# Calculus of Variations

1. Jan 22, 2008

### JukkaVayrynen

Hello everybody.
Sorry, I don't know how to use TeX yet, I couldn't find a testing zone.

Problem:
Let I = \int_0^\infty [(dy/dx)^2 - y^2 + (1/2)y^4]dx, and y(0) = 0, y(\infty) = 1. For I to be extremal, which differential equation does y satisfy?

Solution:
The condition is that \delta I = 0 \Rightarrow \int_0^\infty [2(dy/dx)\delta (dy/dx) -2(y-y^3)\delta y]dx = 0, which results, after partial integration, in y - y^3 + (d^2 y / dx^2) = 0, which I hope is the correct answer.
The question is: why are y(0) = 0 and y(\infty) = 1, mentioned, I didn't use them at all.

2. Jan 22, 2008

### HallsofIvy

Staff Emeritus
you have a second order differential equation which, typically, will involve two "arbitrary constants" in its solution. You the additional conditions to determine the correct solution to the differential equation.

In other words, you don't have, strictly speaking, a "differential equation", you have a "boundary value problem": a differential equation and additional conditions.

3. Jan 22, 2008

### JukkaVayrynen

So the answer should be: y satisfies the boundary value problem $$y - y^3 + \frac{d^2 y}{dx^2} = 0$$, y(0) = 0, $$y(\infty) = 1$$?

Last edited: Jan 22, 2008