Showing that there is a unique solution for the following equation

Question

Let $I = (0, 1)$, $a : H_0^2(I) \times H_0^2(I) \to \mathbb{R}$ a continuous bilinear form defined by $$a(u, v) = \int\limits_I u'' (x) v'' (x) dx.$$

Show that for every $f \in L^2(I)$ there is a unique solution $u \in H_0^2(I)$ of the equation $$\int\limits_I u''(x) v''(x) dx = \int\limits_I f(x) v(x) dx, \;\; \forall v \in H_0^2(I).$$

Can someone, please, give me a hint?

Thank you!

P.S: We can introduce an operator $(A, D(A))$, where $D(A) = H^4(I) \cap H_0^2(I)$ and $A u = u_{xxxx}$. Does it help?

Answer 1

You can show that the bilinear form is coercive in $H^2$. To do so, you have to use the inequality $$ \|u\|_{H^1} \le c \|u''\|_{L^2} \quad \forall u\in H^2_0(\Omega). $$ Then by Lax-Milgram theorem you obtain existence of a unique solution for all $f$.