Prove inequality in Sobolev space

Question

Let $\Omega = [0, 1], \varphi \in L^2(\Omega)$, and $u$ is a solution of a problem

$-u''+u=\varphi$ in $\Omega$, $u(0)=u(1)=0.$

I need to show that $\exists C > 0: \vert u\vert_{H^2(\Omega)} \leq C\vert\vert \varphi\vert\vert_{L^2(\Omega)} $.

The hint suggests to first get an inequality for $\vert\vert\cdot\vert\vert_{H^1(\Omega)}$ by multiplying the differential equation by $u$ and integrating over $\Omega$. This resulted the following:

$-\int_\Omega u''u + \int_\Omega u^2 = \int_\Omega u \varphi$

After partial integration I get

$\int_\Omega u'^2 + \int_\Omega u^2 = \int_\Omega u \varphi$

$\vert\vert u\vert\vert_{H^1(\Omega)} = \int_\Omega u \varphi$

Here I am stuck trying to think of any further estimations. Moreover I don't see, how an inequality for $\vert\vert\cdot\vert\vert_{H^1(\Omega)}$ will give me the required inequality.

Would be grateful for any explanation.

Answer 1

You derived $$\|u'\|_2^2 + \|u\|^2_2 = \|u\phi\|_1$$ which gives the inequalities $$\|u'\|_2^2\leq \|u\phi\|_1\qquad\text{ and }\qquad\|u\|_2^2 \leq \|u\phi\|_1.$$

Apply Hölder's inequality/Cauchy-Schwarz to get a bound of the form $\|u\|_2^2\leq C\|\phi\|_2^2.$ Then, using this bound, get a bound of the form $\|u'\|_2^2\leq C\|\phi\|_2^2.$ Then use the bound $\|u\|_2^2\leq C\|\phi\|_2^2$ and the one piece of information you have about $u''$ to get a bound of the form $\|u''\|_2^2\leq C\|\phi\|_2^2.$ (Each $C$ can be different.)