Differential operator and equivalence

Question

Here is the problem: I have a certain PDE and there is the nonlinear terme $h$, I have as data: $f \in H_0^2(0,L)$,,,$g \in {H^1}(0,L)$ with ${g_x}(0) = {g_x}(L) = 0$ Now on consider the fnction $$h(x) = f'(x)(g'(x) + {(f'(x))^2}){\rm{ }}{\rm{, x}} \in {\rm{[0}}{\rm{,L]}}$$ we notice that $h(0)=h(L)=0$. It is well known that the operator:$$(I - \partial {}_x^2):H_0^1(0,L) \cap {H^2}(0,L) \to {L^2}(0,L)$$ is bounded and invertible, and its inverse defined by: $${(I - \partial {}_x^2)^{ - 1}}:{L^2}(0,L) \to H_0^1(0,L) \cap {H^2}(0,L)$$ the question: have we the inequality $${\left\| {{\partial _x}{{(I - \partial {}_x^2)}^{ - 1}}{\partial _x}h} \right\|_{{L^2}(0,L)}} \le c{\left\| h \right\|_{{L^2}(0,L)}}??$$ ($c$ is a positive constant) and if this is wrong, what conditions can I pose on $f$ and $g$ to satisfy this inequality..? and if this is true, how can I prove it rigorously?

Answer 1

First, we have $$ \|\partial_x(I - \partial_x^2)^{-1}\partial_xh \|_{L^2(0,L)} \le c \|(I - \partial_x^2)^{-1}\partial_xh \|_{H^1(0,L)} . $$ Define $z:=(I - \partial_x^2)^{-1}\partial_xh \in H^1_0(0,L)$. Then it satisfies the weak formulation $$ \int_0^L \partial_x z\partial_x v + zv \ dx = \int_0^L \partial_x h \cdot v\ dx $$ for all $v\in H^1_0(0,L)$. Partial integration on the right-hand side, and setting $v=z$ gives $$ \int_0^L (\partial_x z)^2 + z^2 \ dx = -\int_0^L h \partial_x z\ dx. $$ Hence we get $$ \|z\|_{H^1}^2 \le \|h\|_{L^2(0,L)} \|\partial_xz\|_{L^2(0,L)}\le \|h\|_{L^2(0,L)} \|z\|_{H^1(0,L)}, $$ which shows $\|z\|_{H^1}\le \|h\|_{L^2(0,L)} $. And the claim is proven.

Note that the last step (the estimate of $z$ is nothing else than showing that $\|\partial_xh\|_{(H^1(0,L))^*}\le \|h\|_{L^2(0,L)}$.