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I am trying to solve an example... little stuck in between. Any help would be appreciated.

Consider the quasi-linear partial differential equation $ \frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}=-xu~\\ $ with the initial condition $~u(x,0)=f(x).$

a)Show that $Q=u+x^2/2$ is conserved along characetristics of this equation.

b)Use the method of characteristics to find an implicit (i.e., parametric) solution for u assuming $f(x)>0.$

c) Derive an explicit solution for the special case of $f(x)=-x^2/2$.Show that characteristics starting at $x(0)=\xi<0$ exhibit blow up in finite time. Does this imply blow up in finite time for $u(x,t)$ at a fixed location x? why or why not?

I dont know how to show part (a). for part (b), I solved this way:

$\begin{align} dx/dt&=u => x=ut+\xi\\ du/dt&=-xu =-(ut+\xi)u \end{align}$

I am little stuck here. Please help me..

user142807
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1 Answers1

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Hint:

$\dfrac{\partial u}{\partial t}+u\dfrac{\partial u}{\partial x}=-xu$

$\dfrac{1}{u}\dfrac{\partial u}{\partial t}+\dfrac{\partial u}{\partial x}=-x$

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

$\dfrac{dx}{ds}=1$ , letting $x(0)=0$ , we have $x=s$

$\dfrac{du}{ds}=-x=-s$ , letting $u(0)=\dfrac{u_0^2}{2}$ , we have $u=\dfrac{u_0^2-s^2}{2}=\dfrac{u_0^2-x^2}{2}$

$\dfrac{dt}{ds}=\dfrac{1}{u}=\dfrac{2}{u_0^2-s^2}$ , letting $t(0)=F\left(\dfrac{u_0^2}{2}\right)$ , we have $t=F\left(\dfrac{u_0^2}{2}\right)+\dfrac{2}{u_0}\tanh^{-1}\dfrac{s}{u_0}=F\left(u+\dfrac{x^2}{2}\right)+\dfrac{2}{\sqrt{2u+x^2}}\tanh^{-1}\dfrac{x}{\sqrt{2u+x^2}}=F\left(u+\dfrac{x^2}{2}\right)+\dfrac{2}{\sqrt{2u+x^2}}\sinh^{-1}\dfrac{x}{\sqrt{2u}}$

doraemonpaul
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