Author Topic: TT1-P1  (Read 283 times)

Victor Ivrii

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TT1-P1
« on: October 19, 2016, 10:24:30 PM »
Consider the first order equation:
\begin{equation}
u_t +  xt u_x = - u.
\label{eq-1-1} 
\end{equation}
(a) Find the characteristic curves and sketch them in the $(x,t)$ plane.

(b) Write the general solution.

(c) Solve  equation (\ref{eq-1-1})  with the initial condition $u(x,0)= (x^2+1)^{-1}$.
Explain why the solution is fully  determined by the initial condition.

Roro Sihui Yap

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Re: TT1-P1
« Reply #1 on: October 19, 2016, 10:35:16 PM »
a) Characteristic Equation

\begin{equation}  \frac{dt}{1} = \frac{dx}{xt} = \frac{du}{-u} \end{equation}
From $ \frac{dt}{1} = \frac{dx}{xt} $, $\frac{t^2}{2} + \ln c = \ln x$, thus $ x = ce^\frac{t^2}{2}$

b) General Solution
From $ \frac{dt}{1} = \frac{du}{-u}$, $-t + \ln k = \ln u$ 
So $u = ke^{-t} = \phi(xe^{\frac{-t^2}{2}})e^{-t} $

c) Since $u|_{t=0} = \frac{1}{1+x^2}$,
$\phi(x) =  \frac{1}{1+x^2} $ 
Therefore, \begin{equation} u(x,t) = \frac{1}{1+x^2e^{-t^2}}e^{-t}\end{equation}
 

 

XinYu Zheng

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Re: TT1-P1
« Reply #2 on: October 20, 2016, 12:09:07 AM »
Just to add onto Roro Sihui Yap's solution for (c), the solution is fully determined by condition at $t=0$ because all characteristics intersect the $x$ axis and do not intersect with each other.

Victor Ivrii

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Re: TT1-P1
« Reply #3 on: October 20, 2016, 04:41:03 AM »
 :D