Vrftsjtryk

I'm trying to solve the following partial differential equations:
$$
u_t + frac{1}{3}{u_x}^3 = 0 tag{a}
$$
$$
u_t + frac{1}{3}{u_x}^3 = -cu tag{b}
$$
with the initial value problem
$$
u(x,0)=h(x)=
leftlbrace
begin{aligned}
&e^{x}-1 & &text{for}quad x<0\
&e^{-x}-1 & &text{for}quad x>0
end{aligned}
right.
$$
My idea was to set $v(x,t)=u_x(x,t)$, because then I get
the transport equation in $v$ which I am able to solve:
$v_t + v^2v_x =0$.
But when I do this, my solution for $v$ is
$$
v(x,t)=
leftlbrace
begin{aligned}
&phantom{-}ae^{x-v^2 t} & &text{for}quad x<0\
&{-a}e^{-x+v^2 t} & &text{for}quad x>0
end{aligned}
right.
$$
Can someone help me with this equation? Is
$u_x = a e^{x-u_x^2 t}$ the correct answer?
Or should I maybe do something very different to solve this equation?

Both equations (a) and (b) are Hamilton-Jacobi equations. Indeed, they are derived from a canonical transformation involving a type-2 generating function $u(x,t)$ which makes vanish the new Hamiltonian
$$
K = H(x,u_x) + (partial_t + c), u , .
$$
Here, $H(q,p)=frac{1}{3}p^3$ is the original Hamiltonian and $partial_t + c$ defines the time-differentiation operator. Setting $v=u_x$, we have
$$
v_t = u_{tx} = -left(tfrac{1}{3}{u_x}^3right)_x - cu_x = -v^2v_{x} - cv , .
$$
Thus, we consider the first-order quasilinear PDE $v_t + v^2v_{x} = -cv$ with initial data $v(x,0) = h'(x) = pm e^{pm x}$ for $pm x<0$,
and we apply the method of characteristics:

• 
  $frac{text d t}{text d s} = 1$, letting $t(0)=0$, we know $t=s$.


• 
  $frac{text d v}{text d s} = -cv$, letting $v(0)=h'(x_0)$, we know $v=h'(x_0)e^{-ct}$.


• 
  $frac{text d x}{text d s} = v^2$, letting $x(0)=x_0$, we know $x=frac{1}{2c}h'(x_0)^2(1-e^{-2ct}) + x_0$.

Injecting $h'(x_0) = ve^{ct}$ in the equation of characteristics, one obtains the implicit equation
$$
v = h'!left(x-v^2frac{e^{2ct}-1}{2c}right) e^{-ct}, .
$$
For short times, the previous equation is valid. The method of characteristics breaks down when characteristics intersect (breaking time). We use the fact that $frac{text d x}{text d x_0}$ vanishes at the breaking time
$$
t_B = inf_{x_0in Bbb R} frac{-1}{2 c} lnleft(1 + frac{c}{h'(x_0)h''(x_0)}right) .
$$

If $c=0$, the characteristics are straight lines $x=x_0+h'(x_0)^2t$ along which $v=h'(x_0)$ is constant. A sketch of the $x$-$t$ plane is displayed below:

The breaking time becomes $t_B = inf_{x_0} -(2h'(x_0)h''(x_0))^{-1} = 1/2$.
For short times $t<t_B$, the implicit equation for $v$ reads
$v = h'(x-v^2t)$, i.e. $v = pm e^{pm (x-v^2t)}$ if $pm(x-v^2t)<0$. Its analytic solution
$$
v(x,t) = pmexp! left(pm x- tfrac{1}{2}W(pm 2t e^{pm 2x})right) quadtext{for}quad {pm (}x-t) < 0
$$
involves the Lambert W function, so that $u = int v,text d x$ is hard to express in terms of elementary functions. For larger times $t>t_B$, particular care should be taken when computing weak solutions (shock waves) since the flux $f:vmapsto frac{1}{3}v^3$ is nonconvex.