# Extremal Again

• August 26th 2010, 08:19 AM
funnyinga
Extremal Again
Hi,
I got some help with some worked examples here (thanks Ackbeet).

I'm now trying a question from my text to see if I understand it correctly. Do I have the right idea?

Find the extremal for the following:

$\int_1^2 \ \frac{2t \dot{x} + \dot{x}^2}{t^2} dt$ with $x(1) = 0, x(2) = 4$.

Let $f(t, x, \dot{x})= \frac{2t \dot{x} + \dot{x}^2}{t^2}$

Then Euler-Lagrange eqn is
$
\frac{\partial f}{\partial x} - \frac{d}{dt} \big ( \frac{\partial f}{\partial \dot{x}} \big ) = 0$

So -

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

Integrating $\int \frac{2t + 2 \dot{x}}{t^2} dt$ we get

$- \frac{2}{t} - \frac{\dot{x}}{t^2} = 0$

$\therefore \ \frac{\dot{x}}{t} = C - \frac{2}{t}$

$\dot{x} = Ct - 2$ where C is a constant.

Now - Integrate with respect to $t$ and we get

$x = \frac{1}{2} Ct^2 - 2t + l$

so $x = x(t) = kt^2 - 2t + l$ where $k = \frac{C}{2}$

Using $x(1) = 2$ and $x(2) = 4$ we have

$x(1) = k - 2 + l = 0$ and $x(2) = 4k - 4 + l = 4$

Solving these we get $l = 0, k = 2$

so the extremal is $x = 2t^2 - 4$. Is this right or did I stuff something?
• August 26th 2010, 08:32 AM
Ackbeet
You're fine up to and including this line:

$0 - \frac{d}{dt} \big ( \frac{2t + 2 \dot{x}}{t^2} \big )= 0.$

To integrate, you can just eliminate the derivative symbol, because the LHS is a total derivative. That is,

$\frac{2t + 2 \dot{x}}{t^2}= C.$

At this point, I would solve for $\dot{x}$ and integrate again.
• August 27th 2010, 08:20 PM
funnyinga
This time I got .... (currently editing)
• August 28th 2010, 08:43 AM
Ackbeet
• September 10th 2010, 10:50 PM
funnyinga
I haven't looked at this for a while. Anyway, this time I got

$\frac{2t + 2 \dot{x}}{t^2} = C$

$\dot{x} = \frac{1}{2} \left( -2+Ct \right) t$ where $C$ is a constant.

Integrating again,

$x = \frac{1}{6} C{t}^{3}- \frac{1}{2} {t}^{2} + l$ where $l$ is a constant.

$= kt^2 - \frac{t^2}{2} + l$ where $l = \frac{C}{6}$

$x = t(kt - \frac{t}{2}) + l$

Using $x(1) = 0$ and $x(2) = 4$ we have

$x(1) = t(kt - \frac{t}{2}) + l = 0$ (call this eqn 1)

$x(2) = t(kt - \frac{t}{2}) + l = 4$ (call this eqn 2)

From 1, $x(2) = 4k - 2 + l = 4$

so $l = \frac{1}{2} - k$

Then from (2) $4k - 2 + \frac{1}{2} - k = 4$

$k = \frac{11}{6}$

Sub back into (1)
$
L = \frac{1}{2} - \frac{11}{6} = -\frac{4}{3}$

The extremal is
$x = \frac{11t^2}{6} - \frac{t^2}{2} - \frac{4}{3}$

$= \frac{4}{3} t^2 - \frac{4}{3}$
• September 11th 2010, 03:31 AM
Ackbeet
Again, everything looks good up until this point:

$= kt^2 - \frac{t^2}{2} + l$

You've lost the cubic power! I also think you mean $k=C/6$, not $l=C/6$. Try working those corrections back through and see what you get.

Here's a little hint for doing math in LaTeX: copy and paste. Whenever you're simplifying things, just copy the equation from the previous line, and make whatever changes you want. You'll make fewer mistakes that way than if you re-type out the whole equation. I do this all the time, and it's not only a mistake-saver, it's a real time-saver.
• September 11th 2010, 07:04 PM
funnyinga
Quote:

Originally Posted by Ackbeet
Again, everything looks good up until this point:

$= kt^2 - \frac{t^2}{2} + l$

You've lost the cubic power! I also think you mean $k=C/6$, not $l=C/6$. Try working those corrections back through and see what you get.

Here's a little hint for doing math in LaTeX: copy and paste. Whenever you're simplifying things, just copy the equation from the previous line, and make whatever changes you want. You'll make fewer mistakes that way than if you re-type out the whole equation. I do this all the time, and it's not only a mistake-saver, it's a real time-saver.

Thanks, I really need to check my work before I post things.

$= kt^3 - \frac{t^2}{2} + l$ where $k = \frac{C}{6}$

Using $x(1) = 0$ and $x(2) = 4$ we have

$x(1) = k - \frac{1}{2} + l = 0 \ \Rightarrow \ l = -k + \frac{1}{2}$

Sub into $x(2)$

$x(2) = kt^3 - \frac{t^2}{2} - k + \frac{1}{2} = 4$

$\therefore \ 8k - 2 - k + \frac{1}{2} = 4$

$k = \frac{11}{14}$

Sub back into $x(1)$

$\frac{11}{14} - \frac{1}{2} + l = 0 \ \Rightarrow \ l = \frac{-2}{7}$

and the extremal is (hopefully)

$x = \frac{11t^3}{14} - \frac{t^2}{2} - \frac{2}{7}$
• September 11th 2010, 07:06 PM
Ackbeet
Everything looks good to me!
• September 11th 2010, 07:16 PM
funnyinga
Quote:

Everything looks good to me.
Thanks very much for your help with this.

Quote:

Originally Posted by Ackbeet
You're fine up to and including this line:

$0 - \frac{d}{dt} \big ( \frac{2t + 2 \dot{x}}{t^2} \big )= 0.$

To integrate, you can just eliminate the derivative symbol, because the LHS is a total derivative. That is,

$\frac{2t + 2 \dot{x}}{t^2}= C.$

At this point, I would solve for $\dot{x}$ and integrate again.

Just a question though, why isn't this

$0 - \frac{2t + 2 \dot{x}}{t^2}= C$
• September 11th 2010, 07:29 PM
Ackbeet
Well, you can just absorb the minus sign into the constant. The important thing is to have a constant. In these sorts of situations, you can often re-define the constant as you go along, in appropriate ways.