Toronto Math Forum
APM346-2018S => APM346--Tests => Term Test 1 => Topic started by: Victor Ivrii on February 15, 2018, 08:18:09 PM
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Consider the first order equation:
\begin{equation}
u_t + xt u_x = x te^{-t^2/2}.
\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.
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a) We have that given $ u_t + xtu_x = xte^{-t^2}{2} $
We use: $\frac{dx}{tx} = \frac{dt}{1} = \frac{du}{xte^{\frac{-t^2}{2}}} $
Integrating gives us the relation:
$ c + \frac{t^2}{2} = x $ WRONG. V.I.
This gives us the following characteristic curves:
b) Finding the general solution:
We plug in our value for x into:
$\frac{dt}{1} = \frac{du}{(\frac{t^3 e^{\frac{-t^2}{2}}+ Cte^{\frac{-t^2}{2}}} $|
Integration yields:
$$ u(x,t) = D +\frac{ Ce^{-t^2 / 2} - \frac{Ce^{-t^2 / 2} (t^2 + 1)}{3}}{4}$$
$$ D = \phi(C') = u - \frac{ Ce^{-t^2 / 2} - \frac{Ce^{-t^2 / 2} (t^2 + 1)}{3}}{4} $$
$$ u = \phi( x - \frac{t^2}{2}) + \frac{ ( x - \frac{t^2}{2})e^{-t^2 / 2} - \frac{( x - \frac{t^2}{2})e^{-t^2 / 2} (t^2 + 1)}{3}}{4} $$
Giving us a lovely general solution.
c) Solving (1) with the initial condition of $ u(x,0) = x $ :
$$ u(x,0) = \phi(x) + \frac{x}{3} - \frac{x}{12} = x$$
Therefore $\phi(x) = \frac{3/4} x $
Giving us a solution:
$$ u(x,t) = \frac{3( x - \frac{t^2}{2})}{4}+ \frac{ ( x - \frac{t^2}{2})e^{-t^2 / 2} - \frac{( x - \frac{t^2}{2})e^{-t^2 / 2} (t^2 + 1)}{3}}{4} $$
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Crucial error
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A) Clearly
$$C=xe^{-t^2/2}$$
B)
$$du=e^{t^2/2}dx=C/xdx\implies u=x\ln x e^{-t^2/2}+\varphi(xe^{-t^2/2})$$
C)
$$x=x\ln x + \varphi(x)\implies\varphi(x)=x - x\ln x\implies u = xe^{-t^2/2}+\frac{xt^2}{2}e^{-t^2/2}$$
I am fortunate enough that I didn't write night exam...this I spent more than half an hour.