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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?

DK2AX
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mary
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    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 $\frac {\partial^2 u}{\partial x^2}?$ – Ross Millikan Oct 14 '12 at 00:19
  • Given that the method of characteristics gives $\frac{dy}{dt} = u^2$, I think she means $u^2u_x$. – Michael Albanese Oct 14 '12 at 00:24
  • @Michael, that is correct – mary Oct 14 '12 at 01:02
  • @mary: You say "but not sure if this will help." What happens if you continue to use the method of characteristics? – Michael Albanese Oct 14 '12 at 01:09
  • There becomes a mess in solving the differential equations, i.e., renders useless to the problem – mary Oct 14 '12 at 01:25
  • Why don't you take a look at Witham's Linear and Nonlinear Waves? – Pragabhava Oct 16 '12 at 01:38
  • The characteristic equations you have don't look correct. – timur Oct 16 '12 at 03:10
  • Can you please fix it above? – mary Oct 16 '12 at 05:39
  • I think you're more likely to get people interested in your question by following Ross' advice and link and making the question legible than by setting a bounty on it. – joriki Oct 16 '12 at 07:34
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    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. – Pragabhava Oct 16 '12 at 08:31
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    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. – Pragabhava Oct 16 '12 at 08:40
  • I'm using Strauss's PDE. I will take a look at that – mary Oct 21 '12 at 05:26

1 Answers1

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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 Equations (§1.4)

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 $\xi \rightarrow -2$, $t \rightarrow \infty$.

Characteristic curves

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

$\hskip.75in$Existence

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

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

Pragabhava
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  • Thank you so much for your explanation. – mary Oct 16 '12 at 09:55
  • 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)} $ ? – Infinite_28 Oct 23 '18 at 20:37
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    @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 $\xi = \frac{-(1 + 4t) \pm \sqrt{1 + 4t(2 + x)}}{2 t}$. – Pragabhava Oct 23 '18 at 21:16
  • Now I see it. Thank you very much. – Infinite_28 Oct 23 '18 at 21:48