# Calculus of variations - showing extremal is minimum

#### Naranja

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
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)).

#### Naranja

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
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).

#### Naranja

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
OK then no worries: hope you get what you were looking for!