MHB How to Combine Sturm-Liouville Equations for 4th Derivative?

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The discussion focuses on combining two Sturm-Liouville equations for the fourth derivative, specifically showing that the equations can be expressed in a combined form. Participants explore integration by parts and the implications of differentiating and rearranging terms to achieve the desired result. The integration process involves applying boundary conditions and recognizing the symmetry in the equations. Ultimately, the conclusion is reached that the order of the modes does not affect the outcome, allowing for flexibility in the representation of the combined equation. The thread emphasizes the mathematical rigor required in manipulating Sturm-Liouville equations.
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$$
X^{(4)}_n - \alpha^4X_n = 0\qquad\qquad X^{(4)}_m - \alpha^4X_m = 0
$$
\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:
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dwsmith said:
$$
X^{(4)}_n - \alpha^4X_n = 0\qquad\qquad X^{(4)}_m - \alpha^4X_m = 0
$$
\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 said:
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}
 

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