### Author Topic: TT1-P1  (Read 2939 times)

#### Victor Ivrii

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##### TT1-P1
« on: October 19, 2016, 10:24:30 PM »
Consider the first order equation:

u_t +  xt u_x = - u.
\label{eq-1-1}

(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

$$\frac{dt}{1} = \frac{dx}{xt} = \frac{du}{-u}$$
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, $$u(x,t) = \frac{1}{1+x^2e^{-t^2}}e^{-t}$$

#### 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.