# 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.