I couldn't solve this problem, can you help me please?

The Burgers' equation

$$ u_y + uu_x = 0 $$ $ - \infty < x < \infty $ , $ y > 0 $ , $ u(x,0)=f(x) $

My question;

is there any solution $ f(x)= -x $ for all $ y>0 $ , if not, why?

Thank you for your help.

  • $\begingroup$ You can use the method of characteristics to solve it. $\endgroup$ Jul 29, 2013 at 12:11
  • $\begingroup$ Hi Dear, I tried but it doesn't work. $\endgroup$
    – John
    Jul 29, 2013 at 12:34
  • $\begingroup$ See here. $\endgroup$ Jul 29, 2013 at 13:18

3 Answers 3


I rewrite equation to common view

$$\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}=0$$ There is a solution with each initial value, but this solution exists only for $t\le t_{r}$, where $t_{r}$ is a point where the phenomenon of "rollover" occurs. This phenomenon occurs if some characteristics of this equations intersect. Here is a picture of characteristics: enter image description here

It shows that characteristics of this equations intersect on the moment $t=1$

Before the moment $t=1$ the solution is:

$$u(x,t)=f(x-ut),$$ where $$f(x)=-x $$

After the point $t=1$ the solution does not exist. There you should change your equation to conservation law:

$$\frac{\partial u}{\partial t}+\frac{\partial}{\partial x}\left(\frac{u^2}{2}\right)=0$$

and define the order of this law.

  • $\begingroup$ Here is a mistake. Wait for 1 minute.. $\endgroup$
    – cool
    Jul 29, 2013 at 14:01
  • $\begingroup$ Please,look at the answer,@John $\endgroup$
    – cool
    Jul 29, 2013 at 14:11
  • $\begingroup$ Yes,it's correct. This picture shows that there is no solution for $t>1$, but previous showed that there was, that was incorrect. $\endgroup$
    – cool
    Jul 29, 2013 at 14:19

As discussed in the other answers and this post, the method of characteristics yields the classical solution $u = -(x-uy)$, i.e. $$ u(x,y) = \frac{-x}{1-y} \, . $$ Obviously, the denominator vanishes when $y$ reaches unity: the classical solution breaks down as $y\to 1$. This is reflected by the computation of the breaking time $$ y_b = \inf \frac{-1}{(-\text{id})'(x)} = 1 \, . $$ There, all the characteristic curves intersect simultaneously (cf. picture in answer by @cool), i.e. the solution $u$ takes simultaneously all the values from $−∞$ to $+∞$. It literally blows up! There is no weak solution (e.g. of shock-wave type) to start after this blow-up happened.


Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dy}{dt}=1$ , letting $y(0)=0$ , we have $y=t$

$\dfrac{du}{dt}=0$ , letting $u(0)=u_0$ , we have $u=u_0$

$\dfrac{dx}{dt}=u=u_0$ , letting $x(0)=f(u_0)$ , we have $x=u_0t+f(u_0)=uy+f(u)$ , i.e. $u=F(x-uy)$

$u(x,0)=-x$ :


$\therefore u=-(x-uy)=-x+uy$

Since $u(x,y)$ has perfect explicit form $u(x,y)=\dfrac{x}{y-1}$ , of course it has solution for all $y>0$ .

  • 1
    $\begingroup$ I consider this solution be wrong (partially). The picture higher shows the real situation with characteristics -- they intersect at $y=1$, beginning from that point solution does not exist. $\endgroup$
    – cool
    Jul 30, 2013 at 13:09

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