# Extremals - Euler - Lagrange Equation

• Aug 25th 2010, 05:28 AM
funnyinga
Extremals - Euler - Lagrange Equation
Hello,
I don't know if someone will be able to help me with this, but I'm going to post it anyway. I'm studying a worked example, but I'm confused about a few things and was hoping someone could help me understand some steps.

Question:
Find the extremal for the following fixed-end problem

$\displaystyle \int_1^2 \frac{\dot{x}^2}{t^3} dt$ with $\displaystyle x(1) = 2, x(2) = 17$

Solution (I'm going to number the steps here.

(1) Let $\displaystyle f(t,x,\dot{x}) = \frac{\dot{x}^2}{t^3}$. The Euler-Lagrange equation is
$\displaystyle \frac{\partial f}{\partial x} - \frac{d}{dt} \big ( \frac{\partial f}{\partial \dot{x}} \big ) = 0$

(2) which implies $\displaystyle - \frac{d}{dt} \big ( \frac{2 \dot{x}}{t^3} \big ) = 0$

(3) so $\displaystyle \frac{\dot{x}}{t^3} = C$ where C is a constant.

(4) Then $\displaystyle x = x(t) = kt^4 + l$ where $\displaystyle k = \frac{C}{4}$

(5) Using $\displaystyle x(1) = 2, x(2) = 17$ we have
$\displaystyle k + l = 2, 16k + l = 17$

(6) Solving this linear system yields $\displaystyle k = 1, l = 1$

(7) So the extremal is $\displaystyle x = t^3 + 1$

Firstly, I don't understand how this goes from (2) to (3) i.e where $\displaystyle \frac{\dot{x}}{t^3} = C$, and from (3) to (4).

Another example I have (which I won't post the whole thing for) is

Find the extremal of

(8) $\displaystyle \int_0^{\frac{\pi}{2}} (x^2 - \dot(x)^2 - 2x \sin t) dt$ with $\displaystyle x(0) = 1, x(\frac{\pi}{2}) = 2$

(9) The Euler-Lagrange equation is
$\displaystyle \frac{\partial f}{\partial x} - \frac{d}{dt} \big ( \frac{\partial f}{\partial \dot{x}} \big ) = 0$

(10) which implieds $\displaystyle 2x - 2 \sin t + \frac{d}{dt} (2 \dot{x}) = 0$

(11) This implies $\displaystyle \ddot{x} + x = \sin t$

again, in this example I don't understand how 11 was arrived at from 10. (Note the extremal solution to this second example is

$\displaystyle x(t) = \cos t + 2 \sin t - \frac{1}{2} t \cos t$
• Aug 25th 2010, 05:57 AM
Ackbeet
Quote:

Firstly, I don't understand how this goes from (2) to (3)...
You have the time derivative of an expression being zero, right? That means the expression does not change with respect to time. Therefore it is constant. It's no different from solving the differential equation $\displaystyle \dot{y}(t)=0.$ If you integrate both sides, you're going to get $\displaystyle y(t)=\text{const},$ right? The fact that instead of $\displaystyle y(t)$ on the LHS you have $\displaystyle \frac{2 \dot{x}}{t^3}$ doesn't change how you can integrate the DE. Make sense?

Quote:

...again, in this example I don't understand how 11 was arrived at from 10.
All they did was differentiate the $\displaystyle 2\dot{x}$, cancel all the 2's, and re-arrange. That is:

$\displaystyle 2x - 2 \sin(t) + \frac{d}{dt} (2 \dot{x}) = 0$

$\displaystyle 2x - 2 \sin(t) + 2 \ddot{x} = 0$

$\displaystyle x - \sin(t) + \ddot{x} = 0$

$\displaystyle x + \ddot{x} = \sin(t).$

So there's considerably less there than meets the eye. It's all regular calculus.
• Aug 25th 2010, 06:10 AM
funnyinga
Oh got it. I was looking for something overly complicated......thanks for pointing this out.
• Aug 25th 2010, 06:11 AM
Ackbeet
Going from (3) to (4) is another integration. Take the equation

$\displaystyle \displaystyle{\frac{\dot{x}(t)}{t^{3}}=C}.$ Multiply both sides by $\displaystyle t^{3}:$

$\displaystyle \dot{x}(t)=Ct^{3}.$ Integrating both sides yields

$\displaystyle \displaystyle{x(t)=\frac{Ct^{4}}{4}+l.}$

Define $\displaystyle k=C/4,$ and you have (4).
• Aug 25th 2010, 06:13 AM
Ackbeet
Quote:

Oh got it. I was looking for something overly complicated......thanks for pointing this out.
Hehe. Right. Just because you're studying a complicated subject like Calculus of Variations doesn't mean all the regular calculus or differential equations goes out the window! Remember Occam's razor.