MHB Orthogonality of Eigenfunctions of Mixed Boundary Conditions

Dustinsfl
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$$
\left.(\phi_n\phi_m' - \phi_m\phi_n')\right|_0^L + (\lambda_m^2 - \lambda_n^2)\int_0^L\phi_n\phi_m dx = 0
$$
where $\phi_{n,m}$ and $\lambda_{n,m}$ represent distinct modal eigenfunctions which satisfy mixed boundary conditions at $x = 0,L$ of the form
\begin{alignat*}{3}
a\phi(0) + b\phi'(0) & = & 0\\
c\phi(L) + d\phi'(L) & = & 0
\end{alignat*}
Show that the eigenfunctions are orthogonal.
$$
\int_0^L\phi_m\phi_m dx = 0.
$$
I not sure how to proceed since there are constants a,b,c,d.

If they weren't there, I would proceed as
$$
(\lambda_m^2 - \lambda_n^2)\int_0^L\phi_n\phi_m dx = -\left.(\phi_n\phi_m' - \phi_m\phi_n')\right|_0^L
$$
\begin{alignat*}{3}
\phi(0) &= &-\phi'(0) \\
\phi(L) &=& -\phi'(L)
\end{alignat*}
Therefore,
$$
(\lambda_m^2 - \lambda_n^2)\int_0^L\phi_n\phi_m dx = 0\iff \int_0^L\phi_n\phi_m dx = 0
$$
 
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dwsmith said:
$$
\left.(\phi_n\phi_m' - \phi_m\phi_n')\right|_0^L + (\lambda_m^2 - \lambda_n^2)\int_0^L\phi_n\phi_m dx = 0
$$
where $\phi_{n,m}$ and $\lambda_{n,m}$ represent distinct modal eigenfunctions which satisfy mixed boundary conditions at $x = 0,L$ of the form
\begin{alignat*}{3}
a\phi(0) + b\phi'(0) & = & 0\\
c\phi(L) + d\phi'(L) & = & 0
\end{alignat*}
Show that the eigenfunctions are orthogonal.
$$
\int_0^L\phi_m\phi_m dx = 0.
$$
I not sure how to proceed since there are constants a,b,c,d.

If they weren't there, I would proceed as
$$
(\lambda_m^2 - \lambda_n^2)\int_0^L\phi_n\phi_m dx = -\left.(\phi_n\phi_m' - \phi_m\phi_n')\right|_0^L
$$
\begin{alignat*}{3}
\phi(0) &= &-\phi'(0) \\
\phi(L) &=& -\phi'(L)
\end{alignat*}
Therefore,
$$
(\lambda_m^2 - \lambda_n^2)\int_0^L\phi_n\phi_m dx = 0\iff \int_0^L\phi_n\phi_m dx = 0
$$
You know that $a\phi_m(0) + b\phi'_m(0) = 0$ and $a\phi_n(0) + b\phi'_n(0) = 0$. Therefore $\dfrac{\phi'_m(0)}{\phi_m(0)} = \dfrac{\phi'_n(0)}{\phi_n(0)} = -\dfrac ab.$ It follows that $\phi_n(0)\phi_m'(0) - \phi_m(0)\phi_n'(0) = \phi_n(0)\phi_m(0)\Bigl(\dfrac{\phi'_m(0)}{\phi_m(0)} - \dfrac{\phi'_n(0)}{\phi_n(0)}\Bigr) = \phi_n(0)\phi_m(0)\Bigl(-\dfrac ab + \dfrac ab\Bigr) = 0.$ Similarly at the other endpoint $L$. I leave you to fret about what happens if there are any zeros on the denominators of those fractions.
 
Opalg said:
It follows that $\phi_n(0)\phi_m'(0) - \phi_m(0)\phi_n'(0) = \phi_n(0)\phi_m(0)\Bigl(\dfrac{\phi'_m(0)}{\phi_m(0)} - \dfrac{\phi'_n(0)}{\phi_n(0)}\Bigr) $

How did this come about?
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
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