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?
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$\begingroup$ Make a Fourier transform, the opetator $\partial_x$ becomes the operator $f(t)\to it\cdot f(t)$, and your operator becomes a multiplication by $t^2/(1+t^2)$, which is of course bounded. $\endgroup$– Fedor PetrovCommented May 25, 2017 at 12:12
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$\begingroup$ Hello sir ....in our case we have a finite intervalle not R...How can we make the Fourier transform...? $\endgroup$– GustaveCommented May 25, 2017 at 12:36
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$\begingroup$ We may extend the functions to the whole line (by zero values outside the interval), but at moment I doubt that the inverse of $I-\partial_x^2$ is the same as making Fourier transform, multiplying by $1/(1+t^2)$ and taking inverse Fourier transform... $\endgroup$– Fedor PetrovCommented May 25, 2017 at 12:48
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$\begingroup$ Thanks...so the inequality is true....how can I use the conditions $h(0)=h(L)=0$ ? Can I work without them ? Thanks. there is a second question in my mind: Can I prove the inequality using spectral decomposition ? thanks. $\endgroup$– GustaveCommented May 25, 2017 at 15:34
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2$\begingroup$ Forget this Fourier approach. The operator $A=i\partial_x$ is self-adjoint, thus $A^2(I+A^2)^{-1}$ has norm at most 1 by spectral decomposition. $\endgroup$– Fedor PetrovCommented May 25, 2017 at 16:50
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