# Calculus of Variations/CHAPTER XIV

CHAPTER XIV: THE ISOPERIMETRICAL PROBLEM.

• 190 Statement of the problem.
• 191 A simpler form of the integral that appears.
• 192 The function ${\displaystyle F_{1}}$ for this problem.
• 193 Integration of the differential equation that occurs.
• 194 An immediate consequence is the theorem of Steiner: Those portions of curve that are free to vary, are the arcs of equal circles.
• 195 If there exists a curve, which with a given periphery incloses the greatest surface area, that curve is a circle.
• 196 The admissibility that this property belongs to the circle.

Article 190.
The isoperimetrical problem may be briefly stated as follows :

Determine the curve of given length which maximizes or minimizes a certain definite integral.

For example, it may be asked : Among all curves of a given length joining two points, what is the form of the one which produces a minimum surface of revolution about a definite axis; or, along what arc of given length joining two fixed points does a particle under the influence of gravity descend in the shortest time ?

We shall consider here the Problem V of Chapter I, which may be again stated as follows : Suppose that any portion of the plane is bounded in such a way that one can go from any point in it to any other point without crossing the boundaries. In this portion of plane a line returning into itself is to be so constructed that having a given length it incloses the greatest possible surface area.

Let ${\displaystyle x}$ and ${\displaystyle y}$ be such functions of ${\displaystyle t}$ that for two definite values ${\displaystyle t_{0}}$ and ${\displaystyle t_{1}}$ the corresponding points fall together, and that while ${\displaystyle t}$ goes from the smaller value ${\displaystyle t_{0}}$ to the greater value ${\displaystyle t_{1}}$, the point ${\displaystyle x,y}$ traverses in a positive direction the whole curve from the initial point to the end-point.

The surface area, inclosed by the curve, is expressed by the integral

${\displaystyle 1)\qquad I^{(0)}={\frac {1}{2}}\int _{t_{0}}^{t_{1}}(xy'-yx'){\text{d}}t}$

and its perimeter by

${\displaystyle 2)\qquad I^{(1)}=\int _{t_{0}}^{t_{1}}{\sqrt {x'^{2}+y'^{2}}}{\text{d}}t}$

The problem proposed consists in expressing ${\displaystyle x}$ and ${\displaystyle y}$ as functions of ${\displaystyle t}$ in such a manner that the first integral shall have the greatest possible value, while at the same time the second integral retains a given value.

It makes no difference vihere the origin of coordinates has been chosen ; for by a transformation of the origin the second integral remains unchanged while the first integral is changed only by a constant. This does not alter the maximum property of the integral.

One may also add other conditions ; for example : That the curve go through a certain number of fixed points in a given order, or that it is to include certain portions of curve in a given order, etc. The curve will then contain portions along which the variation is not free.

Article 191.
The function ${\displaystyle F}$ is here

${\displaystyle F={\frac {1}{2}}(xy'-yx')-\lambda {\sqrt {x'^{2}+y'^{2}}}}$

Instead of this function we may substitute another, since

${\displaystyle {\frac {{\text{d}}(xy)}{{\text{d}}t}}=xy'+yx'}$

and consequently.

${\displaystyle {\frac {1}{2}}(xy'-yx')={\frac {1}{2}}{\frac {\text{d}}{{\text{d}}t}}(xy)-yx'}$

Now, if we integrate between the limits ${\displaystyle t_{0}\ldots t_{1}}$, the first term of the right-hand side of the above equation vanishes, since the endpoint and the initial-point of the curve coincide. It follows, then, that

${\displaystyle {\frac {1}{2}}\int _{t_{0}}^{t_{1}}(xy'-yx'){\text{d}}t=-\int _{t_{0}}^{t_{1}}yx'{\text{d}}t}$

We may consequently give the function ${\displaystyle F}$ the value

${\displaystyle 3)\qquad F=-x'y-\lambda {\sqrt {x'^{2}+y'^{2}}}}$

From this we have

${\displaystyle {\frac {\partial F}{\partial x'}}=-y-{\frac {\lambda x'}{\sqrt {x'^{2}+y'^{2}}}}\qquad {\frac {\partial F}{\partial y'}}=-{\frac {\lambda y'}{\sqrt {x'^{2}+y'^{2}}}}}$

But since (Art. 187) ${\displaystyle {\frac {\partial F}{\partial x'}}}$ and ${\displaystyle {\frac {\partial F}{\partial y'}}}$ vary in a continuous manner along the portions of curve that vary freely, since also ${\displaystyle \lambda }$ has the same constant value for the whole curve (Art. 185), and since the quantities that are multiplied by ${\displaystyle \lambda }$ are nothing other than the direction-cosines of the tangent to the curve, it follows that the curve at every point, where the variation is free, changes its direction in a continuous manner.

Article 192.
The function ${\displaystyle F_{1}}$ has the value

${\displaystyle 4)\qquad F_{1}={\frac {-\lambda }{({\sqrt {x'^{2}+y'^{2}}})^{3}}}}$

It is evident that ${\displaystyle F_{1}}$ does not change sign, and since a maximum is to enter and consequently ${\displaystyle F_{1}}$ is to be continuously negative, it follows that ${\displaystyle \lambda }$ must be a positive constant.

Article 193.
In order to find the curve itself, we have to integrate the differential equation ${\displaystyle G^{(0)}-\lambda G^{(1)}=0}$. This equation is equivalent (Art. 79) to the two equations

${\displaystyle {\frac {\text{d}}{{\text{d}}t}}{\frac {\partial F}{\partial x'}}-{\frac {\partial F}{\partial x}}=0\qquad {\frac {\text{d}}{{\text{d}}t}}{\frac {\partial F}{\partial y'}}-{\frac {\partial F}{\partial y}}=0}$

Since ${\displaystyle F}$ does not contain ${\displaystyle x}$ explicitly, the first of these equations gives

${\displaystyle 5)\qquad {\frac {\partial F}{\partial x'}}=~{\text{constant}}~~~{\text{or}}~~~y+{\frac {\lambda x'}{\sqrt {x'^{2}+y'^{2}}}}=b}$

where ${\displaystyle b}$ is an arbitrary constant. Since ${\displaystyle {\frac {\partial F}{\partial x'}}}$ varies in a continuous manner for a portion of curve where there is free variation, it follows that the constant ${\displaystyle b}$ retains the same value throughout such a portion of curve. The curve may, however, consist of separate portions which are free to vary, and for these the constant ${\displaystyle b}$ may have different values.

If we take as the independent variable the arcs of curve measured from the origin, we have from 5),

${\displaystyle 6)\qquad {\frac {{\text{d}}x}{{\text{d}}s}}=-{\frac {1}{\lambda }}(y-b)}$

and consequently, since ${\displaystyle \left({\frac {{\text{d}}x}{{\text{d}}s}}\right)^{2}+\left({\frac {{\text{d}}y}{{\text{d}}s}}\right)^{2}=1}$, it follows that

${\displaystyle \left({\frac {{\text{d}}y}{{\text{d}}s}}\right)^{2}=1-{\frac {1}{\lambda ^{2}}}(y-b)^{2}}$

and

${\displaystyle {\frac {{\text{d}}^{2}y}{{\text{d}}s^{2}}}=-{\frac {1}{\lambda ^{2}}}(y-b)={\frac {1}{\lambda }}{\frac {{\text{d}}x}{{\text{d}}s}}}$

It is seen at once, if we integrate the last equation, that

${\displaystyle 7)\qquad {\frac {{\text{d}}^{2}y}{{\text{d}}s^{2}}}={\frac {1}{\lambda }}(x-a)}$

where ${\displaystyle a}$ is an arbitrary constant ; and consequently the equation of the curve is

${\displaystyle 8)\qquad (x-a)^{2}+(y-b)^{2}=\lambda ^{2}}$

From the nature of the curve it is evident that ${\displaystyle \lambda }$ is a positive constant.

Article 194.
An immediate consequence is the theorem of Steiner, that those portions of the cure, which are free to vary, must be the arcs of equal circles. These circles may have different centers, since ${\displaystyle a}$ and ${\displaystyle b}$ are not determined. Each such arc of the circle may, however, lie on different sides of the chord joining two endpoints ; we have, therefore, to ascertain which of the two arcs is the one required.

The solutions of the differential equation are

${\displaystyle x-a=\lambda \cos {\frac {s-s_{0}}{\lambda }}=\lambda \cos t\qquad t-b=\lambda \sin {\frac {s-s_{0}}{\lambda }}=\lambda \sin t}$

as is seen from equations 6) and 7), when differentiated. Since ${\displaystyle \lambda }$ is positive, ${\displaystyle s}$ increases with ${\displaystyle t}$ and since with increasing ${\displaystyle t}$ the curve is traversed in the positive direction, we must take that arc for which this is also true. Let ${\displaystyle C}$ be the center of the circle, ${\displaystyle A_{1}}$ the initial-point, and ${\displaystyle A_{2}}$ the end-point of the arc. That arc will be the right one which lies on the positive side of ${\displaystyle CA_{1}}$, that is, on the side of the increasing ${\displaystyle t}$'s. For if ${\displaystyle t_{1}}$ is the angle which the radius ${\displaystyle CA_{1}}$ makes with the ${\displaystyle X}$-axis, and if ${\displaystyle x_{1},y_{1}}$ are the coordinates of the point ${\displaystyle A_{1}}$, then we have

${\displaystyle \cos t_{1}={\frac {1}{\lambda }}(x_{1}-a)\qquad \sin t_{1}={\frac {1}{\lambda }}(y_{1}-b)}$

and further the angle, which the tangent ${\displaystyle A_{1}B_{1}}$ drawn to the arc at the point ${\displaystyle A_{1}}$ includes with the ${\displaystyle X}$-axis is ${\displaystyle t_{1}+{\frac {\pi }{2}}}$. Consequently we have

${\displaystyle \cos \left(t_{1}+{\frac {\pi }{2}}\right)=-\sin t_{1}=-{\frac {1}{\lambda }}(y_{1}-b)=\left({\frac {{\text{d}}x}{{\text{d}}s}}\right)_{1}}$
${\displaystyle \sin \left(t_{1}+{\frac {\pi }{2}}\right)=\cos t_{1}={\frac {1}{\lambda }}(x_{1}-a)=\left({\frac {{\text{d}}y}{{\text{d}}s}}\right)_{1}}$

formulae, which have the right signs. This would not be true if we took the other arc and also the tangent which is drawn in the other direction. Hence that arc is always to be taken which, looking out from the center, is traversed in the positive direction.

Article 195.
If no conditions are imposed upon the curve and it is required to find among all isoperimetrical lines that one which offers the greatest surface area, then the question is not of an absolute maximum, since the curve may be shoved anywhere in the plane without an alteration in its shape. The problem may be stated more accurately by saying that the integral which represents the surface area is not to admit of a positive increment, when all possible variations are introduced. The problem thus formulated leads to exactly the same necessary conditions as before, namely that the first variation is to vanish, and consequently we have the same differential equation to solve. We have also the same condition for ${\displaystyle \lambda }$. Since the second variation can never be positive, and consequently ${\displaystyle F_{1}}$ can not change its sign, we conclude as above that ${\displaystyle \lambda }$ is positive. Since the whole curve is free to vary and since ${\displaystyle {\frac {\partial F}{\partial x'}}}$ and ${\displaystyle {\frac {\partial F}{\partial y'}}}$ are continuous functions for the whole trace, the constants ${\displaystyle a}$ and ${\displaystyle b}$ are the same for the whole curve; however, they remain undetermined. We have, consequently, the following result:

If there exists a closed curve which with a given periphery includes the greatest surface area, this curve is a circle.

Article 196.
However, it has not as yet been proved that this property belongs to the circle. The treatment of the second variation is not sufficient, since only such variations have been employed vihere the distance betvieen two corresponding points, and also the difEerence in direction at these points do not exceed certain limits.

The further proof has to be made that every other curve forms the boundary of a smaller surface area. The proof that the circle has this maximum property, ( a proof which is omitted in all previous solutions of the problem), has been considered so difficult that its solution has been denied to be in the province of the Calculus of Variations. We shall, however, in the next Chapter show that in the theorems already treated a means of overcoming this difficulty is offered. It will be seen that without the use of the second variation the desired result is reached in all cases where the function ${\displaystyle F_{1}}$ does not change sign, not only at any point of the curve but also for any direction at any point.