# shock waves characteristics

I'm trying to solve $u_t + u^2u_x = 0$ with $u(x, 0) = 2 + x$.

I'm thinking to proceed by characteristics where we have above that $\frac{dx}{dt} = 1$ and $dy/dt = u^2$, but not sure if this will help. This is from shock waves idea.

Here's what I have

$u_t + u^2 u_x = 0$

$q +u^2 p =0$

$u^2 p +q =0$

$\frac{dx}{u^2} = \frac{dy}{1}$

and $\frac{u^{-1}}{-1} + C = y$

then $u(x,t)^{-1} + C = y$

is this correct?

• Please learn to $\LaTeX$ format your quesitons. There are tutorials on the web and you can see this. Presumably ut is $u_t=\frac {\partial u}{\partial t}$ but I am not sure how to parse u2ux. Is it $\frac {\partial^2 u}{\partial x^2}?$ Oct 14, 2012 at 0:19
• Given that the method of characteristics gives $\frac{dy}{dt} = u^2$, I think she means $u^2u_x$. Oct 14, 2012 at 0:24
• @Michael, that is correct
– mary
Oct 14, 2012 at 1:02
• As timur comments, the characteristic equations are wrong. They should be stated against a parameter, not the involved variables. You are mistaking $t$ with $y$. The construction of the characteristics is based on the supposition that if $x = x(\eta)$ and $t = t(\eta)$, then $$\frac{d}{d\eta}u\big(x(\eta),t(\eta)\big) = u_x x'(\eta) + u_t t'(\eta) = u^2 u_x + u_t = 0$$ and then one says $x'(\eta) = u^2$, $t'(\eta) = 1$, $u'(\eta) = 0$. See my answer for a full analysis. Oct 16, 2012 at 8:31
• Also, if I understand your work, the method you are using is for solving fully nonlinear first order PDE's, and you are using it wrong. Your problem is quasilinear, and there is no need to introduce $p$ and $q$. This are only introduced in the case the derivatives of $u$ are involved nonlinearly in the equation. I strongly suggest you to study the first chapter of John's Partial Differential Equations, as I believe you are very confused. Any doubts, we can try to help. Oct 16, 2012 at 8:40

The quasilinear first order PDE $$a\big(x,y,u(x,y)\big) u_x(x,y) + b\big(x,y,u(x,y)\big)u_y(x,y) = c\big(x,y,u(x,y)\big)$$ where $a,\,b,\,c \in C^1$ with data $\mathcal{C}(\xi) = \big(x(\xi), y(\xi), u(\xi)\big) \in C^1$ and with $$\begin{vmatrix} \frac{dx}{d\xi} & a \\ \frac{dy}{d\xi} & b\end{vmatrix} \neq 0$$ has a unique solution near $\mathcal{C}$ given by \begin{align} \frac{d x}{d \eta} &= a & x\big|_{\eta = 0}&= x(\xi)\\ \frac{d y}{d \eta} &= b & y\big|_{\eta = 0}&= y(\xi)\\ \frac{d u}{d \eta} &= c & u\big|_{\eta = 0}&= u(\xi)\\ \end{align}

For proof and geometrical interpretation, see F. John's Partial Differential Equations1.4)

In your case, $\mathcal{C}(\xi) = \big(\xi,0,\xi+2\big)$. Near $\eta \sim 0$ $$\begin{vmatrix} \frac{dx}{d\xi} & a \\ \frac{dy}{d\xi} & b\end{vmatrix} = \begin{vmatrix} 1 & u^2 \\ 0 & 1\end{vmatrix} = 1$$ and the solution is unique.

The system of ODE's is \begin{align} \frac{d x}{d \eta} &= u^2 & x\big|_{\eta = 0}&= \xi\\ \frac{d t}{d \eta} &= 1 & t\big|_{\eta = 0}&= 0\\ \frac{d u}{d \eta} &= 0 & u\big|_{\eta = 0}&= \xi + 2\\ \end{align} with solution $$t = \eta, \quad u = \xi + 2, \quad x = (\xi + 2)^2 \eta + \xi.$$

The characteristics are $t = \frac{x - \xi}{(\xi + 2)^2}$ hence $\xi = -2$ is a special point. As $\xi \rightarrow \infty$, $t \rightarrow 0$. As $\xi \rightarrow -\infty$, $t \rightarrow 0$. As $\xi \rightarrow -2$, $t \rightarrow \infty$.

This of course, means that there is no solution when the characteristics meet.

A simple explanation for this is that the transformation $$(x,t) \rightarrow (\xi,\eta)$$ is invertible iff $$\begin{vmatrix} \partial_\xi x & \partial_\eta x \\ \partial_\xi t & \partial_\eta t \end{vmatrix} = 1 + 4\eta + 2\xi \eta \neq 0$$ meaning there is no solution when $\xi = -\frac{1 + 4 \eta}{2\eta}$ or, inverting the transformation, when $$t = - \frac{1}{4(x+2)}$$

$\hskip.75in$

Lastly, inverting for $\xi$

$$\xi = \frac{-(1 + 4t) \pm \sqrt{1 + 4t(2 + x)}}{2 t}$$

and

$$u(x,t) = \frac{-1 \pm \sqrt{1 + 4t(2 + x)}}{2 t}.$$

In order to determine the correct sign, we must look at the initial condition. For the minus sign $\lim_{t \rightarrow 0} u(x,t) = -\infty$, while the plus sign gives the correct answer. Hence

$$u(x,t) = \frac{-1 + \sqrt{1 + 4t(2 + x)}}{2 t}.$$

• Thank you so much for your explanation.
– mary
Oct 16, 2012 at 9:55
• Your explanation helped me very much. Could you please explain how you inverted the transformation? How can you see that $t = - \frac{1}{4(x+2)}$ ? Oct 23, 2018 at 20:37
• @Infinite_28 Well, the transformation isn't invertible when $$\tfrac{D(x,t)}{D(\xi,\eta)} = 0,$$ i.e., when $1 + 2\eta(2+\xi) = 0$. The result follows given that $\eta = t$ and $\xi = \frac{-(1 + 4t) \pm \sqrt{1 + 4t(2 + x)}}{2 t}$. Oct 23, 2018 at 21:16
• Now I see it. Thank you very much. Oct 23, 2018 at 21:48