# [SOLVED]orthogonality

#### dwsmith

##### Well-known member
$$\left.(\phi_n\phi_m' - \phi_m\phi_n')\right|_0^L + (\lambda_m^2 - \lambda_n^2)\int_0^L\phi_n\phi_m dx = 0$$
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,
$$(\lambda_m^2 - \lambda_n^2)\int_0^L\phi_n\phi_m dx = 0\iff \int_0^L\phi_n\phi_m dx = 0$$

#### Opalg

##### MHB Oldtimer
Staff member
$$\left.(\phi_n\phi_m' - \phi_m\phi_n')\right|_0^L + (\lambda_m^2 - \lambda_n^2)\int_0^L\phi_n\phi_m dx = 0$$
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.

#### dwsmith

##### Well-known member
It follows that $\phi_n(0)\phi_m'(0) - \phi_m(0)\phi_n'(0) = \phi_n(0)\phi_m(0)\Bigl(\dfrac{\phi'_m(0)}{\phi_m(0)} - \dfrac{\phi'_n(0)}{\phi_n(0)}\Bigr)$