# Shock formation condition in IVP of $u_t + uu_x + \alpha u = 0$

Consider $$u_t + uu_x + \alpha u = 0$$ for $$t > 0$$, all $$x$$ where $$\alpha > 0$$ is a constant. Find the characteristic equations for the equation with initial data $$u(x, 0) = f(x)$$ given. Show that a shock cannot form if $$\alpha \geq \max_{r \in H}|f'(r)|$$ where $$H = \{r : f'(r) < 0\}$$ or if $$H$$ is empty.

So far, I've found the characteristics by parametrizing $$\begin {cases} x_s=u, x(0,r)=r \\ t_s=1,t(0,r)=0 \\ u_s = -\alpha u, u(0,r)=f(r)\end {cases}$$ Then $$\frac{du}{ds}=-\alpha u \Rightarrow u = C_1 e^{-\alpha s}$$. Considering the initial condition, $$u = f(r)e^{-\alpha s}.$$

$$\frac{dt}{ds} = 1 \Rightarrow t = s$$ (since $$t(0,r)=0$$)

$$\frac{dx}{ds}=u=f(r)e^{-\alpha s} \Rightarrow x = -\frac{1}{\alpha}f(r)e^{-\alpha s}+\frac{1}{\alpha}f(r)+r$$ (since $$x(0,r)=r$$), i.e. $$x = -\frac{1}{\alpha}f(r)e^{-\alpha t}+\frac{1}{\alpha}f(r)+r$$

So how do we show that a shock cannot form?

• What do you know about shocks? When do they form? – Mattos Dec 6 '18 at 3:42

This (dissipative) Burgers' equation with relaxation is a typical example of conditional shock formation. The answer follows the steps in this post. The characteristics are the curves $$x = -f(x_0)\frac{e^{-\alpha t} - 1}{\alpha} + x_0$$ along which the solution satisfies $$u = f\left(x - \frac{e^{\alpha t}-1}{\alpha} u\right)e^{-\alpha t} .$$ Computing $$\frac{\text d x}{\text d x_0}$$, we find that this derivative vanishes at a given positive time $$t$$ -i.e. a shock wave forms- if $$-\frac{\ln(1+\alpha/f'(x_0))}{\alpha} = t >0 .$$ For this to be possible, the logarithm should be negative, and thus, $$f'(x_0)<0$$ (in other words, $$x_0\in H$$). However, if $$-\alpha < f'(x_0) < 0$$ for $$x_0$$ in $$H$$, the logarithm becomes complex and no shock occurs. Hence the conclusion: a shock cannot occur if $$\max_{x_0 \in H}|f'(x_0)| \leq \alpha$$ or if $$H = \emptyset$$. Alternatively, this condition may be written $$\inf_{x_0 \in\Bbb R} f'(x_0) \geqslant -\alpha$$.

I am not standing up here to give another answer. My intervention is a comment, but too long to be edited in the comments section.

The analytical solving of the PDE with the specified boundary condition is especially interesting as shown below. $$u_t+uu_x=-\alpha u \tag 1$$ The Charpit-Lagrange equations are : $$\frac{dt}{1}=\frac{dx}{u}=\frac{du}{(-\alpha u)}$$ A first family of characteristic curves comes from $$\frac{dx}{u}=\frac{du}{(-\alpha u)}$$ : $$u+\alpha x =c_1$$ A second family of characteristic curves comes from $$\frac{dt}{1}=\frac{du}{(-\alpha u)}$$ : $$ue^{-\alpha t}=c_2$$ The general solution of the PDE Eq.$$(1)$$ expressed on the form of implicit equation is : $$u+\alpha x=\Phi\left(ue^{-\alpha t}\right) \tag 2$$ where $$\Phi$$ is an arbitrary function (to be determined according to boundary condition).

Condition : $$u(x,0)=f(x)$$ with $$f(x)$$ a known (given) function.

$$f(x)+\alpha x=\Phi\left(f(x)e^{0}\right)=\Phi\left(f(x)\right)$$ Let $$X=f(x)$$ and $$x=f^{-1}(X)$$

$$f^{-1}$$ denotes the inverse function of $$f$$. $$\Phi(X)=X+\alpha f^{-1}(X)$$ So, the function $$\Phi$$ is determined. We put it into Eq.$$(2)$$. $$u+\alpha x=ue^{-\alpha t}+\alpha f^{-1}\left(ue^{-\alpha t}\right)$$ $$f^{-1}\left(ue^{-\alpha t}\right)=x+\frac{u}{\alpha}\left(1-e^{-\alpha t}\right)$$ $$ue^{-\alpha t}=f\left(x+\frac{u}{\alpha}\left(1-e^{-\alpha t}\right)\right) \tag 3$$ Eq.$$(3)$$ is the implicite form of the analytic solution of $$u_t+uu_x+\alpha u=0$$ with condition $$u(x,0)=f(x)$$.

The explicite form of $$u(x,t)$$ requires to solve Eq.$$(3)$$ for $$u$$. The possibility to do it analytically depends on the kind of function $$f$$.

Of course, this is only for information without answering to the OP question as mentioned at first place.