Welcome to our community

Be a part of something great, join today!

[SOLVED] Euler Lagrange equations

dwsmith

Well-known member
Feb 1, 2012
1,673
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*}
 
Last edited:

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
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?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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).
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
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.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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 $y(b) = y_2$?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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:

[HR][/HR]
Maybe I misunderstood the original question though since this is rather intensive.

It original said:
Obtain the E-L eqs associated with extremizing \(\int_a^bFdx\).
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
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 $F$.