/2$ to be chosen so thatTaking $h=1/j$ and using the triangle inequality proves the density of smooth functions in a Sobolev space.

Solved by gpt-5.6-sol high.