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Recently, I found an exercise in Hunter's Applied Analysis(last page in the link), which may be closely related to the question I raised two months ago.

Consider heat flow in a rod with rapidly varying thermal conductivity $a_n(x)=a(nx)$, where $n\in{\mathbb N}$ and $a(y)$ is a strictly positive periodic function with period one assumed continuous for simplicity. If the ends of the rod are held at an equal fixed temperature, and there is a given heat source $f(x)$ per unit length, the temperature $u_n(x)$ satisfies the boundary value problem $$-\frac{d}{dx}\bigg(a_n(x)\frac{d}{dx}u_n\bigg)=f(x),\quad0<x<1,\quad u_n(0)=u_n(1)=0.$$

  • Integrate this ODE and solve for $u_n(x)$.
  • Let $$H_0^1:=\{u:[0,1]\to{\mathbb R}|u,u'\in L^2([0,1]),u(0)=u(1)=0\},$$ with the inner product $$\langle u,v\rangle:=\int_0^1u'v'dx.$$ Show that $u_n\rightharpoonup u$ weakly in $H_0^1([0,1])$, where $u$ is the solution of the homogenized equation $$-\frac{d}{dx}\bigg(a^h\frac{d}{dx}u\bigg)=f(x),\quad 0<x<1,\quad u(0)=u(1)=0,$$ and the effective conductivity $a^h$ is the harmonic mean of the original conductivity, $$\frac{1}{a^h}=\int_0^1\frac{1}{a(y)}dy.$$

Here are my thoughts:
The first problem can be solve with straightforward calculation. For the second problem, one needs to estimate the absolute value of the integral $$\bigg|\int_0^1(u_n'-u')v'dx\bigg|.$$ With Cauchy-Schwarz inequality, one may want to estimate $$\int_0^1(u_n'-u')^2dx$$ A natural idea is that using the original equation, $$u_n'(x)=\frac{1}{a_n(x)}g(x)$$ and $$u'(x)=\frac{1}{a^h}g(x)$$ It seems that things end up with show the convergence $$\frac{1}{a_n(x)}\to\frac{1}{a^h}\quad \text{in}~L^2([0,1])$$ which is the part that may relate to the question whose link is given at the beginning. I get stuck here. Any idea for how to go on?

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I think you should not use Cauchy-Schwarz because then you would need to prove strong convergence. –  timur Jul 9 '11 at 4:36
    
@timur: Hmm, good point.+1. –  Jack Jul 9 '11 at 4:41
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up vote 1 down vote accepted

With

$$b(x)=\frac1{a(x)}-\frac1{a^h}$$

and $w(x)=g(x)v'(x)$, the problem is now to prove

$$\int_0^1 b(nx)w(x)\mathrm{d}x\to0.$$

The function $b$ is periodic and has zero mean. Then you might want to look up (a proof of) the Riemann-Lebesgue lemma.

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