By Stokes' theorem and the results in 5.6.1 and A5.1, the surface-integral for may be expressed as a line integral Verify that the functions

satisfy the identities

14. Note the facts:
(1) the value of the line-integral Q remains unchanged if G is deformed in such a way that it never sweeps over any of the points (-1, 0) or (1,0) during its deformation;
(2) Q = 2p, if G is a small circle around (1, 0) oriented counter-clockwise;
(3) Q =2 p if G is a small circle around (-1, 0) oriented clockwise.

15. Think of C as being a rigid circle made of wire and of G as being a string. Now deform the string G to a new position G ' lying entirely within the plane y = 0. The numbers p and n are not changed during this deformation, and the first formula now follows directly from 14. applied to the curve G ' within the plane y = 0 and the line-segment -1 < x < 1, y = 0, z = 0 of this plane. The factor 4p (instead of 2p, as in the previous example) is due to the fact that the solid angle W increases by 4p along a closed path for which p = 1, n = 0.

One way of carrying out the above deformation of G into G ' analytically is as follows. Assume that G does not meet the z-axis and let

be the parametric equations of G. Consider now the family of curves

depending on the parameter t which decreases from t = I to t = 0. Note that G(l) = G and that G ' = G(0) is a closed curve which lies in the plane y = 0. Note also that (for a fixed value of z) each point P of G(t) rotates about the z-axis as t varies; whence the solid angle W which C subtends at P does not vary with t. This implies that W₁ - W₀ will have the same value for G(0) as for G( l ) = G. In order to prove the second formula, note that

Chapter VI

6.1

1. Use the theorem of the conservation of energy and prove that r ® ¥ as t ® ¥.

2. If (x, h) are the co-ordinates with respect to the axes of the ellipse, then

give the equation of the ellipse; and by the law of areas

[Note that the question ought to read: ". . . the angle P'MP_s , where P' is the point on the auxiliary circle corresponding to P, the position of the planet . . . ".]

3., 4.. Use the theorem of the conservation of energy and the law of areas.

6.2

1.

2. (c) x*—2cx+y*=0 (circles).

3.

and the variables are separated.

4.

5.

6. Introduce l/y as new unknown function; the equation then becomes homogeneous:

6.3

1. Use induction. Suppose that there holds a linear relation . Divide by and differentiate (n_k+1) times, if P_k(x) is of degree n_k. The degree of the coefficients of the other is unchanged, so that they remain different from zero.

2. Multiply both sides of the equation by (1 - n)y ^-n.

3. If we set y = y₁ + u^-1, the equation reduces to the linear equation

4. By equating the right-hand sides of (a) and (b), we obtain the common integral y = x².

5.

In order to draw the graphs of the corresponding family of curves, first plot the two branches of the curve

which divides the plane into two regions where y' < 0 and one region where y' > 0. The two infinite branches of this curve are asymptotic to the two parabolas y = ± x². Show that all the integral curves are asymptotic to these parabolas by proving the two relations

where o(1) denotes a function which tends to zero.

6. Set

Then

so that

Similarly,

Hence,

and similarly, by subtraction,

7. Cf. the relation

in the proof of 6.

Particular solutions of the special equation are

8. The common solution e^x of (a) and (b) is obtained by eliminating y" from the two equations

. 6.4

1. It follows from the fundamental theorem of algebra it follows that f(z) may be written as

where the m_n are positive integers such that

Now,

On differentiating this relation (m_n - 1) times and setting l = a_n in the result, we get by. Leibnitz's rule

So we have n particular solutions

which are linearly independent, by 1. of Exercises 6.3.

2.

3. On substituting in the differential equation, we get

an identity if a₀b₀ = 1, a₀b₁ + a₁b₀ = 0, ··· , from the expansion.The second case reduces to the first if we substitute y' for y.

4.

5.

6. y = e^x(x²/2 + 3x/2 + 7/4) + c₁e^3x + c₂e^2x.

6.5

1. (a) Use the fact that the curvilinear integral

is independent of the path. Integrating between (0, 0) and (x, y) along the broken line (0, 0) ··· (x, 0) ··· (x, y), we get

(b) By inspection, we find the general integral

2. Here dy/dx is a function of y/x alone.

3. x² y — 2xy² — 2cy — 2 = 0 (integrating factor m = 1/y²)

4. The equation is linear in x and its general integral is (xy² +1)² = cy. The identity

displays an integrating factor of the equation.

5.
(c) The differential equation of this family of confocal conics is found to be

which is unaltered if y' is replaced by -l/y'; the family of ellipses (-b²<c<¥ ) is orthogonal to the family of hyperbolas (-a²<c<-b²).
(d) y = log |tan(x/2)| + c and the vertical lines x = kp (k an integer).
(e) The family of curves (tractrix)

and the same family reflected in the x-axis.

6. (a) The family of parabolas y = cx².
(b) The family of hyperbolas xy = c.

7. (a) y= x³, (b) y= -x + x log(-x) (0 > x > -¥ ).

8.

9.

Note that this gives for c = 0 the parabola y = x² - a²/4. What is the geometrical meaning of this result?

10.
which is a family of cycloids and can be expressed in the parametric form x = c + a(j - sin j), y = a(l - cos j). Singular solution y = 2a.

11. and the differential equation is
By the general method this is easily reduced to

The various cases, all of importance in the differential geometry of surfaces (Eisenhart, Differential Geometry, pp. 270-4 (Princeton Press)), are as follows:

12. Eliminate b from the equations obtained by differentiating the equation of the circle twice and three times:

13.

14. If we substitute the power series for cos xt in the expression for J_o(x) in Exercise 4 and interchange summation and integration (why is this permissible?), we get

the value of

as it is found by setting t = sin t and referring to Volume I, 4.4.4. The power series for y(x) and J_o(x) are therefore identical.

6.6

1. Poisson's formula gives a potential function u(r, q) inside the unit circle, with boundary values f(q). Now u( 1/r, q) is also a potential function with the same boundary values, and it is bounded in the region outside the unit circle; thus the expression

is a solution of the problem.

2. The potential is

Since on the ellipsoid

the potential is

the confocal ellipsoids

are equipotential surfaces. The lines of force are the orthogonal trajectories and hence (Exercises 6.5, 5c) are the confocal hyperbolas given by the same equation when 0 £ a £ 1 and the ratio of x to y is constant.

3. Let S be a sphere of radius r and centre (x, y, z), lying inside S. Since D(1/r ) = 0 and Du = 0 in the region bounded by S and S, by Green's theorem ,

where in the first integral n is the outward normal to S and in the second the outward normal to S. Now, we have on the sphere S

hence

since u is a harmonic function; in addition,

and, as r ® 0, this espression obviously tends to u(x, y, z), because it is the mean value of u on S.

6.7

1. (a) u= f(x) + g(y) (f and g are arbitrary functions).
(b) u == f(x, y) + g(x, z) + h(y, z) (f, g, h are arbitrary functions).
(c) The most general solution is obtained from a particular solution by adding the general solution of the homogeneous equation u_xy = 0.

2. Apply the linear transformation

3.

4.

by differentiating the first equation and comparing it with the second one, we have

or

For t

Moreover, for t ³ r,

i.e., . As always a + at ³ 0 and hence g(a + at) = - c, it it follows that

if both x and t are not negative.

5. If then

in addition,

Hence,

where J₀ is the the Bessel function

6. (a) The differential equation yields

As the left hand side does not depend on y, nor the right hand side on x, both sides are equal to a constant (which has to be positive or zero), say c², i.e.,

Hence,

is a solution, where c and b are arbitrary and c² £ 1.
(b) u = f(x)+ g(y) yields

so that

(where a and b are constants).

If u = f(x)/g(y), then

so that in this case

where a, b, c are arbitrary constants.

7. A one-parameter family is obtained from the two-parameter family of solutions z = u(x, y, a, b) by making a and 6 depend in some way on a parameter t:

The envelope of this one-parameter family is obtained by finding t from the equation

and substituting this expression for t in z = u(x, y, f(t), g(t)). The result is again a solution of F(x; y, z, z_x, z_y) = 0, as

and z = u(x, y, a, b) satisfies the equation F(x; y, z, z_x, z_y) = 0.

8.

Chapter VII

7.1

7.2

1.

2. Circle with centre on x-axis.

only depends on the end-points of the curve y = y(x)).

7.

8. Consider F(x, y) for fixed x as a function of y; let this function of y have a minimum for y = . Then, F(x, y) ³ F(x, ) for a certain neighbourhood of and F(x, ) = 0. will depend on the parameter x. i.e., = (x). Then, for any neighbouring y, we have

where (x) satisfies the equation F_y(x, (x)) = 0.

9.

7.3

If y = 1/f(r), then T is given by

Euler's equation for the variable j yields

along a ray. Now, let the polar co-ordinates be chosen in such a way that the plane j = 0 passes through the initial point and the end-point; since j = 0 at both these points, we have for some intermediate point, by the mean value theorem, i.e., C = 0; but then for the whole ray, i.e. j = 0. Hence the whole ray must lie in the plane j = 0.

7.4

The law of conservation of energy gives

hence ds/dt = const = C = initial velocity.

Then Hamilton's principle asserts the stationary character of

a stationary character of Hamilton's integral implies that the length of the path is stationary.

MISCELLANEOUS EXERCISES VII

1. From the differential equations for geodesics, we find that for a cylinder, i.e., if G does not depend on z, dz/dt = const; hence the geodesics on a cylinder make a constant angle with the xy-plane.

from this equation and we have

(b) For any continuous, admissible j, we have

the equality sign holding for j = u.

Chapter VIII

8.1

1. For x ³ 0,

2. Use the principle of comparison.

3. By Volune I, Exercise 1.3, 3b, the coefficient of zⁿ in the expansion of cos² z + sin² z for n > 0 is

4. The series converges if, and only if, |z| < 1. In fact, if |z| = 0 < 1, then

and we may compare it with the geometric series. If |z| > 1, then z^v/(1 - z^v) tends to -1 as n increases, whereas in a convergent series the terms must tend to 0. If |z| = 1, either there are terms in the series which are not defined, or at least its terms are not bounded, since z^v may approach 1 as closely as we please.

8.2

Let f(z) = u + iv, g(z) = u' + iv', fg = p + iq, where p = uu' - vv', q = uv' + vu'. Assume that u, v and also u', v' satisfy the Cauchy-Riemann equations and prove that the same is true of p, q.

8.3

1. The functions (a), (b), (c) are continuous everywhere; (d) is discontinuous at z = 0.

2. None.

3.

Now, for , the difference between the numerator and the denominator is

so that the numerator is larger than the denominator for |z| > 1 and smaller for |z| < 1. If » \ < 1. If the converse is true.

4.

6. First transform by setting z = az + b into the unit circle; then apply the transformation

7. The equation of a circle or straight line in the z-plane has the form

where a and g are real; if we substitute here the expression for z, we get an equation of the same.form for z.

For a fixed point z = z, we hare the quadratic equation

which, in general, has two different solutions. A circle through the fixed points is, as we have just shown, transformed into a circle and must again pass through the fixed points; the family of orthogonal circles transforms into itself, because circles become circles and the transformation is conformal.

8.4

2. The series is absolutely convergent.

8.6

By the Cauchy-Riemann equations, the partial derivatives v_x and v_y are given; a function v with these derivatives does exist, since the condition of integrability u_xx + u_yy = 0 is satisfied; v is uniquely determined, apart from an additive constant c, and is given by the curvilinear integral

It also follows from the Cauchy-Riemann equations that v is a potential function.

8.7

1. lt is easily seen that

is an analytic function of z. By differentiating under the integral sign and using Leibnitz's rule, we find that h^(m)(z) is

Only the terms with m - n £ n differ from zero, as otherwise vanishes. On the other hand, a term with m - n < n vanishes for z = 0; if (m < n, there are no other terms, so that h^(m)(0) = 0. If m ³ 0, there remains only the term with m-n = n so that

2.

where C is the circle of radius r about the origin.

3. is equal to the sum of the residue of f'/f in the interior of C, Now, if f has a zero of order n at z = z₀,

so the residue of at z = z₀ is 2pin.

4. (a) The number of roots of the equation P(z) + qQ(z) = 0, by 3,, is

The denominator differs from zero for every q for which 0 £ q £ 1at any point of C; the entire integral is therefore a continuous function of q. As its value is always an integer, it is constant, and hence the same for q = 0 and q = 1.
(b) If |a| < r⁴ - 1/r, then r > 1; so the equation z⁵ + 1 = 0 has five roots inside the circle |z| = r; if we set P(z) = z⁵ + 1, Q(z)=az, we have on the circle z| = r

5. Cf. proof of 3.

8.8

The left-hand side of the formula is the sum of the residues of the function z^k/f(z) and is therefore equal to taken around a circle enclosing all the roots a_n. But this integral tends to zero as the radius of the circle tends to infinity (the centre remaining fixed).