APM346-2015F > Web Bonus = Nov

Web Bonus Problem to Week 9 (#1)

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Victor Ivrii:
Problem 5
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter6/S6.P.html#problem-6.P.5

Chi Ma:
Substitute the partial derivatives of $u(x,y)=X(x)Y(y)$ into the pde and divide by $X^{(2)}Y^{(2)}$.

\begin{equation}
\frac{\frac{X^{(4)}}{X^{(2)}}}{\frac{Y^{(2)}}{Y}} + 2 + \frac{\frac{Y^{(4)}}{Y^{(2)}}}{\frac{X^{(2)}}{X}}=0
\end{equation}

One set of solutions can be obtained by assuming the following:
\begin{equation}
\frac{X^{(4)}}{X^{(2)}} = \frac{X^{(2)}}{X} = \omega^2 \qquad \frac{Y^{(4)}}{Y^{(2)}} = \frac{Y^{(2)}}{Y} = -\omega^2
\end{equation}

where $\omega > 0$. In this case, the solution is as follows:
\begin{equation}
u(x,y)=X(x)Y(y)=(A\cosh\omega x + B\sinh\omega x) (C\cos\omega y + D\sin\omega y)
\end{equation}

Similarly, another solution is as follows:
\begin{equation}
u(x,y)=X(x)Y(y)=(A\cos\omega x + B\sin\omega x) (C\cosh\omega y + D\sinh\omega y)
\end{equation}

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