1.1 The equation for the determination of the extreme values is
If one writes instead
one can sum the geometric series. Finally, one finds
The locations x = 0 and x = p are not counted due to the denominator in (3). (4) proves the statement.
1.2 The cos series (2.17) yields by integration between 0 and p a sin series and from it the from (2.14) following analogue of Leibnitz's series (1.8). Integration of this series yields a series which advances with 1 - cos x, 1 - cos 3x, ··· and from this series there arises for x = p/2 the next analogue of (2.16) from which one can obtain readily the value of S6. The procedure is readily extended, but does not seem to yield a clear law for the successive analogues.
1.3 The two methods stated yield the series
from which you obtain, for example, for s = 0 a strange representation of 1/2 by a sequence of odd reciprocal numbers.
1.4 In the case a), if one writes t instead of x and performs the integration with respect to x, Equation (4.8) yields
Here |a(w)| is the amplitude of the spectrum of f(x) for the frequency w and is, by (1),
It has its primary maximum of height t/p at w = 0, followed by secondary maxima of decreasing height at the asymptotic distance Dw = 2p/t.
In the case b), where, according to the figure, one is concerned with a function f which is odd in t, one obtains in the same manner
Now, b(w) = 0 for w = 0; the first maximum lies at w ~ 4.7/t ; the following secondary maxima have again successively decreasing heights.
In both cases a) and b), we are dealing with an into infinity running channelled spectrum.
At the start of the theory of X-rays, one wanted to explain these as aperiodic ether pulses of approximately the forms a) and b). From a spectral point of view, which alone is physically justified, i.e., this is not in contradiction to the wave concept, but only a (by the way, arbitrarily specialized) assumption regarding the nature of the X-ray spectrum.
In the case of on both sides fractured wave trains of length 2T = 2nt (Fig. 33c), it is convenient to start from (4.11b) and then find
The principal maximum of b lies, as expected, at the frequency w 0 = 2p/t and has the height
i.e., it grows with increasing length of the line train. It is flanked on both sides by secondary maxima of consecutively decreasing height, which follow each other asymptotically at the distance p/nt ; one has for all of them sin wT ~ 1. We will now seek those two maxima with intensity I=½I0 , for which, by (4),
The difference in the frequencies defines the so-called half-width of the spectral line under consideration. By (5), assuming that ,
Thus, the half-value width decreases with increasing T, as stated in Exercise 1.4. Only for , there arises an absolutely sharp spectral line.
1.5 The function f(t), represented by Fig. 33a) above, is called in the mathematical literature, after specializing t/2 to 1, the Dirichlet discontinuous factor:
If one sets here t = 0, one obtains the already discussed fundamental integral (Fig. 4)
This is readily confirmed by complex integration: Since sin w /w is analytic on the real axis and analytic in its environment, one can bypass the point w = 0, for example, as in Fig. 34a below, and then dissect (2a) into the difference of the two integrals
both taken along the strongly drawn path shown in the figure. It is readily drawn to infinity on the lower half-plane and vanishes there. I must be deformed to infinity in the upper half plane, because eiw only vanishes there, but remains hanging at the pole w = 0. The residue at the pole is 2p i. Hence, we have proved (2) as well as (2a).
We note that one can readily see by formation of majorants that also the proximity of the real axis, indicated in the figure by short arrows, and their broken line extensions do not contribute to I and II.
Complex integration is also suitable to extend the statements of the preceding exercise regarding the wave train, which is bounded on both sides. Obviously, one can conceive it as a superposition of two wave trains, bounded on one side, with opposite phase, one of which extends from t = -T to , the other from t = +T to . However, these two processes cannot be performed in Fourier fashion singly due to the occurring divergence at . For this purpose, one must first decompose in (4) of the preceding exercise the integration path away from the real axis into the complex regime, as is shown in Fig. 34b, and only then proceed with the decomposition. This is explained by the following transformations which start from (4):
where the integral sign without limits indicates the complex path in Fig. 34b.
We claim that I represents the the wave train starting at t = -T, II that starting at t = + T, both continuing to . The proof will have been obtained, when we set, for the sake of simplicity, T = 0 and show that we are faced in
with a sin wave which starts at t = 0 and ends at . This is shown similarly as in the case of (3): For t > 0, one must drag the integration path into the upper half-plane when it yields, due to the poles w=±2p /t , the residues
This is the required proof.
If one starts, instead of from (4), somewhat more simply from
one sees readily that one obtains for an equal choice of the integration path and its equal deformation the same result as above, namely
The interest in this representation occurs in optical dispersion theory. Imagine that the wave train (6) meets perpendicularly at the plane x = 0 a medium filling the half-space x > 0 and decomposes it, according to (5), into partial waves of the form a(w) eiw t. Everyone of them propagates then to the side of increasing x independently of all others and is represented there by a(w)ei[kz -w t]. The wave number k would be kw/e in a dispersion free medium; however, due to the ensuing vibrations of the electrons (their number is /cm³), one has
where ws is the eigen-frequency of the resonating electron (for the sake of simplicity, we disregard here damping). Hence, at infinity in the w-plane:
The half-plane into which we must drag the integration path is determined by the sign of x - ct. This criterion, being independent of w, is the same for all partial waves so that x = ct means the start of the entire light excitation at the location x of the dispersing medium. Hence, the head of our light signal propagates with the velocity in vacuum dx/dt = c, not, as one might assume with the for the dispersing medium characteristic phase velocity V = w/k. This has been interpreted in Volume IV in the way that the electrons are at rest for t < x/c and begin to vibrate at t = x/c. However, they only attain their full amplitude, corresponding to the incoming excitation, at a later time which now again is not determined by the phase velocity v, but by the group velocity U=dw/dk. These oscillations, which precede this time can be called forerunners of the light signal.
1.6 a) Hermite polynomials: Due to the even character of g(x) = e-x² and the interval bounds , these polynomials, like the spherical functions Pn, are even and odd functions of x depending on n being even or odd. Taking this into consideration and using the customary normalization of the Hn, write
and compute a, b, c, d by continuing application of the orthogonality condition [result: a=-2, b=-12, c=-48, d=+12).
b) Laguerre polynomials: Due to their weight g(x) = e-x and their expansion interval , they are not even or odd with n. Taking this into consideration as well as their traditional normalization (Exercise 1.6), write
and compute a, b, ··· , f as in a). (Result: a = -1, b = 1, c = -4, d = -1, e = 9, f = -18).
2.1 The differential equation (7.8), to which this exercise refers, is obtained from the theory of bending of beams as follows: Equation (41.8) in Volume II, differentiated twice with respect to x and u replacing y, is
The bending moment M of the external, static loads on the beam is to be replaced by the dynamic inertial resistance
(r = mass per unit length of the vibrating beam). Let it be clamped at x = 0 and its free end be x=l. At the cross-section x, all cross-sections x < x < l contribute to the bending moment with the lever arm x - x, whence
Substitution in (1) yields (7.8) and for the constant c there (with dimension cm²/sec)
At the free end, by (2), one has This means, according to Volume II,
On the other hand, the clamping demands
With u = Ueiw t, U = ea x, Equation (7.8) yields
Hence, there are 4 particular solutions of the differential equation for U
which are for what follows conveniently combined into
whence the general solution is
By (4) and (5), there exist between the constants A, B, C, D four relations from which there follows the transcendental equation
Its graphical treatment, as in Fig. 7 , yields infinitely many roots which follow each other asymptotically at the distances
In the fundamental oscillation, one has
The differential equation for a tube with air is the same as that for a vibrating string: Equation (7.6) with u the longitudinal velocity of the air, c the velocity of sound. In the case of a tube which is open at both ends and the at x = 0 closed, at x= l open pipe, respectively, the boundary conditions ( means, due to the hydrodynamic continuity equation, the same as i.e., p = p0 = atmospheric pressure, as one will assume approximately)
respectively. Due to (8a, b), u = Ueiw t yields
The value (7a) of k lies between the values (9b) and (9a). The succession of the w for the open and closed pipe is harmonic, in the case of the elastic bar it becomes asymptotic for high overtones only (cf. (7)),
2.2 a) When L(u) has the normal form (10.1), the identities
L is a bilinear form of u,v and their first derivatives. If L is self-adjoint, due to D = E = 0, Equation (10.6) becomes
i.e., symmetric in u and v.
b) If L has the general form (8.1), Equation (1) also applies, but with
If L is self-adjoined, due to (10.6), L has the simpler form
i.e., it becomes also here symmetric in u and v. Moreover, this can still be simplified by shifting the terms
to X and Y on the right hand side of (1) with negative signs and obtain
If one now integrates (1) over a region S with the boundary C, there arises as most general version of the second form of Green's theorem:
c) In order to check whether the solution of a boundary value problem, posed for the differential equation L(u) = 0, is unique, one proceeds as with the equation Du = 0:
Assuming that there exist different solutions u1, u2 for prescribed boundary values of u on the boundary curve C, one sets in (9)
Then, the first term on the left hand side of (9) vanishes (from L(u1) = 0 and L(u2) = 0 follows also, due to the linearity of L, L(u) = 0); in the same way, there vanishes the right hand side of (9) (from u1 = u2 on C follows for all ds on it v = 0 and them, by (6), also X = Y = 0), whence (9) becomes
By the in the problem given restriction of self-adjointness and with the abbreviations , one has
The upper expression leads to a contradiction to (10), if F(x, y) in S is negative throughout; similarly, the lower expression, if the quadratic form Ax ²+ 2Bxh + Ch² is definite and F(x, y) has throughout the opposite sign to it. Both times, one then concludes from (10) that
i.e., the boundary value problem has only one solution.
This is identical with the non-existence of eigen-functions. In particular, there follows for the into normal form transformed, self-adjoint equation Du + Fu = 0, if F = const = , that
has no eigen-functions in contrast to the equation of the vibrating membrane
whenever it is primarily of interest.
The circumstance that, in the case of self-adjoint differential equations of the elliptic type, the question of uniqueness for the boundary value problem and that of the possibility of eigen-functions, respectively, could be decided quite generally and solely on the basis of our second form of Green's theorem explains the preferred role of this differential equation in mathematical physics.
In order to emphasize the physical significance of (13), note that it has in nuclear physics as Yukawa meson equation the same role as the potential equation in electron physics.
2.3 From the conditions a), b) c), one computes readily:
On the other hand, the solution of the posed boundary value problem is obviously
Fig. 35 shows the two curves G and U. Show that, in order for the to one dimension transferred Green function (10.12)
to agree with (2), one employs the values of in (1).
2.4 Since, by (2), v = v0(t) for z = 0 and t > 0 are given, the required Green function G must fulfill the condition
It is obtained from the principal solution V(x, t; x, t), (12.16), by reflection in the straight line x = 0 (cf. also 3.13):
whence follows for x = 0:
This value must be substituted in (12.18) for , which in the present case is simplified as follows: On the right hand side, remove the first term, since, by (4) in the exercise, v = 0 for t = 0. In the second term, the summand relating to drops out so that there only remains the summand which originates from x0 = 0 and is, by (12.18), negative. Here again, due to (1), drops out the component multiplied by V. Hence, due to (2) of the task, one has
If one substitutes instead of t as integration variable one finds the final solution of the problem
For a discussion of (4), expand
Interchanging x, t with x, t, (4) yields
with the abbreviations
In is essentially the well known and tabulated error integral, for which
The corresponding statements for In(x) are
(6) is an expansion suitable for z < 1. One has for
The transition between these limit laws occurs at z = 1, i.e., for
The flow around a plate, examined here, is a useful preparation for more recent studies of turbulence (Volume II, Section 38).
3.1 Following (13.1), start with
f(x) is given, f(-x) to be found. For x = 0,
If one integrates the last integral by parts and lets f(-x) become at x = 0 f(+x), i.e.,
then there disappears the term without an integral sign arising in the integration by parts and there only remains on the right hand side of (3)
The condition for z = 0 in the task is fulfilled when it is met by the two integrands in (2) and (4a). Taking into consideration that for us is like , we write
This differential equation for f(-x) becomes, after multiplication by ehx, integrable and yields after a simple transformation
This expression for f(-x) must eventually be inserted into (1). Verify that there arises the representation of G in (13.15), if one specializes in the thus obtained equation f into a d-function.
3.2 We are concerned with Green's theorem (16.6) by means of which we are to derive the normalization integral (6.3a) by the limiting process . Since it becomes 0/0, one applies L'Hospital's rule by differentiating first of all the numerator and denominator with respect to lm and then setting lm = ln:
The numerator need only be evaluated for x = l, since it vanishes for x = 0 before the limits is taken. By (16.5) and (16.5a), one finds
We have executed this computation in order to be able to use it as an example whenever the normalization integral cannot be performed in an elementary fashion as in the case of the trigonometric functions.
3.3 In the stationary case, (16.11) becomes
Hence, there follows the general solution
A and B are computed from
and, for x = /2,
thus, the variable q in the task becomes
This yields a quadratic equation for el l /2, whence
By (1), we have now also expressed h in terms of q. By (13.5), h is the ratio of the external thermal conductivity (in the notation there 4aT30) to the internal one k. For q = 1, (4) yields h = 0, as it should be according to the task.
3.4 In the stationary state, the energy, taken from the element of the bar (height dx, cross-section q) by heat conduction, must equal the Joule heat developed inside it. If one expresses the current i in terms of the potential difference V, one finds as the differential equation of the stationary state
The integration yields at the bar's ends for the history of the temperature with the boundary condition taken into account a centrally symmetric parabola. a is determined by its maximum U; a yields
Thus, measurement of V and U allows to check the law, empirically found by Wiedemann and Franz, which, according to the theory of metal electrons, states
where T = absolute temperature, k = Boltzmann's constant, e = charge of electron.
4.1 a) In the case of integer n, eir cos w in (19.18) can be expanded into the known power series. One thus finds as coefficient of r k
Expanding under the integral sign binomially, only a single term remains on integration and that happens only for even The result agrees with (19.34).
b) In the case of non-integer n (assuming r to be real), substitute in (19.14)
For example, the assumed rectangular path W transforms then into the loop in Fig. 37a, which comes from , moves around the origin clockwise and returns to according to the scheme
(19.14) then becomes
If one expands now er²/4t, one finds again the series (19.34) if one employs the following, generally valid definition of the G-function
it is readily seen to agree with the elementary definition of the G-function G (x + 1) = x! for integer x by finding the residues for t = 0.
4.2 As supplement to the examination of the real part of ir cos w, one computes for complex r=|r|eiq
For X to become negative at positive imaginary infinity of the w-plane (q > 1), on must have
Hence, the shaded strips in Figs. 18, 19
shifts when Q increases (decreases) from 0 to
i.e., to the right (left) hand side in Fig. 18. Conversely, in the negative imaginary half-plane, where, by (1), the place of (2) is taken by
For 0 < Q < p (positive imaginary r-half-plane), the shaded areas of the upper and lower half-plane meet each other on a finite section of the real axis, so that W1 can be taken all along on shaded territory. Hence, already [without a knowledge of the asymptotic formula (19.55)!], H1 vanishes in the limit in the positive imaginary r-half-plane. As a result of this shift of the shaded region, one also sees that the path W1 in 0 < Q <-p (negative imaginary r-plane) does not necessarily lead over a non- shaded region, so that then H1 becomes infinite in the limit . The inverse applies for both W2 and H2.
Fig. 36a illustrates the effect of this shift during a full circuit of r about its origin with respect to the path W1. The starting point of W1 has shifted by 2p, its end by - 2p. However, W1 is distorted into W1'. However, W1' can be decomposed into three partial paths with the character of the initial W1, W2, according to the symbolic equation
Here, -W1 is the central of the three dotted partial paths of the figure, which differs from the initial path W1 only by the sense in which is its travelled, -W2 is the partial path on its right side; the path on the left hand side originates from W2, if one replaces w by w - 2p , so that, due to the factor einw in the integrand, the integral changes by e-2pin. Thus, one obtains from (4)
For integer n, this becomes
which we can rewrite
This change of H1n by -4In corresponds together with H1 = I + iN to the change of log r by 2p i in (19.47) for N.
Fig. 36b shows the corresponding distorted path W2' of H2 during a full positive circuit of r about r=0. It can be decomposed into 5 partial paths of the kind of W1, W2
W2 is the central of the 5 partial paths, W1 next to it on the left hand side, W1e2pin on the right hand side , etc. Instead of (5), one finds
This yields for integer n
Also, the change (9a) corresponds, together with the relation H2 = I - iN, to the change of log r by 2p i in (19.47).
Equations (5) and (6) are the so-called circulation relations of the Hankel functions for the change of angle DQ = 2p. They correspond to Gauss' relationes inter contigua of the hypergeometric function. Just as (6a), (9a) arise for integer n from (19.47), the general relations (5) and (8) can be derived from (19.31) and (19.30).
These relations are readily generalized from the full circuits DQ = 2pn (n and integer) to half circuits DQ = pn. For the sake of a later application in 6.32, we will explain here still this half circulation relations for n =1 and n =0:
Its proof is obtained from Fig. 36c above. For real r, the path W1 (- · -) leads out of the field -p<p<0 into the field 0<p<+p ; it has shifted at the present argument reip in the upper part by +p, in the lower part by -p, as the arrows there indicate. However, the corresponding path is then identical to the path W2 for H2. But this is what Equation (10) demands.
A relation, analogous to (10), arises if we replace there r by r e-ip, i.e., write
With respect to the factor eipnw, which for arbitrary index n occurs under the integral sign, (10) is readily generalized to
or (interchange of r with r -ip)
Also these half circuit relations can be read off from (19.31) and (19.30), if one takes into consideration the equations, following from (19.34):
These relation become very simple for yn(r) and zn(r), in which the corresponding Hankel and Bessel functions of subscript n + ½ are multiplied by ; in fact, then
4.3 Substitute in the representation (19.22) of H1n(r)
The path W1, assumed for the sake of convenience to be rectangular, then transforms in the x-plane as follows
Thus, we have the contour integral of Fig. 37b, which starts at negative infinity of the real x-axis and returns there after looping the point x = ir ; the sense of passage is controlled by a small parallel displacement of the real branch of W1. Then, (19.22) yields
The integral arises from the combination of the two branches in Fig. 37b, where the initially negative sign of the returning branch reverts after travel around the branch point x = ir. Integration by parts with respect to x yields
After substitution of x - ir into the first term and going to the limit in both terms, one finds
The substitution x = -t converts the last integral into
C and g are identical with the quantities defined in (19.41a) (recalculation on the basis of Lagrange's integral of the G-function). Joining (4) and (3), one finds from (1)
Due to the relations
(5) agrees with (19.48) for N.
4.4 a) 1. Neglect of 1/r in the differential equation (19.11) yields immediately H1n = Aeir (A is the integration constant; the other solution with e-ir corresponds to H2n).
b) One does not view A as a constant, but as a slowly varying function of r, so that one can neglect A", A'/r and A/r². This yields a differential equation for A(r) from which follows A=b/Ör. Obviously, the normalizing constant B cannot be found in this way.
4.5 a)The constants Cm and Amn in the task become, by (1.12) and (22.14), (22.31),
b) The scheme for given in the task follows by multiplication of by
and integration with respect to j0
whence by integration with respect to , due to (22.14), (22.31),
which agrees with the changed notation (n, m instead of n, m) with the expression under a) for Amn apart from the summation sequence (Fig. 38):
For a): The horizontal strips are summed in the vertical direction,
for b), the vertical strips are combined in the horizontal direction. In both cases, the complete area of summation is bounded by the lines the difference is only apparent.
4.6 Fig. 39 below shows the wedge of Fig. 17 with angle 60º once more in a symmetrical position with respect to the horizontal with its five reflected repetitions. The inversion centre C of the wedge is placed conveniently on the horizontal through O and guides also the inversion circle (broken line) through this point. Since infinity is mapped into C and the point of intersection S1 of the inversion circle becomes the line 1, and so does the point O, the position of the circle, into which the line 1, -1 are mapped, is determined by the three points O, C and S1. The two sections of this circle, which correspond to the half-lines 1 and - 1, are also denoted in the figure by 1 and -1. The same applies to the lines 2, -2, which are mapped into an equally large, but located downwards circle, passing through the points O, C and S2. The line 3, -3 becomes a circle with the radius OC, the upper and lower half circle of which correspond to the half-lines -3 and 3.
Verify that the wedge 12 is mapped into the outside of the arc bi-angle CS11O2S2C; this is emphasized, as in Fig. 17, by shading of the edge of the wedge. Moreover, find the images of the reflective repetitions of this wedge in Fig. 39; they are all reproduced by the inside of certain circular arc bi-angles (meniski), for example, the wedge 2 3 by CS22O3C, the wedge -2, -1 by the small line formed section C, -2, O, -1, C at the centre of Fig. 39.
Hitherto, we have described the figures as if they are flat by talking of lines, circles, bi-arc sections, etc. However, there is no reason why the figures should not be reinterpreted three dimensionally and one should talk instead of lines and circles of planes and spheres. The latter lie each with half spheres above and below the plane of the drawing, respectively. Then, there corresponds to the initial wedge the outside of the two intersecting spheres, which in the given manner belong to the two circles 1, -1 and 2, -2; similarly, there corresponds to the wedge repetitions the empty space between two of the spheres 1, 2, 3.
Just as Green's function of the wedge arose from the elementary reflection construction in Fig. 17, we obtain in the case of the potential equation (but only in this case) Green's function of the corresponding wedge or circular region by seeking the electric image point of the given pole at the boundary circles or spheres 1, 2, 3 and providing them with alternating signs. In view of the symmetry of our problem, which is especially obvious in Fig. 17, it is sufficient to use the finite number of only five electric images in order to obtain the condition u = 0 at the boundary of the region under consideration.
.4.7 a) With the inversion centre C, mapping the infinity of all reflected planes and with the in Fig. 40 dashed inversion sphere K, the planes 1 become the two spheres +1 and -1 touching each other, the diameter of which equals the radius a of the inversion sphere. The entire outside of the spheres 1 corresponds to the inside of the plate, the inside of the spheres to the outside of the plate. The planes are mapped into two spheres, which also touch at C and the diameter of which is only a/3. The images of the planes are two spheres, which again lie inside the spheres , have the diameter a/5 and touch each other at C. The space between two spheres of this series corresponds then to reflective repetitions of the initial plate.
Thus, the Green function of potential theory for the outside of two touching spheres (for example, the spheres 1 of Fig. 40) can be derived by inversion from the Green function of the plane-parallel plate. The infinitely many images of the arbitrarily selected image point, which occur in the process, lie in turn in den quoted empty spaces and accumulate at C.
b) If a and 2a are the radii of the two concentric spheres I and II, one can set, for example, the radius of the inversion sphere equal to a and and place its centre C on the sphere I. This sphere then is mapped into a plane EI, the sphere II into a sphere KII of radius 2a/3, the smallest distance of which from the plane becomes a/6. Thus, conversely,EI and KII are mapped into the two concentric spheres I and II.
In the case when the position of the plane E is arbitrary, one can proceed (according to Mr. Carathéodory's friendly information), in a quite elementary manner: Draw from the centre of the sphere K the perpendicular to E, from the foot of the perpendicular F tangents t to K, create with the length of t an auxiliary sphere H about F. If one makes one of the two points of intersection S from L and H into an inversion centre, then L remains a straight line, H becomes a plane perpendicular to L, E and K become two spheres, which are perpendicular to H and L and therefore have their centre at the point of intersection of H and L, i.e., at the inversion centre. The inversion radius remain arbitrary; it determines only the absolute magnitude of the two concentric spheres.
Instead of E and K, one can also consider two arbitrary non- intersecting spheres K and K2 (cf. the last statement of the task). In order to convert them into two concentric spheres, start from the bundle of spheres K1 + lK2 = 0. Such a bundle of spheres arises, for example, from the bi-polar co-ordinates in Fig. 26 of Volume II, by rotating the system of circles r= const about the axis j=0 in this figure. It contains two spheres of radius 0, in fact, the two poles of the bi-polar co-ordinate systems. If one makes one of them into the inversion centre, all spheres of the bundle, i.e., also K1 and K2, are mapped into concentric spheres.
4.8 If u1 and u2 are two linearly independent solutions of the general, linear, second order differential equation
(p and q being any functions of the independent variable r), then with X = u1u2' - u2u1':
a) For the Bessel differential equation (19.11), p = 1/r, i.e.,
If one takes u1 = H1n, u2 = H2n, C is determined most simply from the asymptotic values (19.55), (19.56). One thus finds
Since In = ½(H1n + H2n), (I) in the task, if one understands by the symbol H there H1, half of the preceding X, i.e.,
and its negative, if one understands by H the function H2.
Somewhat more complicated becomes the determination of C, if one starts from r = 0 instead of from .
b) In the differential equation (21.11a), p = 2/r, whence X = C/r ². If one takes u1=z 1n, u2=z 2n, one has, by (21.14), for r ® ¥
Thus, one finds for (II) , depending on whether z = z 1 orz 2: