To find an extremal of the functional $\int_0^1 [(y')^2 + 12 xy] dx$

Question

I have to find extremal of following :

$\int_0^1 [(y')^2 + 12 xy] dx$ with $y(0) = 0$ and $y(1) = 1$.

I applied the Euler's equation $\frac{\partial F}{\partial y} - \frac{d}{dx}(\frac{\partial F}{\partial y'}) = 0$ and I got

$6x - y'' = 0$ and after solving this I get a solution $y = x^3 +cx + d$.

Now applying the condition $y(0) = 0$ gives $d=0$ and thus $y = x^3 +cx$.

Again applying the condition $y(1) = 1$, I get $c=0$ which finally yields the answer

$y=x^3$.

Is my solution right? As I just started this topic, I am not sure about my answer. Thanks for giving time.

Answer 1

Your solution using Euler-Lagrange is correct.

EDIT 1:

Just as for a function you can re-check the maximum point by equating derivative to zero there is a way to check your answer with functionals in variational calculus also. Can be found in variational calculus text books, I have no access at present.

At first you should check your solution is a minimum, but not a maximum.

Next, a crude way to check the minimum is to evaluate definite integral for some

functional variations in the neighborhood like $ y= x^2$ and $ y= x $ for comparing

result with extremizing result $ y= x^3 $.

For $ y= x^3 $, $ \int_{0}^1 21 x^4 dx = 4.25 $

This evaluated definite integral should be less than that obtained for any function

you can think of between (0,0) and (0,1) :

For $ y= x^2 $ , $ \int_{0}^1 (4 x^2+12 x^3)\, dx = 13/3 = 4.333.. $

For $ y= x $ , $ \int_{0}^1 (1 + 12 x^2)\, dx = 5 $

I indicated subject tag as calculus of variations.

Answer 2

Your solution is correct if you search extermals in the space $$\{f \in C^2([0,1]) : f(0)=0, f(1)=1\}. $$ For the space $$\{f \in C^1([0,1]) , f''\in L^1([0,1]) : f(0)=0, f(1)=1\} $$ maybe your approach does not work.