Determine solutions of the Jacobi-Hamilton problem $u_{t}+|u_x|^{2}=0$

Question

How determine solutions of the initial value problem, $$u_{t}+|u_x|^{2}=0\qquad \mbox{in } \mathbb{R}\times(0,\infty)$$

With condition $u=0$ on $\mathbb{R}\times\{t=0\}$. Clearly one solution is $u(x,t)=0$ (as in the answers), but how determine another solution? My teacher say that there exits the following lipschitz continuous solution a.e.

$ u^{*}(x,t):= \begin{cases} 0&\text{if}\, |x|\geq t\\ |x|-t&\text{if}\, |x|\leq t\\ \end{cases} $

But the true is, I don't know how obtain this lipschitz continuous solution, that solves the pde a.e.

So, How I determine the solution $u^{*}$?

Thanks!

Answer 1

For first thing you can search solutions of the type: $u(x,t)= \alpha(x)+\beta(t)$ In this case the equation is

$\beta_t(t)+\alpha^2_x(x)=0$

and so

$\beta_t(t)=-\alpha^2_x(x)$

In particular you can choose, for example, x=0 and you have that

$\beta_t(t)=-\alpha^2_x(0)=cost$

and so

$\beta(t)=at+b$ with a$,b\in\mathbb{R}$

Now you must resolve

$\alpha^2_x(x)=-a$

You can observe that $a\leq 0$ and in this case $\alpha_x(x)=\sqrt(-a)$

And the solution is $\alpha(x)=\sqrt(-a) x+c$

Now for every $x$ you have that $u(x,0)=0$ and you have that $\sqrt(-a) x+c=b $ For x=0 you have that $c=b$ and so for $x\neq0$ you have $ \sqrt(-a)=0$ and so there your solution is $u(x,t)=0$. In other words the unique solution of the form $u(x,t)= \alpha(x)+\beta(t)$ is the constant function zero. You can prove in general that the constant function zero is the unique solution of your problem.

Answer 2

Differentiate the equation w.r.t $x$ and substitute $v = u_x$ to get a first-order problem

$$ v_t + 2vv_x = 0 $$

This can be solved using the method of characteristics to get an implicit solution

$$ v = f(x - 2vt) $$

The initial condition gives

$$ v(x,0) = u_x(x,0) = f(x) = 0 $$

Therefore $u_x\equiv 0$, which implies $u\equiv 0$ is the only solution

Answer 3

The function $u^{\ast}:=\min\left(\left|x\right|-t,0\right)$ is differentiable whenever $\left|x\right|\not=t$ and $x\not=0$. Hence it is differentiable almost everywhere and we have

$$ \partial_{t}u^{\ast}(x,t)=\begin{cases} 0, & \left|x\right|>t\\ -1, & \left|x\right|<t \end{cases},\,\partial_{x}u^{\ast}(x,t)=\begin{cases} 0, & \left|x\right|>t\\ -1, & t<x<0\\ 1, & 0>x>t \end{cases}. $$

These satisfy $\partial_{t}u^{\ast}(x,t) +\left|\partial_{x}u^{\ast}\right(x,t)|^{2}=0$ where they are defined.