Toronto Math Forum
APM3462015F => APM346Home Assignments => HA6 => Topic started by: Yeming Wen on October 28, 2015, 05:17:27 PM

http://www.math.toronto.edu/courses/apm346h1/20159/PDEtextbook/Chapter4/S4.2.P.html#problem4.2.P.2

Just wondering, the boundary condition in this question is $X(0)=0$ and $X'(l)+\beta X(l)=0$?

Just wondering, the boundary condition in this question is $X(0)=0$ and $X'(l)+\beta X(l)=0$?
Indeed: the same as before on the right

This problem follows similar logic to the first problem.
(a) We let $\lambda = \omega^2$. We can then proceed as in the first problem (check that post if you would like to see the steps) to get that: \begin{equation}
X(x) = A\cos(\omega{}x) + B\sin(\omega{}x) \end{equation}
With the derivative being: \begin{equation}
X'(x) = A\omega{}\sin(\omega{}x) + B\omega{}\cos(\omega{}x)
\end{equation}
This time we have a Dirichlet condition that $X(0) = 0$. So plugging this in: \begin{equation}
X(0) = A = 0 \end{equation}
Therefore $A=0$. Thus the solution is: \begin{equation}
X_n = \sin(\omega{}_nx) \end{equation}
We have chosen $B=1$ for convenience.
Edit: the similar tangent relation is, as you can easily show using the boundary condition on the right end, is: \begin{equation}
\tan(\omega{}l) = \frac{\omega{}}{\beta{}} \end{equation}

(b) We let $\lambda =  \gamma^2$. We can again proceed as in the first problem to get that: \begin{equation}
X(x) = A\cosh(\gamma{}x) + B\sinh(\gamma{}x) \end{equation}
With the derivative being: \begin{equation}
X'(x) = A\gamma{}\sinh(\gamma{}x) + B\gamma{}\cosh(\gamma{}x)
\end{equation}
Plugging in the Dirichlet condition as in part (a): \begin{equation}
X(0) = A = 0 \end{equation}
Therefore $A=0$. Thus the solution is: \begin{equation}
X_n = \sinh(\gamma{}_nx) \end{equation}
We have chosen $B=1$ for convenience.
The relation is: \begin{equation}
\tanh(\gamma{}l) = \frac{\gamma{}}{\beta{}} \end{equation}