APM346-2015F > HA6

HA6-P2

(1/1)

Yeming Wen:
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter4/S4.2.P.html#problem-4.2.P.2

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

Victor Ivrii:

--- Quote from: Yeming Wen on October 28, 2015, 05:19:08 PM ---Just wondering, the boundary condition in this question is $X(0)=0$ and $X'(l)+\beta X(l)=0$?

--- End quote ---
Indeed: the same as before on the right

Emily Deibert:
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}

Emily Deibert:
(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}

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