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I'm trying to understand the setup given in Multiscale analytical solutions and homogenization of n-dimensional generalized elliptic equations, pg 41:

Given the boundary value problem:

$$\frac{d}{dx}\left( -K(x)\frac{du}{dx}\right)=0\text{ in } \Omega\equiv(0,1)$$ $$u(0)=0$$ $$u(1)=1$$ where $K(x)$ is a smooth, strictly positive function in $\Omega$.

We can consider the quantity $q^0 \equiv -K(x)\frac{du}{dx}$ to be a constant in $\Omega$, since the equation specifically states that $\frac{d}{dx}\left( q^0\right)=0$. I understand that we can write:

$$\frac{du}{dx}=-\frac{q^0}{K(x)}$$ $$\Rightarrow u=\int_0^x -\frac{q^0}{K(\tau)} d\tau$$.

But the author claims that we can further rewrite this as:

$$u=K^0\int_0^x\frac{d\tau}{K(\tau)}$$ where $$K^0=-q^0=\frac{1}{\int_0^1 \frac{d\tau}{K(\tau)}}$$.

It is not immediately apparent from the given information that we can deduce that $-q^0=\frac{1}{\int_0^1 \frac{d\tau}{K(\tau)}}$. How could I have derived this from the given information?

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up vote 2 down vote accepted

Plug the second boundary condition $u(1)=1$ into your formula $$u=\int_0^x -\frac{q^0}{K(\tau)} d\tau$$

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Oh, I see now... It was staring at me in my face and I didn't even realize it. So, using the derived formula, I can simply apply the condition $u(1)=1$ into the formula and solve for $-q_0$. I can't believe it was that simple! Thanks, PavelM. – Paul Jan 9 '13 at 5:34

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