Find the necessary and sufficient conditions for $f \in L^2(U)$ so that the equation
$-\sum_{i,j=1}^n (a^{ij}u_{x_i})_{x_j}-7u=f$ in $U$ and $u=0$ on $\partial U$.
has a weak solution $u \in H_{0}^1(U)$. Here the matrix $a^{ij}$ is symmetric, bounded, and uniformly elliptic.
I tried to solve this question by solving the bilinear form and Lax Milgram theorem but not able to reach the conclusion. Anyone can suggest some hints?
First of all stating that $u = 0$ on $\partial U$ and stating that a solution has to lie in $H_0^1(U)$ is superfluous (you could either remover the first or state $u \in H^1(U)$). It does not really matter, just a small remark. Second, I will assume that $\partial U$ is (piecewise) Lipschitz/smooth enough that we can use integration by parts.
Then, the bilinear form corresponding to this PDE is equal to $$ B: H_1^0(U) \times H_1^0(U) \rightarrow \mathbb{R}; \quad B[u,v] = \int_U (\nabla u)^TA (\nabla v) - 7 u^2, $$ where $(A)_{ij} = a_{ij}$. Ignoring this $7$ for a moment we would have $B[u,u] \geq \theta \left \lVert u \right \rVert_{H^1(U)}$ (where $\theta$ is the constant from the uniform ellipticity) and we could use Lax-Milgram (even just Riesz representation Theorem as the bilinear form is symmetric). So we would have an answer, any $f$ suffices.
However, due to the $-7u^2$ term we are in a tougher situation. Then the trick I know comes from Evans' book on PDE. In particular, in Evans' PDE book he uses the Fredholm alternative for compact operators to answer this question: $Lu = f $ has a (unique) solution for every $f \in L^2(U)$ if and only if $\langle f, v \rangle_{L^2(U)} = 0$ for all $v \in H_0^1(U)$ such that $Lu = 0$ (This is the second existence theorem, Theorem 4 of $\S$ 6.2.3 in my version, where I used that your A is symmetric). In other words, when $f \perp Ker(\tilde{B}^*) = Ker(\tilde{B})$, where $\tilde{B}:H_0^1(U) \rightarrow (H_0^1(U))^*; u \mapsto B[u,\cdot]$ is the 'weak version of L'.
This is not a very explicit condition of $f$ since it is just a slight rearrangement of the statment that $f \in ran(\tilde{B})$ together with the usual theorem from functional analysis that $ran(A)^\perp = ker(A^*)$ for a bounded linear operator and the fact that $\tilde{B}$ has closed range, and $\tilde{B}$ is symmetric.
A different view of the above can be found using spectral theory. Notice that $ker(\tilde{B}) = \{0\}$ if and only if $0$ is not an element of the spectrum of the function $\tilde{B}$ i.e. if 7 is not an eigenvalue of the function $$ H_0^1(U) \ni \, u \mapsto \int_u (\nabla u)^T A(\nabla \cdot) \, \in (H_0^1(u))^* $$ but this condition would, again, imply that there is no condition necessary for $f$ to lie in the range (i.e. $\tilde{B}$ would be surjective).
My intuition would be that, to describe $ran(\tilde{B})$ (i.e. by explicitely describing $ker(\tilde{B})$), we would need some more information of $B$ so, as daw mentioned in his comment, some more info on the a_{ij}. But there may be some other explicit characterizations of $ker(\tilde{B})$ or $ran(\tilde{B})$ that work in general (for example, possibly using harnack's inequality, that $u \in ker{L}$ implies $\inf u \leq C \sup u$).