# Find the strong form of a PDE from the weak form.

I'm having a little difficulty understanding how to find the strong form of a PDE given the weak form. For example, I have the weak form as:

$\displaystyle\int_\Omega [a(x)\nabla u\cdot\nabla v+b(x)uv] \space dx=-\int_\Omega fv\space dx$

where we have $u,v\in H^1_0(\Omega)$.

I know I am supposed to somehow get the $\nabla$ off of the $v$ in the first integral so that I can, effectively, write

$\int_\Omega F[u]v\space dx=-\int_\Omega fv\space dx$

for some function(al?) on $u$ from which I can extract my strong form PDE, but I'm not sure if (1) I'm thinking about this the correct way and if I am (2) how to go about doing this.

Any help/hints/tricks/suggestions are highly appreciated. Thanks in advance!

Let $A(x)$ be the $n\times n$ matrix of coefficients $$A_{ij}(x)= \delta_{ij}a(x),$$ where $\delta_{ij}$ is the Kronecker delta; that is, $A(x)$ is a diagonal matrix with diagonal entries all equal to $a(x)$. (I think I'm interpreting the notation in the problem correctly, but the exposition that follows can be modified appropriately if I'm not.) Write $$a(x)\nabla u\cdot\nabla v = \sum_i A_{ij}(x)u_{x_i}v_{x_i},$$ Let $L$ be the differential operator so that the given weak form of the PDE is the weak formulation of $$Lu = f.$$ Write $L = G + H$, where $G$ is the second-order differential term and $H$ is the zero-th order term. Then $G$ is the second-order operator $$Gu = -\sum_{i=1}^n \left(A_{ii}(x)u_{x_i}\right)_{x_i}.$$ This is because if we formally integrate by parts in the weak formulation, we obtain $$\int \sum_{i=1}^nA_{ii}u_{x_i}v_{x_i} = -\int \sum_{i=1}^n\left(A_{ii}u_{x_i}\right)_{x_i} v.$$ And in fact, for a strong solution this is exactly what we would expect for all test functions $v$. The advantage to using $\nabla u\cdot\nabla v$ is that we can put weaker regularity assumptions on the $u$ and $v$ used in the weak formulation (which is exactly the point of a weak formulation).
So to wrap up the story, $L$ is given by $$Lu = -\sum_{i=1}^n \left(A_{ii}(x)u_{x_i}\right)_{x_i} - b(x)u.$$ Recalling that $A_ii(x) = a(x)$, we can express this as $$Lu = -\sum_{i=1}^n a(x)u_{x_ix_i} - \sum_{i=1}^n a_{x_i}(x)u_{x_i} - b(x)u.$$ So the strong formulation of the PDE is $$-\sum_{i=1}^n a(x)u_{x_ix_i} - \sum_{i=1}^n a_{x_i}(x)u_{x_i} - b(x)u = f.$$