I have a pde of the following form: \begin{align} &P(x,D)u = f \text{ on } \Omega, \\ &P(x,D) = \sum\limits_{|\alpha|=2m}a_\alpha(x)D^{\alpha}, \end{align} where one can assume that $f$ and $a_\alpha$ belong to $C_0^{\infty}(\Omega)$ ($\Omega$ is open in $\mathbb{R}^2$). It is important that differential operator consists only of highest order derivatives. Also, I assume that: $$ P(x,\xi) \geq C|\xi|^{2m}, $$ which is known as a strong ellipticity condition. I wonder do these properties imply that such equation is solvable for any $f$ (I'm interested in weak solvability as well as in strong solvability)?

The point is that, when the elliptic pde is rewritten in a variational form, one associates to it a quadratic form. The uniform positivity of this form is sufficient for solvability by Lax-Milgram theorem (for example for the Dirichlet boundary problem with zero on some far border). The problem is that the form must have the coefficients under derivatives (or what is called - have the divergence form), e.g., for $m = 1$ we have: \begin{align} &P(x,D)u = \sum\limits_{k,j}\partial_k a_{k,j}\partial_ju + \text{lower order terms},\\ &\Phi_\Omega(u,v) = \sum\limits_{k,j}a_{k,j}\partial_ju \partial_k\overline{v} + \text{lower order terms}. \end{align} Clearly, for the aforementioned $P(x,D)$ lower-order terms will appear in the form, which will not appear in the ellipticity condition.

I believe that that the answer should be positive, but cannot see it directly.