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In $L^2([0;1])$: $u(x)$ is given smooth (piecewise linear) function, s.t.$u: [0;1]\to \mathbb{R}$. Function $\rho: [0;1] \to \mathbb{R}$, s.t.

$ \rho(x) = \begin{cases} 1, & \text{if $x\in[0;{1\over2})$;} \\ 2, & \text{if $x\in[{1\over2};1]$.} \end{cases} $.

Then $$c = {1\over{\int_{0}^1{1\over{\rho(x)}}dx}} = {4\over3}.$$

We should find a sequence of functions $u_n$, $u_n$ $\Gamma$-converges to $u$, s.t. $$\lim_{n\to\inf}\int_0^1\rho(nx)(u_n^{'}(x))^2dx \to \int_0^1{4\over3}(u^{'}(x))^2dx.$$

Moreover, I know that if $u = {{b-x}\over{b-a}}$, then $u_n$ is a solution of $$\left\{ \begin{aligned} {d\over{dx}}\Big{[}\rho(nx){d\over{dx}}u_n(x)\Big{]} &= 0 \\ u(a) &= 1 \\ u(b) &= 0 \end{aligned} \right.$$ I don't know how to find any solution even a partial one for $u = {{b-x}\over{b-a}}$.

Can someone help me, please?

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  • $\begingroup$ for your function $\rho(x)$ you should clarify the boundaries of $[0,\frac{1}{2}]$ and $[\frac{1}{2}, 1]$ because at $x = \frac{1}{2}$ it is in both. It should be $[0 , \frac{1}{2}) $ or $(\frac{1}{2}, 1] $ not both. $\endgroup$ – Shogun Jun 13 at 21:10
  • $\begingroup$ @Shogun, yes, sorry. Let it be $[0;{1\over2})$ $\endgroup$ – Дарья Романова Jun 14 at 6:32

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