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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?

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    Please learn to $\LaTeX$ format your quesitons. There are tutorials on the web and you can see [this](http://meta.math.stackexchange.com/questions/1773/do-we-have-an-equation-editing-howto). 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}?$2012-10-14
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    Given that the method of characteristics gives $\frac{dy}{dt} = u^2$, I think she means $u^2u_x$.2012-10-14
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    @Michael, that is correct2012-10-14
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    @mary: You say "but not sure if this will help." What happens if you continue to use the method of characteristics?2012-10-14
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    There becomes a mess in solving the differential equations, i.e., renders useless to the problem2012-10-14
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    Why don't you take a look at Witham's _Linear and Nonlinear Waves_?2012-10-16
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    The characteristic equations you have don't look correct.2012-10-16
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    Can you please fix it above?2012-10-16
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    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.2012-10-16
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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.2012-10-16
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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.2012-10-16
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    I'm using Strauss's PDE. I will take a look at that2012-10-21

1 Answers 1

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

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

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

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    Thank you so much for your explanation.2012-10-16
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    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)} $ ?2018-10-23
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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 $\xi = \frac{-(1 + 4t) \pm \sqrt{1 + 4t(2 + x)}}{2 t}$.2018-10-23
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    Now I see it. Thank you very much.2018-10-23