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.



Positive form for a homogeneous elliptic pde
Unfortunately uniform ellipticity (as it is written above) does not imply solvability in general. The reason for that comes from the properties of the associated quadratic form $\Phi\_\Omega(u,u)$. For simplicity, I will consider only the case of $m=1$. 


The property of uniform ellipticity is used to establish what is called the **Garding inequality**:
\begin{equation}
\|u\|^2\_{H^1\_0(\Omega)} \leq C\_1 \Phi\_{\Omega}(u,u) + C\_2\|u\|^2\_{H^0}(\Omega).
\end{equation}
Classical solvability theorems require a stronger version of this inequality to hold (for example, for a homogeneous boundary valued Dirichlet or Neumann problem to be uniquely solvable). More precisely, the requirement asks for aforementioned inequality to hold with $C\_2=0$. Such property of $\Phi\_\Omega$ is called **the strong coercivity of** $\Phi\_\Omega$. You can find it in all books about elliptic pdes.


I say again, that if you have the strong coercivity property then your problem has a unique solution for any right part (modulo writing correct indices for Sobolev spaces). 


However , if you have only coercivity, then you can say that your differential operator is Fredholm in $H\_0^1(\Omega)$. This means that operator $P(x,D)$ may have a nonempty finite-dimensional cokernel. At this point the best one can do is to apply is to apply Fredholm theory:
\begin{align}
&P(x,D)u = f \text{ is solvable in }H^{1}(\Omega) \text{ if and only if }\\
&(f, v)\_{\Omega} = 0 \text{ for all } v\in \mathrm{Ker}(P^\*(x,D)),
\end{align}
where $P^\*(x,D)$ is the formally adjoint operator to $P(x,D)$.


