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I am trying to solve the following equation with initial condition: \begin{equation} u_t + uu_x = -cu, \end{equation} $$u(t=0) = v_0(x)$$

In addition to that, I know the solution of equation $u_t + uu_x = 0$, which is $v(x,t)$, with the same initial condition. So, by using method of charachteristics: $$u = dx/dt, \ \ \ cdt = -\frac{du}{u}$$ Integrating the second equation to get $$u = v_0(x)e^{-ct},$$ where I have used initial condition. After that I am kind of stuck, I think we can say that the general solutuion will be $$u = v(x,t)e^{-ct},$$ but why? Is it correct that first charachteristic equation will kind of give us as the solution some function $u = u(x,t)$, which we happened to know already, because it's the same charachteristic as for equation $u_t + uu_x = 0$, solution of which is $v(x,t)$, which we used to replace $v_0(x)$ in the equation above?

Alej
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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 $t=s$

$\dfrac{du}{ds}=-cu$ , letting $u(0)=u_0$ , we have $u=u_0e^{-cs}=u_0e^{-ct}$

$\dfrac{dx}{ds}=u=u_0e^{-cs}$ , letting $x(0)=f(u_0)$ , we have $x=\dfrac{u_0(1-e^{-cs})}{c}+f(u_0)=\dfrac{u(e^{ct}-1)}{c}+f(ue^{ct})$ , i.e. $u=e^{-ct}F\left(x-\dfrac{u(e^{ct}-1)}{c}\right)$

$u(x,0)=v_0(x)$ :

$F(x)=v_0(x)$

$\therefore u=e^{-ct}v_0\left(x-\dfrac{u(e^{ct}-1)}{c}\right)$

doraemonpaul
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  • Thank you for answer!! So, in the answer $v_0$ function was taken from the initial condition? What about $v(x,t)$ function, which is the solution of $u_t + uu_x = 0$ ? Just we kind of know it. – Alej Jan 07 '16 at 23:16