Lax Milgram, prove continuity for $a : {W^{2,2}(I)} \times {W^{2,2}(I)} \rightarrow\mathbb{R}$

Question

Let $I = (0,1)$ and $b>0$. Let $f\in L^2(I)$. I need to show that there exists a unique $u \in W^{2,2} (I)$ such that

$$a(u,\phi) = \int^1_0 (u''\phi'' + bu'\phi' + u\phi) dx = \int^1_0 f\phi \ dx, \ \forall \phi \in W^{2,2}(I)$$.

I want to use Lax-Milgram so I need to show continuity:

$$|a(u,\phi)| \leq C_0\|u\|_{W^{2,2}(I)} \ \|v\|_{W^{2,2}(I)}$$

Recall $$\|u\|^2_{W^{2,2}(I)} = \|u\|^2_{L^{2}(I)} + \|u'\|^2_{L^{2}(I)} + \|u''\|^2_{L^{2}(I)} $$

What I did :

\begin{align}|a(u,\phi)| &= |\int^1_0 (u''\phi'' + bu'\phi' + u\phi) dx|\\ &\leq \int^1_0 |u''\phi'' + bu'\phi' + u\phi| dx \\ &\leq \int^1_0 |u''\phi''| + b|u'\phi'| + |u\phi| dx \\ &= \int^1_0 |u''\phi''|dx + b\int^1_0|u'\phi'|dx + \int^1_0|u\phi| dx \\ &= \|u''\phi'\|_{L^1(I)} + b\|u'\phi'\|_{L^1(I)} + \|u\phi\|_{L^1(I)} \\ &\overset{\rm Hölder}{\leq}\|u''\|_{L^2(I)}\|\phi''\|_{L^2(I)} + b\|u'\|_{L^2(I)}\|\phi'\|_{L^2(I)} +\|u\|_{L^2(I)}\|\phi\|_{L^2(I)} \end{align}

I don't see what the next step should be to relate the last line to the ${W^{2,2}(I)}$ norms of $u$ and $\phi$. Thanks !

Answer 1

$$\|u''\|_{L^2(I)}\|\phi''\|_{L^2(I)} + b\|u'\|_{L^2(I)}\|\phi'\|_{L^2(I)} +\|u\|_{L^2(I)}\|\phi\|_{L^2(I)}$$ $$ \leq \max{\{1,b\}}\left(\|u''\|_{L^2(I)}\|\phi''\|_{L^2(I)} + \|u'\|_{L^2(I)}\|\phi'\|_{L^2(I)} +\|u\|_{L^2(I)}\|\phi\|_{L^2(I)}\right)$$ $$\underbrace{\leq}_{\text{Cauchy-Schwarz in } \mathbb R^3 } $$ $$\max{\{1,b\}} \sqrt{\|u''\|_{L^2(I)}^2+\|u'\|_{L^2(I)}^2+\|u\|_{L^2(I)}^2}\sqrt{\|\phi''\|_{L^2(I)}^2+\|\phi'\|_{L^2(I)}^2+\|\phi\|_{L^2(I)}^2}$$ $$= \max{\{1,b\}} \|u\|_{W^{2,2}(I)}\|\phi\|_{W^{2,2}(I)}$$

Note that you also need to show coercivity of the bilinear form.