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Source
IM Gelfand, SV Fomin. 1963. Calculus of Variations. Mineola, NY: Dover. Tr. RA Silverman. pp. 26-27.
Problem
“Starting from the point \( P = (a, A) \), a heavy particle slides down a curve in the vertical plane. Find the curve such that the particle reaches the vertical line \( x = b \) in the shortest time” (p. 26).
Although not explicitly assumed in the setup, it is assumed that the particle is under the influence of Earth gravity, and that it slides without friction.
I go through nearly the entire problem, because there was a lot in it that was not immediately evident to me. The only thing I absolutely could not do was derive the family of cycloids from the Euler equation, so at that point I just confirmed that the former is indeed the general solution.
Walkthrough
Part I: Getting from \( v = \frac{ds}{dt} = \sqrt{1 + y’^2}\frac{dx}{dt} \) to \( dt = \frac{\sqrt{1+ y’^2}}{\sqrt{2gy}}dx \).
We start with
\( v = \sqrt{1 + y’^2}\frac{dx}{dt} \tag{1} \)
I set this aside for the moment to rederive some physics that Gelfand and Fomin assume. Let \( (b, B) \) be the point at which the particle reaches \( x = b \). Since only gravity is acting on the particle, conservation of energy implies that the kinetic energy of the particle when it reaches \( (b, B) \) will be the same as the kinetic energy the particle would have after freefall from \( (a, A) \) to \( (a, B) \). Since kinetic energy is given by \( \frac{1}{2}mv^2 \), and mass is presumed to be constant in this example, then \( v \) will have the same value in the two cases. \( \vec{v} \) will point in different directions in the two cases, but we need only the magnitudes up ahead, so there’s nothing to worry about. The punchline of this paragraph is that anything that holds for \( v_y = \frac{dy}{dt} \) under freefall holds equally for \( v = \frac{ds}{dt} \) in the problem we are considering, so we can work with the former for a while, which we will want to do since the trajectory of the particle in that case is already known.
Under freefall, we have
\( g = \frac{d^2y}{dt^2} =_{\textsf{df}} \frac{d}{dt}v_y \tag{2} \)
Integrate both sides to get
\( gt = v_y + C_1 \tag{3} \)
Since \( gt \) should equal the change in velocity over the time interval from \( t_0 \) to \( t \), we know \( C_1 \) must be interpreted in terms of the initial velocity, but in our case the initial velocity is zero, so we have
\( gt = v_y \tag{4} \)
Integrate both sides again to get
\( \frac{1}{2} gt^2 = y + C_2 \tag{5} \)
Since \( \frac{1}{2} gt^2 \) should equal the displacement over the time interval from \(t_0\) to \( t \) (hopefully this is familiar from first-year physics) we know \( C_2 \) must be interpreted in terms of the initial position, but in our case the initial position is zero, so we have
\( \frac{1}{2} gt^2 = y \tag{6} \)
Now, solve for \( t \)—multiply both sides by \( \frac{2}{g} \) and then take the square root of both sides:
\( t = \sqrt{\frac{2y}{g}} \tag{7} \)
Substitute 7 into 4 and do some algebra:
\( v_y = g\sqrt{\frac{2y}{g}} = \sqrt{\frac{2yg^2}{g}} = \sqrt{2gy} \tag{8} \)
Remember that whatever holds for \( v_y \) in freefall holds for \( v \) in Gelfand and Fomin’s scenario, so we can substitute 8 back into 1:
\( \sqrt{2gy} = \sqrt{1 + y’^2}\frac{dx}{dt} \tag{9} \)
And multiplying both sides by \(\frac{dt}{\sqrt{2gy}}\) gives us our target:
\( dt = \frac{\sqrt{1+ y’^2}}{\sqrt{2gy}}dx \tag{10} \)
Part II: Using the Euler equation.
The transit time \( T \) is the sum of all of the little bits of time \( dt \), so from 10 we get
\( \displaystyle T = \int \frac{\sqrt{1 + y’^2}}{\sqrt{2gy}}dx \tag{11} \)
Since the integrand does not contain \( x \), this problem falls into Gelfand and Fomin’s Case 2 (see p. 18 of the book), which means that the Euler equation has a first integral of
\( F – y’F_{y’} = C \tag{12} \)
Let’s work with that.
\( F \) is just the integrand of \( T \), so we plug that in, calculating \( F_{y’} \) along the way:
\( \frac{\sqrt{1 + y’^2}}{\sqrt{2gy}} – y’\left[\frac{1}{\sqrt{2gy}}\left(\frac{1}{2}\right)\frac{1}{\sqrt{1+y’^2}}2y’\right] = C \tag{13} \)
I clean up 13 a bit here, consolidating and canceling a couple of things:
\( \frac{\sqrt{1 + y’^2}}{\sqrt{2gy}} – \frac{y’^2}{\sqrt{2gy}\sqrt{1+y’^2}} = C \tag{14} \)
Multiply both sides by \( \sqrt{1 + y’^2} \):
\( \frac{1 + y’^2}{\sqrt{2gy}} – \frac{y’^2}{\sqrt{2gy}} = C \sqrt{1 + y’^2} \tag{15} \)
Swap the sides of 15 and consolidate:
\( C \sqrt{1 + y’^2} = \frac{1}{\sqrt{2gy}} \tag{16} \)
Square both sides:
\( C^2(1 + y’^2) = \frac{1}{2gy} \tag{17} \)
We want to isolate \( y’ \), so we first distribute \( C^2 \) in the left term, then subtract \( C^2 \) from both sides, and then divide through by \( C^2 \):
\( y’^2 = \frac{1}{2gC^2y} – 1 \tag{18} \)
Gelfand and Fomin seem to take it for granted that readers will immediately recognize this differential equation, and will know that its general solution is a family of cycloids. Figuring this out from first principles eludes me for now, so I have confined myself in the next part to verifying it.
Part III: Verifying the general cycloid solution
We start with the general equation for a cycloid in the next two lines:
\( x = r (\theta – \sin \theta ) + c \tag{19} \)
\( y = r(1 – \cos \theta ) \tag{20} \)
Take the first derivatives of 19 and 20 to get:
\( \frac{dx}{d\theta} = r(1 – \cos \theta) \tag{21} \)
\( \frac{dy}{d\theta} = r \sin \theta \tag {22} \)
Divide 22 by 21:
\( \frac{dy}{dx} = \frac{r \sin \theta}{r(1 – \cos \theta)} \tag{23} \)
Substitute 20 into 23 and, just to make it consistent with Gelfand and Fomin’s format, relabel \( \frac{dy}{dx} \) as \( y’ \):
\( y’ = \frac{r \sin \theta}{y} \tag{24} \)
Square both sides:
\( y’^2 = \frac{r^2 \sin^2\theta}{y^2} \tag{25} \)
Using the relation \( \sin^2\theta + \cos^2\theta =1 \), substitute \(1 – \cos^2 \theta \) into 25:
\( y’^2 = \frac{r^2 (1 – cos^2\theta)}{y^2} \tag{26} \)
Our target doesn’t have any explicit trignometric functions in it, so we want to find something we can substitute into 26 to get rid of the \( \cos^2 \theta \). If we look back through the information we have, we see that 20 gives us what we need. Divide both sides of 20 by \( r \), then add \( \cos \theta – \frac{y}{r} \) to both sides:
\( \cos \theta = 1 – \frac{y}{r} \tag{27} \)
Plug 27 into 26, and then do some algebra:
\( \displaystyle y’^2 = \frac{r^2 \left(1 – \left(1 – \frac{y}{r}\right)^2\right)}{y^2} = \frac{r^2\left(1 – \left(1 – \frac{2y}{r} + \frac{y^2}{r^2}\right)\right)}{y^2} \\ = \frac{r^2\left( \frac{2y}{r} – \frac{y^2}{r^2} \right)}{y^2} = \frac{2yr – y^2}{y^2} = \frac{2r}{y} – 1 \tag{28}\)
Compare the final result with 18 and you will see that 18 has this form with \( r = \frac{1}{4gC^2} \). So the family of cylcoids is indeed the general solution.
Part IV: Finding \( r \).
We still haven’t determined the exact value of \( r \) in the problem, because \( C \)—not to be confused with \( c \) in 19—is as yet undetermined. Instead of trying to use the constraints on the problem to determine \( C \), Gelfand and Fomin find it expedient just to dump \( g \) and \( C \) and all that, and instead figure out directly what \( c \)—not to be confused with \( C \)—and \( r \) are.
Their explanation of why \( c = 0 \) seems clear enough. The only part of their discussion that I initially found mysterious was how they got \( r = \frac{b}{\pi} \) from the fact that \( y’ \) must equal zero when \( x = b \), a move they offer with no further comment. So, I go over that here. We start with
\( x = b \textsf{ implies } y’ = 0 \tag{29} \)
Substitute 24 into 29:
\( x = b \textsf{ implies } \frac{r \sin \theta}{y} = 0 \tag{30} \)
The consequent is true only if the numerator is equal to 0, so we have:
\( x = b \textsf{ implies } r \sin \theta = 0 \tag{31} \)
We know that \( r \) is nonzero since the particle is not stationary, so we have:
\( x = b \textsf{ implies } \sin \theta = 0 \tag{32} \)
Hence,
\(x = b \textsf{ implies } \theta = n\pi \tag{33} \)
Since we want the first time when \( x = b \), we can set \( n = 1 \):
\(x = b \textsf{ implies } \theta = \pi \tag{34} \)
Now, take 19, plug in \( c = 0 \), let \( x = b \), and then combine the rest with the information in 31 and 34 to get
\( b = r \pi \tag{35} \)
So
\( r = \frac{b}{\pi} \tag{36} \)
Put this and \( c = 0 \) back into 19 and 20, and you get the exact solution you are looking for:
\( x = \frac{b}{\pi}(\theta – \sin \theta ) \tag{37} \)
\( y = \frac{b}{\pi}(1 – \cos \theta ) \tag{38} \)
And there you have it.