Calculus of Variations/CHAPTER VII

CHAPTER VII: REMOVAL OF CERTAIN LIMITATIONS THAT HAVE BEEN MADE. INTEGRATION OF THE DIFFERENTIAL EQUATION $G=0$ FOR THE PROBLEMS OF CHAPTER I.

• 96 Instead of a sing-le regular trace, the curve may consist of a finite number of such traces.
• 97 The first derivatives of $F$ with respect to $x'$ and $y'$ vary in a continuous manner for the curve $G=0$, even if there are sudden changes in the direction of this curve.
• 98 Explanation of the result given in the preceding article.
• 99 Summary.
• 100 Solution of the differential equation for Problem I of Chapter I.
• 101,102 The discontinuous solution.
• 103 The equation $G=0$ solved for Problem II, Article 9.
• 104 The two fixed points must lie on the same loop of the cycloid.
• 105 Through two points may be drawn one and only one cycloidal-loop, which does not include a cusp.
• 106 Problem III. Problem of the shortest line on a surface.
• 107 The same result derived in a different manner.
• 108 Problem IV. Surface of rotation which offers the least resistance.
• 109,110 Solution of the equation $G=0$ for Problem IV of Chapter I.

Article 96.
In the derivation of the formulae of Chapter V, it was presupposed that the portion of curve under consideration changed its direction in a continuous manner throughout its whole trace ; that is, $x'$, $y'$ varied in a continuous manner. We shall now assume only that the curve is composed of regular portions of curve ; so that, therefore, the tangent need not vary continuously at every point of the curve. Then it may be shown as follows that each portion of curve must satisfy the differential equation $G=0$. For if the curve consists of two regular portions $AC$ and $CB$, then among all possible variations of $AB$ there exist those in which $CB$ remains unchanged and only $AC$ is subjected to variation.

As above, we conclude that this portion of curve must satisfy the differential equation $G=0$. The same is true of $CB$.

We may now do away with the restriction that the curve consists of one regular trace, and assume that it consists of a finite number of regular traces.

Article 97.
Suppose that the function $F$ does not contain explicitly the variable $x$, and consequently ${\frac {\partial F}{\partial x}}=0$. Instead of the equation $G=0$, let us take $G_{1}=0$, or

${\frac {\partial F}{\partial x}}-{\frac {\text{d}}{{\text{d}}t}}{\frac {\partial F}{\partial x'}}=0$.

It follows that

${\frac {\partial F}{\partial x'}}={\text{ constant}}$,

the constant being independent of $t$; a priori, however, we do not know that ${\frac {\partial F}{\partial x'}}$ does not undergo a sudden change at points of discontinuity of $x'$ and $y'$. Consequently, the more important is the following theorem for the integration of the differential equation $G=0$:

Even if $x'$, $y'$, and thereby also the direction of the curve, suffer at certain points sudden changes, nevertheless, the quantities ${\frac {\partial F}{\partial x'}}$, ${\frac {\partial F}{\partial y'}}$ vary in a continuous manner throughout the whole curve for which $G=0$.

If $t'$ is a point of discontinuity in the curve, then on both sides of $t'$ we take the points $\tau$ and $\tau '$ in such a manner that within the portions $\tau \ldots t'$ and $t'\ldots \tau '$ there is no other discontinuity in the direction of the curve. Then a possible variation of the curve is also the one by which $t_{0}\ldots \tau$ and $\tau '\ldots t_{1}$ remain unaltered and $\tau \ldots \tau '$ b only the portion $\tau \ldots \tau '$ is varied. Here the points $\tau$ and $\tau '$ are supposed to remain fixed, while $t'$ is subjected to any kind of sliding.

The variation of the integral

$I=\int _{t_{0}}^{t_{1}}F(x,y,x',y')~{\text{d}}t$,

then depends only upon the variations of the sum of the integrals

$\int _{\tau }^{t'}F(x,y,x',y')~{\text{d}}t+\int _{t'}^{\tau '}F(x,y,x',y')~{\text{d}}t$.

Since the first variation of this expression must vanish, we necessarily have (Art. 79)

$0=\int _{\tau }^{t'}G(y'\xi -x'\eta )~{\text{d}}t+\int _{t'}^{\tau '}G(y'\xi -x'\eta )~{\text{d}}t+\left[{\frac {\partial F}{\partial x'}}\xi +{\frac {\partial F}{\partial y'}}\eta \right]_{\tau }^{t'}+\left[{\frac {\partial F}{\partial x'}}\xi +{\frac {\partial F}{\partial y'}}\eta \right]_{t'}^{\tau }$.

Since $G=0$ along the whole curve, it follows that

$\left[{\frac {\partial F}{\partial x'}}\xi +{\frac {\partial F}{\partial y'}}\eta \right]_{\tau }^{t'}+\left[{\frac {\partial F}{\partial x'}}\xi +{\frac {\partial F}{\partial y'}}\eta \right]_{t'}^{\tau }=0$.

The quantities $\xi$ and $\eta$ are both zero at the fixed points $\tau$ and $\tau '$; and, if we denote the values that may belong to the quantity ${\frac {\partial F}{\partial x'}}$, according as we approach the point $t'$ from the points $\tau$ or $\tau '$ by

$\left[{\frac {\partial F}{\partial x'}}\right]_{t'}^{-}$ and $\left[{\frac {\partial F}{\partial x'}}\right]_{t'}^{+}$,

the above expression becomes

$\left(\left[{\frac {\partial F}{\partial x'}}\right]_{t'}^{-}-\left[{\frac {\partial F}{\partial x'}}\right]_{t'}^{+}\right)(\xi )'+\left(\left[{\frac {\partial F}{\partial y'}}\right]_{t'}^{-}-\left[{\frac {\partial F}{\partial y'}}\right]_{t'}^{+}\right)(\eta )'=0$,

where $(\xi )'$ and $(\eta )'$ are the values of $\xi$ and $\eta$ at the point $t'$. Since the quantities $(\xi )'$ and $(\eta )'$ are quite arbitrary, it follows that their coefficients in the above expression must respectively vanish, so that

$\left[{\frac {\partial F}{\partial x'}}\right]_{t'}^{-}=\left[{\frac {\partial F}{\partial x'}}\right]_{t'}^{+}$ and $\left[{\frac {\partial F}{\partial y'}}\right]_{t'}^{-}=\left[{\frac {\partial F}{\partial y'}}\right]_{t'}^{+}$;

that is, the quantities ${\frac {\partial F}{\partial x'}}$ and ${\frac {\partial F}{\partial y'}}$ vary in a continuous manner by the transition from one regular part of the curve to the other, even if $x'$ and $y'$ at this point suffer sudden changes.

This is a new necessary condition for the existence of a maximum or a minimum of the integral $I$, which does not depend upon the nature of the differential equation $G=0$.

Article 98.
The question naturally arises: How is it possible that the functions ${\frac {\partial F}{\partial x'}}$, ${\frac {\partial F}{\partial y'}}$, which depend upon $x'$ and $y'$, vary in a continuous manner, even when $x'$ and $y'$ experience discontinuities? To answer this question we may say that the composition of these functions is of a peculiar nature, viz., the terms which contain $x'$, $y'$ are multiplied by functions which vanish at the points considered. This is illustrated more clearly in the example treated in Art. 100. The theorem is of the greatest importance in the determination of the constant. In the special dp case of the preceding article, where ${\frac {\partial F}{\partial x'}}=$ constant it is clear that this constant must have the same value for all points of the curve. The theorem may also be used in many cases to prove that the direction of the curve nowhere changes in a discontinuous manner, and consequently does not consist of several regular portions but of one single regular trace. This is also illustrated in the examples which follow (Arts. 100 et seq.).

Article 99.
We may give here a summary of what has been obtained through the vanishing of the first variation as necessary conditions for the existence of a maximum or a minimum of the integral $I$:

1) The curve offering the maximum or minimum must satisfy the differential equation

$G\equiv {\frac {\partial ^{2}F}{\partial x\partial y'}}-{\frac {\partial ^{2}F}{\partial y\partial x'}}+F_{1}\left(x'{\frac {{\text{d}}y'}{{\text{d}}t}}-y'{\frac {{\text{d}}x'}{{\text{d}}t}}\right)=0$,

or, what is the same thing, the two equations

$G_{2}\equiv {\frac {\partial F}{\partial x}}-{\frac {\text{d}}{{\text{d}}t}}\left({\frac {\partial F}{\partial x'}}\right)=0$, $G_{2}\equiv {\frac {\partial F}{\partial y}}-{\frac {\text{d}}{{\text{d}}t}}\left({\frac {\partial F}{\partial y'}}\right)=0$;

2) The two derivatives of the function $F$ with respect to $x'$ and $y'$ must vary in a continuous m,anner even at the points where the direction of the curve does not vary continuously.

In order to establish the criteria by means of which it may be ascertained whether the curve determined through the equation $G=0$ offers a maximum or a minimum, we must investigate the terms of the second dimension in $\Delta I$ of Chapter V. First, however, to make clear what has already been written, we may apply our deductions to some of the problems already proposed.

\begin{center}SOLUTION OF THE DIFFERENTIAL EQUATION G=0 FOR THE PROBLEMS OF CHAPTER I.\end{center}

Article 100.
Let us consider Problem I of Art. 7. The integral which we have to minimize is

${\frac {S}{2\pi }}=\int _{t_{0}}^{t_{1}}y{\sqrt {x'^{2}+y'^{2}}}~{\text{d}}t$. $\qquad {\text{[1]}}$

Hence

$F=y{\sqrt {x'^{2}+y'^{2}}}$, $\qquad {\text{[2]}}$

and consequently

${\frac {\partial F}{\partial x'}}={\frac {yx'}{\sqrt {x'^{2}+y'^{2}}}}$; $\qquad {\frac {\partial F}{\partial y'}}={\frac {yy'}{\sqrt {x'^{2}+y'^{2}}}}$. $\qquad {\text{[3]}}$

From this it is seen that ${\frac {\partial F}{\partial x'}}$ and ${\frac {\partial F}{\partial y'}}$ are proportional to the direction cosines of the tangent to the curve at any point $x(t)$, $y(t)$; and, since ${\frac {\partial F}{\partial x'}}$ and ${\frac {\partial F}{\partial y'}}$ must vary everywhere in a continuous manner, it follows also that the direction of the curve varies everywhere in a continuous manner except for the case where $y=0$. But the quantity ${\sqrt {x'^{2}+y'^{2}}}$ varies in a discontinuous manner if $x'$ and $y'$ are discontinuous; at the same time, however, $y$ is equal to zero, as is more clearly seen in the figure below.

Since $F$ does not contain $x$ explicitly, we may use the equation

$G_{1}=0$, or ${\frac {\partial F}{\partial x'}}={\frac {yx'}{\sqrt {x'^{2}+y'^{2}}}}=\beta$, $\qquad {\text{[4]}}$

where $\beta$ is the constant of integration. Hence

$y^{2}\left({\frac {{\text{d}}x}{{\text{d}}t}}\right)^{2}=\beta ^{2}\left[\left({\frac {{\text{d}}x}{{\text{d}}t}}\right)^{2}+\left({\frac {{\text{d}}y}{{\text{d}}t}}\right)^{2}\right]$. $\qquad {\text{[5]}}$

The solution of this equation is the catenary:

$x=\alpha +\beta t$, $\qquad y={\frac {\beta }{2}}(e^{t}+e^{-t}),$ $\qquad {\text{[6]}}$

where $\alpha$ is a second arbitrary constant.

Article 101.
A discontinuous solution. If we take the arc $s$ as independent variable instead of the variable $t$, the differential equation of the curve is

$y{\frac {{\text{d}}x}{{\text{d}}s}}=\beta$.

Suppose that $\beta =0$, which value it must retain within the whole interval $t_{0}\ldots t_{1}$. Further, since $y\neq 0$ at the point $P_{0}$. it dx follows that ${\frac {{\text{d}}x}{{\text{d}}s}}=\cos(\phi )=0$ (where $\phi$ is the angle which the tanas gent makes with the $X$-axis), and that $\cos(\phi )$ must remain zero until $y=0$; that is the point which describes the curve must move along the ordinate $P_{0}M_{0}$ to the point $M_{0}$. At this point ${\frac {{\text{d}}x}{{\text{d}}s}}$ cannot, and must not, equal zero if the point is to move to $P_{1}$. Hence, at $M_{0}$ there is a sudden change in the direction of the curve, as there is again at the point $M_{1}$. The curve giving the minimum surface of revolution is consequently, in this case, offered by the irregular trace $P_{0}M_{0}M_{1}P_{1}$. The case where $\beta =0$ may be regarded as an exceptional case. The unconstrained lines $P_{0}M_{0}$ and $P_{1}M_{1}$, i.e., $x=x_{0}$ and $x=x_{1}$ satisfy the condition $G=0$, since $y'G=G_{1}$, and for these values $G_{1}=0$; also for these lines, $y\neq 0$. But $G\neq 0$ for the restricted portion $M_{0}M_{1}$ and is, in fact, equal to 1.

Article 102.
We may prove as follows that the two ordinates and the section of the $X$-axis give a minimum. This is seen at once when we have shown that the first variation for all allowable deformations is positive. The problem is a particular case of Art. 79.

The first variation may be decomposed into several parts (cf. Arts. 79 and 81):

$\delta I=-\int _{M_{0}}^{P_{0}}Gw_{N}~{\text{d}}s+\left[{\frac {\partial F}{\partial x'}}\xi +{\frac {\partial F}{\partial y'}}\eta \right]_{M_{0}}^{P_{0}}-\int _{M_{1}}^{M_{0}}Gw_{N}~{\text{d}}s+\left[{\frac {\partial F}{\partial x'}}\xi +{\frac {\partial F}{\partial y'}}\eta \right]_{M_{1}}^{M_{0}}-\int _{P-1}^{M_{1}}Gw_{N}~{\text{d}}s+\left[{\frac {\partial F}{\partial x'}}\xi +{\frac {\partial F}{\partial y'}}\eta \right]_{P_{1}}^{M_{1}}$.

Now all the boundary terms are zero, since

${\frac {\partial F}{\partial x'}}={\frac {y'x}{\sqrt {x'^{2}+y'^{2}}}}$, $\qquad {\frac {\partial F}{\partial y'}}={\frac {yy'}{\sqrt {x'^{2}+y'^{2}}}}$,

and therefore both are zero at the points $M_{0}$ and $M_{1}$, while $\xi$ and $\eta$ are zero at $P_{0}$ and $P_{1}$. In the first and third integrals $G=0$; in the second this function equals unity, and if we reverse the limits, ${\text{d}}s$ is positive, as is also $w_{N}$. Hence the first variation $\delta I$ is always positive.

When the arbitrary constant $<\beta \neq 0$, the curve consists of one regular trace that lies wholly above the $X$-axis. Further investigation is necessary to determine when this curve offers in reality a minimum.

Article 103.
In the second problem (Art. 9) we have for the time of falling the integral

$T=\int _{t_{0}}^{t_{1}}{\frac {\sqrt {x'^{2}+y'^{2}}}{\sqrt {4gy+\alpha ^{2}}}}~{\text{d}}t$. $\qquad {\text{[1]}}$

That this expression may, in reality, express the time of falling (the time and, therefore, also the increment ${\text{d}}t$ being essentially a positive quantity), the two roots that appear under the integral sign must always have the same sign. Since ${\sqrt {4gy+\alpha ^{2}}}$ can always be chosen positive, it follows that ${\sqrt {x'^{2}+y'^{2}}}$ must be positive within the interval $T_{0}\ldots t_{1}$.

It might happen, however, if we express $x$ and $y$ in terms of $t$, that $x'$ and $y'$ might both vanish for a value of $t$ within the interval $t_{0}\ldots t_{1}$. In this case the curve has at the point $x$,$y$, which belongs to this value of $t$, a singular point, at which the velocity of the moving point is zero.

Suppose that this is the case for $t=t'$, and that the corresponding point is $x_{0}$, $y_{0}$ so that we have

$x=x_{0}+a(t-t')^{m}+\cdots$, $\qquad y=y_{0}+(t-t')^{m}+\cdots$,

where $m\geq 2$, and at least one of the two quantities $a$ and $b$ is different from zero.

Then is

$x'^{2}+y'^{2}=m^{2}(a^{2}+b^{2})(t-t')^{2(m-1)}+\cdots$,

and

${\sqrt {x'^{2}+y'^{2}}}=m{\sqrt {a^{2}+b^{2}}}(t-t')^{m-1}+\cdots$.

Here we may suppose ${\sqrt {a^{2}+b^{2}}}$ positive.

If now $m$ is odd, then for small values of $t-t'$, the expression on the right is positive, and hence ${\sqrt {x'^{2}+y'^{2}}}$ always has a positive sign.

If on the contrary $m$ is even, equal to 2, say, then the curve has at the point $x_{0}$,$y_{0}$ a cusp, since here ${\sqrt {x'^{2}+y'^{2}}}$ has a positive or a negative value according as $t>t'$ or $t<t'$.

If therefore the above integral is to express the time, ${\sqrt {x'^{2}+y'^{2}}}$ cannot always be put equal to the same series of $t$, but must after passing the cusp be put equal to the opposite value of the series. We therefore limit ourselves to the consideration of a portion of the curve which is free from singular points.

Such limitations must often be made in problems, since otherwise the integrals have no definite meaning. Hence with this supposition ${\sqrt {x'^{2}+y'^{2}}}$ will never equal zero.

We may then write:

$F={\frac {\sqrt {x'^{2}+y'^{2}}}{\sqrt {4gy+\alpha ^{2}}}}$, $\qquad {\text{[2]}}$

and consequently

${\frac {\partial F}{\partial x'}}={\frac {1}{\sqrt {4gy+\alpha ^{2}}}}{\frac {x'}{\sqrt {x'^{2}+y'^{2}}}}$, $\qquad {\frac {\partial F}{\partial y'}}={\frac {1}{\sqrt {4gy+\alpha ^{2}}}}{\frac {y'}{\sqrt {x'^{2}+y'^{2}}}}$. $\qquad {\text{[3]}}$

From this we may conclude, in a similar manner as in the first example, that ${\frac {\partial F}{\partial x'}}$,${\frac {\partial F}{\partial y'}}$ are proportional to the direction cosines of the tangent of the curve at the point $x$, $y$. Since now ${\frac {\partial F}{\partial x'}}$,${\frac {\partial F}{\partial y'}}$ vary in a continuous manner along the whole curve, and since, further, ${\sqrt {4gy+\alpha ^{2}}}$ has a definite value which is different from zero, it follows also that the direction of the required curve varies in a continuous manner, or the curve must consist of one single trace.

Also here $F$ is independent of $x$, and consequently we employ the differential equation $G_{1}=0$, from which we have

${\frac {\partial F}{\partial x'}}={\frac {1}{\sqrt {4gy+\alpha ^{2}}}}{\frac {x'}{\sqrt {x'^{2}+y'^{2}}}}=C$, $\qquad {\text{[4]}}$

where $C$ is an arbitrary constant.

If $C$ is equal to zero, then in the whole extent of the curve $C$ must equal zero; and consequently, since ${\sqrt {4gy+\alpha ^{2}}}$ is neither 0 nor $\infty$, ${\frac {x'}{\sqrt {x'^{2}+y'^{2}}}}=\cos(\alpha )$ must always equal zero; that is, the curve must be a vertical line. Neglecting this self-evident case, $C$ must have a definite value which is always the same for the whole curve and different from zero.

From [4], it follows that

${\text{d}}x^{2}=C^{2}(4gy+\alpha ^{2})({\text{d}}x^{2}+{\text{d}}y^{2})$,

or, if we absorb $4g$ in the arbitrary constant and write

${\frac {\alpha ^{2}}{4g}}=a$, and $4gC^{2}=c^{2}$,

we have

${\text{d}}x^{2}=c^{2}(y+a)({\text{d}}x^{2}+{\text{d}}y^{2})$;

whence

${\text{d}}x={\frac {c(y+a){\text{d}}y}{\sqrt {(y+a)[1-c^{2}(y+1)]]}}}$. $\qquad {\text{[5]}}$

In order to perform this last integration, write

${\text{d}}\phi ={\frac {c{\text{d}}y}{\sqrt {(y+a)[1-c^{2}(y+a)]}}}$; $\qquad {\text{[6]}}$

therefore

${\text{d}}x=(y+a){\text{d}}\phi$. $\qquad [5^{\text{a}}]$

In the expression for ${\text{d}}\phi$, write

$2c^{2}(y+a)=1-\xi$. $\qquad [7]$

Then is

$2[1-c^{2}(y+a)]=1+\xi$, $\qquad [8]$

and

$2c^{2}{\text{d}}y=-{\text{d}}\xi$. $\qquad [9]$

Therefore

${\text{d}}\phi =-{\frac {{\text{d}}\xi }{\sqrt {1-\xi ^{2}}}}$, $\qquad [10]$

and hence

$\xi =\cos(\phi )$. $\qquad [11]$

Here the constant of integration may be omitted, since $\phi$ itself is fully arbitrary.

Hence,

$y+a={\frac {1}{2c^{2}}}(1-\cos(\phi ))$, and, from [$5^{\text{a}}$], $x+x_{0}={\frac {1}{2c^{2}}}(\phi -\sin(\phi ))$; $\qquad [12]$

${\displaystyle }$

equations, which represent a cycloid.

The constants of integration $x_{0}$, $c$ are determined from the condition that the curve is to go through the two points $A$ and $B$. Now develop $x$ and $y$ in powers of $\phi$: then in $y$ the lowest power is $\phi ^{2}$, and in $x$ it is $\phi$; so that the curve has in reality a cusp for $\phi =0$, and this is repeated for $\phi =2\pi ,4\pi ,\ldots$.

$A$ and $B$ must lie between two consecutive cusps (Art. 104).

The curve may be constructed, if we draw a horizontal line through the point $-x_{0}$, $-a$, and construct on the under side of this line a circle with radius $1/(2c^{2})$, which touches the horizontal line at the point $-x_{0}$, $-a$. Let this circle roll in the positive $X$-direction on the horizontal line, then the original point of contact describes a cycloid which goes through $A$ and B and which satisfies the differential equation.

Article 104.
That the points $A$ and $B$ cannot lie upon different loops of a cycloid may be seen as follows: For simplicity, let the initial velocity $\alpha$ be zero and shift the origin of coordinates so as to get rid of the constants.

The equation of the cycloid is then

$x=r(\phi -\sin(\phi ))$, $\qquad yr(1-\cos(\phi ))$,

where we have written $r$ in the place of $1/(2c^{2})$.

The cycloidal arc is seen from the accompanying figure. Take j two points lying upon different loops very near and symmetrically situated with respect to an apex, and let us compare the time it would take to travel from one of these points to the other by the way of the apex with the time taken over a straight line joining them. The parameters of the two points may be expressed by

$\phi _{0}=2\pi -\psi _{0}$, $\qquad \phi _{1}=2\pi +\psi _{0}$.

The time required to go by the way of the apex is

$T={\frac {1}{\sqrt {2g}}}\int _{t_{0}}^{t_{1}}{\sqrt {\frac {x'^{2}+y'^{2}}{y}}}~{\text{d}}t={\frac {1}{\sqrt {2g}}}\int _{s_{0}}^{s_{1}}{\frac {{\text{d}}s}{\sqrt {y}}}$.

Now

${\text{d}}x=r(1-\cos(\phi ))~{\text{d}}\phi$,

and

${\text{d}}y=r\sin(\phi )~{\text{d}}\phi$,

so that

${\text{d}}s={\sqrt {{\text{d}}x^{2}+{\text{d}}y^{2}}}=2r\sin(\phi /2)~{\text{d}}\phi$,

and consequently

$T={\frac {1}{\sqrt {2g}}}\int _{\phi _{0}}^{\phi _{1}}{\frac {2r\sin(\phi /2)}{{\sqrt {r}}{\sqrt {1-\cos(\phi )}}}}{\text{d}}\phi ={\sqrt {\frac {r}{g}}}\int _{\phi _{0}}^{\phi _{1}}{\text{d}}\phi$

${\sqrt {\frac {r}{g}}}(\phi _{1}-\phi _{0})={\sqrt {\frac {r}{g}}}[2\pi +\psi _{0}-2\pi +\psi _{0}]=2{\sqrt {\frac {r}{g}}}\psi _{0}$.

The component of velocity across the horizontal line from $\phi _{0}$ to $\phi _{1}$ is $\left[v{\frac {{\text{d}}x}{{\text{d}}s}}\right]$ or, since ${\frac {{\text{d}}x}{{\text{d}}s}}=\sin {\frac {\phi }{2}}$ and $v^{2}=2gy$, this component is equal to

$\left[{\sqrt {2gr}}{\sqrt {1-\cos(\phi )}}\sin {\frac {\phi }{2}}\right]_{\phi _{0}}^{\phi _{1}}=2{\sqrt {gr}}\sin ^{2}{\frac {\psi _{0}}{2}}$.

The length of the line to be traversed from $\phi _{0}$ to $\phi _{1}$ is

$x_{1}-x_{0}=r[\phi _{1}-\phi _{0}-\sin \phi _{1}+\sin \phi _{0}]=2r[\psi _{0}-\sin \psi _{0}]$.

Hence, the time required is

$T_{1}={\frac {2r(\psi _{0}-\sin \psi _{0})}{2{\sqrt {gr}}\sin ^{2}(\psi _{0}/2)}}$,

and consequently

${\frac {T_{1}}{T}}={\frac {2r(\psi _{0}-\sin \psi _{0})}{2{\sqrt {gr}}\sin ^{2}{\frac {\psi _{0}}{2}}\cdot 2{\sqrt {\frac {r}{g}}}\psi _{0}}}={\frac {\psi _{0}-\sin \psi _{0}}{2\psi _{0}\sin ^{2}{\frac {\psi _{0}}{2}}}}={\frac {{\frac {\psi _{0}^{3}}{3!}}-{\frac {\psi _{0}^{5}}{5!}}+\cdots }{2\psi _{0}\left[{\frac {\psi _{0}}{2}}-{\frac {1}{3!}}\left({\frac {\psi _{0}}{2}}\right)^{3}+\cdots \right]^{2}}}$.

Hence,

${\frac {T_{1}}{T}}<{\frac {\frac {\psi _{0}^{3}}{3!}}{2\psi _{0}\left[{\frac {\psi _{0}}{2}}-{\frac {1}{6}}\left({\frac {\psi _{0}}{2}}\right)^{3}+\cdots \right]^{2}}}$,

or

${\frac {T_{1}}{T}}<{\frac {1}{3}}{\frac {1}{\left(1-{\frac {\psi _{0}^{2}}{24}}+\cdots \right)^{3}}}$.

It follows, therefore, for small values of $\psi _{0}$ that

$T_{1}<T$.

From this it is evident that a path of the particle including an apex cannot give a minimum.

Article 105.
Corresponding to the two constants that are contained in the general solution of the differential equation $G=0$ of Art. 103, it is seen that we have all the curves of the family $G=0$, if we vary $r$ and slide the cycloid along the $X$-axis.

We shall now show that only one of these cycloids can contain the two points $A$ and $B$ on the same loop. Suppose that the ordinates of the points $A$ and $B$ to be such that $DB>AC$, and consider any other cycloid with the same parameter $r$ described about the horizontal $X$-axis with the origin at $O$. Through $O$ draw a chord parallel to $AB$ and move this chord through parallel positions until it leaves the curve. We note that in these positions the ordinate $A'C'$ increases continuously, since it can never reach the lowest point of the cycloid, and that the arc $A'B'$ continuously diminishes. Consequently the ratio $A'B':A'C'$ continuously diminishes. When $A'$ coincides with the origin this ratio is infinite, and is zero when the chord becomes tangent to the curve.

Then for some one position we must have

${\frac {A'B'}{A'C'}}={\frac {AB}{AC}}$.

Since the points $A$ and $B$ are fixed, the length $AB$ and the direction $AB$ are both determined.

If $A'B'=AB$, then $A'C'=AC$, and a cycloid can be drawn through $A$ and $B$ as required. But if $A'B'\neq AB$, then our cycloid does not fulfill the required conditions.

Next choose a quantity $r'$ such that

$r:r'=AC:A'C'$.

With $O$ as the center of similitude increase the coordinates of our cycloid parameters in the ratio $r:r'$. These coordinates then become

$x=r'(\phi -\sin \phi )$, $\qquad y=r'(1-\cos \phi )$,

which are the coordinates of a new cycloid.

The latter cycloid is similar to the first, since the transformation moves the ordinate $A'C'$ and the chord $A'B'$ parallel to themselves. Their transformed lengths are respectively

${\frac {r'}{r}}A'C'=AC\qquad$ and $\qquad {\frac {r'}{r}}A'B'=AB$,

giving us a cycloid with the requisite lengths for the ordinate $AC$ and the chord $AB$.

Further, there is but one cycloid which answers the required conditions. For, if we already had $A'B'=AB$ and $A'C'=AC$, the only value of $r'$ which could then make ${\frac {r'}{r}}A'B'=AB$ is $r'=r$. Hence through the two points $A$ and $B$ there can be constructed one and only one cycloid-loop with respect to the $X$-axis.

Article 106.
Problem III. Problem of the shortest line on a surface. This problem cannot in general be solved, since the variables in the differential equation cannot be separated and the integration cannot be performed. Only in a few instances has one succeeded in carrying out the integration and thus represented the curve which satisfies the differential equation.

This, for example, has been done in the case of the plane, the sphere and all the other surfaces of the second degree.

As a simple example, we will take the problem of the shortest line between two points on the surface of a sphere. The radius of the sphere is put equal to 1, and the equation of the sphere is given in the form

$x^{2}+y^{2}+z^{2}=1$.

Now writing:

$x=\cos u$, $y=\sin u\cos v$, $z=\in u\sin v$, $\qquad [1]$

then $u=$ constant and $v=$ constant are the equations of the parallel circles and of the meridians respectively.

The element of arc is

${\text{d}}s={\sqrt {{\text{d}}u^{2}+\sin ^{2}u{\text{d}}v^{2}}}$, $\qquad [2]$

and consequently the integral which is to be made a minimum is

$L=\int _{t_{0}}^{t_{1}}{\sqrt {u'^{2}+v'^{2}\sin ^{2}u}}~{\text{d}}t$; $\qquad [3]$

so that here we have

$F={\sqrt {u'^{2}+v'^{2}\sin ^{2}u}}$

and

${\frac {\partial F}{\partial u'}}={\frac {u'}{\sqrt {u'^{2}+v'^{2}\sin ^{2}u}}}$, $\qquad {\frac {\partial F}{\partial v'}}={\frac {v'\sin ^{2}u}{\sqrt {u'^{2}+v'^{2}\sin ^{2}u}}}$. $\qquad [5]$

Since $F$ does not contain the quantity $v$, we will use the equation $G_{1}=0$, and have:

${\frac {\partial F}{\partial v'}}={\frac {v'\sin ^{2}u}{\sqrt {u'^{2}+v'^{2}\sin ^{2}u}}}=c$,

where $c$ is an arbitrary constant, which has the same value along the whole curve.

If for the initial point $A$ of the curve $u\neq 0$, and consequently, therefore not the north pole of the sphere, then $c$ will be everywhere equal to zero, only if $v'=0$. We must therefore have $v$ constant. It follows as a solution of the problem that $A$ and $B$ must lie on the same meridian.

If this is not the case, then always $c\neq 0$. It is easy to see that $c<1$; we may therefore write $\sin c$ instead of $c$, and have

${\frac {v'\sin ^{2}u}{\sqrt {u'^{2}+v'^{2}\sin ^{2}u}}}=\sin c$, $\qquad [6]$

or

${\text{d}}v={\frac {\sin c~{\text{d}}u}{\sin u{\sqrt {\sin ^{2}u-\sin c}}}}$. $\qquad [7]$

If we write

$\cos u=\cos c\cos w$, $\qquad [8]$

then is

${\text{d}}v={\frac {\sin c~{\text{d}}w}{1-\cos ^{2}c\cos ^{2}w}}$;

since 1 may be replaced by $\sin ^{2}w+\cos ^{2}w$, we have

${\text{d}}v={\frac {\sin c~{\text{d}}w}{\sin ^{2}w+\cos ^{2}w\sin ^{2}c}}={\frac {\sin c{\frac {{\text{d}}w}{\cos ^{2}w}}}{\sin ^{2}c+\tan ^{2}w}}={\frac {{\text{d}}{\frac {\tan w}{\sin c}}}{1+{\frac {\tan ^{2}w}{\sin ^{2}c}}}}$.

Therefore

$v-\beta =\tan ^{-1}\left({\frac {\tan w}{\sin c}}\right)$,

where $\beta$ represents an arbitrary constant.

It follows that

$\tan(v-\beta )={\frac {\tan w}{\sin c}}$. $\qquad [9]$

Eliminating $w$ by means of [8], we have

$\tan u\cos(v-\beta )=\tan c$. $\qquad [10]$

This is the equation of the curve which we are seeking, expressed in the spherical coordinates $u$, $v$.

In order to study their meaning more closely, we may express $u$, $v$ separately through the arc $s$, where $s$ is measured from the intersection of the zero meridian with the shortest line.

Through [7] the expression [2] goes into

${\text{d}}s={\frac {\sin u~{\text{d}}u}{\sqrt {\sin ^{2}u-\sin ^{2}c}}}$,

and this, owing to the substitution [8], becomes

${\text{d}}s={\text{d}}w$,

and, therefore, if $b$ is a new constant,

$s-b=w$. $\qquad [11]$

Hence, from equations [8] and [9] we have the following equations:

$\cos u=\cos c\cos(s-b)$, $\qquad \cot(v-\beta )=\sin c\cot(s-b)$. $\qquad [12]$

But these are relations which exist among the sides and the angles of a right-angled spherical triangle.

If we consider the meridian drawn from the north pole which cuts at right angles the curve we are seeking, then this meridian forms with the curve, and any other meridian, a triangle to which the above relations may be applied.

Therefore, the curve which satisfies the differential equation must itself be the arc of a great circle. The constants of integration $c$, $b$, $\beta$ are determined from the conditions that the curve is to pass through the two points $A$ and $B$.

The geometrical interpretation is: that $c$ is the length of the geodetic normal from the point $u=0$ to the shortest line; $s-b$, the arc from the foot of this normal to any point of the curve, that is, the difference of length between the end-points of this arc; and $v-\beta$, the angle opposite this arc.

If we therefore assume that the zero meridian passes through $A$, then $b$ is the length of arc of the shortest line from $A$ to the normal, and $\beta$ the geographical longitude of the foot of this normal.

Article 107.
We may derive the same results by considering the differential equation $G=0$.

Since

$F_{1}u'^{2}={\frac {\partial ^{2}F}{\partial v'^{2}}}$,

we have

$F_{1}={\frac {\sin ^{2}u}{(u'^{2}+v'^{2}\sin ^{2}u)^{3/2}}}$.

This value, substituted in

$G\equiv {\frac {\partial ^{2}F}{\partial v\partial u'}}-{\frac {\partial ^{2}F}{\partial u\partial v'}}+F_{1}(v'u''-u'v'')=0$,

causes this expression to become

$-[2\cos uu'^{2}v'+\sin ^{2}u\cos uv'^{3}]+\sin u(v'u''-u'v'')=0$,

or

${\frac {{\text{d}}v}{{\text{d}}u}}\left[2+\sin ^{2}u\left({\frac {{\text{d}}v}{{\text{d}}u}}\right)^{2}\right]\cos u+\sin u{\frac {{\text{d}}^{2}v}{{\text{d}}u^{2}}}=0$.

In this equation write

$1)\qquad w=\sin u{\frac {{\text{d}}v}{{\text{d}}u}}$,

and we have

$\cot u(w+w^{3})+{\frac {{\text{d}}w}{{\text{d}}u}}=0$,

or

$2)\qquad {\frac {{\text{d}}w}{w+w^{3}}}+\cot u~{\text{d}}u=0$.

Integrating the last equation, it follows that

$\ln \left({\frac {w}{\sqrt {1+w^{2}}}}\sin u\right)=c$,

and consequently

$3)\qquad w^{2}\sin ^{2}u=C^{2}(1+w^{2})$.

Suppose that $A$ is the north pole of the sphere, $u$ the angular distance measured from $A$ along the arc of a great circle, and $v$ the angle which the plane of this great circle makes with the plane of a great circle through the point $B$.

Hence for all curves of the family $G=0$ that pass through $A$, we must have $C=0$, since $\sin u=0$ for $u=0$. It follows also that $w=0$, and consequently,

$\sin u{\frac {{\text{d}}v}{{\text{d}}u}}=0$,

or

$v=$ constant.

Hence, as above, $A$ and $B$ must lie on the arc of a great circle. Next, if $A$ is not taken as the pole, then always $C\neq 0$, and is less than unity. It follows then at once from equations 1) and 3) that

${\text{d}}s^{2}={\text{d}}u^{2}+\sin ^{2}u{\text{d}}v^{2}={\text{d}}u^{2}[1+w^{2}]$,

or

${\text{d}}s=\pm {\frac {\sin u~{\text{d}}u}{\sqrt {\sin ^{2}u-\sin ^{2}C}}}$, (where we have written $\sin ^{2}C$ for $C^{2}$),

and

${\text{d}}v={\frac {\sin C~{\text{d}}u}{\sin(u){\sqrt {\sin ^{2}u-\sin ^{2}C}}}}$.

Writing $\cos u=\cos C\cos t$, these two equations when integrated become, as in the last article,

$\cos u=\cos C\cos(s-b)$, $\qquad \cot(v-\beta )=\sin C\cot(s-b)$.

Article 108.
Problem IV. Surface of rotation which offers the least resistance. To solve this problem we saw (Art 12) that the integral

$I=\int _{t_{0}}^{t_{1}}{\frac {xx'^{3}}{x'^{2}+y'^{2}}}~{\text{d}}t$

must be a minimum.

We have here

$F={\frac {xx'^{3}}{x'^{2}+y'^{2}}}$

and we see that $F$ is a rational function of the arguments $x'$ and $y'$. For such functions Weierstrass has shown that there can never be a maximum or a minimum value of the integral. But leaving the general problem for a later discussion (Art. 173), we shall confine our attention to the problem before us.

We may determine the function $F-1$ from the relation

${\frac {\partial ^{2}F}{\partial x'\partial y'}}=-x'y'F_{1}$.

It is seen that

$F_{1}={\frac {2xx'(3y'^{2}-x'^{2})}{(x'^{2}+y'^{2})^{3}}}$.

We may take $x$ positive, and also confine our attention to a portion of curve along which $x$ increases with $t$, so that $x'$ is also positive.

Consequently $F_{1}$ has the same sign as $3y'^{2}-x'^{2}$, or of $3\sin ^{2}\lambda -\cos ^{2}\lambda$, where $\lambda$ is the angle that the tangent to the curve at the point in question makes with the $X$-axis.

$F_{1}$ is therefore positive, if $|\tan \lambda |>{\frac {1}{\sqrt {3}}}$, and is negative, if $|\tan \lambda |<{\frac {1}{\sqrt {3}}}$, for the portion of curve considered.

We shall see later (Art. 117) that $F_{1}$ must have a positive sign in order that the integral be a minimum. Hence, for the present problem, $|\tan \lambda |$ must be greater than ${\frac {1}{\sqrt {3}}}$ for the portion of curve considered; and as this must be true for all points of the curve at which $x'$ has a positive sign, the tangent at any of these points cannot make an angle greater than $30^{\circ }$ with the $X$-axis (see Todhunter, Researches in the Calculus of Variations p. 168).

Article 109.
We shall next consider the difEerential equation $G=0$ of the problem.

Since $F$ does not contain explicitly the variable $y$, we may best employ the equation

$-x'G=G_{2}\equiv {\frac {\partial F}{\partial y}}-{\frac {\text{d}}{{\text{d}}t}}{\frac {\partial F}{\partial y'}}=0$.

We have at once

${\frac {\partial F}{\partial y'}}=$ constant,

or

$-{\frac {2xx'^{3}y'}{(x'^{2}+y'^{2})^{2}}}=C$.

Now, if there is any portion of the surface offering resistance, which lies indefinitely near the axis of rotation, then the constant must be zero, since $x=0$ makes $C=0$.

If $C=0$, we have

$x'^{3}y'=0$,

and consequently

$x'=0$ or $y'=0$.

From this we derive

$x=$ constant or $y=$ constant.

In the first case, the surface would be a cylinder of indefinite length, with the $Y$-axis as the axis of rotation, and with an indefinitely small radius (since by hypothesis a portion of the surface lies indefinitely near the $Y$-axis); in the second case, the resistingsurface would be a disc of indefinitely large diameter. These solutions being without significance may be neglected, and we may therefore, suppose that the surface offering resistance has no points in the neighborhood of the $Y$-axis. This disproves the notion once held that the body was egg-shaped.

Article 110.
We consider next the differential equation

$-{\frac {2xx'^{3}y'}{(x'^{2}+y'^{2})^{2}}}=C$,

where $C$ is different from zero. We may take $x$ positive, and as the constant $C$ must always retain the same sign (Art. 97), it follows that the product $x'y'$ cannot change sign.

Instead of retaining the variable $t$, let us write

$t=-y$,

and

${\frac {{\text{d}}x}{{\text{d}}y}}=u$.

The differential equation is then

${\frac {-2xu^{3}}{(u^{2}+1)^{2}}}=C$.

That we may write $-y$ in the place of $t$, is seen from the fact that $x'y'$ cannot change sign, and consequently either $x$ is continuously increasing with increasing $y$, or is continuously decreasing when $y$ is increased. Hence, corresponding to a given value of $y$ there is one value of $x$.

We have then

$x=-{\frac {C(u^{2}+1)^{2}}{2u^{3}}}=-{\frac {C}{2}}(u+2u^{-1}+u^{-3})$,

and

${\frac {{\text{d}}x}{{\text{d}}u}}={\frac {{\text{d}}x}{{\text{d}}y}}{\frac {{\text{d}}y}{{\text{d}}u}}=u{\frac {{\text{d}}y}{{\text{d}}u}}=-{\frac {C}{2}}(1-2u^{-2}-3u^{-4})$,

or

${\frac {{\text{d}}y}{{\text{d}}u}}=-{\frac {C}{2}}(u^{-1}-2u^{-3}-4u^{-5})$;

consequently

$y=-{\frac {C}{2}}\left[\ln u+u^{-2}+{\frac {3}{4}}u^{-4}\right]+C_{1}$.

The equations

$x=-{\frac {C}{2}}(u+2u^{-1}+u^{-3})$, $\qquad y=-{\frac {C}{2}}\left[\ln u+u^{-2}+{\frac {3}{4}}u^{-4}\right]+C_{1}$,

determine a family of curves, one of which is the arc, which generates the surface of revolution that gives a minimum value, if such a minimum exists. For such a curve we have all the real points if we give to $u$ all real values from 0 to $+\infty$. Among these values is ${\sqrt {3}}$, and as we saw above, it is necessary that

${\frac {{\text{d}}y}{{\text{d}}x}}={\frac {1}{u}}>{\frac {1}{\sqrt {3}}}$ continuously,

or

${\frac {1}{u}}<{\frac {1}{\sqrt {3}}}$ continuously.

In other words, if the acute angle which the tangent at any point of the arc makes with the $X$-axis is less than $30^{\circ }$, it must continue less than $30^{\circ }$ for the other points of the arc, and if it is greater than $30^{\circ }$ for any point of the arc it must remain greater than $30^{\circ }$ for all points of the arc. Hence, if $P$ is the point at which the inclination of the tangent with the $X$-axis is $30^{\circ }$, we shall have on one side of $P$ that portion of curve for which the inclination is less than $30^{\circ }$, and on the other side the portion of curve for which the inclination is greater than $30^{\circ }$. The arc in question must belong entirely to one of the two portions.