Cauchy Problem for inviscid Burgers' equation

Question

Consider the Cauchy Problem of finding $u(x,t)$ such that $$\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}=0,x\in\mathbb{R},t>0$$ $$u(x,0) = u_0(x), x\in\mathbb{R}$$ Which choices of the following functions for $u_{0}$ yield a $C^{1}$ solution $u(x,t)$ for all $x\in\mathbb{R},t>0$

1. $u_{0}(x)=\frac{1}{1+x^{2}}$

2. $u_{0}(x)=x$

3. $u_{0}(x)=1+x^{2}$

4. $u_{0}(x)=1+2x$.

If I use the existence and uniqueness theorem for Cauchy problem i get the corresponding determinant is non zero so all are true according to me. But in answer key only option 2nd and 4th is given. Please help me to solve the problem. Thanks a lot.

Answer 1

In the present case of the inviscid Burgers' equation, the method of characteristics gives the unique solution, for short times and small smooth initial data. However, the method of characteristics fails when reaching the breaking time $t_b$ given by (see e.g. these posts (1) (2)) $$ t_b = -\frac{1}{\min u_0'(x)} \, . $$ At this time, the characteristics cross, and a shock wave (discontinuity of the solution) is generated. Here, we compute the derivatives of $u_0$ and draw the characteristics in $x$-$t$ plane for each case:

1. $\min u_0'(x) = -\frac{3\sqrt{3}}{8}$. A shock occurs at $t_b \approx 1.54$ s.

2. $\min u_0'(x) = 1$. No shock occurs for $t>0$.

3. $\min u_0'(x) = -\infty$. Characteristics intersect at $t_b = 0$.

4. $\min u_0'(x) = 2$. No shock occurs for $t>0$.