Chapter 16

Calculus of Variations

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Problem 5961

Find sufficient conditions for a strong minimum of the integral

I=∫x1x2f(x,y) dx I = \int _{x_{1}}^{x_{2}} f(x, y) \, dx

given that the endpoints of the minimizing curve lie respectively on the lines x=x1x = x_{1} and x=x2x = x_{2}.

Examples:

  1. I=∫−1+1(y3−3x4y) dxI = \int_{-1}^{+1} (y^{3} - 3x^{4}y) \, dx;

  2. I=∫01y(y+x) dxI = \int_{0}^{1} y(y+x) \, dx;

  3. I=∫−a+a(x2y+13y3−a2y) dx; a>0I = \int_{-a}^{+a} \left(x^{2}y + \frac{1}{3}y^{3} - a^{2}y\right) \, dx; \ a > 0;

  4. I=∫0x1(6ax−y2)y dx; a>0, x1>0I = \int_{0}^{x_{1}} (6ax - y^{2})y \, dx; \ a > 0, \ x_{1} > 0.

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Problem 5962

Find the minimum of the integral

I=∫(x1,y1)(x2,y2)(y2+2xyy′) dx. I = \int _{(x_{1},y_{1})}^{(x_{2},y_{2})} (y^{2} + 2xyy') \, dx.
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Problem 5963

Find the extremum of the integral

I=∫(x1,y1)(x2,y2)(xy+y2−2y2−2y2y′) dx. I = \int _{(x_{1},y_{1})}^{(x_{2},y_{2})} (xy + y^{2} - 2y^{2} - 2y^{2}y') \, dx.
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Problem 5964

Find the extremum of the integral

I=∫(x1,y1)(x2,y2)(x2+y2+yy′) dx. I = \int _{(x_{1},y_{1})}^{(x_{2},y_{2})} (x^{2} + y^{2} + yy') \, dx.
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Problem 5965

Find the extremal of the integral

I=∫AB1+y′2y dx, I = \int _{A}^{B} \frac{\sqrt{1+y'^{2}}}{y} \, dx,

passing through given points AA and BB, lying in the upper half-plane y>0y > 0.

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Problem 5966

Find the extremals of the integral

I=∫(x1,y1)(x2,y2)y(1+y′2) dx,y1>0,y2>0. I = \int _{(x_{1},y_{1})}^{(x_{2},y_{2})} \sqrt{y(1+y'^{2})} \, dx, \quad y_{1} > 0, \quad y_{2} > 0.
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Problem 5967

Find in quadratures the extremals of the integral

I=∫M1M2yn1+y′2 dx. I = \int _{M_{1}}^{M_{2}} y^{n} \sqrt{1+y'^{2}} \, dx.
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Problem 5968

Find the extremals of the integral

I=∫M1M2rn ds I = \int _{M_{1}}^{M_{2}} r^{n} \, ds

(in polar coordinates rr, θ\theta), expressing the nn-th moment of the arc M1M2M_{1}M_{2} relative to the origin.

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Problem 5969

A material point moves in a plane so that its speed vv is inversely proportional to the distance to the origin, and ∫v ds\int v \, ds, taken along the trajectory between any two of its points, is an extremum. Find the direction of the force acting on the point, and the orbit of the point.

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Problem 5970

A plane curve is the orbit of a material point moving freely under the action of a central force, and at the same time is a brachistochrone (see problem 5989) for a different motion, this time under a different central force. Show that if vv and v′v' are the speeds of the point in these two motions at some point of the trajectory, then for all points of the trajectory the product vv′vv' remains constant.

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Problem 5971

Among all curves joining the origin to the line x=ax = a, determine the one which, together with the axis OxOx and the given line, encloses an area of given magnitude SS, and has the smallest length.

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Problem 5972

On the sides OAOA and OBOB of an angle with vertex OO are given points AA and BB; it is required to join them by the shortest line ACBACB, under the condition that the area OACBOOACBO has a given magnitude SS.

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Problem 5973

Find the surface of revolution about the axis OXOX, having a given area, and enclosing the largest volume, given that the surface intersects the axis of revolution exactly twice.

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Problem 5974

Find all integrals

I=∫(x1,y1)(x2,y2)F(x,y,y′) dx, I = \int _{(x_{1},y_{1})}^{(x_{2},y_{2})} F(x, y, y') \, dx,

for which straight lines serve as extremals.

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Problem 5975

Find all integrals

I=∫x1x2F(x,y,y′) dx, I = \int _{x_{1}}^{x_{2}} F(x, y, y') \, dx,

for which the extremals are the parabolas y=ax+bx2y = ax + bx^{2}.

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Problem 5976

Among the curves joining the points A(1,3)A(1,3) and B(2,5)B(2,5), find the one that gives the minimum of the integral

I=∫12y′(1+x2y′) dx. I = \int _{1}^{2} y'(1+x^{2}y') \, dx.
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Problem 5977

Among the lines joining the points A(0,1)A(0,1) and B(1,4)B(1,4), find the one that gives the minimum of the integral

I=∫01(12xy+yy′+y′2) dx. I = \int _{0}^{1} (12xy + yy' + y'^{2}) \, dx.
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Problem 5978

Among the lines joining the points A(1,0)A(1,0) and B(2,1)B(2,1), find the one that gives the minimum of the integral

I=∫12(xy′2−y) dx. I = \int _{1}^{2} (xy'^{2} - y) \, dx.

Find the character of the minimum.

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Problem 5979

Find the minimum of the integral

I=∫(1,0)(2,1)(xy′4−2yy′3) dx I = \int _{(1,0)}^{(2,1)} (xy'^{4} - 2yy'^{3}) \, dx

and determine its character.

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Problem 5980

Show that if the integrand of the integral

I=∫(A)(B)F(x,y′) dx I = \int _{(A)}^{(B)} F(x, y') \, dx

does not contain yy explicitly, then every extremal can always be surrounded by a field, and for a strong minimum it suffices that the condition ∂2F(x,y′)∂y′2>0\frac{\partial^{2} F(x,y')}{\partial y'^{2}} > 0 hold at every point of the field for any direction.

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Problem 5981

Find the minimum of the integral

I=∫(0,0)(x1,y1)ex1+y′2 dx. I = \int _{(0,0)}^{(x_{1},y_{1})} e^{x} \sqrt{1+y'^{2}} \, dx.
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Problem 5982

Show that the line y=0y = 0 gives only a weak minimum of the integral

I=∫(0,0)(1,0)(y′2−yy′3) dx, I = \int _{(0,0)}^{(1,0)} (y'^{2} - yy'^{3}) \, dx,

although Legendre's condition ∂2F∂y′2>0\frac{\partial^{2} F}{\partial y'^{2}} > 0 holds at points of the extremal y=0y = 0 for all y′y'.

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Problem 5983

Find the critical conjugate points on the extremals of the integral

I=∫ABy1+y′2 dx I = \int _{A}^{B} y\sqrt{1+y'^{2}} \, dx

by means of Jacobi's differential equation.

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Problem 5984

Find the minimum of the integral

I=∫(x1,y1)(x2,y2)(y′2+2y′y−9y2) dx. I = \int _{(x_{1},y_{1})}^{(x_{2},y_{2})} (y'^{2} + 2y'y - 9y^{2}) \, dx.
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Problem 5985

Find the minimum of the integral

I=∫(x1,y1)(x2,y2)1+y2y′2 dx. I = \int _{(x_{1},y_{1})}^{(x_{2},y_{2})} \frac{1+y^{2}}{y'^{2}} \, dx.

Find the extremals, check the conditions of Legendre, Jacobi, and Weierstrass, and find the character of the minimum.

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Problem 5986

Find the minimum of the integral

I=∫(x1,y1)(x2,y2)yy′2 dx. I = \int _{(x_{1},y_{1})}^{(x_{2},y_{2})} \frac{y}{y'^{2}} \, dx.

Find the extremals, check the conditions of Legendre, Jacobi, and Weierstrass, and find the character of the minimum.

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Problem 5987

Show that the extremals of the integral

I=∫(x0,y0)(x1,y1)ymG(y′) dx I = \int _{(x_{0},y_{0})}^{(x_{1},y_{1})} y^{m}G(y') \, dx

are obtained from one another by a similarity transformation and translation parallel to the axis OXOX. Here the tangents at conjugate critical points intersect on the axis OXOX.

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Problem 5988

Among the curves joining the points M1(−a,b)M_{1}(-a,b) and M2(a,b)M_{2}(a,b), a>0a > 0, b>0b > 0, find the one that, when rotated about OXOX, gives the smallest surface. Here:

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(a)

find the extremals;

(b)

investigate the question of the possibility of drawing an extremal through two given points;

(c)

in the case where an extremal passing through two given points exists, determine which of them gives the minimum, and find the character of this minimum.

Problem 5989

Among the curves lying in the plane XOYXOY (the axis OYOY directed vertically downward), find the one along which, sliding down, a heavy point will descend most quickly from the point O(0,0)O(0,0) to the point M(a,b)M(a,b). Here the initial velocity of the point is zero.

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Problem 5990

In the vertical plane XOYXOY, find the curve, moving along which a heavy material point will arrive in the shortest time from the point O(0,0)O(0,0) to the point M(a,b)M(a,b); here it is given that the initial velocity at the point OO equals 2gh\sqrt{2gh}.

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Problem 5991

Find the shape that must be given to a body of revolution so that it experiences the smallest resistance under uniform motion along the axis of revolution. Here it is assumed that at speed vv the element dσd\sigma experiences, in the direction of the axis OXOX, a resistance kv2cos⁡3φ dσkv^{2}\cos^{3}\varphi \, d\sigma. Here φ\varphi is the angle of the normal to dσd\sigma with OXOX. It is required to find the extremals, consider Jacobi's condition, and investigate the character of the minimum.

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(a)
(b)
(c)
Problem 5992

Among the lines joining the points A(−1,−1)A(-1,-1) and B(1,1)B(1,1), find the one that gives the minimum of the integral

I=∫(−1,−1)(1,1)(x2y′2+12y2) dx. I = \int _{(-1,-1)}^{(1,1)} (x^{2}y'^{2} + 12y^{2}) \, dx.

Investigate the total increment of the integral ΔI\Delta I.

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Problem 5993

Find the minimum of the integral

I=∫(0,0)(1,1)xy′2 dx. I = \int _{(0,0)}^{(1,1)} xy'^{2} \, dx.
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Problem 5994

Show that for the integral

I=∫(−1,0)(1,0)(x2y′2+xy′3) dx I = \int _{(-1,0)}^{(1,0)} (x^{2}y'^{2} + xy'^{3}) \, dx

OXOX is an extremal for which the second variation is positive, but the integral nevertheless has no strong minimum.

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Problem 5995

Show that there is no continuous function y=φ(x)y = \varphi (x) giving the minimum of the integral

I=∫−11x2y′2 dx I = \int _{-1}^{1} x^{2}y'^{2} \, dx

and satisfying the conditions: φ(−1)=−1\varphi (-1) = -1; φ(1)=1\varphi (1) = 1. (Weierstrass's example.)

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Problem 5996

Find the minimum of the integral

∫(0,0)(π,1)y′2cos⁡2x dx. \int _{(0,0)}^{(\pi ,1)} y'^{2}\cos ^{2}x \, dx.
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Problem 5997

Find the minimum of the integral

∫(−1,−1)(1,1)x23y′2 dx. \int _{(-1,-1)}^{(1,1)} x^{\frac{2}{3}}y'^{2} \, dx.
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Problem 5998

Find the maximum of the integral

I=∫(x1,y1)(x2,y2)(12(xy′−yx′)−Rx′2+y′2) dτ,R>0. I = \int _{(x_{1},y_{1})}^{(x_{2},y_{2})} \left(\frac{1}{2}(xy'-yx') - R\sqrt{x'^{2}+y'^{2}}\right) \, d\tau , \quad R > 0.
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Problem 5999

Find the geodesic lines on the surface of revolution r2=2f(z)r^{2} = 2f(z) (rr, φ\varphi, zz are cylindrical coordinates). Determine for which arcs M1M2M_{1}M_{2} the integral ∫M1M2ds\int_{M_{1}}^{M_{2}} ds is indeed a minimum.

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Problem 6000

Find the geodesic lines of the circular cylinder r=Rr = R.

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Problem 6001

Find the geodesic lines of the cone r=ztg⁡αr = z \operatorname {tg} \alpha.

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Problem 6002

Investigate the geodesic lines on a sphere of radius RR, find the extremals, investigate Jacobi's condition, and determine the character of the minimum of the integral ∫ds\int ds.

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Problem 6003

Find the geodesic lines on the paraboloid of revolution 2z=r22z = r^{2}. Under what condition will the geodesic line between two points on the same meridian be the shortest distance?

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Problem 6004

For the oblate ellipsoid of revolution

x2+z2a2+y2b2=1;a>b \frac{x^{2}+z^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1; \quad a > b

the equator (y=0y=0, x2+z2=a2x^{2}+z^{2}=a^{2}) is a geodesic line. Show that the critical point A′A' conjugate to the point AA on this line is at a distance πb\pi b from AA.

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Problem 6005

We have two circles of radius rr with centers at the points M1(−a,b)M_{1}(-a,b) and M2(a,b)M_{2}(a,b). A curve lies in the upper half-plane, ending on the given circles, and when rotated about OXOX gives a surface of the smallest area. At what points and at what angle does it intersect the circles?

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Problem 6006

Write down the transversality condition for the integral

I=∫M1M2(12(xy′−x′y)−Rx′2+y′2) dτ. I = \int _{M_{1}}^{M_{2}} \left(\frac{1}{2}(xy'-x'y) - R\sqrt{x'^{2}+y'^{2}}\right) \, d\tau .
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Problem 6007

For which integrals of the form ∫F(x,y,y′) dx\int F(x,y,y') \, dx does transversality reduce to orthogonality?

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Problem 6008

Find all integrals ∫f(x,y,y′) dx\int f(x,y,y') \, dx for which the transversals intersect the extremals at an angle of 45°45°.

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Problem 6009

Among the lines joining the point O(0,0)O(0,0) with the curve y3=2−xy^{3} = 2-x, find the one that gives the minimum of the integral

I=12∫(0,0)(x1,y1)(y′2−y2) dx. I = \frac{1}{2} \int _{(0,0)}^{(x_{1},y_{1})} (y'^{2}-y^{2}) \, dx.
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Problem 6010

In the vertical plane XOYXOY with the axis OYOY directed downward, find the curve, moving along which a heavy material point will arrive in the shortest time from the point OO to a given line y=(x−c)tg⁡βy = (x-c)\operatorname {tg}\beta; 0<β<π0 < \beta < \pi. Here the initial velocity is zero.

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Problem 6011

Find the field function p(x,y)p(x,y) and the field of extremals itself, passing through the origin, of the integral

∫(0,0)(x,y)1+y′2y dx,y>0. \int _{(0,0)}^{(x,y)} \frac{\sqrt{1+y'^{2}}}{y} \, dx, \quad y > 0.
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Problem 6012

It is given that a family of extremals passes through the point (x0,y0)(x_{0},y_{0}). Denoting by (x1,y1)(x_{1},y_{1}) the point of intersection of these extremals with some transversal, show that the integral

∫(x0,y0)(x1,y1)f(x,y,y′) dx \int _{(x_{0},y_{0})}^{(x_{1},y_{1})} f(x,y,y') \, dx

taken along the extremal has a constant value.

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Problem 6013

A one-parameter family of extremals is intersected by two transversals. Prove that the integral ∫f(x,y,y′) dx\int f(x,y,y') \, dx, taken along any extremal between the points of intersection with the transversals, has a constant value.

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Problem 6014

Among the lines joining the line x=0x=0 with the point A(x1,y1)A(x_{1},y_{1}), where x1>0x_{1} > 0, y1>0y_{1} > 0, find the one that gives the minimum of the integral

∫0x1+y′2y dx,y>0. \int _{0}^{x} \frac{\sqrt{1+y'^{2}}}{y} \, dx, \quad y > 0.

Also form the Hamilton--Jacobi equation and find the equidistant curves for the extremals transversally intersecting the line x=0x=0.

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Problem 6015

Find the Hamilton--Jacobi equation and the transversals for the field of extremals y=axy=ax of the integral ∫y′2 dx\int y'^{2} \, dx.

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Problem 6016

Find the transversals for the field of extremals y=axy=ax of the integral

I=∫G(y′) dx. I = \int G(y') \, dx.
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Problem 6017

Find the transversals for the field of extremals y=x+ay=x+a of the integral

I=∫(xy′4−2yy′3) dx. I = \int (xy'^{4} - 2yy'^{3}) \, dx.
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Problem 6018

Find the transversals for the field of extremals y=x−x2ay = x - \frac{x^{2}}{a} of the integral

∫y(1−y′2) dx,a>0, x>0; y>0. \int \sqrt{y(1-y'^{2})} \, dx, \quad a > 0, \ x > 0; \ y > 0.
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Problem 6019

Find the extremals of the integral

I=∫M1M2G(y)1+y′2 dx I = \int _{M_{1}}^{M_{2}} G(y)\sqrt{1+y'^{2}} \, dx

by solving the Hamilton--Jacobi equation, and conversely --- find the solution of the Hamilton--Jacobi equation, starting from the first integral of Euler's equation.

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Problem 6020

Find the extremals of the integral

I=∫x0x1x2+y21+y′2 dx I = \int _{x_{0}}^{x_{1}} \sqrt{x^{2}+y^{2}} \sqrt{1+y'^{2}} \, dx

by means of solving the Hamilton--Jacobi equation.

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Problem 6021

Knowing that the paths of light rays are curves along which light propagates in the shortest time, and knowing that the speed of light is inversely proportional to the refractive index of the medium nn, show that the brachistochrone (problem 5989) is the path of light in a medium in which the refractive index nn is expressed by the formula

n=12gy. n = \frac{1}{\sqrt{2gy}}.

In connection with this, find the paths of light rays in such a medium and the equation of the wave surfaces. For the brachistochrones, find the synchrones, i.e., the curves reached simultaneously by points moving from the point (0,0)(0,0) along different brachistochrones.

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Problem 6022

Find the trajectory of a material point moving under the influence of gravity, starting from the principle of least action, according to which for the trajectory of the point the first variation of the "action" integral vanishes:

I=∫2(U+h) ds. I = \int \sqrt{2(U+h)} \, ds.

Here UU is the potential; in this case U=−gyU = -gy, if the axis OYOY is directed upward.

Form the Hamilton--Jacobi equation and, having solved it, find the extremals, in particular those passing through the origin. Investigate the character of the minimum of the integral.

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Problem 6023

Find the trajectory for the motion of a point in a plane under the action of a repulsive force from the axis OXOX, proportional to the distance of the point from this line and directed parallel to the axis OYOY, under the condition that the energy integral has the form v22−y22=0\frac{v^{2}}{2} - \frac{y^{2}}{2} = 0, starting from the "action" integral

I=∫x0x1y1+y′2 dx,y>0. I = \int _{x_{0}}^{x_{1}} y\sqrt{1+y'^{2}} \, dx, \quad y > 0.
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Problem 6024

A material point describes the circle r=2Rcos⁡θr = 2R\cos \theta (in polar coordinates rr and θ\theta) of radius RR under the action of a central force kr5\frac{k}{r^{5}}, inversely proportional to the fifth power of the distance from the center of force, lying at the origin. Show that the "action" integral, taken along any arc of this circle: −π2<θ1≤θ≤θ2<π2-\frac{\pi }{2} < \theta_{1} \leq \theta \leq \theta_{2} < \frac{\pi }{2}, is a strong minimum.

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Problem 6025

Study the motion of a material point under the action of a central force proportional to the distance from the center OO of the force, starting from the principle of least action and applying the Jacobi--Hamilton method.

Show that at the conjugate critical points of any trajectory of the point, regarded as an extremal of the "action" integral, the tangents are mutually perpendicular.

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Problem 6026

Solving the Hamilton--Jacobi equation, find the geodesic lines on a Liouville surface, for which the line element has the form

ds2=(f(x)+g(y))(dx2+dy2). ds^{2} = (f(x)+g(y))(dx^{2}+dy^{2}).
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Problem 6027

Prove that for surfaces whose line element is determined by the equality

ds2=E(u,v) du2+2F(u,v) du dv+G(u,v) dv2, ds^{2} = E(u,v) \, du^{2} + 2F(u,v) \, du \, dv + G(u,v) \, dv^{2},

the Jacobi--Hamilton equation for the problem of the geodesic line has the form

G(∂w∂u)2−2F∂w∂u∂w∂v+E(∂w∂v)2=EG−F2. G\left(\frac{\partial w}{\partial u}\right)^{2} - 2F\frac{\partial w}{\partial u}\frac{\partial w}{\partial v} + E\left(\frac{\partial w}{\partial v}\right)^{2} = EG - F^{2}.
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Problem 6028

Find the shortest distance between a given point and the parabola y2=2pxy^{2} = 2px.

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Problem 6029

Find the minimum of the integral

∫01(12y′2+yy′+y′+y) dx, \int _{0}^{1} \left(\frac{1}{2}y'^{2} + yy' + y' + y\right) \, dx,

if the values at the endpoints of the interval are not given.

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Problem 6030

Find the shortest distance between two circles C1C_{1} and C2C_{2}, lying in one plane.

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Problem 6031

For the problem of finding the minimum of the integral

I=∫f(x,y,y′) dx I = \int f(x,y,y') \, dx

the Hamilton--Jacobi equation is given: wx=H(x,y,wy)w_{x} = H(x,y,w_{y}). Find the function f(x,y,y′)f(x,y,y').

Example. (∂w∂x)2+(∂w∂y)2=1\left(\frac{\partial w}{\partial x}\right)^{2} + \left(\frac{\partial w}{\partial y}\right)^{2} = 1.

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Problem 6032

The Hamilton--Jacobi equation for the problem of the minimum of the integral ∫f(x,y,y′) dx\int f(x,y,y') \, dx has the form

(∂w∂x)2+(∂w∂y)2=x2+y2. \left(\frac{\partial w}{\partial x}\right)^{2} + \left(\frac{\partial w}{\partial y}\right)^{2} = x^{2}+y^{2}.

Find the function f(x,y,y′)f(x,y,y').

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Problem 6033

For the problem of the minimum of the integral

∫x1x2f(x,y,y′,y′′) dx \int _{x_{1}}^{x_{2}} f(x,y,y',y'') \, dx

find Euler's equation and the boundary conditions, given that only the coordinates of the endpoints are specified.

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Problem 6034

Show that Euler's equation for the conditions of problem 6033 has a first integral:

∂f∂y′−ddx(∂f∂y′′)=const, \frac{\partial f}{\partial y'} - \frac{d}{dx}\left(\frac{\partial f}{\partial y''}\right) = \mathrm{const},

if ff does not contain yy. In the case where ff does not contain xx, the first integral can be written in the form:

f−y′(∂f∂y′−ddx(∂f∂y′′))−y′′∂f∂y′′=const. f - y'\left(\frac{\partial f}{\partial y'} - \frac{d}{dx}\left(\frac{\partial f}{\partial y''}\right)\right) - y''\frac{\partial f}{\partial y''} = \mathrm{const}.

Finally, if ff contains neither xx nor yy, then Euler's equation has a second integral:

f−y′′∂f∂y′′=C1+C2y′. f - y''\frac{\partial f}{\partial y''} = C_{1} + C_{2}y'.
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Problem 6035

Among the lines passing through the points A(0,0)A(0,0) and B(1,0)B(1,0), find the one that minimizes the integral ∫01y′′2 dx\int_{0}^{1} y''^{2} \, dx in two cases:

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(a)

if y′(0)=ay'(0)=a, y′(1)=by'(1)=b;

(b)

if no other conditions are given.

Problem 6036

Find the extremals of the integral ∫(y′′)n dx\int (y'')^{n} \, dx.

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Problem 6037

The minimum is sought of the integral

∫x1x2f(x,y,y′,y′′) dx, \int _{x_{1}}^{x_{2}} f(x,y,y',y'') \, dx,

with the condition given at the variable upper limit: φ(x,y,y′)=0\varphi (x,y,y')=0. Find the condition at this end, generalizing the transversality condition.

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Problem 6038

The same problem, but at the upper limit the value y′(x2)y'(x_{2}) and the curve h(x,y)=0h(x,y)=0 are given, on which the point (x2,y2)(x_{2},y_{2}) must lie.

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Problem 6039

Find a plane curve such that the area bounded by the arc M1M2M_{1}M_{2}, the normals M1C1M_{1}C_{1} and M2C2M_{2}C_{2} at the points M1M_{1} and M2M_{2}, and the arc of the evolute C1C2C_{1}C_{2}, is a minimum. Here the quantities x1x_{1}, y1y_{1}, y1′y'_{1} and x2x_{2}, y2y_{2}, y2′y'_{2}, i.e., the positions and directions of the ends of M1M2M_{1}M_{2}, are given.

Example. x1=y1=0x_{1}=y_{1}=0, y1′=1y'_{1}=1; x2=1x_{2}=1, y2=0y_{2}=0; y2′=−1y'_{2}=-1.

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Problem 6040

The previous problem, but without the directions at the points M1M_{1} and M2M_{2} being given. Here M1(0,0)M_{1}(0,0) and M2(2π,0)M_{2}(2\pi ,0).

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Problem 6041

Find a necessary and sufficient condition for the function f(x,y,y′,…,y(n))f(x,y,y',\ldots ,y^{(n)}), for any choice of the function yy, to be the total derivative with respect to xx of some function

φ(x,y,y′,…,y(n−1)). \varphi (x,y,y',\ldots ,y^{(n-1)}).
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Problem 6042

Is the function

v=x2y′′′+xy′′+(2xy−1)y′+y2+2x v = x^{2}y''' + xy'' + (2xy-1)y' + y^{2} + 2x

a total derivative of some function u(x,y,y′,y′′)u(x,y,y',y''), and if so, find the function uu.

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Problem 6043

A vertical rod is clamped at its ends OO and AA. The lower end OO is fixed, the upper end can move horizontally. On the upper end AA a force PP acts, directed downward. Determine the critical load P0P_{0} at which the rectilinear shape ceases to be stable. The modulus of elasticity is EE. The moment of inertia of the cross-section relative to the axis passing through the center of gravity of the section perpendicular to the plane of bending equals II.

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Problem 6044

The same question, but under the condition that the lower end is clamped and the upper end is free, i.e., at the lower end θ=0\theta = 0, and nothing is specified at the upper end.

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Problem 6045

The same problem under the condition of simply supported ends.

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Problem 6046

The same problem, but under the condition that both ends are clamped, i.e., at them v=0v=0, v′=0v'=0.

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Problem 6047

The same problem, but the lower end is clamped and the upper end is simply supported, i.e., at the upper end v=0v=0, and v′v' is not specified.

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Problem 6048

A horizontal rod ABAB of length ll is freely supported at one end, and rests on an elastic support at the other end, and is compressed by two horizontal forces PP. Determine the critical load P0P_{0}, given that under a vertical displacement vv the elastic support provides a resistance equal to −kv-kv.

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Problem 6049

If the integral I=∫f(x,y,y′,y′′) dxI = \int f(x,y,y',y'') \, dx is transformed by Weierstrass into I=∫F(x,x′,x′′,y,y′,y′′) dtI = \int F(x,x',x'',y,y',y'') \, dt, then the function FF will identically satisfy the equalities:

F=x′Fx′+y′Fy′+2x′′Fx′′+2y′′Fy′′;0=x′Fx′′+y′Fy′′. F = x'F_{x'} + y'F_{y'} + 2x''F_{x''} + 2y''F_{y''}; \quad 0 = x'F_{x''} + y'F_{y''}.
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Problem 6050

Show that Euler's equations for the integral

∫f(x,y,y′,z,z′) dx \int f(x,y,y',z,z') \, dx

admit the following first integrals:

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(a)

∂f∂y′=C\frac{\partial f}{\partial y'} = C, if ff does not contain yy;

(b)

f−y′∂f∂y′−z′∂f∂z′=Cf - y'\frac{\partial f}{\partial y'} - z'\frac{\partial f}{\partial z'} = C, if ff does not contain xx.

Problem 6051

Show that Euler's equations for the integral

∫f(x,y,y′,z,z′) dx \int f(x,y,y',z,z') \, dx

can be represented in the form:

dydx=∂H∂η;dzdx=∂H∂ζ;dηdx=−∂H∂y;dζdx=−∂H∂z; \frac{dy}{dx} = \frac{\partial H}{\partial \eta }; \quad \frac{dz}{dx} = \frac{\partial H}{\partial \zeta }; \quad \frac{d\eta }{dx} = -\frac{\partial H}{\partial y}; \quad \frac{d\zeta }{dx} = -\frac{\partial H}{\partial z}; H(x,y,z,η,ζ)=y′∂f∂y′+η′∂f∂z′−f;η=∂f∂y′;ζ=∂f∂z′. H(x,y,z,\eta ,\zeta ) = y'\frac{\partial f}{\partial y'} + \eta '\frac{\partial f}{\partial z'} - f; \quad \eta = \frac{\partial f}{\partial y'}; \quad \zeta = \frac{\partial f}{\partial z'}.
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Problem 6052

Find the transversality condition for the integral

∫f(x,y,y′,z,z′) dx, \int f(x,y,y',z,z') \, dx,

if the endpoint lies on the surface S(x,y,z)=0S(x,y,z)=0.

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Problem 6053

The same problem, but the endpoint lies on the curve

y=F(x),z=G(x). y = F(x), \quad z = G(x).
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Problem 6054

Show that for integrals ∫f(x,y,z) ds\int f(x,y,z) \, ds, and only for these, the transversality conditions on a surface or curve coincide with the conditions of orthogonality.

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Problem 6055

The minimum is sought of the integral

I=∫M1M2μ(x,y,z) ds, I = \int _{M_{1}}^{M_{2}} \mu (x,y,z) \, ds,

where μ(x,y,z)\mu (x,y,z) is a positive function, continuous together with its derivatives. Show that:

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(a)

At the point where the extremal intersects the level surface μ=C\mu = C, the osculating plane of the extremal contains the normal to the level surface.

(b)

If ψ\psi is the angle between the tangent to the extremal and the normal to the level surface, and the surface passes through the point of tangency, then μsin⁡ψ=C\mu \sin \psi = C along the whole extremal.

(c)

The Weierstrass condition is satisfied.

Problem 6056

A flexible string is under the action of a force FF, having a potential UU, so that

X=−∂U∂x,Y=−∂U∂y,Z=−∂U∂z. X = -\frac{\partial U}{\partial x}, \quad Y = -\frac{\partial U}{\partial y}, \quad Z = -\frac{\partial U}{\partial z}.

Here the force FF is referred to a unit length of the string. The equilibrium equations of the string have the form:

dds(Tdxds)=∂U∂x;dds(Tdyds)=∂U∂y;dds(Tdzds)=∂U∂z. \frac{d}{ds}\left(T\frac{dx}{ds}\right) = \frac{\partial U}{\partial x}; \quad \frac{d}{ds}\left(T\frac{dy}{ds}\right) = \frac{\partial U}{\partial y}; \quad \frac{d}{ds}\left(T\frac{dz}{ds}\right) = \frac{\partial U}{\partial z}.

Here TT is the tension of the string, ds2=dx2+dy2+dz2ds^{2} = dx^{2}+dy^{2}+dz^{2}. From these equations it follows that T=U+hT = U+h, where hh is a constant.

Prove that the equilibrium conditions of the string are necessary conditions for the minimum of the integral ∫φ(x,y,z) ds\int \varphi (x,y,z) \, ds, where φ=U+h\varphi = U+h.

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Problem 6057

Find the differential equations of the geodesic lines in an nn-dimensional Riemannian space, i.e., the extremals of the integral defining the length of a curve between two points:

I=∫τ1τ2∑gik(x1,x2,…,xn)xi′xk′ dτ. I = \int _{\tau _{1}}^{\tau _{2}} \sqrt{\sum g_{ik}(x_{1},x_{2},\ldots ,x_{n})x'_{i}x'_{k}} \, d\tau .

Here the quadratic form dS2=∑gik dxi dxkdS^{2} = \sum g_{ik} \, dx_{i} \, dx_{k} is symmetric (i.e., gik=gkig_{ik}=g_{ki}) and positive.

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Problem 6058

Find the curve between the points M1(x1,y1,z1)M_{1}(x_{1},y_{1},z_{1}) and M2(x2,y2,z2)M_{2}(x_{2},y_{2},z_{2}) such that its linear moment relative to the axis OXOX, i.e., the integral

∫y2+z2⋅1+y′2+z′2 dx \int \sqrt{y^{2}+z^{2}} \cdot \sqrt{1+y'^{2}+z'^{2}} \, dx

is a minimum.

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Problem 6059

Find the minimum of the integral ∫h−x2−y2−z2 ds\int \sqrt{h-x^{2}-y^{2}-z^{2}} \, ds, giving the "action" in the case where a material point is attracted to a fixed center by a force proportional to the distance.

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(a)

Find the extremals directly and by the Hamilton--Jacobi method.

(b)

Find the conjugate critical points.

Problem 6060

Find the shortest distance between the circle

x2+y2=a2,z=0 x^{2}+y^{2}=a^{2}, \quad z=0

and the hyperbola

z2−y2=c2,x=0. z^{2}-y^{2}=c^{2}, \quad x=0.
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Problem 6061

Find the discontinuous solutions with a corner point for the problem of the minimum of the integral

∫(0,0)(2,1)y′2(y′−1)2 dx. \int _{(0,0)}^{(2,1)} y'^{2}(y'-1)^{2} \, dx.
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Problem 6062

Among the lines going from the point A(0,0)A(0,0) to the point B(2,0)B(2,0), find the line with a corner point along which the integral

∫AB(y′4−6y′2) dx \int _{A}^{B} (y'^{4} - 6y'^{2}) \, dx

attains its minimum value.

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Problem 6063

Find the discontinuous solutions with a corner point for the problem of the minimum of the integral

∫M1M2(y′4−2y′2) dx. \int _{M_{1}}^{M_{2}} (y'^{4} - 2y'^{2}) \, dx.
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Problem 6064

For the problem of the extremum of the integral

∫(0,0)(x1,y1)sin⁡y′ dx \int _{(0,0)}^{(x_{1},y_{1})} \sin y' \, dx

find the continuous solutions, as well as the solutions with a corner point.

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Problem 6065

Among all cylindrical surfaces with generators parallel to OZOZ, find the one on which the plane z=0z=0, the cone z=x2+y2tg⁡az = \sqrt{x^{2}+y^{2}}\operatorname {tg}a, and the two generators passing through the points A(x1,y1)A(x_{1},y_{1}) and B(x2,y2)B(x_{2},y_{2}) cut off the smallest area.

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Problem 6066

In the plane XOYXOY, find the curve joining two points M1(r1,θ1)M_{1}(r_{1},\theta_{1}) and M2(r2,θ2)M_{2}(r_{2},\theta_{2}) and having the smallest moment of inertia relative to the point OO. Find the continuous and discontinuous solutions.

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Problem 6067

Among the lines lying in the upper half-plane and joining the points AA and BB, find the one that, when rotated about OXOX, gives the smallest surface.

Find the discontinuous solutions (see problem 5988).

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Problem 6068

Find the curve, lying in the region y≥k>0y \geq k > 0, joining the points M1(−a,b)M_{1}(-a,b) and M2(a,b)M_{2}(a,b) and forming, when rotated about OXOX, the smallest surface.

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Problem 6069

Among the lines joining the given points A(x0,y0)A(x_{0},y_{0}) and B(x1,y1)B(x_{1},y_{1}), find the discontinuous solution of the problem of the minimum of the integral

I=∫AB(h−gy)(x′2+y′2) dτ. I = \int _{A}^{B} \sqrt{(h-gy)(x'^{2}+y'^{2})} \, d\tau .
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Problem 6070

Find the shortest path from the point A(−2,3)A(-2,3) to the point B(2,3)B(2,3), lying in the region y≤x2y \leq x^{2}.

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Problem 6071

Among the curves joining the points A(0,0)A(0,0) and B(b,0)B(b,0) and lying in the region y≥0y \geq 0, find the one that gives the maximum of the integral

∫AB(y2−ay1+y′2) dx \int _{A}^{B} \left(y^{2} - ay\sqrt{1+y'^{2}}\right) \, dx

under the condition b≥2a>0b \geq 2a > 0.

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Problem 6072

The minimum is sought of the integral

I=∫abU(x,y)+h1+y′2 dx I = \int _{a}^{b} \sqrt{U(x,y)+h}\sqrt{1+y'^{2}} \, dx

in the region y≥f(x)y \geq f(x), under the condition that at one limit the values of the function and its derivative (a,ya,ya′)(a, y_{a}, y'_{a}) are given, while at the other limit only the number bb itself is given. Show that if the extremal meets the boundary curve y=f(x)y=f(x), then its further continuation makes with the boundary curve the same angle, i.e., the angle of incidence equals the angle of reflection.

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Problem 6073

Inside a convex region BB two points M1M_{1} and M2M_{2} are given. Find the shortest path from M1M_{1} to M2M_{2} that touches the boundary of BB.

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Problem 6074

Find the minimum of the integral

∫(0,0)(x1,y1)y⋅1−y′2 dx \int _{(0,0)}^{(x_{1},y_{1})} \sqrt{y} \cdot \sqrt{1-y'^{2}} \, dx

under the conditions y≥0y \geq 0; 1−y′2≥01-y'^{2} \geq 0.

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Problem 6075

Among the curves joining the points A(a,ya)A(a,y_{a}) and B(b,yb)B(b,y_{b}), find the one that gives an extremal value of the integral

∫aby1−y2y′2 dx \int _{a}^{b} y\sqrt{1-y^{2}y'^{2}} \, dx

under the condition y≥0y \geq 0, 1−y2y′2≥01-y^{2}y'^{2} \geq 0.

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Problem 6076

On the surface z=φ(x,y)z = \varphi (x,y) the points M1M_{1} and M2M_{2} are given.

Draw on the surface the shortest line from M1M_{1} to M2M_{2}, under the condition that the slope of the curve, i.e., dzds\frac{dz}{ds}, does not exceed a given number kk (mountain railway).

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Problem 6077

Join the points AA and BB by the shortest path, leaving AA in a given direction and having curvature not exceeding a given value 1R\frac{1}{R}. The point BB lies outside the circles of radius RR tangent to the given direction at the point AA.

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Problem 6078

Light travels along a path such that the first variation of the integral

∫(x0,y0)(x1,y1)n(x,y)x′2+y′2 dt \int _{(x_{0},y_{0})}^{(x_{1},y_{1})} n(x,y)\sqrt{x'^{2}+y'^{2}} \, dt

along the path equals zero. Here n(x,y)n(x,y) is the refractive index of the medium.

If a ray passes from one medium into another, then n(x,y)n(x,y) may undergo a sudden change. Derive the condition for the refraction of a light ray when the ray passes from a medium with refractive index n1n_{1} into a medium with refractive index n2n_{2} (see problem 6055).

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Problem 6079

Form Euler's equation for the integral ∬ν dS\iint \nu \, dS, where ν\nu is the segment of the normal between the surface and the plane xyxy.

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Problem 6080

Euler's equation for the integral

∬Sf(x,y,z,p,q) dx dy, \iint _{S} f(x,y,z,p,q) \, dx \, dy,

where ff takes given values on the contour, is capable of becoming an identity. Prove that a necessary and sufficient condition for this is the equality:

f=L(x,y,z)+M(x,y,z)p+N(x,y,z)q;∂L∂z=∂M∂x+∂N∂y. f = L(x,y,z) + M(x,y,z)p + N(x,y,z)q; \quad \frac{\partial L}{\partial z} = \frac{\partial M}{\partial x} + \frac{\partial N}{\partial y}.

(On the other hand, a necessary and sufficient condition is the independence of the value of the integral from the shape of the surface SS.)

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Problem 6081

Find Euler's equation for the problem of the minimum of the integral

∬((∂u∂x)2+(∂u∂y)2+2uf(x,y)) dx dy \iint \left(\left(\frac{\partial u}{\partial x}\right)^{2} + \left(\frac{\partial u}{\partial y}\right)^{2} + 2uf(x,y)\right) \, dx \, dy

for given values of uu on the contour of the region.

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Problem 6082

Find the minimum of the integral

∬S(p2+q2) dx dy, \iint _{S} (p^{2}+q^{2}) \, dx \, dy,

if the values of the function uu are given on the contour. Here:

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(a)

form Euler's equation;

(b)

show that every solution of Euler's equation, satisfying the conditions on the contour, gives a strong minimum.

Problem 6083

Find Euler's equation for the problem of the extremum of the integral

I=∬((∂φ∂x)2−(∂φ∂y)2) dx dy. I = \iint \left(\left(\frac{\partial \varphi }{\partial x}\right)^{2} - \left(\frac{\partial \varphi }{\partial y}\right)^{2}\right) \, dx \, dy.
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Problem 6084

Obtain the differential equation of minimal surfaces, i.e., among all surfaces z=z(x,y)z=z(x,y), resting on a given contour CC (or several contours), find the one having the smallest area. In particular, find the minimal surfaces whose equation has the form z=φ(x)+ψ(y)z = \varphi (x)+\psi (y).

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Problem 6085

Show that if we have a minimal surface z=z(x,y)z=z(x,y), resting on a given contour CC, then this surface gives a strong minimum of the integral

I=∬S1+p2+q2 dx dy. I = \iint _{S} \sqrt{1+p^{2}+q^{2}} \, dx \, dy.
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Problem 6086

Show that the problem of the minimum of the integral

∬G(u2+v2) dx dy, \iint G\left(\sqrt{u^{2}+v^{2}}\right) \, dx \, dy,

where

u=∂φ∂x,v=∂φ∂y, u = \frac{\partial \varphi }{\partial x}, \quad v = \frac{\partial \varphi }{\partial y},

leads to Euler's equation

(c2−u2)∂2φ∂x2−2uv∂2φ∂x∂y+(c2−v2)∂2φ∂y2=0, (c^{2}-u^{2})\frac{\partial ^{2}\varphi }{\partial x^{2}} - 2uv\frac{\partial ^{2}\varphi }{\partial x \partial y} + (c^{2}-v^{2})\frac{\partial ^{2}\varphi }{\partial y^{2}} = 0,

where

c2=−q2G′(q)qG′′(q)−G′(q) c^{2} = \frac{-q^{2}G'(q)}{qG''(q)-G'(q)}

(an equation of this kind arises from the problem of the plane steady adiabatic motion of a compressible fluid with velocity potential φ\varphi, if we set:

G(q)=(2α2+(γ−1)(U2−q2))γγ−1; G(q) = \left(2\alpha ^{2}+(\gamma -1)(U^{2}-q^{2})\right)^{\frac{\gamma }{\gamma -1}};

α\alpha, γ\gamma, UU are constants).

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Problem 6087

Let ξ(x,y)\xi (x,y), η(x,y)\eta (x,y) be curvilinear coordinates, and let:

ds2=dx2+dy2=e(ξ,η) dξ2+2f(ξ,η) dξ dη+g(ξ,η) dη2. ds^{2} = dx^{2}+dy^{2} = e(\xi ,\eta ) \, d\xi ^{2} + 2f(\xi ,\eta ) \, d\xi \, d\eta + g(\xi ,\eta ) \, d\eta ^{2}.

Starting from the first variation of the integral

J=∬((∂u∂x)2+(∂u∂y)2) dx dy, J = \iint \left(\left(\frac{\partial u}{\partial x}\right)^{2} + \left(\frac{\partial u}{\partial y}\right)^{2}\right) \, dx \, dy,

show that the Laplace operator Δu\Delta u transforms to the new curvilinear coordinates by the formula

Δu=∂2u∂x2+∂2u∂y2=1eg−f2(∂∂ξ(guξ−fuηeg−f2)+∂∂η(euη−fuξeg−f2)). \Delta u = \frac{\partial ^{2}u}{\partial x^{2}} + \frac{\partial ^{2}u}{\partial y^{2}} = \frac{1}{\sqrt{eg-f^{2}}}\left(\frac{\partial }{\partial \xi }\left(\frac{gu_{\xi }-fu_{\eta }}{\sqrt{eg-f^{2}}}\right) + \frac{\partial }{\partial \eta }\left(\frac{eu_{\eta }-fu_{\xi }}{\sqrt{eg-f^{2}}}\right)\right).

Example. Polar coordinates, where ds2=dr2+r2dφ2ds^{2} = dr^{2} + r^{2}d\varphi^{2}.

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Problem 6088

The values of the function uu are given on the boundary surface of the volume VV.

Find the necessary condition for the minimum of the integral

∭V((∂u∂x)2+(∂u∂y)2+(∂u∂z)2) dx dy dz. \iiint _{V} \left(\left(\frac{\partial u}{\partial x}\right)^{2} + \left(\frac{\partial u}{\partial y}\right)^{2} + \left(\frac{\partial u}{\partial z}\right)^{2}\right) \, dx \, dy \, dz.
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Problem 6089

The same question applied to the integral

∭V1+(∂u∂x)2+(∂u∂y)2+(∂u∂z)2 dx dy dz. \iiint _{V} \sqrt{1+\left(\frac{\partial u}{\partial x}\right)^{2} + \left(\frac{\partial u}{\partial y}\right)^{2} + \left(\frac{\partial u}{\partial z}\right)^{2}} \, dx \, dy \, dz.
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Problem 6090

Let ξ1\xi_{1}, ξ2\xi_{2}, ξ3\xi_{3} be curvilinear orthogonal coordinates, defined by the formulas:

x=x(ξ1,ξ2,ξ3),y=y(ξ1,ξ2,ξ3),z=z(ξ1,ξ2,ξ3). x = x(\xi _{1},\xi _{2},\xi _{3}), \quad y = y(\xi _{1},\xi _{2},\xi _{3}), \quad z = z(\xi _{1},\xi _{2},\xi _{3}).

Let also

ds2=dx2+dy2+dz2=H12dξ12+H22dξ22+H32dξ32. ds^{2} = dx^{2}+dy^{2}+dz^{2} = H_{1}^{2}d\xi _{1}^{2} + H_{2}^{2}d\xi _{2}^{2} + H_{3}^{2}d\xi _{3}^{2}.

Using the first variation of the integral

∭((∂u∂x)2+(∂u∂y)2+(∂u∂z)2) dx dy dz, \iiint \left(\left(\frac{\partial u}{\partial x}\right)^{2} + \left(\frac{\partial u}{\partial y}\right)^{2} + \left(\frac{\partial u}{\partial z}\right)^{2}\right) \, dx \, dy \, dz,

show that the Laplace operator Δu\Delta u in the new coordinates has the form

Δu=1H1H2H3(∂∂ξ1(H2H3H1∂u∂ξ1)+∂∂ξ2(H3H1H2∂u∂ξ2)+∂∂ξ3(H1H2H3∂u∂ξ3)). \Delta u = \frac{1}{H_{1}H_{2}H_{3}}\left(\frac{\partial }{\partial \xi _{1}}\left(\frac{H_{2}H_{3}}{H_{1}}\frac{\partial u}{\partial \xi _{1}}\right) + \frac{\partial }{\partial \xi _{2}}\left(\frac{H_{3}H_{1}}{H_{2}}\frac{\partial u}{\partial \xi _{2}}\right) + \frac{\partial }{\partial \xi _{3}}\left(\frac{H_{1}H_{2}}{H_{3}}\frac{\partial u}{\partial \xi _{3}}\right)\right).

Example. Spherical coordinates, where ds2=dr2+r2sin⁡2θ dφ2+r2dθ2ds^{2} = dr^{2} + r^{2}\sin^{2}\theta \, d\varphi^{2} + r^{2}d\theta^{2}.

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Problem 6091

Show that for the problem of the minimum of the integral

I=12\iiiint(Hx2+Hy2+Hz2−Ex2−Ey2−Ez2) dx dy dz dt, I = \frac{1}{2}\iiiint (H_{x}^{2}+H_{y}^{2}+H_{z}^{2}-E_{x}^{2}-E_{y}^{2}-E_{z}^{2}) \, dx \, dy \, dz \, dt,

where

Hx=∂Az∂y−∂Ay∂z;Hy=∂Ax∂z−∂Az∂x;Hz=∂Ay∂x−∂Ax∂y; H_{x} = \frac{\partial A_{z}}{\partial y} - \frac{\partial A_{y}}{\partial z}; \qquad H_{y} = \frac{\partial A_{x}}{\partial z} - \frac{\partial A_{z}}{\partial x}; \qquad H_{z} = \frac{\partial A_{y}}{\partial x} - \frac{\partial A_{x}}{\partial y}; Ex=−∂Ax∂t−∂Φ∂x;Ey=−∂Ay∂t−∂Φ∂y;Ez=−∂Az∂t−∂Φ∂z, E_{x} = -\frac{\partial A_{x}}{\partial t} - \frac{\partial \Phi }{\partial x}; \qquad E_{y} = -\frac{\partial A_{y}}{\partial t} - \frac{\partial \Phi }{\partial y}; \qquad E_{z} = -\frac{\partial A_{z}}{\partial t} - \frac{\partial \Phi }{\partial z},

Euler's equations coincide with Maxwell's equations for the propagation of electromagnetic waves:

∂Hz∂y−∂Hy∂z=∂Ex∂t;∂Hx∂z−∂Hz∂x=∂Ey∂t; \frac{\partial H_{z}}{\partial y} - \frac{\partial H_{y}}{\partial z} = \frac{\partial E_{x}}{\partial t}; \qquad \frac{\partial H_{x}}{\partial z} - \frac{\partial H_{z}}{\partial x} = \frac{\partial E_{y}}{\partial t}; ∂Hy∂x−∂Hx∂y=∂Ez∂t;∂Ex∂x+∂Ey∂y+∂Ez∂z=0. \frac{\partial H_{y}}{\partial x} - \frac{\partial H_{x}}{\partial y} = \frac{\partial E_{z}}{\partial t}; \qquad \frac{\partial E_{x}}{\partial x} + \frac{\partial E_{y}}{\partial y} + \frac{\partial E_{z}}{\partial z} = 0.
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Problem 6092

Form the differential equation of the longitudinal vibrations of a rod, starting from Hamilton's principle.

The rod is located along the axis OXOX between the points (0,0,0)(0,0,0) and (l,0,0)(l,0,0) and works under longitudinal compression and tension. Notation: u(x,t)u(x,t) --- displacement; ω(x)\omega (x) --- cross-sectional area; ρ(x)\rho (x) --- density of the material; EE --- modulus of elasticity; σ\sigma --- Poisson's ratio; I(x)I(x) --- polar moment of the cross-section relative to the point (x,0,0)(x,0,0); F(x)F(x) --- external force per unit mass. Determine what boundary conditions can be considered at the ends.

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Problem 6093

Form the differential equation of small transverse vibrations of a rod, starting from Hamilton's principle. The rod of length ll is located along the axis OXOX and works under bending. The vibrations occur in the plane XOYXOY. Notation: v(x,t)v(x,t) --- displacement; ω(x)\omega (x) --- cross-section; ρ(x)\rho (x) --- density of the material; EE --- modulus of elasticity; I(x)I(x) --- moment of inertia of the cross-section relative to the axis passing through its center of gravity perpendicular to XOYXOY; F(x)F(x) --- external force, per unit mass. Also find the form of the limiting conditions that must be satisfied in various cases.

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Problem 6094

Starting from Hamilton's principle, form the differential equation of small transverse vibrations of a membrane, located in the plane XOYXOY and working only under tension. Notation: ζ(x,y,t)\zeta (x,y,t) --- displacement, perpendicular to XOYXOY; ρ(x,y)\rho (x,y) --- surface density; S0S_{0} --- area of the membrane in the equilibrium position; PP --- tension of the membrane, constant, giving a force acting in the plane XOYXOY per unit length of the contour bounding any element of area; F(x,y)F(x,y) --- external force, perpendicular to XOYXOY, per unit mass; φ\varphi --- external force, perpendicular to XOYXOY, acting on the contour, per unit length of the contour. Also find the form of the limiting condition on the contour CC of the membrane, when it is not fixed (if it is fixed, then the limiting condition will be ζ=φ(c,t)\zeta = \varphi (c,t) --- a given function of the position of the point on the contour).

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Problem 6095

Form the differential equation of equilibrium of a membrane and determine the boundary conditions in the case where an external force acts on it only on the contour, equal to k(s)ζ(s)k(s)\zeta (s), where kk is a coefficient of proportionality, and ζ(s)\zeta (s) is the transverse displacement of a point of the membrane. The force is directed opposite to the displacement.

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Problem 6096

Form the exact differential equation of equilibrium of a membrane serving as the bottom of a cylindrical vessel in which a liquid of density ρ0\rho_{0}, of depth hh (relative to the original position of the membrane), is poured. Notation: ζ(x,y,t)\zeta (x,y,t) --- displacement of a point of the membrane in the vertical direction OZOZ (positive direction downward), PP --- constant tension of the membrane.

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Problem 6097

Starting from Hamilton's principle, form the equation of small transverse vibrations of a thin plate. The plate works under tension and bending, its thickness is 2h2h. In the equilibrium position the midplane of the plate lies in the plane XOYXOY and is bounded by the contour CC. Here: ζ(x,y,t)\zeta (x,y,t) --- displacement of a point along OZOZ, ρ\rho --- constant density of the plate, EE --- modulus of elasticity, σ\sigma --- Poisson's ratio, there are no external forces. Determine the character of the limiting conditions.

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Problem 6098

Derive the equations of equilibrium of an elastic isotropic body, starting from the principle of virtual displacements, i.e., from the condition:

δ∭Π dx dy dz=0, \delta \iiint \Pi \, dx \, dy \, dz = 0,

where

Π=A−W,W=ρ(uX+vY+wZ), \Pi = A - W, \quad W = \rho (uX+vY+wZ), 2A=(λ+2μ)(ux+vy+wz)2+μ((wy+vz)2+(vx+uy)2+(uz+wx)2−4vywz−4wzux−4uxvy), 2A = (\lambda +2\mu )(u_{x}+v_{y}+w_{z})^{2} + \mu \left((w_{y}+v_{z})^{2}+(v_{x}+u_{y})^{2}+(u_{z}+w_{x})^{2} - 4v_{y}w_{z} - 4w_{z}u_{x} - 4u_{x}v_{y}\right),

λ\lambda and μ\mu are elastic constants, AA is the strain energy per unit volume, XX, YY, ZZ are the components of the external force per unit mass, uu, vv, ww are the displacements of points of the elastic body, given on the surface; uxu_{x} denotes ∂u∂x\frac{\partial u}{\partial x}, and so on.

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Problem 6099

Derive the equations of motion of an elastic isotropic body, starting from Hamilton's principle:

δ∫t0t1(T−Π) dt=0, \delta \int _{t_{0}}^{t_{1}} (T-\Pi ) \, dt = 0,

where

T=12∭ρ((∂u∂t)2+(∂v∂t)2+(∂w∂t)2) dx dy dz, T = \frac{1}{2}\iiint \rho \left(\left(\frac{\partial u}{\partial t}\right)^{2}+\left(\frac{\partial v}{\partial t}\right)^{2}+\left(\frac{\partial w}{\partial t}\right)^{2}\right) \, dx \, dy \, dz,

and the quantity Π\Pi has the same meaning as in problem 6098.

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Problem 6100

Find the shape of a body of revolution about the axis OXOX, homogeneous, having a given volume VV and exerting the greatest Newtonian attraction on a point lying at the intersection of the body with OXOX.

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Problem 6101

Find the homogeneous body of revolution about the axis OXOX with given volume VV and with the smallest moment of inertia relative to the axis OZOZ.

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Problem 6102

The cost of land depends on its location and can be characterized by giving a positive function ρ(x,y)\rho (x,y). Among the curves of length ll joining the points M1M_{1} and M2M_{2}, find the one that, together with the chord M1M2M_{1}M_{2}, would bound the area of greatest cost.

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Problem 6103

Of the curves of length ll joining the points M1(−a,0)M_{1}(-a,0), M2(a,0)M_{2}(a,0), determine the one that, together with OXOX, bounds the largest area. Here l>2al > 2a.

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Problem 6104

Join two unknown points M1M_{1} and M2M_{2} on the axis OXOX by a curve of least length, enclosing with OXOX an area of given magnitude SS.

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Problem 6105

Find the closed curve of given length ll, bounding an area of the greatest magnitude.

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Problem 6106

On two given ordinates AA1AA_{1} (x=ax=a) and BB1BB_{1} (x=bx=b), find such points MM and NN and such a curve joining them, that the area AMNBAMNB is maximal. Here the sum of the lengths of the arc MNMN and the ordinates AMAM and BNBN is given.

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Problem 6107

Among all curves drawn between the ordinates x=x0x=x_{0} and x=x1x=x_{1} and forming, together with them and OXOX, an area of given magnitude, find the one for which the center of gravity of this area occupies the lowest position.

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Problem 6108

Join the given point (0,b)(0,b) on the axis OYOY to OXOX by a curve enclosing, together with the axes OXOX and OYOY, a given area SS and forming, when rotated about OXOX, the smallest surface.

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Problem 6109

From the origin of the plane XOYXOY, draw a curve OAOA of length ll, ending on the line y=hy=h and forming, together with the ordinate of the point AA and the axis OXOX, the largest area.

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Problem 6110

An angle with vertex at the origin is given. Join the given point M1M_{1} on one side of the angle with an unknown point M2M_{2} on the other side by a curve of length ll, so that the area between the curve and the sides of the angle is the largest.

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Problem 6111

Among the curves of length ll joining the points M1(x1,y1)M_{1}(x_{1},y_{1}) and M2(x2,y2)M_{2}(x_{2},y_{2}), find the one whose center of gravity lies lowest.

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Problem 6112

Find the shape of a heavy homogeneous string, one end of which is fixed at the point B(x1,y1)B(x_{1},y_{1}), and the other end lies on the axis OYOY. The length of the string is ll.

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Problem 6113

Among the curves drawn from the periphery of the circle x2+y2=R2x^{2}+y^{2} = R^{2} to the point M(a,0)M(a,0), and for which the areas of the sector with vertex at the center of the circle have a given value, find the one for which the potential of the arc at the point OO has the smallest value. The density is constant, the attraction follows Newton's law.

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Problem 6114

Among the curves of length ll passing through the points M1(x1,y1)M_{1}(x_{1},y_{1}) and M2(x2,y2)M_{2}(x_{2},y_{2}) and having at these points given slope angles θ1\theta_{1} and θ2\theta_{2}, determine the one that, together with the normals M1N1M_{1}N_{1} and M2N2M_{2}N_{2} and the arc of the evolute N1N2N_{1}N_{2}, encloses the smallest area.

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Problem 6115

Find the position of equilibrium of an elastic rod clamped at its ends.

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Problem 6116

An airplane has its own speed uu. The wind blows parallel to the axis OXOX with speed v<uv < u. The airplane has a supply of fuel for a definite flight time. Determine the maximum area of the curve that the airplane can trace while flying with a return to the point of departure.

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Problem 6117

Find the minimum of the integral ∫0πy′2 dx\int_{0}^{\pi } y'^{2} \, dx, if ∫0πy2 dx=1\int_{0}^{\pi } y^{2} \, dx = 1, y(0)=y(π)=0y(0)=y(\pi )=0. Find the extremals and check Jacobi's condition.

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Problem 6118

Find the minimum of the integral ∫0πy′′2 dx\int_{0}^{\pi } y''^{2} \, dx under the conditions:

∫0πy′2 dx=1;y(0)=y(π)=0. \int _{0}^{\pi } y'^{2} \, dx = 1; \quad y(0)=y(\pi )=0.
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Problem 6119

Find the minimum of ∫0πy′′2 dx\int_{0}^{\pi } y''^{2} \, dx, if ∫0πy2 dx=1\int_{0}^{\pi } y^{2} \, dx = 1, y(0)=y(π)=0y(0)=y(\pi )=0.

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Problem 6120

Among the lines going from A(0,0)A(0,0) to B(1,14)B\left(1,\frac{1}{4}\right), find the one that gives the extremum of the integral ∫01y′2 dx\int_{0}^{1} y'^{2} \, dx under the condition:

∫01(y−y′2) dx=112. \int _{0}^{1} (y-y'^{2}) \, dx = \frac{1}{12}.
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Problem 6121

Among the lines from the point A(0,0,0)A(0,0,0) to B(1,1,1)B(1,1,1) for which

∫AB(y′2−xy′−z′2) dx=2, \int _{A}^{B} (y'^{2}-xy'-z'^{2}) \, dx = 2,

find the one for which

∫AB(y′2+z′2−4xz′−4z) dx \int _{A}^{B} (y'^{2}+z'^{2}-4xz'-4z) \, dx

is a minimum. Consider the same problem under the condition that all that is known about the point AA is that it lies in the plane YOZYOZ.

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Problem 6122

Among all surfaces z=z(x,y)z=z(x,y), resting on a given contour LL and bounding, together with the cylinder projecting LL onto XOYXOY and the base of this cylinder, a given volume vv, find the one having the smallest area. Here form Euler's equation and determine its geometric meaning.

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Problem 6123

Among surfaces with given area SS, resting on the plane XOYXOY and bounding a given volume VV, find the one for which the center of gravity of this volume would have the lowest position. Here determine the geometric meaning of Euler's equation and find what boundary conditions must hold on the line of intersection of the surface with the plane XOYXOY.

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Problem 6124

Among all surfaces z=z(x,y)z=z(x,y) of given area and resting on a given contour, find the one for which the center of gravity lies lowest. Form Euler's equation and determine its geometric meaning.

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Problem 6125

Among closed surfaces of given area, find the one bounding the largest volume. Form Euler's equation, determine its geometric meaning, and show that the spherical surface satisfies Euler's equation.

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Problem 6126

A system of material points M1,M2,…,MnM_{1}, M_{2}, \ldots , M_{n} is given. The mass of the point MνM_{\nu } equals mνm_{\nu }, its coordinates are xν,yν,zνx_{\nu }, y_{\nu }, z_{\nu }, and the force acting on it has components Xν,Yν,ZνX_{\nu }, Y_{\nu }, Z_{\nu }. Suppose there exists a function V(xν,yν,zν,t)V(x_{\nu },y_{\nu },z_{\nu },t) --- the potential energy --- such that

Xν=−∂V∂xν,Yν=−∂V∂yν,Zν=−∂V∂zν. X_{\nu } = -\frac{\partial V}{\partial x_{\nu }}, \quad Y_{\nu } = -\frac{\partial V}{\partial y_{\nu }}, \quad Z_{\nu } = -\frac{\partial V}{\partial z_{\nu }}.

Denote by TT the kinetic energy:

T=12∑ν=1nmν(x˙ν2+y˙ν2+z˙ν2); T = \frac{1}{2}\sum _{\nu =1}^{n} m_{\nu }(\dot{x}_{\nu }^{2}+\dot{y}_{\nu }^{2}+\dot{z}_{\nu }^{2});

here the dot denotes differentiation with respect to tt. Suppose mm conditions of the form φa(xν,yν,zν,t)=0\varphi_{a}(x_{\nu },y_{\nu },z_{\nu },t)=0, a=1,2,…,ma=1,2,\ldots ,m, are given.

Show that finding the extremum of the Hamilton integral

I=∫t1t2(T−V) dt I = \int _{t_{1}}^{t_{2}} (T-V) \, dt

for given xν,yν,zνx_{\nu }, y_{\nu }, z_{\nu } at t=t1t=t_{1} and t=t2t=t_{2}, leads to the Lagrange differential equations of the first kind:

mνd2xνdt2=Xν+∑a=1mλa∂φa∂xν. m_{\nu }\frac{d^{2}x_{\nu }}{dt^{2}} = X_{\nu } + \sum _{a=1}^{m} \lambda _{a}\frac{\partial \varphi _{a}}{\partial x_{\nu }}.

If, on the other hand, the coordinates are expressed in terms of r=3n−mr=3n-m independent variables q1,q2,…,qrq_{1}, q_{2}, \ldots , q_{r} and tt, then we arrive at the Lagrange differential equations of the second kind:

ddt∂T∂q˙ν−∂V∂qν+∂T∂qν=0,ν=1,2,…,r. \frac{d}{dt}\frac{\partial T}{\partial \dot{q}_{\nu }} - \frac{\partial V}{\partial q_{\nu }} + \frac{\partial T}{\partial q_{\nu }} = 0, \quad \nu = 1,2,\ldots ,r.
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Problem 6127

With the notation of the previous problem, suppose that the functions φa\varphi_{a} and V=−UV=-U do not contain tt. Call a trajectory the curve in 3n3n-dimensional space:

xν=xν(τ),yν=yν(τ),zν=zν(τ), x_{\nu } = x_{\nu }(\tau ), \qquad y_{\nu } = y_{\nu }(\tau ), \qquad z_{\nu } = z_{\nu }(\tau ),

satisfying all the conditions φa(xν,yν,zν)=0\varphi_{a}(x_{\nu },y_{\nu },z_{\nu })=0.

To each motion of the points under the action of the given forces there corresponds its own trajectory. Show that this trajectory is an extremal for the problem of the minimum of the "action" integral

I=∫τ0τ12(U+h)∑mν(xν′2+yν′2+zν′2) dτ. I = \int _{\tau _{0}}^{\tau _{1}} \sqrt{2(U+h)}\sqrt{\sum m_{\nu }(x'^{2}_{\nu }+y'^{2}_{\nu }+z'^{2}_{\nu })} \, d\tau .

Here hh is the total energy, T=U+hT=U+h, and differentiation is carried out with respect to τ\tau; the values of the variables at the ends are given.

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Problem 6128

The functions y(x)y(x) and z(x)z(x) satisfy the equation

y2+y′2=z′. y^{2}+y'^{2} = z'.

Moreover, it is given that

y(0)=a+b,y(n)=aen+be−n,z(0)=a2−b2. y(0) = a+b, \quad y(n) = ae^{n}+be^{-n}, \quad z(0) = a^{2}-b^{2}.

Find yy and zz from the condition that z(n)z(n) is a maximum.

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Problem 6129

Consider the problem of the shortest distance between two points M1M_{1} and M2M_{2} in space as a Mayer problem.

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Problem 6130

Among all curves joining the given points M1(x1,y1)M_{1}(x_{1},y_{1}) and M2(x2,y2)M_{2}(x_{2},y_{2}), lying in the vertical plane XOYXOY (OYOY --- downward), find the one along which, moving, a heavy point will arrive from M1M_{1} to M2M_{2} in the shortest time. The initial speed is v1v_{1}; besides gravity, the point is acted on by a resistance R=kvnR=kv^{n}.

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Problem 6131

Among the curves joining the given points M1(x1,y1)M_{1}(x_{1},y_{1}) and M2(x2,y2)M_{2}(x_{2},y_{2}), lying in the plane XOYXOY (OYOY --- downward), find the one along which, moving, a point will arrive at M2M_{2} with the greatest speed. Here the speed of the point at M1M_{1} equals v1v_{1}; the point is acted on by gravity and by a resistance force R=kvnR=kv^{n}. Compare the normal reaction of the curve with the centrifugal force.

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Problem 6132

Join the points M1(0,0,0)M_{1}(0,0,0) and M2(a,0,0)M_{2}(a,0,0) by a curve having given curvature 1ρ\frac{1}{\rho } and the greatest length.

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Problem 6133

An airplane departs from the point A(x1,y1)A(x_{1},y_{1}) at the moment t=0t=0, intending to reach the point B(x2,y2)B(x_{2},y_{2}) in the shortest time. Find how the airplane must be directed for this, given that the distribution of wind in space and time is known, i.e., the components of the wind velocity ww along the axes OXOX and OYOY are known: wx=u(x,y,t)w_{x}=u(x,y,t), wy=v(x,y,t)w_{y}=v(x,y,t). The airplane's own speed is VV, and w2=u2+v2<V2w^{2}=u^{2}+v^{2} < V^{2}.

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Problem 6134

An airplane wing of length ll is located along the axis OXOX and moves parallel to the axis OYOY with speed v0v_{0} (the plane XOYXOY is horizontal). Denote by f(x) dxf(x) \, dx the lift of an element of the wing of length dxdx. In aerodynamics it is shown, under certain assumptions, that under these conditions a vortex of intensity dΓ=1ρv0f′(x) dxd\Gamma = \frac{1}{\rho v_{0}}f'(x) \, dx is shed from the element of the wing. Here ρ\rho is the density of air, taken to be constant. (This formula is a consequence of the basic formula of N.E.Zhukovsky.) These vortices induce at every point ξ\xi of the wing a downward air velocity

w(ξ)=14πρv0∫−l/2l/2f′(x)ξ−x dx. w(\xi ) = \frac{1}{4\pi \rho v_{0}} \int _{-l/2}^{l/2} \frac{f'(x)}{\xi -x} \, dx.

Here the axis OZOZ is directed downward, and the integral is understood everywhere, here and below, in the principal-value sense. The induced drag of the wing is determined by the formula

B=∫−l/2l/2ω(ξ)v0f(ξ) dξ=14πρv02∫−l/2l/2∫−l/2l/2f′(x)f(ξ)ξ−x dx dξ. B = \int _{-l/2}^{l/2} \frac{\omega (\xi )}{v_{0}}f(\xi ) \, d\xi = \frac{1}{4\pi \rho v_{0}^{2}} \int _{-l/2}^{l/2}\int _{-l/2}^{l/2} \frac{f'(x)f(\xi )}{\xi -x} \, dx \, d\xi .

It is required to determine, for a given total lift of the wing

A=∫−l/2l/2f(x) dx A = \int _{-l/2}^{l/2} f(x) \, dx

the distribution of lift, i.e., the function f(x)f(x), for which the induced drag is smallest. Here it should be taken into account that, because of the absence of vortices outside the interval −l2≤x≤l2-\frac{l}{2} \leq x \leq \frac{l}{2}, the condition f(±l2)=0f\left(\pm \frac{l}{2}\right) = 0 must hold.

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Problem 6135

Two equations are given

u′=f(x,y,y′,z,u),z′=g(x,y,y′,z,u). u' = f(x,y,y',z,u), \quad z' = g(x,y,y',z,u).

Find yy as a function of xx under the condition that the values of xx, yy, zz, and uu are given at x=x0x=x_{0}, and yy is also given at x=x1x=x_{1}, while u(x1)u(x_{1}) must be an extremum. Example:

u(x1)=∫x0x1f(x,y,y′,z) dx;z=∫x0x1+y′2 dx. u(x_{1}) = \int _{x_{0}}^{x_{1}} f(x,y,y',z) \, dx; \quad z = \int _{x_{0}}^{x} \sqrt{1+y'^{2}} \, dx.

Form Euler's differential equations.

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Problem 6136

Find the extremum of the function W=f(u,v)W = f(u,v), where

u=∫t1t2F(x,y,x′,y′) dt,v=∫t1t2G(x,y,x′,y′) dt u = \int _{t_{1}}^{t_{2}} F(x,y,x',y') \, dt, \quad v = \int _{t_{1}}^{t_{2}} G(x,y,x',y') \, dt

for given values of xx and yy at the ends of the interval (t1,t2)(t_{1},t_{2}).

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Problem 6137

Find the form of the extremals for the problem of the minimum of the integral

∫ABy dx⋅∫AB1+y′2 dx, \int _{A}^{B} y \, dx \cdot \int _{A}^{B} \sqrt{1+y'^{2}} \, dx,

where A(x1,y1)A(x_{1},y_{1}) and B(x2,y2)B(x_{2},y_{2}) are given points.

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