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I've confused myself when thinking about the following variational problem: \begin{equation} \min_{f} \int_0^T \left([f'(x)]^2 + [f(x)]^2\right)\,dx \qquad f(0) = 1, f'(0) = 0. \tag{*} \end{equation}

Intuitively I expect the solution to be some kind of sigmoid-like decay towards zero, but applying the calculus of variations I get that the Euler-Lagrange equation is simply $f'' = f$ giving me the stationary solution $$f(x) = \frac{e^x + e^{-x}}{2},$$ which is obviously not a minimizer of (*).

What is going on? Does the minimum of (*) not exist? Is there a family of functions $f(x)$ whose energy gets arbitrarily close to zero (or some other infimum)?

Gonçalo
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user7530
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1 Answers1

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You are technically correct that based on your conditions you have no minimum. But there is a function which minimizes your integral if you ignore the condition that $f'(0)=0$. The problem is that you have basically solved the ODE

$$\frac{\partial L}{\partial f}-\frac{d}{dx}\frac{\partial L}{\partial f'}$$

where $L=f^2+f'^2$ are your initial conditions. Of course this will only give you a stationary solution. However, your mistake is trying to apply the condition $f'(0)=0$ when it actually does not matter. Observe: let $g(x)$ be any function such that $g(0)=1$ and let $h(x)$ be the cubic function such that $h(0)=1$, $h'(0)=0$, $h(\epsilon)=g(\epsilon)$, and $h'(\epsilon)=g'(\epsilon)$. Then the function defined

$$f(x)=\left\{\begin{array}{ll} h(x) & x\in[0,\epsilon) \\ g(x) & x\in [\epsilon,\infty) \\ \end{array} \right. $$

clearly satisfies the conditions in the question but

$$\lim_{\epsilon\to 0^+} \int_0^T f'(x)^2+f(x)^2=\int_0^T g'(x)^2+g(x)^2$$

since

$$0\leq \int_0^\epsilon h'(x)^2+h(x)^2\leq \epsilon \cdot\text{max}_{x\in [0,1]}[ h'(x)^2+h(x)^2]$$

To properly show this last statement it would be necessary to actually write out and define the cubic $h(x)$ but that is not too hard with a good computer software tool.

But now we can actually solve your problem: solving the functional

$$\left[\frac{\partial}{\partial f}-\frac{d}{dx}\frac{\partial}{\partial f'}\right](f^2+f'^2)=0$$

with the conditions $f(0)=1$ and $f(T)=s$. We will solve for $s$ later. Now, using these conditions we get the equation

$$f(x)=(1-b)e^x+be^{-x}$$

$$b=\frac{e^T-s}{e^T-e^{-T}}$$

The integral isn't too difficult, and we get that

$$\int_0^T f(x)^2+f'(x)^2=(1 + s^2) \coth(T) - 2 s \text{csch}(T)$$

At this point, we can find $s$ as well by normal calculus methods (differentiate and set to $0$). Solving gives

$$s=\text{sech}(T)$$

for a final minimum solution of

$$f(x)=(1-b)e^x+be^{-x}$$

$$b=\frac{e^T-\text{sech}(T)}{e^T-e^{-T}}$$

Of course, as you noted this function does not satisfy $f'(0)=0$ and can only be thought of as a global infimum in this minimization problem.

QC_QAOA
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