Find the extremals of the functional
$$\text{J}(y)= y^2(1) + \int_0^1 y'^2(x)dx , \ \ y(0)=1.$$
Answer: $y(x)=1-\frac{1}{2}x$
My solution:
$ F (x,y,y')=y'^2(x)$
After solving the Euler Lagrange equation we get $\frac{\mathrm{d}}{\mathrm{d}x}(2y')=0$
Which implies that $y=\frac{a}{2}x+b$ , using initial value condition we get $b=1$. Could you please help me find value of $a$?
Best Answer
Note that we can write $J$ as
$$J(y) = \int_0^1 \left(2 y(x)y'(x) + y'(x)^2\right) \ dx \, + \, y(0)^2$$
Setting aside the constant $y(0)^2 = 1$ for now, we have $J$ as the integral of a function $F(y,y',x)$ where $F(y,y',x) = 2yy' + y'^2$.
You can now deploy your box of tricks on $F$ to find $y$. That is, the Euler-Lagrange equation:
$$0 = \frac{d \ }{dx} \frac{\partial F}{\partial y'} - \frac{\partial F}{\partial y} = \ ... $$