# Weierstass's Excess Function - minimizing curve

• October 10th 2010, 08:46 PM
BrooketheChook
Weierstass's Excess Function - minimizing curve
Using the Weierstass condition, find the strongly minimizing curve and the value of $J_{min}$ for the cases

$\displaystyle \int_1^2 (t^3 \dot{x}^2 + 3tx^2) \ dt$

where $x(1) = 0, x(2) = 3$.

I've gotten to a point where I have tried to continue, but I can't seem to get anything that resembles a correct solution.

Let $f = t^3 \dot{x} ^2 + 3tx^2$

Using the Euler-Largrange eqn, I get

$6tx - 2t^3 \ddot{x} = 0$ so $\ddot{x} t^2 - 3x = 0$ ...(1)

which is a Euler's type DE which can be solved by changing variable $t = e^s$.

Let $x(s) = x(t) = x(e^s)$. By the chain rule and noting that $\frac{dt}{ds} = e^s = t$ :

$\displaystyle \frac{dx}{ds} = \frac{dx}{dt} \frac{dt}{ds} = e^s \frac{dx}{dt} = t \frac{dx}{dt}$ ........(2)

and

$\displaystyle \frac{d^2 x}{d^2 s} = \frac{d}{ds} \big( t \frac{dx}{dt} \big) = t \frac{dx}{dt} + t \frac{d}{ds} \frac{dx}{dt} = t \frac{dx}{dt} + t^2 \frac{d^2 x}{dt^2}$ ....(3)

Subtracting (2) and (3) I get,

$\displaystyle \frac{d^2 x}{d^2 s} - \frac{dx}{ds} = t^2 \frac{d^2 x}{dt^2} = 3x$

implying $\displaystyle \frac{d^2 x}{d^2 s} - \frac{dx}{ds} - 3x = 0$

which is equivalent to equation (1).

The extremal is therefore

$\displaystyle x = x(t) = x(s) = Ae^{\frac{1}{2}(1 + \sqrt{13}) s} + Be^{\frac{-1}{2} (-1 + \sqrt{13})s}$

Now, when using $x(1) = 0, x(2) = 3$ I don't get a solution that appears to be correct.....and I don't know how to continue. Any help would be greatly appreciated.
Thank-you.
• October 11th 2010, 11:52 AM
zzzoak
Euler-Largrange equation is

$
\displaystyle{\frac{\partial L}{\partial x} - \frac{d}{d t}\frac{\partial L}{\partial x \dot}=0}
$

$
\displaystyle{L=t^3 \dot{x}^2 + 3tx^2}
$

$
\displaystyle{\frac{\partial L}{\partial x}=6tx}
$

$
\displaystyle{\frac{\partial L}{\partial x \dot}=2t^3 x \dot}
$

$
\displaystyle{
6tx-\frac{d}{d t}(2t^3 {x \dot})=0
}
$
• October 11th 2010, 04:40 PM
BrooketheChook
Quote:

$\displaystyle{
6tx-\frac{d}{d t}(2t^3 {x \dot})=0
}$

Maybe I did this incorrectly.

I get $6tx - 2t^3 \ddot{x} = 0$ Maybe this line of calculation is wrong? In my first post there was a typo here which has now been fixed.

$\therefore 6x = 2t^2 \ddot{x}$

$3x = \ddot{x} t^2$

so $\ddot{x} t^2 - 3x = 0$
• October 12th 2010, 04:51 AM
Ackbeet
As zzzoak mentioned, you must compute

$\displaystyle \frac{d}{dt}(2t^{3}\dot{x}).$

Now, you're viewing $x=x(t),$ which means that the derivative in question must be computed using the product rule (it's a total derivative, not a partial derivative!). I get

$2(3t^{2}\dot{x}+t^{3}\ddot{x})=6t^{2}\dot{x}+2t^{3 }\ddot{x}.$

Another puzzle: you're asked to use the Weierstrass condition. Is that the Weierstrass-Erdmann Corner Condition?
• October 12th 2010, 07:39 AM
BrooketheChook
I messed that up....twice.

Now I have $t^2 \ddot{x} - 3t \dot{x} - 3x = 0$

There is a hint to this question that I forgot to mention earlier, which says "Look for solutions of the Euler-Lagrange equation of the form $t^{\alpha}$.

That doesn't really make sense to me...how can I get the extremal from here?
• October 12th 2010, 07:45 AM
Ackbeet
I agree with your DE now.

Your hint is hinting at the fact that the DE you now have to solve is a Cauchy-Euler equation. To solve Cauchy-Euler equations, your ansatz looks like $t^{\alpha}.$ So plug in $x(t)=t^{\alpha}$ and solve for $\alpha.$ What do you get?
• October 12th 2010, 08:04 AM
BrooketheChook
Assume $x(t) = t^{\alpha}$

Differentiating

$\dot{x} = \alpha t ^{\alpha - 1}$

and

$\ddot{x} = \alpha(\alpha - 1) t^{\alpha - 2}$

Substituting back into the DE,

$t^2 (\alpha(\alpha - 1)t^{\alpha - 2} - 3t(\alpha t ^{\alpha - 1}) - 3t^{\alpha} = 0$
• October 12th 2010, 08:07 AM
Ackbeet
Looking good so far... What next?
• October 12th 2010, 08:25 AM
BrooketheChook
which I think (just quickly) is

$\alpha(\alpha - 1) t^{\alpha} - 3 \alpha t^{\alpha} - 3t^{\alpha}$

so

$(\alpha (\alpha - 1) - 3 \alpha - 3) t^{\alpha}$

So, either $x = 0$ (trivial) or $x^{\alpha}$ = 0. Solving the quadratic yields

$\alpha = 2 \pm \ \sqrt{7}$

So general solution is

$x(t) = At^{2 + \sqrt{7}} + Bt^{2 - \sqrt{7}}$
• October 12th 2010, 08:29 AM
Ackbeet
Whoops. I think there was a sign error way back in post # 5. The double derivative and single derivative terms should have the same sign.
• October 12th 2010, 09:18 AM
BrooketheChook
Thanks for picking that up.....

but the solving the quadratic yields $x = -3, 1$

so the general solution is

$x(t) = At + \frac{B}{t^3}$

Then

$x(1) = 0 = A + B \Rightarrow \ B = - A$

$x(2) = 3 = 2A - \frac{A}{8} \Rightarrow \ A = \frac{8}{5}$

$B = - \frac{8}{5}$

Then $x = x(t) = \frac{8t}{5} - \frac{8}{5t^3}$

Im not sure how to calculate $J_{min}$ for this though....
• October 12th 2010, 09:21 AM
Ackbeet
That all looks correct.

Isn't

$\displaystyle J=\int_1^2 (t^3 \dot{x}^2 + 3tx^2) \ dt?$

Then, if you've found the minimizing function $x(t),$ can't you just plug that into the expression for $J$ and plug and chug?
• October 12th 2010, 09:48 AM
BrooketheChook
Quote:

Then, if you've found the minimizing function $x(t),$ can't you just plug that into the expression for $J$ and plug and chug?
The way you worded this really made things clear.

I get $\frac{228}{5}$
• October 12th 2010, 09:51 AM
Ackbeet
Looks good to me.

Glad I was able to word things in a clear fashion. That's something I strive for constantly!

Have a good one!