hyperbolic-equationspartial differential equations
Solve Burgers' equation $$u_t + uu_x =0,$$
with $u=u(x,t)$ and the side condition $u(x,0)=x$.
I am not sure on how to find the solution $u(x,t)$. I have learned the method of characteristics. I am neither sure on how to use the side condition in Burgers' equation.
Here is a sketch of the characteristic curves in the $x$-$t$ plane, which equation is given by $$ \begin{aligned} x'(t) &= u(x(t),t) \\ &= u(x(0),0) \end{aligned} $$ Those curves intersect already at time zero:
Thus, a shock wave arises.
The Burgers' equation is rewritten in the conservative form $u_t + f(u)_x = 0$, where $f(u) = \frac{1}{2}u^2$. The shock wave with speed $s$, left state $u_L = 1$ and right state $u_R = 0$ writes $$ u(x,t) = \left\lbrace \begin{aligned} &1 &&\text{if}\quad x<st \\ &0 &&\text{if}\quad st<x \, . \end{aligned} \right. $$ The speed of shock must satisfy the Rankine-Hugoniot condition $s = \frac{f(u_R) - f(u_L)}{u_R - u_L}$, i.e. $s = \frac{1}{2}$. Here is a modified sketch of the $x$-$t$ plane, which accounts for the shock:
The initial condition is discontinuous, so the shock is there from the very beginning, and if $u_l>u_r$ it remains.
I will also note that it's not true that the border between area where $u = u_l$ and area where $u=u_r$ is given by equation $x-t=0$. If $u_l>u_r$ the equation of the discontinuity line is given by $x = \frac{u_l+u_r}{2}t $ (which can be derived from Rankine-Hugoniot condition), and the solution is $$ u(x,t) =\left\{\begin{array}{ll} u_l & \text{for } x< \frac{u_l+u_r}{2}t\\ u_r & \text{for } x >\frac{u_l+u_r}{2}t\end{array}\right.$$ If $u_l<u_r$, then there's no shockwave, but rarefaction appears instead: $$ u(x,t) =\left\{\begin{array}{ll} u_l & \text{for } x \le u_l t\\ \frac{x}{t} & \text{for } u_lt \le x \le u_rt \\u_r & \text{for } x \ge u_r t\end{array}\right.$$
Best Answer
This PDE is called the (inviscid) Burgers' equation. The characteristic equations are the system of ODEs $${dx\over dt}=u, \quad {du\over dt}=0,$$ which has solution $$u=C_1, \quad x=ut+C_2,$$ where $C_1,C_2$ are constants. But $C_1$ is a function of $C_2$, so $$u=C_1\implies u=C_1(C_2)=C_1(x-ut).$$ To determine $C_1$, note that the side condition requires $$u(x,0)=C_1(x-u(x,0)\cdot 0)=x\implies C_1(x)=x.$$ Thus, $u=x-ut$ and we conclude $$u(x,t)={x\over 1+t}.$$