IVP for nonlinear PDE $u_t + \frac{1}{3}{u_x}^3 = -cu$

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)= \left\lbrace \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)= \left\lbrace \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 $$v_t = u_{tx} = -\left(\tfrac{1}{3}{u_x}^3\right)_x - cu_x = -v^2v_{x} - cv \, .$$ 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 $$v = h'\!\left(x-v^2\frac{e^{2ct}-1}{2c}\right) e^{-ct}\, .$$ Along the same characteristic curves, we have

• $$\frac{\text d u}{\text d s} = \tfrac23 v^3 - c u$$, letting $$u(0) = h(x_0)$$, we know $$u = h(x_0) e^{-ct} + \frac23\! \int_0^t e^{-c(t-s)} v(s)^3 \text d s$$.

Thus, we get $$u = \left(h\!\left(x-v^2\frac{e^{2ct}-1}{2c}\right) + h'\!\left(x-v^2\frac{e^{2ct}-1}{2c}\right)^3 \frac{1-e^{-2ct}}{3c} \right) e^{-ct} \, ,$$ where the link between $$x_0$$ and $$v$$ along characteristics has been used. For short times, the previous solution 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 $$t_B = \inf_{x_0\in \Bbb R} \frac{-1}{2 c} \ln\left(1 + \frac{c}{h'(x_0)h''(x_0)}\right) .$$ However, it seems pointless to look further for full analytical expressions in the general case.

If $$c=0$$, the characteristics are straight lines $$x=x_0+v^2t$$ along which $$v=h'(x_0)$$ is constant, and along which $$u = h(x_0) + \frac23 v^3 t$$. 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 $$t, the implicit equation for $$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) = \pm\exp\! \left(\pm x- \tfrac{1}{2}W(\pm 2t e^{\pm 2x})\right) \quad\text{for}\quad {\pm (}x-t) < 0$$ involves the Lambert W function. The expression of $$u$$ is deduced from $$u = h(x-v^2 t) + \frac23 v^3 t$$. For larger times $$t>t_B$$, particular care should be taken when computing weak solutions (shock waves) since the flux $$f:v\mapsto \frac{1}{3}v^3$$ is nonconvex.

• Thank you a lot for your answer! So with this method, you get a formula for $v$, but not for $u$? But if I want to compute u, for $c=0$ I could calculate $u$ with the formula for the Hamilton-Jacobi equation. Is there any way to compute $u$ directly if $c \neq 0$? – Infinite_28 Oct 25 '18 at 21:54
• Okay, thanks. So there is no theory how to solve a non homogeneous Hamilton-Jacobi equation and get an explicit formula of the function, is there? – Infinite_28 Oct 26 '18 at 16:46
• @Infinite_28 Maybe you were hoping that there is a Lax-Hopf formula for this non-homogeneous HJE... I don't know if such a thing exists, but at least we can find $v=u_x$ via the method of characteristics. – Harry49 Dec 11 '18 at 21:53