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Calculus of Variations |
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| Oct14-05, 12:56 AM | #1 |
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Calculus of Variations
I am facing a difficult integral here for calculus of variations. The question reads:
Find the extremum to the integral: [tex] I[y(x)] = \int_{Q}^{P} (dy/dx)^2(1+dy/dx)^2 dx [/tex] [tex] where [/tex][tex] P = (0,0) [/tex] [tex] and [/tex] [tex]Q = (1,2)[/tex] |
| Oct14-05, 04:34 AM | #2 |
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How about using Euler's equation:
[tex]\frac{\partial F}{\partial y}-\frac{d}{dx}\left(\frac{\partial F}{\partial y^{'}}\right)=0[/tex] where: [tex]F(x,y,y^{'})=\left(y^{'}\right)^2(1+y^{'})^2[/tex] However, that's usually on an interval and not from point to point as you have it defined above. This of course results in a highly non-linear ODE. Push come to shove, I'd solve it numerically. |
| Oct14-05, 04:43 AM | #3 |
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| Oct14-05, 05:16 AM | #4 |
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Calculus of Variations |
| Oct14-05, 10:57 AM | #5 |
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[tex] 2y'^3 + 3y'^2 + y' = constant [/tex] |
| Oct14-05, 12:56 PM | #6 |
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That's just a cubic equation in y' -- it factors into (y' - a)(y' - b)(y' - c) = 0, does it not?
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| Oct14-05, 05:50 PM | #7 |
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[tex]6(y^{'})^2+6y^{'}+1=0[/tex] Can someone rectify this discrepancy please? |
| Oct14-05, 07:35 PM | #8 |
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Let say for example, [tex] (y'+1)(y'+2) = 0 [/tex] Then, [tex] y'+1 = 0 [/tex] or [tex] y'+2 = 0 [/tex] That means to say that y is a function whereby its first derivative can take only the value -1 or -2 ? What's the next step? |
| Oct14-05, 08:05 PM | #9 |
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Well, there aren't many functions whose derivative is always either -1 or -2, are there? (Even fewer if you assume the derivative is continuous...)
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| Oct14-05, 09:34 PM | #10 |
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IMO, the only function whose derivative is either -1 or -2, are somewhat like stepwise functions with straight lines. I suppose you are saying that this applies to the original question. Are there any other functions? |
| Oct15-05, 04:52 AM | #11 |
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[tex]y(x)=-x+b[/tex] or: [tex]y(x)=-2x+b[/tex] Thus, being a variational problem, I'd substitute them both into the integral, evaluate it, and see which one yields the extremum I seek. Or should I say, "is that what I SHOULD do?" |
| Oct15-05, 07:02 AM | #12 |
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But in the original question, there are boundary conditions, ie, P=(0,0) and Q=(1,2) If it is a straight line, obviously we can get the function y straight away from just the end points P and Q. What is your opinion? |
| Oct15-05, 07:34 AM | #13 |
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Straight lines, even splits, et cetera are often solutions to extremal problems, so I'm not surprised.
![]() I'm pretty sure my approach is valid -- if you've proven that the derivative must satisfy that cubic, then at all points it must be equal to one of its roots, and if the derivative is continuous, it must be a constant. I don't know if you should assume the derivative is continuous or not. If it's not, then the simpler solutions for y are piecewise-linear... I'm not sure if there are pathological continuous solutions. |
| Oct15-05, 08:18 PM | #14 |
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For the record I wish to correct a conclusion I made above:
Solving the Euler equation, I obtained: [tex]2(1+6y^{'}+6(y^{'})^2)y^{''}=0[/tex] Now, either the quantity in paranthesis is zero or the second derivative. In the former case: [tex]1+6y^{'}+6(y^{'})^2=0[/tex] I obtain: [tex]y^{'}=\frac{-1\pm\sqrt{1-2/3}}{2}[/tex] Note that the slope is negative in both instances so this solution cannot meet the boundary conditions above. In the second case: [tex]y^{''}=0[/tex] The solution is: [tex]y(x)=ax+b[/tex] Using the boundary conditions: y(0)=0 and y(1)=2, I obtain: [tex]y(x)=2x[/tex] Toqura, I assume you're finished with this problem by now right? Edit: Oh yea, can someone tell me how to prove y(x) is a solution to this functional equation? |
| Oct16-05, 06:53 AM | #15 |
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Thanks.
I think I will write assumptions whether the first derivative is continuous or not, first case and second case. And then, go on to solve it by factorising like what saltydog did. |
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