# Finding a sequence of functions which Г-converges to a given smooth function in $L^2([0;1])$

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 $$u = {{b-x}\over{b-a}}$$.

• 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. – Shogun Jun 13 at 21:10
• @Shogun, yes, sorry. Let it be $[0;{1\over2})$ – Дарья Романова Jun 14 at 6:32