
#1
Oct1405, 12:56 AM

P: 284

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

Sci Advisor
HW Helper
P: 1,593

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: 284





#4
Oct1405, 05:16 AM

Sci Advisor
HW Helper
P: 1,593

Calculus of Variations 



#5
Oct1405, 10:57 AM

P: 284

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



#6
Oct1405, 12:56 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

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

Sci Advisor
HW Helper
P: 1,593

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



#8
Oct1405, 07:35 PM

P: 284

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

Emeritus
Sci Advisor
PF Gold
P: 16,101

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: 284

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

Sci Advisor
HW Helper
P: 1,593

[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: 284

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

Emeritus
Sci Advisor
PF Gold
P: 16,101

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

Sci Advisor
HW Helper
P: 1,593

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: 284

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. 


Register to reply 
Related Discussions  
Calculus of Variations  Calculus & Beyond Homework  2  
Calculus of Variations.  Calculus  2  
Calculus of Variations  Calculus & Beyond Homework  3  
Calculus of Variations II  Calculus & Beyond Homework  8  
Calculus of variations  Introductory Physics Homework  4 