# Quasi-linear PDE with Cauchy conditions:

Am attempting to solve the partial differential equation given by $$u_t + uu_x = 0$$ for some wave function $u(x, t)$, subject to the conditions $$u(x, 0) = \phi(x) = 1-x^2$$ I started by using the method of characteristics, and hence obtaining the system $$dt = \frac{dx}{u} = \frac{du}{0}$$

And I can hence solve a system of ordinary differential equations given by $$\frac{du}{dt} = 0$$ and $$\frac{dx}{dt} = u$$ If the first ode tells us that $u$ is constant in $t$, then we can integrate the second ode such that $$x = ut + c$$ which, considering a characteristic curve emerging from the $x$-axis at $(\xi, 0)$ we can then get $$x = ut + \xi$$ So now I figure I need to use those Cauchy conditions to solve the equation. If it's true that $$u(x, t) = \phi(\xi)$$ then I get that $$x = (1-\xi^2)t + \xi$$

I am now slightly unclear on the best method to proceed. Factoring the final equation I have written for $\xi$? If I do that, however, I will be still left with a $\xi$ in the remaining expression, by virtue of the presence of the $\xi^2$ term.

Any thoughts are much appreciated.

$$\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}=0$$ $du=\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial t}dt$

Change of variable : $x=f(u,t)$

$dx=\frac{\partial f}{\partial u}du+\frac{\partial f}{\partial t}dt=\frac{\partial f}{\partial u}\left(\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial t}dt \right)+\frac{\partial f}{\partial t}dt$

$\begin{cases} 1=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}\\ 0=\frac{\partial f}{\partial u}\frac{\partial u}{\partial t}+\frac{\partial f}{\partial t} \end{cases} \rightarrow \begin{cases} \frac{\partial u}{\partial x}=\frac{1}{\frac{\partial f}{\partial u}}\\ \frac{\partial u}{\partial t}=-\frac{\frac{\partial f}{\partial t}}{\frac{\partial f}{\partial u}} \end{cases} \rightarrow \frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}=-\frac{\frac{\partial f}{\partial t}}{\frac{\partial f}{\partial u}}+u \frac{1}{\frac{\partial f}{\partial u}}=0$

$\frac{\partial f}{\partial t}=u \rightarrow f=ut+F(u)$ any derivable function $F$

$f(u,t)=x=ut+F(u)\rightarrow F(u)=x-ut$

or, with any derivable function $G$ : $$u=G(x-ut)$$ Condition $u(x,0)=\phi(x) \rightarrow G(x)=\phi(x)=1-x^2$

$G(x-ut)=1-(x-ut)^2$ $$u=1-(x-ut)^2$$ $t^2u^2+(1-2tx)u-1+x^2=0$ $$u=\frac{1}{2t^2}\left(2tx-1\pm\sqrt{1-4tx+4t^2} \right)$$

This is consistent with the method of characteristics :

with your notations : $$u(x,t)=\phi(\xi)=\phi(x-ut)=1-(x-ut)^2$$

• Comprehensive! I really appreciate the time that went into this, thank you. Jun 3, 2015 at 10:28