Calculus of variations - showing extremal is minimum

Apr 2013
13
2
N/A
Hi,

J[y] = \(\displaystyle \int_0^1{e^y(y')^2 dx\)
y(0) = 1, y(1) = 2ln(2).

I need to find the extremal and show whether it provides a maximum or minimum.

In this case I only have problems with the second part. I have the extremal as y(x) = 2ln(x(2-\(\displaystyle \sqrt{e}\))+\(\displaystyle \sqrt{e}\)).

To find whether its a maximum or minimum the process I follow is; consider v(x) such that v(0)=0 and v(1)=0, and find the sign of:

J[y+v] - J[y] = \(\displaystyle \int_0^1{e^{y+v}((y+v)')^2 dx\) - \(\displaystyle \int_0^1{e^y(y')^2 dx\).

I have that this should come out as positive in my notes and thus y provides a minimum - but I don't know how to get there. Squaring out the J[y+v] integral bracket gives me terms I don't know how to integrate, and integration by parts has proved unable to alleviate this problem. I fear I may be going wrong with my approach. Does all of what I've written make sense? Can someone offer any pointers?

Thanks!
 

chiro

MHF Helper
Sep 2012
6,608
1,263
Australia
Hey Naranja.

Subtituting u = y + v gives e^u * u' for J[v+y]. You can then integrate by parts to get e^u * u which is solve-able. The limits will be y(0) + v(0) to y(1) + v(1) for this transformation (since we are using u(x) = y(x)+ v(x)).
 
Apr 2013
13
2
N/A
Thanks for your reply,

I should make clear though that I am not actually looking to do any integration here. y and v are functions of x - I can't integrate these with respect to x. I am just looking to manipulate the entire expression to show that it is positive. I.e. by cancelling as many terms as possible.
 

chiro

MHF Helper
Sep 2012
6,608
1,263
Australia
You can evaluate the integral if its in terms of u since you are in u-space rather than x-space (as is done by doing a substitution). [Remember that the point of the substitution is to go from x to u(x) just like we usually go from say x to u in a normal substitution).
 
Apr 2013
13
2
N/A
Hi,

As I said in the previous post - I am not looking to do any integration. That may be an option but it is not the intended route for this problem, and not the method I am required to learn. This is purely manipulation of the integral to show it must be positive.
 

chiro

MHF Helper
Sep 2012
6,608
1,263
Australia
OK then no worries: hope you get what you were looking for!