# Non-homogeneous Burgers' equation $u_t + u u_x = -\sin x$

While studying for an exam, I found this question from a previous exam:

Consider the forced Burgers' equation for $u(x,t)$ on the periodic domain $x \in [-\pi, \pi]$. $$u_t + uu_x = -\sin(x)$$ (a) Show that the equation conserves total mass $M = \int_{-\pi}^{\pi} u(x,t)\,dx$.

(b) Find explicitly the smooth steady solutions. Show that for these solutions $|M| > C$, and find the constant $C$. Show that the limiting solution when $|M| = C$ has a "corner" at $x = \pi$.

So part (a) was pretty straight forward after integrating with respect to $x$ from $[-\pi,\pi]$. I'm stuck on part (b). Maybe I'm not sure what they mean by steady state solution? Does it mean that $u_t = 0$? Otherwise, I've tried:

Let $z(t) = u(x(t),t)$, so $z'(t) = x'(t)u_x + u_t = -\sin(x)$ provided that $x'(t) = u(x(t),t)$. So I have a system:

$$z'(t) = -\sin(x(t))$$ $$x'(t) = u(x(t),t) = z(t)$$

From these we can conclude:

$$x''(t) + \sin(x(t)) = 0$$

Which is an ODE you get when studying pendulums, and possibly a bit to complicated for this problem. The corresponding ODE for $z$ was even worse.

steady state solution? Does it mean that $u_t=0$?
Yes, this is exactly what it means. The equation simplifies to $(u^2)_x=-2\sin x$, which integrates to $u=\pm \sqrt{2\cos x+B}$ for some constant $B$. For this to make sense, we need $B\ge 2$. This leads to a lower bound on $\int |u|$, attained when $B=2$.
Show that the limiting solution when $|M|=C$ has a "corner" at $x=\pi$.
Indeed, because when $B=2$, the expression under the square root drops to $0$ at $\pm \pi$. The solution in this case is $\sqrt{2(1+\cos x)}=4\cos (x/2)$, $|x|\le \pi$, if I got trigonometry right.