Shortest distance between two points via calculus of variations

Question

This problem might be trivial but when solving it using calculus of variations it's not so stupid.

Suppose we have a fixed boundary condition $f(a) = f(b) = 0$ and we want to find the shortest distance between two points, so we choose $f$ to minimize the functional $$ I(f) = \int_a^b \sqrt{1+f'(x)^2} dx $$ we solved this in class and got $f'' = 0$ which is a straight line. However, could we have just solved the problem $$ \min I(f) = \int_a^b f'(x)^2 dx $$ since it is a monotonic transformation of the original function? I know this works in calculus but not sure about when dealing with functionals. It gives me the same answer ($f'' = 0$) but is this just coincidental? The second problem involves significantly less work. Thanks!

Answer 1

This is an extended comment, not an answer, just to share my thoughts.

You mention that monotone transformations of functions yield equivalent optimization problems. This is true because, taking $g(x)$ to be monotone increasing so that $g'(x)>0$ for all $x$ in the domain, $$\frac{d}{dx}g(f(x))=g'(f(x))f'(x)=0\iff f'(x)=0$$ since $g'(x)$ is never zero.

But I simply cannot find a way to extend this argument to the case of variational problems. If we have a solution to the transformed problem with $g(f(x,y,y'))$ (in your case $g(t)=t^2-1$ and the domain is $t\ge 0$) then we know that $$\frac{\partial }{\partial y}g\big(f(x,y,y')\big)=\frac{d}{dx}\frac{\partial }{\partial y'}g\big(f(x,y,y')\big)$$ which expands to $$g'(f)\frac{\partial f}{\partial y}=g''(f)\frac{df}{dx}\frac{\partial f}{\partial y'}+g'(f)\frac{d}{dx}\frac{\partial f}{\partial y'}.$$ We want this to imply that $$\frac{\partial f}{\partial y}=\frac{d}{dx}\frac{\partial f}{\partial y'}$$ so that we can conclude (under suitable conditions) that we have optimized the original problem. But I can find no way to realize that implication, given only that $g'(t)>0$.

Clearly the proof is done if we have $g''(t)\equiv 0$, that is, if $g$ is a linear function. Unfortunately yours is quadratic.

Answer 2

I would say that the result is coincidental.

Consider minimization of the functional: $$ I(f)=\int_a^b\sqrt{{f(x)}^2+{f'(x)}^2}\;dx.\tag1 $$

If I correctly understand your point, the alternative way in this case would be the minimization of $$ I^*(f)=\int_a^b\left({f(x)}^2+{f'(x)}^2\right)\;dx.\tag2 $$ But these two functionals lead to completely different Euler-Lagrange equations: $$\begin{align} f(f^2+2f'^2-f''f)=0;\tag{1'}\\ f-f''=0,\tag{2'} \end{align}$$ which have (except for the case $f(a)=f(b)=0$) completely different solutions.

Answer 3

Consider the general functional

$$\int_a^b\Phi(f'(x))\,dx.$$

The Euler-Lagrange equations read

$$\frac{d}{dx}\frac\partial{\partial f'}\Phi(f')=0$$

and

$$\frac{d}{dx}\Phi'(f')=f''\Phi''(f')=0.$$

Hence $f''(x)=0$ is always a solution, as is $f'(x)=\Phi''^{-1}(0)$, a constant.

Answer 4

Yes, it is expected, not coincidental. That is how it should be as per calculus of variations concepts verifiable using Euler-Lagrange. The effect is the same in the two cases you mention. However the cause is different, it is not from monotonic changes from the functional. It is a class (or type) of functionals dealt with Euler-Lagrange through this EL uniform procedure. Let us see how.

Application of Euler-Lagrange on functionals

$$ \sqrt{1+ f^{'2}(x)}, f^{'2}(x),\dfrac{1- f^{'2}(x)}{\sin(1+ f^{'7}(x))},\;...$$

all lead to the very same solution viz., $ y'=$ a constant, the straight lines.

This happens due to the appearance of a single derivative variable $y'$ dependent pure functional.

For example when you take another hypothetical

$$ \int y' (1- y^{'2})dx$$

after extremization the solution should be the same again, you can verify with EL Equn.

For that matter in general any $f(y')$ pure functional (without x) would lead to the same straight lines for a solution, viz., $ y^{'}=const,\;y^{''}=0.$

I am supplying the proof here that any functional dependent purely on $y'$ would lead to straight line solutions.

$$ L= \int f(y') dx \tag1$$ Beltrami's formula when devoid of $x$ or its functions gives:

$$ f(y') -y'\cdot \dfrac{df(y')}{dy'}=C_1 \tag2 $$

(Full derivative as for single variable, $y$ terms treated constant when differentiating wrt $y'$).

For symbolic brevity let $u= y'$ operating as independent variable.

$$ f(u)-u\cdot\dfrac{d f(u)}{du}=C_1\tag3$$

$$\dfrac{df(u)}{f(u)-C_1}=\dfrac{du}{u} \tag4$$

Integrating, $$ log [f(u)-C_1]= log \;u + log\;C_2 \tag5 $$

$$ f(u)= C_2 \;u +C_1\tag6 $$

$$ f(y')= C_2\; y'+ C_1\tag7 $$

Differentiating with respect to $x$

$$\dfrac{df(y')}{dx}= C_2\; y^{''} \tag8$$

The LHS should vanish as we began with the premise/assumption that $f'(u)$ does not involve (is independent of) $x$ explicitly.

$$ y''=0 \rightarrow y= C_3 x + C_4 \tag9$$ are straight lines in the plane for all such functionals as a class.