Welcome to our community

Be a part of something great, join today!

[SOLVED] Sturm-Liouiville 4th derivative

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
X^{(4)}_n - \alpha^4X_n = 0\qquad\qquad X^{(4)}_m - \alpha^4X_m = 0
$$
\begin{alignat*}{7}
X(0) & = & 0 &\qquad\qquad & X'(0) & = & 0\\
X(L) & = & 0 &\qquad\qquad & X''(L) & = & 0
\end{alignat*}
Using a Sturm-Liouville methodology:
Show that the these two equations can be combined in the following manner
$$
X^{(4)}_nX_m - X^{(4)}_mX_n + (\alpha^4_n - \alpha_m^4)X_nX_m = 0
$$
I am not sure how to start this.
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
X^{(4)}_n - \alpha^4X_n = 0\qquad\qquad X^{(4)}_m - \alpha^4X_m = 0
$$
\begin{alignat*}{7}
X(0) & = & 0 &\qquad\qquad & X'(0) & = & 0\\
X(L) & = & 0 &\qquad\qquad & X''(L) & = & 0
\end{alignat*}
Using a Sturm-Liouville methodology:
Show that the these two equations can be combined in the following manner
$$
X^{(4)}_nX_m - X^{(4)}_mX_n + (\alpha^4_n - \alpha_m^4)X_nX_m = 0
$$
I am not sure how to start this.
So I have something that makes sense and fits the idea but not entirely.
Let $\phi,\psi$ be sufficiently differentiable functions on $[0,L]$.
Then
$$
\int_0^L\phi^{(4)}\psi dx
$$
after integration by parts 4 times we have
$$
\int_0^L\phi^{(4)}\psi dx - \int_0^L\phi\psi^{(4)} dx =\left[\phi'''\psi - \phi''\psi' + \phi'\psi''-\phi\psi'''\right|_0^L
$$
If $\phi = X_n$ and $\psi = X_m$ and $X^{(4)} = \alpha^4X$, then
$$
\int_0^LX_n^{(4)}X_m dx - \int_0^LX_nX_m^{(4)} dx =(\alpha^4_n-\alpha^4_m)\int_0^LX_nX_m =\left[\phi'''\psi - \phi''\psi' + \phi'\psi''-\phi\psi'''\right|_0^L
$$
So how do I get from here to where I am trying to go?
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
So I have something that makes sense and fits the idea but not entirely.
Let $\phi,\psi$ be sufficiently differentiable functions on $[0,L]$.
Then
$$
\int_0^L\phi^{(4)}\psi dx
$$
after integration by parts 4 times we have
$$
\int_0^L\phi^{(4)}\psi dx - \int_0^L\phi\psi^{(4)} dx =\left[\phi'''\psi - \phi''\psi' + \phi'\psi''-\phi\psi'''\right|_0^L
$$
If $\phi = X_n$ and $\psi = X_m$ and $X^{(4)} = \alpha^4X$, then
$$
\int_0^LX_n^{(4)}X_m dx - \int_0^LX_nX_m^{(4)} dx =(\alpha^4_n-\alpha^4_m)\int_0^LX_nX_m =\left[\phi'''\psi - \phi''\psi' + \phi'\psi''-\phi\psi'''\right|_0^L
$$
So how do I get from here to where I am trying to go?
So here is what I think. Is this correct?
Differentiating and multiplying through by negative yields
$$
X_nX_m^{(4)} - X_n^{(4)}X_m + (\alpha^4_m - \alpha^4_n)X_nX_m = 0.
$$
Since the order of $n^{\text{th}}$ and the $m^{\text{th}}$ mode is irrelevant, the result is the same. Thus, we can just switch the subscript.
\begin{alignat}{3}
X_mX_n^{(4)} - X_m^{(4)}X_n + (\alpha^4_n - \alpha^4_m)X_nX_m & = & 0.
\end{alignat}