# [SOLVED]Boundary Conditions of the Third Kind

#### dwsmith

##### Well-known member
\begin{align}
\varphi''+\lambda\varphi &= 0, & \quad 0< x < L\\
\varphi'(0) &= 0 &\\
\end{align}
$$\varphi = A\cos x\sqrt{\lambda} + B\frac{\sin x\sqrt{\lambda}}{\sqrt{\lambda}}$$
Since $\varphi'(0) = 0$, $\varphi = A\cos x\sqrt{\lambda}$. Then
$$-\sqrt{\lambda}\sin L\sqrt{\lambda}+h\cos L\sqrt{\lambda} = 0 \Leftrightarrow \sqrt{\lambda}\tan L\sqrt{\lambda} = h$$
Multiplying be $L$.
$$L\sqrt{\lambda}\tan L\sqrt{\lambda} = Lh$$
Let $L\sqrt{\lambda} = s$. Then $s\tan s = Lh$. By definition, $\tan (-s) = -\tan s$ and $\lambda = \left(\dfrac{s}{L}\right)^2 = \left(\dfrac{-s}{L}\right)^2$.
\begin{align}
y_1 &= \tan s, & \quad 0\leq s <\infty\\
y_2 &= \frac{Lh}{s}, & \quad 0\leq s <\infty
\end{align}

Why does $Lh = \dfrac{\pi}{2}\mbox{?}$

#### Ackbach

##### Indicium Physicus
Staff member
\begin{align}
\varphi''+\lambda\varphi &= 0, & \quad 0< x < L\\
\varphi'(0) &= 0 &\\
\end{align}
$$\varphi = A\cos x\sqrt{\lambda} + B\frac{\sin x\sqrt{\lambda}}{\sqrt{\lambda}}$$
Since $\varphi'(0) = 0$, $\varphi = A\cos x\sqrt{\lambda}$. Then
$$-\sqrt{\lambda}\sin L\sqrt{\lambda}+h\cos L\sqrt{\lambda} = 0 \Leftrightarrow \sqrt{\lambda}\tan L\sqrt{\lambda} = h$$
Multiplying be $L$.
$$L\sqrt{\lambda}\tan L\sqrt{\lambda} = Lh$$
Let $L\sqrt{\lambda} = s$. Then $s\tan s = Lh$. By definition, $\tan (-s) = -\tan s$ and $\lambda = \left(\dfrac{s}{L}\right)^2 = \left(\dfrac{-s}{L}\right)^2$.
\begin{align}
y_1 &= \tan s, & \quad 0\leq s <\infty\\
y_2 &= \frac{Lh}{s}, & \quad 0\leq s <\infty
\end{align}

Why does $Lh = \dfrac{\pi}{2}\mbox{?}$
I don't think it does. From this plot, you can see that the solutions are going to get closer and closer to integer multiples of $\pi$, not $\pi/2$.