I've used Euler-Lagrange and can't seem to get the right answer, please help if you can.

Determine the extremal for the functional

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

with \(\displaystyle y(0)=0, y(1)=2\)

Using Euler-Lagrange I get,

\(\displaystyle \frac{\partial f}{\partial y}+\frac{\mathrm{d}}{\mathrm{d} x}(\frac{\partial f}{\partial y'})\)

Actually, this should be

\(\displaystyle \frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'},\)

but you seem to have computed the correct expression here:

\(\displaystyle =x+2y-4yy'+4yy'=x+2y=0.\)

On the face of it, it doesn't surprise me that the term with \(\displaystyle y'\) drops out. After all,

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

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

So the problem reduces down to finding the extremal of

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

\(\displaystyle \frac{\partial L}{\partial y}=0,\) which implies

\(\displaystyle x+2y=0,\) or \(\displaystyle y=-x/2,\) 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 \(\displaystyle y'\) 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.