# How to prove that a strong solution of the homogeneous Dirichlet boundary problem really exists?

For homogeneous Dirichlet boundary problem \left\{\!\! \begin{aligned} &-\Delta u+c(x)u=f(x),~\text{in}~\Omega,~~~~~~~~~~~~~~(I)\\ &u=0,~\text{on}~ \partial\Omega \end{aligned} \right. I know that the strong (classical) solution is defined as the function $$u\in C^2(\bar{\Omega})$$ satisfying the problem (I) above in the usual sense. On the other hand, the weak solution is the function $$u\in H^1_0(\Omega)$$ satisfying the following:
$$\int_\Omega\left(\sum_{i=1}^n\frac{\partial u}{\partial x_i} \frac{\partial v}{\partial x_i}+c(x)uv\right)\,dx=\int_\Omega fv\,dx~~~~~~~(II)$$ for every $$v\in H_0^1(\Omega)$$.

My question are the following:

(a) How do I prove/show that such a strong solution $$u\in C^2(\Omega)$$ really exists? In other words, how do I solve for this $$u$$?

(b) If I want to obtain a weak solution, do I have to solve (II) explicitly to obtain $$u$$? How is it easily solved? Any reference, guide?

This is a question of regularity i.e. you know that a weak solution $$u \in H^1_0(\Omega)$$ exists so now you must prove that any weak solution of (I) is actually an element of $$C^2(\Omega) \cap C(\overline{\Omega})$$. (Note: you didn't mention it but for $$u$$ to be a classical solution you also need $$u \in C(\overline{\Omega})$$ for the boundary condition to make sense).
To answer your second question about 'explicitly solving' (II) it is practically impossible. To put it into perspective if $$u$$ satisfies (II) with $$c=0$$, $$f \in C^\infty(\Omega)$$ and $$\Omega$$ with $$C^\infty$$ boundary then standard regularity theory implies that $$u \in C^\infty(\overline{\Omega})$$. Hence, $$u$$ satisfies $$-\Delta u =f$$ in $$\Omega$$ and $$u=0$$ on $$\partial \Omega$$. We know then that we can write $$u$$ in terms of a Green's function, but only in very very specific cases of $$\Omega$$ can we actually work out that Green's function. If $$c \neq 0$$ there are still 'Green's functions' but these are even more challenging to calculate than before! This is why mathematicians gave up on trying to 'solve' PDE a long time ago - it is much easier (and often more useful) to understand the underlying theory.