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Let us define $$ \mathbb{H}^{1} = H^{1}(-L,0) \times H^{1}(0,L) \ \ \text{and} \ \ \mathbb{L}^{2} = L^{2}(-L,0)\times L^{2}(0,L), $$ where $H^{1}(I) = \big\lbrace u \in L^{2}(I) \ \text{and} \ u_{x} \in L^{2}(I); I = (a,b) \big\rbrace$.

Besides these, $$ \mathbb{M} = \big\lbrace (u,v) \in \mathbb{H}^{1}; u(-L) = v(L) = 0 \ \text{and} \ u(0) = v(0) \big\rbrace . $$ Under the above conditions, we have that the phase space is given by $$ \mathcal{H} = \mathbb{M} \times \mathbb{L}^{2}. $$ Note that this space equipped with the inner product $$ \langle (u_{1},v_{1},w_{1},z_{1}), (u_{2},v_{2},w_{2},z_{2}) \rangle = \int_{-L}^{0}u_{1_{x}}\overline{u}_{2_{x}} + \int_{0}^{L}v_{1_{x}}\overline{v}_{2_{x}} + \int_{-L}^{0}w_{1}\overline{w}_{2} + \int_{0}^{L}z_{1}\overline{z}_{2} $$ is a Hilbert space.

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  • $\begingroup$ Why is the inequality deleted? $\endgroup$
    – Zerox
    Nov 13 at 16:50

1 Answer 1

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Just take a derivative, and you get

$$ i \lambda u_x = f_x - w_x $$

So taking the $L^2(-L,0)$ norms on both sides, you get

$$ \lambda^2 \int |u_x|^2 \leq \int |f_x - w_x|^2 $$

The RHS can be expanded and estimated using AM-GM to be

$$ \lambda^2 \int |u_x|^2 \leq 2 \int |f_x|^2 + |w_x|^2 $$

The first term is bounded by $\|F\|^2$, and the second, by your assumption on the $L^2$ bound of $w_x$, is bounded by $\|U\| \|F\|$.

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  • $\begingroup$ why the condition of $|\lambda |> 1$? of your answer, it is valid for all $\lambda \in \mathbb{R}$. $\endgroup$
    – user253963
    Sep 6 at 13:57
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    $\begingroup$ Nothing in your question requires $|\lambda | > 1$; there may be other consideration in whatever book/paper you are reading. But since you didn't provide that info, I cannot answer. $\endgroup$ Sep 6 at 15:46
  • $\begingroup$ In the thesis of the question he affirms the inequality for $|\lambda |> 1$. I wrote this information. But thanks for the reply. $\endgroup$
    – user253963
    Sep 6 at 16:10

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