# Find the extremal of this functional with given b.c.s

• Nov 6th 2011, 04:39 AM
featherbox
Find the extremal of this functional with given b.c.s
Determine the extremal for the functional
$\int_{0}^{1}(xy+y^2-2y^2y') dx$
with $y(0)=0, y(1)=2$

Using Euler-Lagrange I get,
$\frac{\partial f}{\partial y}+\frac{\mathrm{d}}{\mathrm{d} x}(\frac{\partial f}{\partial y'})=x+2y-4yy'+4yy'=x+2y=0$
but that isn't consistent with the given b.c.s

• Nov 7th 2011, 04:11 AM
Ackbeet
Re: Find the extremal of this functional with given b.c.s
Quote:

Originally Posted by featherbox
Determine the extremal for the functional
$\int_{0}^{1}(xy+y^2-2y^2y') dx$
with $y(0)=0, y(1)=2$

Using Euler-Lagrange I get,
$\frac{\partial f}{\partial y}+\frac{\mathrm{d}}{\mathrm{d} x}(\frac{\partial f}{\partial y'})$

Actually, this should be

$\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'},$

but you seem to have computed the correct expression here:

Quote:

$=x+2y-4yy'+4yy'=x+2y=0.$
On the face of it, it doesn't surprise me that the term with $y'$ drops out. After all,

$\int_{0}^{1}(xy+y^{2}-2y^{2}y')\,dx=\int_{0}^{1}(xy+y^{2})\,dx-2\int_{0}^{1}y^{2}y'\,dx$

$=\int_{0}^{1}(xy+y^{2})\,dx-2\int_{0}^{2}y^{2}\,dy=\int_{0}^{1}(xy+y^{2})\,dx-\frac{16}{3}.$

So the problem reduces down to finding the extremal of

$\int_{0}^{1}(xy+y^{2})\,dx,$ subject to the boundary conditions. Since there is now no $y'$ term, the Euler-Lagrange equation simplifies down to setting

$\frac{\partial L}{\partial y}=0,$ which implies

$x+2y=0,$ or $y=-x/2,$ as before. And, as you've noted, this function does not satisfy the boundary conditions.

Question: what is the domain of functions over which you're searching for a solution? Continuous? Differentiable? (I would assume probably differentiable, since you have a $y'$ in the integrand; however, you might be interpreting that derivative in a weak sense, or in some other similarly exotic fashion.)

If you require a differentiable function as your solution, then I would say your problem has no solution.