Author Topic: P1  (Read 456 times)

Victor Ivrii

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P1
« on: February 15, 2018, 06:56:34 PM »
Consider the first order equation:
\begin{equation}
u_t +  3(t^2-1)  u_x =  6t^2.
\tag{1} 
\end{equation}

(a)  Find the characteristic curves and sketch them in the $(x,t)$ plane.

(b)  Write the general solution.

(c) Solve  equation (1)  with the initial condition $u(x,0)= x$. Explain why the solution is fully  determined by the initial condition.


Jilong Bi

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Re: P1
« Reply #1 on: February 15, 2018, 09:23:12 PM »
Characteristic Equation :$$\frac{dt}{1} = \frac{dx}{3(t^2-1)} = \frac{du}{6t^2} $$
a). From $$\frac{dt}{1} = \frac{dx}{3(t^2-1)} \\ (3t^2-3)dt = dx \\ \Rightarrow t^3 -3t +D =x  \\ \Rightarrow D = x -t^3 -3t $$ b).From :$$\frac{dt}{1} =  \frac{du}{6t^2}  \\ 6t^2dt = du \\ \Rightarrow 2t^3+A = u\\ \Rightarrow u(x,t) = 2t^3+\phi(x-t^3-3t)$$ c). $$ u(x,0) = \phi(x) = x \\ \Rightarrow u(x,t) = 2t^3+x-t^3-3t  \\ \Rightarrow u(x,t) = t^3+x-3t$$

Victor Ivrii

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Re: P1
« Reply #2 on: February 22, 2018, 06:33:57 AM »
Jilong
In the last line of (a) should be $D=x-t^3 + 3t$. Correct from here

Zhongnan Wu

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Re: P1
« Reply #3 on: April 07, 2018, 09:41:44 PM »
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