Consider the following 1-dimension problem: $$ -a(x)u_{xx}+b(x)u_x+c(x)u=f,0<x<L,u(0)=u(L)=0, $$ where $$ 0<c_1\le a(x)\le c_2. $$ Show that a weak solution to the problem exists if $b$ is bounded and $c\ge 0$.
I want to use the following theorem on Existence of Weak Solutions to Elliptic Equations:
For the equation $Lu=a(x)u_{xx}+b(x)u_x+c(x)u=f$, assume the coefficients of $L$ satisfy $0<c_1\le a(x)\le c_2$ and $b$, $c$ are bounded. Then for each $\varphi\in H^1(U)$ and $f\in L^2(U)$, there exists a unique weak solution $u\in H^1(U)$ of the equation $Lu=f$ in $U$ and $u=\varphi$ on $\partial U$.
So I need to rewrite the original equation such that the coefficient of $u$ is bounded. I tried to put the equation into divergence form by letting $\widetilde{a}=e^{-\int \frac{b(x)}{a(x)}\ dx}$, $\widetilde{c}=\frac{c(x)}{a(x)} e^{-\int \frac{b(x)}{a(x)}\ dx}$ and $\widetilde{f}=\frac{f(x)}{a(x)} e^{-\int \frac{b(x)}{a(x)}\ dx}$, which yields $-(\widetilde{a}u_x)_x+\widetilde{c}u=\widetilde{f}$. But for $c>0$, $\widetilde{c}$ is still not bounded necessarily.
So my question is: Is it possible to put the original equation into a particular form where the coefficient corresponding to $u$ is bounded?