PROOF OF THEOREM 1 Uniqueness¹

Assuming that the problem consisting of the ODE

and the two initial conditions

has two solutions y₁(x) and y₂(x) on the interval I in the theorem, we show that their difference

is identically zero on I; then y₁ ≡ y₂ on I, which implies uniqueness.

Since (1) is homogeneous and linear, y is a solution of that ODE on I, and since y₁ and y₂ satisfy the same initial conditions, y satisfies the conditions

We consider the function

and its derivative

From the ODE we have

By substituting this in the expression for z′ we obtain

Now, since y and y′ are real,

From this and the definition of z we obtain the two inequalities

From (13b) we have 2yy′ −z. Together, |2yy′| z. For the last term in (12) we now obtain

Using this result as well as −p |p| and applying (13a) to the term 2yy′ in (12), we find

Since y′² y² + y′² = z, from this we obtain

or, denoting the function in parentheses by h,

Similarly, from (12) and (13) it follows that

The inequalities (14a) and (14b) are equivalent to the inequalities

Integrating factors for the two expressions on the left are

The integrals in the exponents exist because h is continuous. Since F₁ and F₂ are positive, we thus have from (15)

This means that F₁z is nonincreasing and F₂z is nondecreasing on I. Since z(x₀) = 0 by (11), when x x₀ we thus obtain

and similarly, when x x₀,

Dividing by F₁ and F₂ and noting that these functions are positive, we altogether have

This implies that z =y² + y ′² ≡ 0 on I. Hence y ≡ 0 or y₁ ≡ y₂ on I.

PROOF OF THEOREM 2 Frobenius Method. Basis of Solutions. Three Cases

The formula numbers in this proof are the same as in the text of Sec. 5.3. An additional formula not appearing in Sec. 5.3 will be called (A) (see below).

The ODE in Theorem 2 is

where b(x) and c(x) are analytic functions. We can write it

The indicial equation of (1) is

The roots r₁, r₂ of this quadratic equation determine the general form of a basis of solutions of (1), and there are three possible cases as follows.

Case 1. Distinct Roots Not Differing by an Integer. A first solution of (1) is of the form

and can be determined as in the power series method. For a proof that in this case, the ODE (1) has a second independent solution of the form

see Ref. [A11] listed in App. 1.

Case 2. Double Root. The indicial equation (4) has a double root r if and only if (b₀ − 1)² − 4c₀ = 0, and then A first solution

can be determined as in Case 1. We show that a second independent solution is of the form

We use the method of reduction of order (see Sec. 2.1), that is, we determine u(x) such that y₂(x) = u(x)y₁(x) is a solution of (1). By inserting this and the derivatives

into the ODE (1′) we obtain

Since y₁ is a solution of (1′), the sum of the terms involving u is zero, and this equation reduces to

By dividing by x²y₁ and inserting the power series for b we obtain

Here, and in the following, the dots designate terms that are constant or involve positive powers of x. Now, from (7), it follows that

Hence the previous equation can be written

Since r = (1 − b₀)/2, the term (2r + b₀)/x equals 1/x, and by dividing by u′ we thus have

By integration we obtain ln u′ = −ln x + …, hence u′ = (1/x) e^(…). Expanding the exponential function in powers of x and integrating once more, we see that u is of the form

Inserting this into y₂ =uy₁, we obtain for y₂ a representation of the form (8).

Case 3. Roots Differing by an Integer. We write r₁ = r and r₂ = r − p where p is a positive integer. A first solution

can be determined as in Cases 1 and 2. We show that a second independent solution is of the form

where we may have k ≠ 0 or k = 0. As in Case 2 we set y₂ = uy₁. The first steps are literally as in Case 2 and give Eq. (A),

Now by elementary algebra, the coefficient b₀ − 1 of r in (4) equals minus the sum of the roots,

Hence 2r + b₀ = p + 1, and division by u′ gives

The further steps are as in Case 2. Integrating, we find

where dots stand for some series of nonnegative integer powers of x. By expanding the exponential function as before we obtain a series of the form

We integrate once more. Writing the resulting logarithmic term first, we get

Hence, by (9) we get for y₂ = uy₁ the formula

But this is of the form (10) with k = k_p since r₁ − p = r₂ and the product of the two series involves nonnegative integer powers of x only.

THEOREM Determinants

The definition of a determinant

as given in Sec. 7.7 is unambiguous, that is, it yields the same value of D no matter which rows or columns we choose in the development.

PROOF

In this proof we shall use formula numbers not yet used in Sec. 7.7.

We shall prove first that the same value is obtained no matter whichrow is chosen.

The proof is by induction. The statement is true for a second-order determinant, for which the developments by the first row a₁₁a₂₂ + a₁₂(−a₂₁) and by the second row a₂₁(−a₁₂) + a₂₂a₁₁ give the same value a₁₁a₂₂ −a₁₂a₂₁. Assuming the statement to be true for an (n − 1) st-order determinant, we prove that it is true for an nth-order determinant.

For this purpose we expand D in terms of each of two arbitrary rows, say, the ith and the jth, and compare the results. Without loss of generality let us assume i < j.

First Expansion. We expand D by the ith row. A typical term in this expansion is

The minor M_ik of a_ik in D is an (n − 1)st-order determinant. By the induction hypothesis we may expand it by any row. We expand it by the row corresponding to the jth row of D. This row contains the entries a_jl (l ≠ k). It is the (j − 1) st row of M_ik, because M_ik does not contain entries of the ith row of D, and i < j. We have to distinguish between two cases as follows.

Case I. If l < k, then the entry a_ij belongs to the lth column of M_ik (see Fig. 561). Hence the term involving a_jl in this expansion is

where M_ikjl is the minor of a_jl in M_ik. Since this minor is obtained from M_ik by deleting the row and column of a_jl, it is obtained from D by deleting the ith and jth rows and the kth and lth columns of D. We insert the expansions of the M_ik into that of D. Then it follows from (19) and (20) that the terms of the resulting representation of D are of the form

where

Case II. If l > k, the only difference is that then a_jl belongs to the (l − 1)st column of M_ik, because M_ik does not contain entries of the kth column of D, and k < l. This causes an additional minus sign in (20), and, instead of (21a), we therefore obtain

where b is the same as before.

Fig. 561. Cases I and II of the two expansions of D

Second Expansion. We now expand D at first by the jth row. A typical term in this expansion is

By the induction hypothesis we may expand the minor M_jl of a_jl in D by its ith row, which corresponds to the ith row of D, since j > i.

Case I. If k > l, the entry a_jk in that row belongs to the (k − 1)st column of M_jl, because M_jl does not contain entries of the lth column of D, and l < k (see Fig. 561). Hence the term involving a_p in this expansion is

where the minor M_ikjl of a_jk in M_jl is obtained by deleting the ith and jth rows and the kth and lth columns of D [and is, therefore, identical with M_ikjl in (20), so that our notation is consistent]. We insert the expansions of the M_jl into that of D. It follows from (22) and (23) that this yields a representation whose terms are identical with those given by (21a) when l <k.

Case II. If k < l, then a_ik belongs to the kth column of M_jl, we obtain an additional minus sign, and the result agrees with that characterized by (21b).

We have shown that the two expansions of D consist of the same terms, and this proves our statement concerning rows.

The proof of the statement concerning columns is quite similar; if we expand D in terms of two arbitrary columns, say, the kth and the lth, we find that the general term involving a_jla_ik is exactly the same as before. This proves that not only all column expansions of D yield the same value, but also that their common value is equal to the common value of the row expansions of D.

This completes the proof and shows that our definition of an nth-order determinant is unambiguous.

PROOF OF FORMULA (2)

We prove that in right-handed Cartesian coordinates, the vector product

has the components

We need only consider the case v ≠ 0. Since v is perpendicular to both a and b, Theorem 1 in Sec. 9.2 gives a · v = 0 and b · v = 0; in components [see (2), Sec. 9.2],

Multiplying the first equation by b₃, the last by a₃, and subtracting, we obtain

Multiplying the first equation by b₁, the last by a₁, and subtracting, we obtain

We can easily verify that these two equations are satisfied by

where c is a constant. The reader may verify, by inserting, that (4) also satisfies (3). Now each of the equations in (3) represents a plane through the origin in v₁v₂v₃-space. The vectors a and b are normal vectors of these planes (see Example 6 in Sec. 9.2). Since v ≠ 0, these vectors are not parallel and the two planes do not coincide. Hence their intersection is a straight line L through the origin. Since (4) is a solution of (3) and, for varying c, represents a straight line, we conclude that (4) represents L, and every solution of (3) must be of the form (4). In particular, the components of v must be of this form, where c is to be determined. From (4) we obtain

This can be written

as can be verified by performing the indicated multiplications in both formulas and comparing. Using (2) in Sec. 9.2, we thus have

By comparing this with formula (12) in Prob. 4 of Problem Set 9.3 we conclude that c = ±1.

We show that c = +1. This can be done as follows.

If we change the lengths and directions of a and b continuously and so that at the end a = i and b = j (Fig. 188a in Sec. 9.3), then v will change its length and direction continuously, and at the end, v = i × j = k. Obviously we may effect the change so that both a and b remain different from the zero vector and are not parallel at any instant. Then v is never equal to the zero vector, and since the change is continuous and c can only assume the values +1 or −1, it follows that at the end c must have the same value as before. Now at the end a = i, b = j, v = k and, therefore, a₁ = 1, b₂ = 1, v₃ = 1, and the other components in (4) are zero. Hence from (4) we see that v₃ =c = +1. This proves Theorem 1.

For a left-handed coordinate system, i × j = − k (see Fig. 188b in Sec. 9.3), resulting in c = −1. This proves the statement right after formula (2).

THEOREM A Transformation Law for Vector Components

For any vector v the components v₁, v₂, v₃ and in any two systems of Cartesian coordinates x₁, x₂, x₃ and , respectively, are related by

and conversely

with coefficients

satisfying

where the Kronecker delta²is given by

and i, j, k and i*, j*, k* denote the unit vectors in the positive and -directions, respectively.

PROOF

The representation of v in the two systems are

Since i* · i* = 1, i* · j* = 0, i* · k* = 0, we get from (5b) simply i* · v = v₁* and from this and (5a)

Because of (3), this is the first formula in (1), and the other two formulas are obtained similarly, by considering j* · v and then k* · v. Formula (2) follows by the same idea, taking i · v = v₁ from (5a) and then from (5b) and (3)

and similarly for the other two components.

We prove (4). We can write (1) and (2) briefly as

Substituting v_j into , we get

where k = 1, 2, 3. Taking k = 1, we have

For this to hold for every vector v, the first sum must be 1 and the other two sums 0. This proves (4) with k = 1 for m = 1, 2, 3. Taking k = 2 and then k = 3, we obtain (4) with k = 2 and 3, for m = 1, 2, 3.

THEOREM B Transformation Law for Cartesian Coordinates

The transformation of any Cartesian x₁x₂x₃-coordinate system into any other Cartesian -coordinate system is of the form

with coefficients (3) and constants b₁, b₂, b₃; conversely,

PROOF OF THE INVARIANCE OF THE CURL

We write again x₁, x₂, x₃ instead of x, y, z, and similarly for other Cartesian coordinates, assuming that both systems are right-handed. Let a₁, a₂, a₃ denote the components of curl v in the x₁x₂x₃-coordinates, as given by (1), Sec. 9.9, with

Similarly, let denote the components of curl v in the -coordinate system. We prove that the length and direction of curlv are independent of the particular choice of Cartesian coordinates, as asserted. We do this by showing that the components of curl v satisfy the transformation law (2), which is characteristic of vector components. We consider a₁. We use (6a), and then the chain rule for functions of several variables (Sec. 9.6). This gives

From this and (7) we obtain

Note what we did. The double sum had 3 × 3 = 9 terms, 3 of which were zero (when m = j), and the remaining 6 terms we combined in pairs as we needed them in getting .

We now use (3), Lagrange's identity (see Formula (15) in Team Project 24 in Problem Set 9.3) and k^* × j^* −i* and k × j = −i. Then

Hence . This is of the form of the first formula in (2) in Theorem A, and the other two formulas of the form (2) are obtained similarly. This proves the theorem for right-handed systems. If the x₁x₂x₃-coordinates are left-handed, then k × j = +i, but then there is a minus sign in front of the determinant in (1), Sec. 9.9.

PROOF OF THEOREM 1, PART (b) We prove that if

with continuous F₁, F₂, F₃in a domain D is independent of path in D, then F = grad f in D for some f; in components

We choose any fixed A: (x₀, y₀, z₀) in D and any B: (x, y, z) in D and define f by

with any constant f₀ and any path from A to B in D. Since A is fixed and we have independence of path, the integral depends only on the coordinates x, y, z, so that (3) defines a function f(x, y, z) in D. We show that F = grad f with this f, beginning with the first of the three relations (2′). Because of independence of path we may integrate from A to B₁: (x₁, y, z) and then parallel to the x-axis along the segment B₁B in Fig. 562 with B₁ chosen so that the whole segment lies in D. Then

We now take the partial derivative with respect to x on both sides. On the left we get δf/δx. We show that on the right we get F₁. The derivative of the first integral is zero because A: (x₀, y₀, z₀) and B₁: (x₁, y, z) do not depend on x. We consider the second integral. Since on the segment B₁B, both y and z are constant, the terms F₂dy* and F₃dz * do not contribute to the derivative of the integral. The remaining part can be written as a definite integral,

Fig. 562. Proof of Theorem 1

Hence its partial derivative with respect to x is F₁(x, y, z), and the first of the relations (2′) is proved. The other two formulas in (2) follow by the same argument.

THEOREM Reality of Eigenvalues

If p, q, r, and p′ in the Sturm–Liouville equation (1) of Sec. 11.5 are real-valued and continuous on the interval a x b and r (x) > 0 throughout that interval (or r (x) 0 throughout that interval), then all the eigenvalues of the Sturm–Liouville problem (1), (2), Sec. 11.5, are real.

PROOF

Let λ = α + iβ be an eigenvalue of the problem and let

be a corresponding eigenfunction; here α, β, u, and v are real. Substituting this into (1), Sec. 11.5, we have

This complex equation is equivalent to the following pair of equations for the real and the imaginary parts:

Multiplying the first equation by v, the second by −u and adding, we get

The expression in brackets is continuous on a x b, for reasons similar to those in the proof of Theorem 1, Sec. 11.5. Integrating over x from a to b, we thus obtain

Because of the boundary conditions, the right side is zero; this is as in that proof. Since y is an eigenfunction, u² + v² 0. Since y and r are continuous and r > (or r < 0) on the interval a x b, the integral on the left is not zero. Hence, β = 0, which means that λ = α is real. This completes the proof.

PROOF OF THEOREM 2 Cauchy–Riemann Equations

We prove that Cauchy–Riemann equations

are sufficient for a complex function f(z) = u (x, y) + iv (x, y) to be analytic; precisely, if the real part u and the imaginary part v of f(z) satisfy (1) in a domain D in the complex plane and if the partial derivatives in (1) are continuous in D, then f(z) is analytic in D.

In this proof we write Δz = Δx + iΔy and Δf = f(z + Δz) − f(z). The idea of proof is as follows.

(a) We express Δf in terms of first partial derivatives of u and v, by applying the mean value theorem of Sec. 9.6.

(b) We get rid of partial derivatives with respect to y by applying the Cauchy–Riemann equations.

(c) We let Δz approach zero and show that then Δf/Δz, as obtained, approaches a limit, which is equal to u_x + iv_x, the right side of (4) in Sec. 13.4, regardless of the way of approach to zero.

(a) Let P: (x, y) be any fixed point in D. Since D is a domain, it contains a neighborhood of P. We can choose a point Q: (x + Δx, y + Δy) in this neighborhood such that the straight-line segment PQ is in D. Because of our continuity assumptions we may apply the mean value theorem in Sec. 9.6. This yields

where M₁ and M₂ (≠ M₁ in general!) are suitable points on that segment. The first line is Re Δf and the second is Im Δf, so that

(b) u_y = −v_x and v_y = u_x by the Cauchy–Riemann equations, so that

Also Δz = Δx + iΔy, so that we can write Δx = Δz −iΔy in the first term and Δy = (Δz − Δx)/i = −i (Δz − Δx) in the second term. This gives

By performing the multiplications and reordering we obtain

Division by Δz now yields

(c) We finally let Δz approach zero and note that |Δy/Δz| 1 and |Δx/Δz| 1 in (A). Then Q: (x + Δx, y + Δy) approaches P: (x, y), so that M₁ and M₂ must approach P. Also, since the partial derivatives in (A) are assumed to be continuous, they approach their value at P. In particular, the differences in the braces {…} in (A) approach zero. Hence the limit of the right side of (A) exists and is independent of the path along which Δz → 0. We see that this limit equals the right side of (4) in Sec. 13.4. This means that f(z) is analytic at every point z in D, and the proof is complete.

Section 14.2, pages 653–654

GOURSAT'S PROOF OF CAUCHY'S INTEGRAL THEOREM

Goursat proved Cauchy's integral theorem without assuming that f′(z) is continuous, as follows.

We start with the case when C is the boundary of a triangle. We orient C counterclockwise. By joining the midpoints of the sides we subdivide the triangle into four congruent triangles (Fig. 563). Let C_I, C_II, C_III, C_IV denote their boundaries. We claim that (see Fig. 563).

Indeed, on the right we integrate along each of the three segments of subdivision in both possible directions (Fig. 563), so that the corresponding integrals cancel out in pairs, and the sum of the integrals on the right equals the integral on the left. We now pick an integral on the right that is biggest in absolute value and call its path C₁. Then, by the triangle inequality (Sec. 13.2),

We now subdivide the triangle bounded by C₁ as before and select a triangle of subdivision with boundary C₂ for which

Fig. 563. Proof of Cauchy's integral theorem

Continuing in this fashion, we obtain a sequence of triangles T₁, T₂, … with boundaries C₁, C₂, … that are similar and such that T_n lies in T_m when n > m, and

Let z₀ be the point that belongs to all these triangles. Since f is differentiable at z = z₀, the derivative f′(z₀) exists. Let

Solving this algebraically for f(z) we have

Integrating this over the boundary C_n of the triangle T_n gives

Since f(z₀) and f′(z₀) are constants and C_n is a closed path, the first two integrals on the right are zero, as follows from Cauchy's proof, which is applicable because the integrands do have continuous derivatives (0 and const, respectively). We thus have

Since f′(z₀) is the limit of the difference quotient in (3), for given > 0 we can find a δ > 0 such that

We may now take n so large that the triangle T_n lies in the disk |z − z₀|< δ. Let L_n be the length of C_n. Then |z − z₀| < L_n for all z on C_n and z₀ in T_n. From this and (4) we have |h(z)(z − z₀)| < L_n. The ML-inequality in Sec. 14.1 now gives

Now denote the length of C by L. Then the path C₁ has the length L₁ = L/2, the path C₂ has the length L₂ = L₁/2 = L/4, etc., and C_n has the length L_n = L/2ⁿ. Hence . From (2) and (5) we thus obtain

By choosing (> 0) sufficiently small we can make the expression on the right as small as we please, while the expression on the left is the definite value of an integral. Consequently, this value must be zero, and the proof is complete.

The proof for the case in which C is the boundary of a polygon follows from the previous proof by subdividing the polygon into triangles (Fig. 564). The integral corresponding to each such triangle is zero. The sum of these integrals is equal to the integral over C, because we integrate along each segment of subdivision in both directions, the corresponding integrals cancel out in pairs, and we are left with the integral over C.

The case of a general simple closed path C can be reduced to the preceding one by inscribing in C a closed polygon P of chords, which approximates C “sufficiently accurately,” and it can be shown that there is a polygon P such that the integral over P differs from that over C by less than any preassigned positive real number , no matter how small. The details of this proof are somewhat involved and can be found in Ref. [D6] listed in App. 1.

PROOF OF THEOREM 4 Cauchy's Convergence Principle for Series

(a) In this proof we need two concepts and a theorem, which we list first.

1. A bounded sequence s₁, s₂, … is a sequence whose terms all lie in a disk of (sufficiently large, finite) radius K with center at the origin; thus |s_n| < K for all n.

2. A limit point a of a sequence s₁, s₂, … is a point such that, given an > 0, there are infinitely many terms satisfying |s_n − a| > . (Note that this does not imply convergence, since there may still be infinitely many terms that do not lie within that circle of radius and center a.)

3. A bounded sequence in the complex plane has at least one limit point. (Bolzano–Weierstrass theorem; proof below. Recall that “sequence” always means infinite sequence.)

(b) We now turn to the actual proof that z₁ + z₂ + … converges if and only if, for every > 0, we can find an N such that

Here, by the definition of partial sums,

Writing n + p = r, we see from this that (1) is equivalent to

Suppose that s₁, s₂, … converges. Denote its limit by s. Then for a given > 0 we can find an N such that

Hence, if r > N and n > N, then by the triangle inequality (Sec. 13.2),

that is, (1*) holds.

(c) Conversely, assume that s₁, s₂, … satisfies (1*). We first prove that then the sequence must be bounded. Indeed, choose a fixed and a fixed n = n₀ > N in (1*). Then (1*) implies that all s_r with r > N lie in the disk of radius and center s_n0 and only finitely many terms s₁, …, s_N may not lie in this disk. Clearly, we can now find a circle so large that this disk and these finitely many terms all lie within this new circle. Hence the sequence is bounded. By the Bolzano–Weierstrass theorem, it has at least one limit point, call it s.

We now show that the sequence is convergent with the limit s. Let > 0 be given. Then there is an N* such that |s_r − s_n|</2 for all r > N* and n > N*, by (1*). Also, by the definition of a limit point, |s_n − s| < /2 for infinitely many n, so that we can find and fix an n > N* such that |s_n − s|/2. Together, for every r > N *,

that is, the sequence s₁, s₂, … is convergent with the limit s.

THEOREM Bolzano–Weierstrass Theorem³

A bounded infinite sequence z₁, z₂, z₃, … in the complex plane has at least one limit point.

PROOF

It is obvious that we need both conditions: a finite sequence cannot have a limit point, and the sequence 1, 2, 3, …, which is infinite but not bounded, has no limit point. To prove the theorem, consider a bounded infinite sequence z₁, z₂, … and let K be such that |z_n| < K for all n. If only finitely many values of the z_n are different, then, since the sequence is infinite, some number z must occur infinitely many times in the sequence, and, by definition, this number is a limit point of the sequence.

We may now turn to the case when the sequence contains infinitely many different terms. We draw a large square Q₀ that contains all z_n. We subdivide Q₀ into four congruent squares, which we number 1, 2, 3, 4. Clearly, at least one of these squares (each taken with its complete boundary) must contain infinitely many terms of the sequence. The square of this type with the lowest number (1, 2, 3, or 4) will be denoted by Q₁. This is the first step. In the next step we subdivide Q₁ into four congruent squares and select a square Q₂ by the same rule, and so on. This yields an infinite sequence of squares Q₀, Q₁, Q₂, …, Q_n, … with the property that the side of Q_n approaches zero as n approaches infinity, and Q_m contains all Q_n with n > m. It is not difficult to see that the number which belongs to all these squares,⁴ call it z = a, is a limit point of the sequence. In fact, given an > 0, we can choose an N so large that the side of the square Q_N is less than and, since Q_N contains infinitely many z_n, we have |z_n − a| < for infinitely many n. This completes the proof.

PART (b) OF THE PROOF OF THEOREM 5

We have to show that

thus,

If we set z + Δz = b and z = a, thus Δz = b − a, this becomes simply

where A_n is the expression in the brackets on the right,

thus, A₂ = 1, A₃ = b + 2a, etc. We prove (7) by induction. When n = 2, then (7) holds, since then

Assuming that (7) holds for n = k, we show that it holds for n = k + 1. By adding and subtracting a term in the numerator and then dividing we first obtain

By the induction hypothesis, the right side equals b[(b − a)A_k + ka^k−1] + a^k. Direct calculation shows that this is equal to

From (7b) with n = k we see that the expression in the braces {…} equals

Hence our result is

Taking the last term to the left, we obtain (7) with n = k + 1. This proves (7) for any integer n 2 and completes the proof.

ANOTHER PROOF OF THEOREM 1 without the use of a harmonic conjugate

We show that if w = u + iv = f(z) is analytic and maps a domain D conformally onto a domain D* and Φ*(u, v) is harmonic in D*, then

is harmonic in D, that is, ∇²Φ = 0 in D. We make no use of a harmonic conjugate of Φ*, but use straightforward differentiation. By the chain rule,

We apply the chain rule again, underscoring the terms that will drop out when we form ∇²Φ:

Φ_yy is the same with each x replaced by y. We form the sum ∇²Φ. In it, is multiplied by

which is 0 by the Cauchy–Riemann equations. Also ∇²u = 0 and ∇²v = 0. There remains

By the Cauchy–Riemann equations this becomes

and is 0 since Φ* is harmonic.