I came across this question while studying for an exam:
Consider the equation $$u_t = u u_{xx} + uu_{x} $$ to be solved for $t > 0$, $0 < x < 1$, with $u(0,t) = u(1,t)$, $u_x(0,t) = u_x(1,t)$, and $u(x,0) = \phi(x)$ is smooth and periodic with period 1. What restriction on $\phi(x)$ do you need in order to obtain a convergent finite difference scheme? Given a $\phi$ that satisfies this restriction, give a convergent finite difference scheme for the equation.
I've never encountered a PDE of this form. It resembles Burger's Equation, but has a nonlinear viscosity term. Is there some sort of transformation we can use to simplify the PDE? Any suggestions would be very much appreciated.
Edit: I believe Jeff's answer in this thread explains how I should go about solving this problem. I believe this is correct, but confirmation would be great.
Propose the scheme $$\frac{u_j^{n+1}-u_j^n}{k} = u_{j}^n \frac{u_{j+1}^n-2u_j^n+u_{j-1}^n}{h^2} + u_j^n \frac{u_{j+1}^n-u_{j-1}^n}{2h},$$ which can be rewritten in the form $$ u_j^{n+1} = \left( 1-2 \mu u_j^n \right)u_j^n+ \left(\mu u_j^n + \frac{\lambda u_j^n}{2} \right) u_{j+1}^n + \left( \mu u_j^n - \frac{\lambda u_j^n}{2} \right)u_{j-1}^n,$$ where $\lambda = \frac{k}{h}$, $\mu = \frac{k}{h^2}$. Noting that the coefficients add up to $1$, we want to find conditions on $\mu$, $\lambda$, and $u_j^n$, so that each coefficient is positive, which would give us a monotone method. The second coefficient tells us that $u_j^n \geq 0$, which can be enforced by having $\phi(x) \geq 0$. This is because the PDE satisfies the maximum principle, which is easily checked. The third coefficient tells us that $\mu \geq \lambda/2$, which implies that $h \leq 2$. Finally, the first coefficient tells us \begin{align*} u_j^n &\leq \frac{1}{2 \mu} \\ & \leq \frac{1}{\lambda} \\ &= \frac{h}{k} \\ & \leq \frac{2}{k} \end{align*} Since $u(x,t) \leq \sup_{x \in [0,1]} \phi(x) = M $, this condition will be satisfied for $k$ sufficiently small. Therefore let's impose the condition on $k$ that $M \leq \frac{1}{k}$. Now we just need to verify that the method is in fact monotone. By construction, $\frac{\partial u_{j}^{n+1}}{\partial u_{j-1}^n}$, $\frac{\partial u_j^{n+1}}{\partial u_{j+1}^n} \geq 0$. Finally, \begin{align*} \frac{\partial u_j^{n+1}}{ \partial u_j^n} &= 1 - 4 \mu u_j^n + \left( \mu + \frac{\lambda}{2} \right)u_{j+1}^{n} + \left( \mu - \frac{\lambda}{2} \right) u_{j-1}^{n} \\ &\geq 1 - 4 \mu M \\ &\geq 0, \end{align*} for $\mu$ small enough. Provided that we have $M \leq \max\left\{\frac{1}{4 \mu}, \frac{1}{k}\right\}$, the method should be monotone, thus stable. Consistency follows easily.