Thread: Calculus of Variations (Eulers Equation)

1. Calculus of Variations (Eulers Equation)

Hello, was wondering if you could help me with a question im stuck on...

The functional I is defined by

I[y] = integral from 1 to 0 of (1/2 y''^2) dx + (y(1))^2 + y'(1),

where y(0) = y'(0) = 0 and y belongs to C^4[0,1].
Show from first principles that a extremum y0(x) of I satisfies the euler equation y0'''' = 0
together with the transversality conditions y0''(1) = -1, y0'''(1) = 2y0(1).
show that y0= -1/10x^2(2+x).

Thanks : ) x

2. May be this helps:

$
\displaystyle
I(y)= \frac{1}{2} \int_0^1 y'' \ ^2 dx+ (y(1))^2 + y'(1)
$

$
\displaystyle
\delta I=I(y+\delta y)-I(y)= \frac{1}{2} \int_0^1 (y+\delta y)'' \ ^2 dx-\frac{1}{2} \int_0^1 (y)'' \ ^2 dx=\int_0^1 y'' \ \delta y'' \ dx \ .
$

Where

$
\delta y(0)=\delta y(1)=0 \ .
$

I don't know if this is true:

$
\delta y'(0)=\delta y'(1)=0 \ .
$

Then we may integrate two times by parts.