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Calculus of Variations 
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#1
Oct1405, 12:56 AM

P: 282

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] 


#2
Oct1405, 04:34 AM

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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 nonlinear ODE. Push come to shove, I'd solve it numerically. 


#3
Oct1405, 04:43 AM

P: 282




#4
Oct1405, 05:16 AM

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Calculus of Variations



#5
Oct1405, 10:57 AM

P: 282

[tex] 2y'^3 + 3y'^2 + y' = constant [/tex] 


#6
Oct1405, 12:56 PM

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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?



#7
Oct1405, 05:50 PM

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[tex]6(y^{'})^2+6y^{'}+1=0[/tex] Can someone rectify this discrepancy please? 


#8
Oct1405, 07:35 PM

P: 282

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? 


#9
Oct1405, 08:05 PM

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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...)



#10
Oct1405, 09:34 PM

P: 282

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? 


#11
Oct1505, 04:52 AM

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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?" 


#12
Oct1505, 07:02 AM

P: 282

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? 


#13
Oct1505, 07:34 AM

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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 piecewiselinear... I'm not sure if there are pathological continuous solutions. 


#14
Oct1505, 08:18 PM

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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{12/3}}{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? 


#15
Oct1605, 06:53 AM

P: 282

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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