Check the conditions of Lax-Milgram Lemma - proving coercivity. Check the conditions of the Lax-Milgram Lemma for the problem
$$a(u,v) = \left< F,v \right>, v \in H_0^1(\Omega)$$
where $\Omega$ is bounded and
$$a(u,v) = \int_\Omega \left[ \sum_{i,j=1}^d (i+j)^2 \frac{\partial u}{\partial x_j} \frac{\partial v}{\partial x_i} + (1+x^2)uv \right] ~dx$$
$$\left< F,v \right> = \int_\Omega 2v ~dx$$
What I've managed to do:
$a$ is obviously bilinear.
Functional $F$ is bounded:
$$| \left< F,v \right>| \leq 2 \int_\Omega |v| dx \leq 2 |\Omega| \left( \int_\Omega |v|^2 ~dx \right)^{1/2} \leq C \| v \|_{H^1(\Omega)} $$
$a$ is continuous:
$$|a(u,v)| \leq \sum_{i,j=1}^d \int_\Omega (i+j)^2 \left| \frac{\partial u}{\partial x_j} \right| \left| \frac{\partial v}{\partial x_i} \right| + (1+x^2) |u| |v| ~dx \\ \leq d^2 n \cdot d^2 |u|_{H^1{\Omega}} |v|_{H^1(\Omega)} + (1+M^2) \| u \|_{L^2(\Omega)} \| v \|_{L^2(\Omega)} \\ \leq max\{ n d^4, 1+M^2 \} \left[ |u|_{H^1(\Omega)} |v|_{H^1(\Omega)} + \| u \|_{L^2} \| v \|_{L^2} \right] \\ \leq D \cdot \| u \|_{H^1} \| v \|_{H^1}$$
where $M$ is such that $1+x^2 \leq 1+M^2, \forall x \in \Omega.$
What I can't prove is coercivity of $a.$ I don't even know if it holds.
Any hints would be appreciated!
 A: Let $A$ be the leading matrix. For simplicity, we assume that $A$ is diagonal and $A_{11} < 0$ (this can be obtained after some rotation). Moreover, we assume that $0$ belongs to the interior of the domain $\Omega$.
Now, consider some $\psi \in C_c^\infty(\Omega)$, $\psi \ne 0$, such that the support of $\psi$ is a ball centered at $0$ (again, for simplicity).
We define
$$
\psi_n(x) := \psi(n \, x_1, x_2, \ldots, x_d).$$
Then, one can check
$$
\int_\Omega \frac{\partial^2}{\partial x_1^2} \psi_n(x) \, \mathrm dx
=
\int_\Omega n^2 \, \frac{\partial^2}{\partial x_1^2} \psi(n\,x_1,x_2,\ldots) \, \mathrm dx
=
n \, \int_\Omega \frac{\partial^2}{\partial x_1^2} \, \psi(x) \, \mathrm dx
=: n \, c_1.
$$
Similarly,
$$
\int_\Omega \frac{\partial^2}{\partial x_i^2} \psi_n(x) \, \mathrm dx
=
\frac1n \, \int_\Omega \frac{\partial^2}{\partial x_i^2} \psi(x) \, \mathrm dx
=: \frac1n \, c_i
$$
for $i > 1$.
Thus,
$$
\int_\Omega \nabla \psi_n^\top \, A \, \nabla\psi_n \, \mathrm dx
=
n \, c_1 \, A_{11} + \frac1n \, \sum_{i=2}^d c_i \, A_{ii}
$$
and this will be negative for $n$ large enough.
