APM346-2018S > Term Test 1

P1 Night

(1/1)

Victor Ivrii:
 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.

Tristan Fraser:
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:

$$ u(x,t) = \frac{3( x - \frac{t^2}{2})}{4}+ \frac{ ( x - \frac{t^2}{2})e^{-t^2 / 2} - \frac{( x - \frac{t^2}{2})e^{-t^2 / 2} (t^2 + 1)}{3}}{4} $$

Victor Ivrii:
Crucial error

Jingxuan Zhang:
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.

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