Classical solution of $u_t + (u^4)_x = 0$ (characteristics)

Question

Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be an odd, continuous, increasing function on real line. Let us consider an equation: $$u_t + (u^4)_x = 0, \qquad (t,x) \in \mathbb{R}_{+} \times \mathbb{R}$$ $$u(0,x) = g(x), \qquad x \in \mathbb{R} .$$

a) Show that there exists classic solution on $\mathbb{R}_{+} \times \mathbb{R}$,

b) Find $\lim_{t \rightarrow \infty} u(t,x)$ for fixed $x \in \mathbb{R}$,

c) Show that for every $t>0$ there exist $\lim_{x \rightarrow +\infty}u(t,x)$ and $\lim_{x \rightarrow -\infty}u(t,x)$.

I was trying to do part a) by characteristics method, and here is what I managed to do:

Let $t=t(s)$, $x=x(s)$, $p(s)= \nabla u(x(s))$, $z(s) = u(x(s))$.

Let us consider a function:

$F(x, z, p) = p_1(s) + 4z^3(s)\cdot p_2(s)$.

We have $F_p(x,z,p) = [1, 4z^3(s)]$, so:

$t_s = 1$

$x_s = 4z^3(s)$

$z_s = [p_1, p_2] \cdot F_p = p_1 + p_2 \cdot 4z^3(s) = 0$.

And we have boundary conditions: $t(0) = 0$, $x(0) = \alpha$, $z(0) = g(\alpha)$. After solving this system of ordinary differential equations, we obtain:

$t(s) = s$,

$z(s) = g(\alpha)$,

$x(s) = 4g^3(\alpha)\cdot s + \alpha$.

Now we want to express $(s, \alpha)$ in terms of $(t,x)$. And here comes the problem. Expressing $s$ in terms of $t$ is trivial, because we have $t(s) = s$. But when we want to express $\alpha$, then we have an equation for $x(s)$, where both $\alpha$ and $g(\alpha)$ appear. Can we obtain a formula for this solution? Or is it the end of the proof, because we know, that there exists $g^{-1}$ and it is possible to obtain a formula for alpha, although not directly?

Answer 1

a) Rewriting the PDE in quasi-linear form, $u_t + 4u^3 u_x = 0$, the solution obtained by the method of characteristics reads implicitly as $$ u = g(x - 4 u^3 t)\, . $$ The derivation in OP looks correct, since it leads to the same implicit formula. Following one of the methods in this post, we can show that the solution becomes singular unless $s\mapsto 4g(s)^3$ is a non-decreasing function of $s$. This property is true, as a consequence of the properties of $g$ and of the cubic polynomial function. Thus, the classical solution is defined for all $t\geq 0$.

b) Taking the limit as $t\to+\infty$ in the implicit formula yields $$\lim_{t\to+\infty} u(t,x) = 0\qquad\text{at}\qquad x =0 \, . $$ For other abscissas, I have no clue how to prove it, but I'd say $u\to 0$ too.

c) If $g$ is bounded, taking the limit in the implicit formula yields the result, and the values of $u$ at $x\to\pm\infty$ are the limits of $g$. If $g$ is unbounded, I've no clue how to prove the result.