Calculus of Variation - Minimize a Functional

Question

I am struggling with a basic question from a homework assignment regarding Calculus of Variation. I would love if somebody could get me started in the right direction, as I am pretty lost and the textbook does not cover problems this simple. I have never dealt with Functionals before, so I am a little short on intuition. Here is the problem:


We consider the functional $F:X\to\mathbb{R}$ defined by $$F(y)=\int_{0}^1 (y(x)^2-1)^2dx$$ Find the minimum and all the minimizers of the functional $F$ over the set $X$ in the following cases:

a. $X=\{\text{continuous functions}\ y: [0,1]\to\mathbb{R}\}$

b. $X=\{\text{continuous functions}\ y: [0,1]\to\mathbb{R} \; \text{such that} \; y(0)=1\}$


In short, I really don't know where to start. I looked online, and it seems like most of these types of problems are supposed to be solved using Euler's formula or some other method. However, my professor has not introduced any formulas or any explicit method that we can use to solve such a problem.

The professor has only given us one similar example in whiched he solved $F(y)=\int_{0}^1 (y(x)-1)^2dx$ instead of the above problem. However, he solved it by simply claiming that $F$ has a minimum of $0$, and then proving that. He did not show us how he determined the minimum in the first place.

Any tips in the right direction would be very helpful. Thank you in advance!

Answer 1

No matter the what the function $y$ is, the function $(y^2(t)-1)^2$ is always nonnegative, so the smallest that $$F(y)=\int_0^1 (y^2(t)-1)^2\,dt$$ could possibly be is zero. Is there any continuous function $y:[0,1]\to\mathbb R$ which makes $F(y)$ equal to zero? If so, then there is your answer to part (a). For any of the answer(s) to part(a), does $y(0)=1$? If so, then there is your answer to (b) as well.

Answer 2

If $F \geq 0$ on $\Omega$, then $\int_{\Omega}F \geq 0$ and $\int_{\Omega}F = 0 \Leftrightarrow F\equiv 0$. Hence, $\int_{\Omega}F$ has a global minimum of $0$ when $F\equiv 0$ (except for countably many points, but if $F$ is continuous then this can't happen).

In regards to the difference between the two sets, notice that the second one implies that $y>0$ for some $x\in [0,1]$. In this case, it implies that the integrand is $0$ at that point, which means that $y \equiv 1$ is the only minimizer. For the first set, however, we could also have $y \equiv -1$.

In the hypothetical case where the integrand was just $y^2$, these two sets' differences are highlighted. The first set would simply have the solution $y\equiv 0$, as expected. However, the second set has that $y>0$ for some $x\in [0,1]$. Since $y$ is continuous, $\int_{[0,1]}y^2 > 0$ (note the strict inequality). This is a fact worth proving on your own. This means that there is actually no minimum (also worth proving), though the infimum is $0$.