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Solve the following lst order $pde$ using the method of characteristics:

$ u_t + 3t^2 u_x = -u$ With $ u(x,0) = \begin{cases} e^{-1} & \quad x<0\\ e^{x-1} & \quad 0<x<1\\ e^{-(x-1)} & \quad x>1\\ \end{cases} $

I found $u(x,t)= f(x_•)e^{-t}$ But i get stuck with inatial condition .

Thanks for help.

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  • $\begingroup$ Notice that $f(x_•) = u(x,0)$ $\endgroup$
    – Sara
    Jun 5, 2015 at 10:41
  • $\begingroup$ How i can write the solution as form of inatial condition . $\endgroup$
    – Sara
    Jun 5, 2015 at 10:44

1 Answer 1

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

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

$\dfrac{dx}{ds}=3t^2=3s^2$ , letting $x(0)=x_0$ , we have $x=x_0+s^3=x_0+t^3$

$\dfrac{du}{ds}=-u$ , letting $u(0)=f(x_0)$ , we have $u(x,t)=f(x_0)e^{-s}=f(x-t^3)e^{-t}$

$u(x,0)=\begin{cases}e^{-1}&x<0\\e^{x-1}&0<x<1\\e^{-(x-1)}&x>1\end{cases}$ :

$f(x)=\begin{cases}e^{-1}&x<0\\e^{x-1}&0<x<1\\e^{-(x-1)}&x>1\end{cases}$

$\therefore u(x,t)=\begin{cases}e^{-t}&x-t^3<0\\e^{x-t^3-1-t}&0<x-t^3<1\\e^{-(x-t^3-1)-t}&x-t^3>1\end{cases}=\begin{cases}e^{-t}&x<t^3\\e^{x-t^3-t-1}&t^3<x<t^3+1\\e^{-x+t^3-t+1}&x>t^3+1\end{cases}$

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  • $\begingroup$ Thank you vary much , after 1 hour i have final exam , and this topic is included. $\endgroup$
    – Sara
    Jun 6, 2015 at 4:46

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