Non-divergence form to divergence form.

Question

Consider 1-dimensional boundary-value problem $$ \begin{equation} \left\{ \begin{array}{lr} -a(x)u_{xx} + b(x) u_x + c(x) u = f, 0<x<L \\ u(0) = u(L) = 0 \end{array} \right. \end{equation} $$ where $0<c_1 \le a(x) \le c_2$.
Can it be transformed in to the divergence form as follows? $$ \begin{equation} \left\{ \begin{array}{lr} -(\tilde{a}(x)u_{x})_x + \tilde{c}(x) u = \tilde{f}, 0<x<L \\ u(0) = u(L) = 0 \end{array} \right. \end{equation} $$ I don't understand how come the term $u_x$ can vanish.

Answer 1

$u_x$ did not "vanish". Note from the product rule:

$$ -(\bar{a}u_x)_x + \bar{c}u = -(\bar{a}u_{xx} + \bar{a}_xu_x) + \bar{c}u $$

Thus we can choose an integrating factor $\mu(x)$ (refer back to first-order ODEs for derivations) such that

$$ -(\mu a u_{xx} - \mu bu_x) + \mu cu = -(\bar{a}u_{xx} + \bar{a}_xu_x) + \bar{c}u $$

For this to hold, we require $$ (\mu a)_x = -\mu b $$

Solving the first-order ODE for $\mu$ gives $$ \mu_x a + \mu (a_x + b) = 0 $$ $$ \mu = -\int \frac{a_x+b}{a} dx $$