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$\pmb{Problem} :$

Given the differential equation :

$(E) : -u''(x) + tanh(x). u(x) = -\lambda.u(x)$

Let $f_n(x) = C_n.cos(\sqrt{\lambda-1}. x). \varphi(\frac{x-n^2}{n})$ where $\varphi \in C^{\infty}(\mathbb{R})$ and $$\varphi(x) = \left\{ \begin{array}{c} 1 & if \ \ x \in [-1, 1]\\ 0 & if \ \ \lvert x\rvert \geq 2 \\ g(x) &else \end{array} \right. $$ where $g \in C^{\infty}$ (That's all we know about the function g)

Let $\alpha_n = \int_{-\infty}^\infty f_n^2(x) .dx$

Determine $C_n$ such that $\displaystyle\lim_{n\to\infty} \alpha_n \neq 0$ $and \int_{-\infty}^\infty L^2(f_n)(x).dx \to 0$ when $n \to +\infty$

where $L(f_n)(x) = -f''_n(x) + tanh(x).f_n(x) - \lambda.f_n(x)$

$\pmb{Discussion :}$

For instance, I started by calculating $f_n ^2(x)$ to evalute $\alpha_n$ in terms of n. I also tried to solve the differential equation $(E)$ but I couldn't do that due to the hyperbolic tangent (non constant coefficients). In addition, I calculated the first derivative of $f_n$ for the hope of finding a simpler expression of $f_n$ after integrating but all of this didn't get me anything. So I literally can't find where to begin to address this problem.

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