APM346-2015F > Web Bonus = Sept

Web bonus problem : Week 3 (#5)

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Victor Ivrii:
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.4.P.html#problem-2.4.P.3 Problem 3

Zaihao Zhou:
Part A
\begin{equation}
u_{xy} = u_{x}u_{y}u^{-1}
\end{equation}

Rearrange the function, we have: \begin{equation}uu_{xy} = u_{x}u_{y}\end{equation}
Divide both sides by \begin{equation}uu_{x}\end{equation}, we have \begin{equation}\frac{u_{xy}}{u_{x}} = \frac{u_{y}}{u}\end{equation}

which is \begin{equation} (lnu_{x})_{y} = (lnu)_{y}\end{equation}.
Integrate and do some arrangements we have \begin{equation} lnu = \int{e^{f(x)}dx}+g(y) \end{equation}
Thus \begin{equation} u = e^{\int{e^{f(x)}dx}}e^{g(y)} \end{equation}


Part B arrange we have \begin{equation}(lnu_{x})_{y} = u_{y} \end{equation}
Integrate and arrange we have \begin{equation} -e^{-u} = \int{e^{f(x)}dx}+g(y) \end{equation}
thus we have \begin{equation} u = -ln[-(\int{e^{f(x)}dx}+g(y))] \end{equation}

Part C

Victor Ivrii:
Part A -- after you got $\bigl(\ln (u_x)\bigr)_y = \bigl(\ln (u)\bigr)_y$ integration gives $\ln (u_x)= \ln (u)  +\ln \phi(x)$ (we look for a simple form) $\implies u_x= \phi(x) u\implies (\ln (u))_x =\phi(x)\implies \ln (u)= f(x)+g(y)$ with $f$ primitive of of $\phi$.

So: $u =F(x)G(y)$ where $F=e^f, G=e^g$ are arbitrary functions. You did correct but you need to jump to simple forms all the time

Part B -- ditto, but again $u=-\ln (F(x)+G(y))$ is a simpler form.




Write \ln (and so on)

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