MHB What is the integral of f(x) and q(x) in the Euler Lagrange equations?

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The discussion revolves around finding the integrals of functions \(f(x)\) and \(q(x)\) within the context of the Euler-Lagrange equations applied to the functional \(F = p(x)y'^2 - q(x)y^2 + 2f(x)y\). Participants clarify that the question about the integrals is somewhat ambiguous, particularly regarding the treatment of \(y\) in the integrals. They suggest that the correct formulation involves separating the integrals of \(f(x)\) and \(yq(x)\) without pulling \(y\) out of the integral. The conversation also touches on solving the resulting integro-differential equation and proposes alternative approaches to simplify the problem, including transforming it into a linear differential equation. Ultimately, the focus is on correctly applying the Euler-Lagrange equations to derive the necessary differential equation for \(y\).
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Given this \(F = p(x)y^{'2}-q(x)y^2+2f(x)y\). What would be the integral of \(f(x)\) and \(q(x)\)?
\begin{align*}
f(x) - q(x)y - \frac{d}{dx}\left[p(x)y'\right] &= 0\\
\frac{d}{dx}\left[p(x)y'\right] &= f(x) - q(x)y\\
y'p(x) &= \int f(x)dx - y\int q(x)dx
\end{align*}
 
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Assuming you meant to apply the EL equations to $F$, you've started out correctly. I'm not sure you can pull $y$ out of the last integral, though. If $y=y(x)$, then I think you have to have
$$y'p(x) = \int f(x) \,dx - \int y \, q(x) \,dx.$$
The question, "What would be the integral of $f(x)$ and $q(x)$?" is a bit unclear. You have sort-of found it, I suppose, although the $y$ under the $q$ integral is a bit troubling. Is that the exact wording of the original question?
 
Ackbach said:
Assuming you meant to apply the EL equations to $F$, you've started out correctly. I'm not sure you can pull $y$ out of the last integral, though. If $y=y(x)$, then I think you have to have
$$y'p(x) = \int f(x) \,dx - \int y \, q(x) \,dx.$$
The question, "What would be the integral of $f(x)$ and $q(x)$?" is a bit unclear. You have sort-of found it, I suppose, although the $y$ under the $q$ integral is a bit troubling. Is that the exact wording of the original question?
How can I continue to solve the problem is the real question since I have this integrals. I suppose, for f, I could say \int f = F but then I still have the issue of the other integral of yq(x).
 
Yes, you have an integro-differential equation for $y$. Not so nice. I might actually peal back your integral and try to solve
$$f(x) - q(x)y - p'(x)y' -p(x)y''= 0,$$
or
$$p(x)y''+p'(x)y'+q(x)y=f(x).$$
It's at least linear in $y$. The homogeneous equation, if I'm not mistaken, is Sturm-Liouville. That might help you some.
 
This may be helpful then. I know F is integrable on from a to b.
$$
\int_a^b Fdx
$$
So could one say then $y(a) = y_1$ and $y(b) = y_2$?
 
If we solve that DE, here is what I came up with using variation of parameters
\begin{align*}
y'' + \frac{p'}{p}y' + \frac{q}{p}y &= \frac{f}{p}\\
y'' + Ay' + By &= C
\end{align*}
Then we have \(m^2 + Am + B = 0\) where \(m = \frac{-A\pm\sqrt{A^2-4B}}{2}\).
Then we have 3 cases:
If \(A^2-4B = 0\), then
$$
y_c = c_1\exp\left[-\frac{A}{2}x\right] + c_2x\exp\left[-\frac{A}{2}x\right]
$$
Then the Wronskian is
$$
W = \exp(-Ax),
$$
\(u_1 = \frac{2C}{A^2}\exp\left[\frac{A}{2}x\right](2-Ax)\), and \(u_2 = \frac{2C}{A}\exp\left[\frac{A}{2}x\right]\).
$$
y = c_1\exp\left[-\frac{A}{2}x\right] + c_2x\exp\left[-\frac{A}{2}x\right] + \frac{2C}{A^2}\exp\left[\frac{A}{2}x\right](2-Ax) + \frac{2C}{A}\exp\left[\frac{A}{2}x\right]
$$

If \(A^2-4B < 0\), then
$$
y_c = \exp\left[-\frac{A}{2}x\right]\left(c_1\cos\left(-\frac{\sqrt{A^2-4B}}{2}x\right) + c_2\sin\left(-\frac{\sqrt{A^2-4B}}{2}x\right)\right)
$$
Repeat first case here:

If \(A^2-4B > 0\), then
$$
y_c = c_1\exp\left[\frac{-A+\sqrt{A^2-4B}}{2}\right] + c_2\exp\left[\frac{-A-\sqrt{A^2-4B}}{2}\right]
$$
Repeat first case here:


Maybe I misunderstood the original question though since this is rather intensive.

It original said:
Obtain the E-L eqs associated with extremizing \(\int_a^bFdx\).
 
I'm afraid your solution to the DE is not valid. The guess-and-check methods associated with the exponential only work with constant coefficients.

Given your problem statement, I think you could stop here:
$$p(x)y''+p'(x)y'+q(x)y=f(x).$$
This DE is the result of applying the EL equation to $F$.
 
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