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

#### featherbox

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

#### Ackbeet

MHF Hall of Honor
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.