Appendix I11
Solutions of the Exercises
Note: Solutions immediate by merely following the directions given in their statements are omitted from this list. 1.2.1 1.2.2
Use ( I . 2  5 ) 3 , Axiom B3, and the definition of meet to prove the first implication. Use (1.25)4 and the definition of join to prove the second. Adopting the first of the two possible hypotheses, we set
By the definition of meet, 9 3
Thus 9
3
49,
9 3 4
U,
9 3
49.
is a part of .C%, U, and 9. Now suppose that
94.C%, 3 4 U ,
949.
Then, again by the definition of meet, 3491,
9  4 9 2 ,
9493.
Thus any part 9 of 9 , U, and 9 is a part of 343
9 3 .
By the definition
344
APPENDIX
111. SOLUTIONS OF EXERCISES
of the meet of three bodies, then,
From this last, we see that
Suppose now that
Since 9 A
V A 9 exists,
The definition of meet shows that
1.2.3 1.2.4 1.3.1
1.4.1 1.5.1 1.5.3
1.8.1 1.8.2 1.8.3
1.9.1
A similar proof holds if we assume that 9 A (h4 A 9) exists. Use (1.28)l and (1.212)~. Use (1.28)1, (I.216)1, and (1.212)~. Let 91 and 9 2 be bodies of 0 0 . Then 9 j= intcloaj, i = 1,2. We know also that int clo(91 U 9 2 ) belongs to 0 0 .Consider any 9 E Q suchthat91 c 9 a n d 9 z c 9 ; t h e n 9 = i n t c l o 9 a n d 9 1 U 9 z c 9 , and so intclo(91 U92) c 9. Since 9 is arbitrary, the conclusion follows. Use (1.238) to write 9 V h4 as the join of separate bodies. Then use Axiom M3, (1.43), and Axiom M1 (if necessary). Substitute (1.516) into (1.526), then use (1.520). Expand f ( 9 V V, 9 v $7) with the aid of Axioms FE2 and FE3, then similarly expand the results. (1.86)l follows simply by differentiating (1.84). To derive (I.86)z , use the definition of M,, so as to get M,, ,and adjust the terms. Write F ( 9 , ge) F ( 9 , 9)explicitly, and use (1.526). In (1.86) take x, for x l , and use (1.829). The second term in (1.830) is the rate of change of rotational momentum with respect to the fixed place x,, of a masspoint located at the center of mass of ~ ( t) 9, and endowed with the linear momentum of 9. Let 4 and V’ be two ndimensional innerproduct spaces over 9. Let h: @ t Y be an isometry. Let Q: 4 t Y be defined by Q(u) =
+
CHAPTER
I
345
h(u)  h(0) for each u in @. Then Q(0) = 0, and we conclude that (i) IQ(u)ly = 1 ~ 1 % for each u in 42. Let u and v be vectors in @. By multiplying out both sides of IQ(u) Q(v)& = Ju and using (i) to simplify the resulting expression, we obtain (ii) (Q(u), Q ( v ) ) = ~ (uyv)%for every u and v in @. By (i) and (ii), it is easy to show that (iii) IQ(u +v)  (Q(u)+Q(v)) l’y = 0 and (iv) IQ(W XQ(u)l’y =.O, for each scalar A in 9 and all vectors u and v in @. It follows from (111) and (iv) that Q is linear; (ii) states that Q is orthogonal. Moreover, by definition Q satisfies the equation Qu = h(u)  h(0) for each u in @. The uniqueness of such a Q is easy to prove. 1.9.2 X* =
G(t)+ Q ( N x  XO),
+ Q(t)(x  % ( t ) + Z o ( t )  XO), = g(t)+ Q(t)(%(t)  XO) + Q(t)(x  %o(t))y = & ( t ) + Q(t)(x  Mt)), =$(t)
1.9.3
1.9.4
+
say. Since a transformation of this kind, along with t* = t II, preserves the metrics in 8 and 9,it defines a change of frame. To prove the group property, use the fact that the orthogonal tensors form a group. Differentiate Z = YYT and use (1.916) and (1.915). A solution of (1.917) is furnished by Z ( t ) = 1; by uniqueness it is the only solution satisfying Z ( t o ) = 1. A* is formed from the tensor Q* that enters the inverse of (1.94). Since Q* = QT, (1.918) follows. To derive (1.919), write out the equations for the three changes of frame. A3 = Q3Qi ; simplification by (1.919)l yields (I.919)2. When and i2coincide at some instant, Q2 = QS = 1 at that instant. If e is a unit vector on the axis of rotation, Qe = fe. Hence Qe Qe = f e ; that is,
i3
1.9.5
+
Thus if e = 0, it follows that Ae = 0. The relation between o and b can be proved by use of an explicit representation of the orthonormal components of rotations over a threedimensional space: cos 8
[Q]= sin 8 0
sin 8 0 cos 0 0
0 , 1
346
APPENDIX
111. SOLUTIONS OF EXERCISES
where 8 is the angle of rotation and the basis vector e3 lies in the axis of rotation. 1.9.6
tr[kRT(R  R')] = tr[R + (kRT)TRT], = 2trR.
In a space of 3 dimensions tr R = 2(sin
e)e.
Also, by appeal to the explicit representation given in Section App. IIA.14,
')] = 2(sin 8)AKLcLKpeP, tr[kRT(R  R = 4(sin
8)o me.
If 8 = 0, no relation between w e and 4 can hold, because e can be any unit vector. If 8 = ?r, the foregoing argument delivers nothing, but w e is a continuous function of e, and so we may infer (1.920) by a passage to the limit, using the conclusion established for values of 8 near r. The remark just preceding the exercise solves half of it, for in any frame that gives rise to a spin W having the same axis as W at each t the points on that axis will maintain their mutual distances. Conversely, suppose that

1.10.1
Then
Thus W  W is a skew tensor such that
(W W)(x  y) = 0
if
x E x(a,t )
and
Y E x(@, 0 .
For any given x and y there are infinitely many skew tensors S such that S(x  y) = 0. Such S belong in common to all x and y that lie upon the same straight line. If x(@, t) contains three not collinear
CHAPTER
1.10.2 1.10.4
1.10.5 1.10.7 1.10.8
1.11.1
1.11.2 1.11.3
341
1
places, W  W is a skew tensor whose nullspace contains two distinct straight lines; since the nullspace of a skew tensor other than 0 is ldimensional, W  W = 0. (1.101)2 shows that & = c. Therefore p i = Wp,, i = 1, 2. Compute d a(p,.p2), and use W = WT. Use (1.82)l , (1.101)2, (1.105), and the fact that c is a function of t only. To obtain (I. 108)2, note that Q(a A b)QT = Qa A Qb, and use (1.103) and (1.107). To obtain (1.109), use Guo's formula in Section App.IIA. 12. Use W = QTAQ and (1.915). If e = 0, then We = 0 if and only if (W W2)e = 0. Use (1.83), (1.101), and (1.104) to get (1.1015). If ~ ( tis )chosen as directed, then &(t) = c, and p = 0. (1.1016) follows with the aid of a little manipulation in the third term on the righthand side of (1.1015). Let TT: Y'T + Y'T be the linear transformation in question. Then T = (D$,)TT(D$~)*, and T* = (D$;)TT(D$;)'. Thence
+
Since under galilean transformations X* = QX and M* = M , (1.85)l shows that m* = Qm. The first statement follows at once from the chain rule. The fourth statement is proved as follows from (I. 112): det T* = det(QTQT)= (det Q)(det T)(det QT), = det T.
This being so, det(T rl) is frameindifferent, no matter what be the number r. Therefore, T' and T have the same latent roots. Consequently they have the same trace and the same proper numbers. If e is a proper vector corresponding to the proper number r, then Te = re, so that
(Q'T'Q)e
= re,
348
APPENDIX
111. SOLUTIONS OF
EXERCISES
and hence T*(Qe) = r(Qe). 1.11.4
1.12.1 1.12.2
Thus Qe is the corresponding proper vector of T*. A given smooth surface can be embedded in a smooth family of surfaces f = const., where f is a frameindifferent scalar. Then, by conclusions of Exercise I. 11.3, both Vf and )OfI are frame).A better proof can be constructed by indifferent, and n = Of/[Of writing an equation for a single surface as
where a and b are parameters. For each a and b, the lefthand side is a frameindifferent vector, and so the vectorvalued function f is frameindifferent. Accordingly, a, g and a b g are frameindifferent. They span the tangent plane at ( a , b). The line normal to the tangent plane contains exactly two unit vectors. Their construction as above shows that both are frameindifferent. Use Axiom A2 and conclusions from Exercise I. 11.3. = Qx, and so In (1.913) suppose that x' = 0, Q = 0. Then I* W *= W implies that
1
.d f i e =
.d h e ,
1
whence Q'dfie = dfge. 1.13.1
1.13.2
We note that (1.522) holds as long as the system of forces is balanced. Since the axioms of inertia as applied to analytical dynamics respect the requirement that the forces be balanced, they do not alter the requirement (1.522). Note that (f hg)' = 0. The spectral decomposition of EG is
EL.
= Ele 8
e
+ E2f 8f+ E 3 g 8 g .
It is easy to show that E;,A~  A~E;, = 0,
349
CHAPTER
and
1.14.1
Thus (I. 1322) reduces to (I. 1324). Invariance under translation is equivalent to a representation
Wqk being a frame indifferent function of vectors. A theorem of CAUCHY [NFTM, p. 291 tells us that
for all v and all changes of frame if and only if Wqk(v)= Vqk(v.v). Calculation of fqk and Vqk yields
1.15.1 11.1.1
The conclusion about f i follows by a similar argument. The heating Q obeys the identity (1.52). If d U is of class C ' , by use of the divergence theorem it follows from the definition of perimeter that per( U ) 5 A ( d U ) .
If d u is of class c*,one can easily arrive at the inequality per(U) 2 A ( d U )
11.5.1
by using the divergence theorem to compute J div g dV for g E Cb(S, Y') such that lgl 5 1 and gl3g = ng . Comparison of the two inequalities delivers the desired conclusion if d U is of class C 2 . If d g is merely of class C ' , the same conclusion follows from a rather more delicate argument which is sketched on p. 157 of the book by & HUDJAEV, cited in Footnote 1 on p. 88. VOL'PERT From the definition of a determinant, or by use of the characteristic polynomial, it follows that for a tensor A det( 1
+ A) = 1 + tr A + o(A)
as A
+ 0.
350
APPENDIX
111. SOLUTIONS
OF EXERCISES
Thus if L is a fixed invertible tensor
+ U) = det L det( 1 + LIU) = det L[l + tr(UL')] + o(U)
det(L
as U + 0.
If an invertible tensor F is a differentiable function of a parameter, we may put F for L and EFfor U and so obtain
det(F
Divide by
E
+ EF) detF = det F(l + E tr FFI)  detF + O ( E ) = E(detF)trFFl + o(E). and then let E + 0 to conclude that (det F)' = (det F) tr(FFl).
11.5.2
Interpret F as being the transplacement gradient and use the chain rule = Vxy or simply use (11.115) and (11.117). to show that (An easier problem of this kind is given below as Exercise 11.6.3.) Consider the linear partial differential equation
m'
where P O ,P1 ,. . . , P , , R are given functions of X O , X I , . . . , x n . The chamcteristics of (L) are the integral curves of the system
A chamcteristic integral is a function f i ( X 0 , XI ,. . . , x n ,Z ) such that f i = const. on every characteristic curve. The formal statement of LAGRANGE'S theorem is that if f 1 ,. . . fn are any n functionally independent, characteristic integrals of (C), then the general solution of (L) is
To treat (11.5.7) in n dimensions, let xo = 1 , write x for (xl, . . , x n ) , Z := log p . Then Po = 1, Pi = , t i , and, by (11.56),
XZ,.
CHAPTER
II
35 1
R =  J / J . Hence n members of (C) can be written in the form dx = x d t , and so n families of characteristic curves are provided by the pathlines of the substantial points. Thus x;’ denotes n characteristic integrals. An ( n lyt integral can be obtained by integrating dt = d Z / R = d log p / d log J, the resulting integral being p J . Thus the general solution of (11.57) is
+
11.5.3
and this is (11.54). Note: The method of characteristics for linear partial differential equations of first order was invented by LAGRANGE on the basis of this example and the one in Exercise 11.6.3, both of these having arisen in hydrodynamics. The trivial generalization of the particular case (11.5it is the only 6) from 3 dimensions to n was obtained by LIOUVILLE; one of the several statements physicists call “Liouville’s theorem in statistical mechanics” that has any connection with LIOUVILLE. A rigorous treatment of LAGRANGE’S theory in the large is intricate. Most modem books on partialdifferential equations omit it. A clear and precise treatment of the local theory may be found in Chapter 2 of P. R. GARABEDIAN’S Partial Differential Equations, New York, John Wiley & Sons, 1964, reprinted New York, Chelsea Publications, 1986. A simple treatment of characteristics is given by C.C. WANG in the appendix to his Mathematical Principles of Mechanics and Electromagnetism, Part A , N.Y. & London, Plenum, 1979. By the theorem of integral calculus used to derive (11.26),the volume of x1(9, t ) is given by
JdV.
11.5.4
The condition of isochoric motion is therefore locally equivalent to J = 1. To complete the exercise, use (11.56) and (11.57). For a plane motion (11.58)becomes
axx +ayy = 0, in which x , y are Cartesian coordinates and x ,y are the corresponding components of the velocity field. This is a necessary and sufficient condition that in each simply connected region there be a function q such that
x = &q,
y =&q,
352
11.5.5
APPENDIX

dA(X) = c k p q dXp dXQ = ckpqFP,Fq8dX" d X a .
(An interpretation for this transformation law is given in Section 11.13.) The conclusion follows by comparing both sides of (11.512). As (11.611) suggests, take V / p for Y! in (11.69), then use (11.57), (11.63), and the divergence theorem. The value of the lefthand side of (11.610) is the time derivative of J 4 dV for a given part 9 of 97; the operation denoted by a prime is the time derivative of J 4 dV obtained if, neglecting the motion x , we confuse 9 with its present shape ~ ( t). 9, The difference between these is explained and evaluated by the third term, which gives the rate of increase of J 4 dV for 9 effected by the motion of substantial points out of or into the present shape of 9.To complete the exercise, refer to the definitions (1.81) and (1.82), and take for 4 first px and then (X
11.6.2
SOLUTIONS OF EXERCISES
and this is (11.511). For the extension to multiply connected domains see CFT, Section 161. Clearly x. V q = 0, and so the stream lines are normal to the normals of the curves q( , t) = const. nk
11.6.1
UI.
 Xg) A p X .
For a moving surface Y , choose a particular parametric representation:
and think of A as being attached permanently to a point on Y as it progresses. Then the velocity u at A is given by
Of course the field u so defined on Y depends upon the particular parametrization used to describe Y . Now suppose the parameter A to have been eliminated, so that an equation for Y is f ( x , t ) = 0. All the infinitely many different parametrizations of Y will lead to one and the same set of points satisfying a relation of this kind, and this relation characterizes Y over an interval of time:
CHAPTER
11
353
for each fixed A in any parametrization. If h(x, t) = 0 is another equation for Y , then h is an invertible function off. Differentiation of (1) yields
f ' + (grad f ).u
= 0.
(2)
Now the unit normal to Y in the direction of increasing f is given by
Therefore (2) asserts that
u.n=
11.6.3 11.6.4
lgrad f I
 ~h'  lgrad h
1'
h being any differentiable function off. Because the righthand side is independent of the parametrization, so is the lefthand side. Thus what we have defined as the speed of displacement S, is in fact the common normal speed of advance of all possible assignments of velocity to points on Y. For the method of characteristics, see Exercise 11.5.2. In the present instance R = 0, and f is the unknown function. Note that g ( x , t) = g(x,(X, t), t) := G(X, t) = 0, and
n, =
11.9.1
f'
~
GradG(X, t )  FTgrad g lGradG(X, t)l lGradG(X, t)l'
A little simplification gives (11.621). If Y Kis not a substantial surface, then at different times different substantial points will lie upon it. Of course (11.622) is merely an application of (11.616). To get (11.623), use (II.63)l and (11.616). The common proof starts from the assumption FFT = 1 and by differentiating it and using the fact that Fk,b = F $ , a concludes that F = const. GURTIN & WILLIAMS have found an elegant proof that does not require F to be differentiable. Let f be a differentiable function of place z in some open, connected set Y o n which (Vf)( = 1. Then det V f = f 1. If ~0 E F, there is an open ball Y such that xo E Y c Y a n d that f is invertible in Y. If x E Y and y E 9, let V be the line segment from y to x . Then
f ( x )  f(y) =
s,
Vf(z)dz.
354
APPENDIX
m.SOLUTIONS OF EXERCISES
Because lQul = lul for any orthogonal tensor Q and any vector u,
Just the same argument applies to fI:
Comparison of these two inequalities yields
Therefore f preserves distances in Y.Since F i s connected, the assertion follows. If U = U,then grad(& o x;') = FF  mT, which must be constant in virtue of the preceding. C(7) = FT(7)F(7).Use (11.87), and simplify. The principal stretches u are the roots of det(B  u21) = 0. B = RCR', and
'
11.9.2 11.9.3
cos 8
sin 8 0
[R] = sin 8 cos 8 0 0 11.9.4
11.9.5
1
since the principal axes are rotated about the ~3axis.
Calculate the physical components of B by employing Bkm =
11.9.6 11.11.3
0
m B k'd
(no summation),
and compare with (11.913)'. Use (11.65) and (II.95)4 to get (11.919). EULERproved the statement by first differentiating the component & WILLIAMS equations xk, + x m , k = 0.The elegant proof of GURTIN does not require that G be differentiable. Let the notations be as in
355 Exercise 11.9.1. Then
[f(x)  f(y)]. [x  y] =
s,
[x  yl.Vf(z)dz.
Since V is a straight line, dz is parallel to x  y, and so the integrand is 0 if Vf is skew. Therefore
This condition is equivalent to (I. 10.1) in the present notation:
f =c
+ W(x  xg),
WT = W = const.
(B)
Indeed, that (B) +(A) is immediate. Conversely, by differentiating (A) with respect to x we obtain
Vf(x)T(x y)
+ f(x)  f(Y) = 0.
Differentiation with respect to y yields
 Vf(x)T  Vf(y) = 0. 11.11.5 11.11.6 11.11.7
Thus Vf is both constant and skew. Show first that divx = E trG2. The preceding exercise shows that in a rigid motion divx = lWI2. Use (11.111 1) and (11.1122) to obtain (11.1123), and then use Section App .IIA .15 . G = duF(~)F'(t)Ju,,. Use the polar decompositions of F(u) and F(t), and then carry out the indicated differentiation to obtain
+
G = RR'
+ RUU~R'.
This equation can be written as
D + W = kFt' + qR(UU'  U'U)RT
+ iR(UU' + U'U)RT. Use uniqueness of the additive decomposition of a tensor into symmetric and skew parts to get (11.1126)~.3 To get (II.1126)l , start
.
356
11.11.8 11.11.9
APPENDIX
m.SOLUTIONS OF EXERCISES
from (11.1126)3, and use C = U2 and the polar decomposition theorem. (In fact, (11.1126)' is easy to derive directly, but from it as a starting point there seems to be no obvious way to reach (11.1 126)2.) The last relation follqws from B = FFT and FIF=I= G. C = (FF 1)' = FF +F(F')', (FFI)' = 0 = F(FI) +F(F')'; hence (F'). = FIFF . To get (11.1132), use Leibniz's rule to differentiate Ft(7)TFt(7).To get (11.1133), first prove that (n)
C =FTA,F.
(A prescription for proving this formula is given in the text of Section 11.14, where it is listed as (11.1416).) Hence
= FTA,F
+ F T ~ +, ,F~A,,F. ~
Now use (11.115). 11.11.10 A formula for the derivative of the determinant of an invertible tensor
is given in Exercise 11.5.1. Differentiating it yields
+ (det L)' tr(LLI),  3LL'LL' + 2(LL')3]
(det L)" = (det L) tr[LL'  (LL')2] (det L)"' = (det L) tr[LL'
+ (detL)'(...) + (detL)"(...),
etc. If det L = 1 always, these relations reduce to tr[LL' tr[LLl  ~LLILL'

(LL')*I = 0,
+ ~ ( L L  ' ) ~=I 0,
etc. In an isochoric motion, we may substitute C t ( u )for L. Putting u for t, followed by use of the definition (11.1131), yields (11.1134)2,3 . The term involving the time derivative of highest order in the formula
('" )
for (detL)(") is (detL)tr LL' , and so the general assertion of the exercise follows. 11.11.11 Let x be a point of Y , and let k be a vector in the tangent plane of Y at x. Then there are points y ( h ) on Y such that y(h) =x
+ hk + o(h)
as h
+
0.
CHAPTER
11
357
Therefore,
Gk := lim
X(X
+ hk)  X(X) h
h 0
9
If x vanishes on Y , the difference quotient on
L :. righthand side vanishes, and so (IT.1138) follows. Now in (11.1113)replace e by n, a unit normal to the tangent plane at x; then denote by e a unit vector in the axis of W, so that W = iwn A f , We = 0. Because of (11.1138), Ge = Gf = 0, and so
De = 0 ,
Df = Wf = iwn,
the last equation being a consequence of (11.1114)~. Now since
it follows that
Dn = (n.Dn)n  i w f , E = n.Dn. Hence (11.1139)follows. Because De = 0, e is a principal axis of stretching, the corresponding principal stretching is 0, and det D = 0. The second principal invariant of D is  bw2. Thus the characteristic equation of D is
D(D2  E D

:w2)= 0,
the solutions of which are (11.1140). 11.11.12 From (11.1140)we see that
B2= 1/[1 + (E/w)23.
358
APPENDIX
m.SOLUTIONS OF EXERCISES
+
(FTF). = FTF a symmetric tensor. By choosing n successively as 1 and 2 in (11.1128)show that (FT(V%)F)’= FT(Vx)F+a symmetric tensor. Take the skew part of this relation to get (11.1142). 11.11.14 By (11.114)
11.11.13
w=o 11.11.15
Use (11.1128)to show that
11.11.16
The skew part is (11.1148). By (11.1148)
*
w,=o.
(;(Wl2)’ = W.(W,  D W WD). If dim Y = 3, then for any skew tensor W and any symmetric tensor D
W.(DW
11.11.18
11.12.1
+ WD) = IWI2(trD  nDn),
in which n is either unit vector in the nullspace of W. Use of (11.1115) yields (11.1149),from which the conclusion of the exercise is obvious. In the proof of the theorem of Kelvin and Helmholtz replace the assumption of steady density by the general equation (11.66)2 and so obtain
generalizing (11.1 158). Under the conditions stated in the exercise the surface integral vanishes. In unbounded domains the condition (11.1159)suffices to make the surface integral vanish. (The isochoric instance is more fruitful because the student has at his disposition the developed discipline called “potential theory”, while conditions at 00 for a mass density that depends upon x and t are difficult to ascertain in practice.) The ellipsoid in ~(97) is swept out by the termini of vectors m, such that
.
const. = Im l2 = IFm,. Fm, I = m, Cm,
.
CHAPTER
11
359
Let el , e2, e3 be an orthonormal set of unit proper vectors of C , so that Ce, = u,?e,,where u, is the principal stretch corresponding to e, . Let the coordinates of m, with respect to this basis be mk . Then the above equation for the ellipsoid assumes in Cartesian coordinates the form 3
C ( m k ) 2 u z= const. k=l
Therefore, the principal axes of the ellipsoid are the principal axes of strain at X, and the lengths of the semiaxes are inversely proportional to the corresponding squared principal stretches. The extremal properties of the principal stretches correspond inversely to the extremal properties of the lengths of vectors to points on the ellipsoid. That the principal axes are not sheared, is the same as the statement cos O(ei , e j ) = &, , which is an immediate consequence of (11.126). Since (11.121) can be written in the form
11.12.3
11.13.1 11.13.2
the last statement follows immediately by aid of (11.94). Differentiate (11.126) after writing it as
J x ( u , l )f  d x =
JK(u)f(X, t).F(X, t ) d X . Now on the righthand side differentiation can be performed under the integral sign. The volume V of a tetrahedron whose vertices are ~ ( tand ) the termini of p 1 , p 2 , and p 3 , is given in terms of the components p,k as follows:
Hence
11.13.3
Put dx = t d s , f = f t in (11.131).
360 11.13.4
APPENDIX
IU.
SOLUTIONS OF EXERCISES
By (11.137) and (11.118)
ec = e  (D + W)e. Since erne = 0 and W is skew,
e.ec = e.De. Substitutingthis formula into (11.1314) and then putting the outcome into (A) yields (11.1315). If De = de, (11.1315) reduces to
e =We. C.' the discussion of rigid motion in Section I. 10. For any vector m we obtain from (11.1315)
m.e = m.De + m.We  (e.De)(m.e).
11.13.5
Hence (11.1316) follows. All these conclusions apply to the position vectors p in a homogeneous motion because pc = 0. Since mn = cos 8(,,,"), (II.1316)l reduces to (11.1210) if mn = 0 at the instant in question. Likewise (11.1316)~reduces to (11.1215). Let the vector field f be tangent to a vector line of S at the time t o . Then f is tangent to a vector line of S for all t 2 to if and only if (Sf)' = 0.
Hence by use of (11.134) and (11.137) SCf+ sfc = 0.
11.13.7
By (11.1312), there is a scalar field 01 such that fc = af. Thus, f too, has to be tangent to a vector line of Sc: SCf = 0. Now recall that two nonnull skew tensors have one and the same vector lines if and only if they commute. The argument is phrased in terms of vortex tubes. These are surfaces swept out by the vortex lines through the points of some circuit nowhere tangent to the axes of spin. The flux of spin has the same value, at a given instant, for all likeoriented surfaces bounded by
36 1
CHAPTER
circuits embracing the tube just once (HELMHOLTZ’S First Vorticargument, a classic example of conceptual ity Theorem). KELVIN’S mathematics, may be found in LAMB’S treatise, in Section 128 of CFT, and elsewhere. P O I N C A RTheorem ~S makes the hypothesis equivalent to WW, = 11.13.8 W,W. Exercise 11.1 1.12 makes the differential relation equivalent to WW, = 0. If neither W nor W, vanishes, the two requirements are incompatible, for one requires the axes of the two tensors to coincide and the other requires that they be perpendicular to each other. If W = 0, then W, = 0, as is shown in Exercise 11.11.12. Thus W, = 0 is the only possibility, and clearly it is sufficient that the two conditions be compatible. 11.13.9 Take x for c in MASOTTI’S formula (Section App.IIC.6) to obtain for w  R x a vector which for a screw motion must vanish because w and x are collinear. 11.13.10 Put x for c in the second formula for R in Section App.IIC.6, then use (11.1323). 11.13.11 Take the curl of (11.1323); then the inner product of the result and 6,and use (11.1323). 11.13.12 Inspect (11.119). 11.13.13 As was remarked just after (11.66), in a motion with steady density div (px) = 0. Using (11.1323) delivers div [(p/R)w]= 0, and so w . grad (p/R) = 0. Thus x is the tangent to one of the surfaces p/R = const. 11.13.14 If a screw motion preserves circulation, then taking the screw part of the gradient of (11.119) yields w’ = 0. The conclusion follows from (11.1323) and (11.1324). 11.13.15 Begin as in Exercise 11.13.11. For the first statement the condition grad R x x = 0 is necessary and sufficient. The others follow by taking curls of the preceding. Referring to (1.914), for A(x*  $) write w x p* and note that 11.14.1
= w.l;p* x
ds*.
For a plane circuit V the vector delivered by the latter integral is normal to the plane, and its magnitude is the area of the region bounded by V+. .
362 11.14.2
APPENDIX
m.SOLUTIONS OF EXERCISES
By (11.87) and (11.146),
Hence
111.1.1
111.1.2
The conclusion (11.1420) follows by the uniqueness of a polar decomposition.
by (1.828). By (III.150)2 and (111.146),
J,t y d A 111.2.1
=
L

The Lebesgue differentiation theorem gives t y = t+ a.e. Expand p ( x  v) @ (x  v)n and p A [ p ( x  v) @ (x  v)n]; integrate over the shape of L8;use the divergence theorem; note that div(pP Q x) = p x
111.3.1
tydA.
+ p div(px);
use (11.66) for a motion with steady density, and note that x.n = 0 on the obstacle. Choose Ar such that A ( A 4 (= A ( A d ' ) A ( A d * ) .Then (111.311) and (III.39)2 yield (111.312).Hence
+
+
A(d A.9) = A ( A d ) A ( A d * )+A(&''), V(A.9) = o ( A r 3 )
as A r
+
0.
CHAPTER
111.3.2
III
363
The student would be well advised to draw a figure. Let be the tangent plane to Y and F a t x. With respect to Cartesian coordinates ( x , y, z) with xaxis and yaxis in y , let z = f ( x , y) and z = g(x, y ) be the representations of Y and F n e a r x. Choose Ar such that when x 2 y 2 5 A r 2 , Y and Y l i e entirely between two paraboloids z = f K ( x 2 + y 2 ) , where
+
111.4.1
Follow the same procedure as before. Let C be a cube, and let two of its faces be normal to k. Then
l,t
dA = 2( V (C))2’3k,
and
but lim
V(C)0
111.4.2 111.6.1
K:,I
V(C)
= +m,
and so (111.158) is violated, while (111.159) is not. Immediate from (111.41) and the definition of the transpose. Prove the identity div(v 8 S) = v 61div S + ( Vv)ST,
111.6.2
take the skew part, set v = xxo ,and apply the divergence theorem. Hold x fixed, and drop it from the notation; do not assume that TT= T. Trivially (111.610) @ (111.611), and (111.61 1)
+
(111.68) 8t (111.69).
Write (111.69) in the form n.Tn = p for all unit vectors n; let n = cos 8 nl +sin 8 n2 , nl and n2 being unit vectors, and show that n1.Tnz = n2.Tnl . Hence conclude that (111.69)
+
T = pl + S ,
364
APPENDIX
111. SOLUTIONS OF EXERCISES
S being a skew tensor. If T is symmetric, (111.611) follows. Otherwise it does not. (111.68) requires that (t(n)I2= n.TTTn= p2. The conclusion just reached in regard to (111.69) shows that TTT= p21 a skew tensor, but this latter is 0 because TTT is symmetric. If p = 0, (111.611) holds trivially; if p 0, we have shown that p  ' T is an orthogonal tensor, say  Q. Then
+
+
n.t(n) = pn.Qn. If R is the rotation such that Q = f R, show that
nRn = 1  2 n t sin2 $9,
n l being the magnitude of the component of n normal to the axis of R, and 0 being the angle of R. In order that n.Rn > 0 Vn, it is necessary and sufficient that 1  2 sin2 $I > 0. Thus (111.68) and
111.6.3
are equivalent. If T is symmetric, R = RT, and hence 6 = 0 or T . Since the latter alternative is excluded by the conclusion just drawn, R = 1. Thus (111.611) follows from (111.68) if T is symmetric. Otherwise it does not. The statement is really an instance of NOLL'Stheorem in Section I. 12 but is more than a century older. For an independent proof, hold t fixed and consider the rigid transplacement defined as follows by a constant vector vo ,the position vector p, and a constant skew tensor WO: v = vo Wop. Since a. Sb = S . (a Q b) for any skew tensor S,
+
In order that P = 0 for all choices of vo and W O , it is necessary
CHAPTER
111.6.4 111.6.5 111.6.6
111.7.1
111
365
and sufficient that the first bracket vanish and the second bracket be symmetric. Use (I. 141),(111.612), the divergence theorem, (111.6l), and (11.118). In a rigid motion D = 0. In an isochoric motion t r D = 0. Substitute from (111.17) and (111.615) and (111.612) and use (11.69) to obtain (111.617).Then (111.618)and (111.619) are easy to obtain. The existence of h , the influx of heating, first demonstrated by STOKES, follows by use of arguments parallel to those that deliver CAUCHY’S Fundamental Theorem (Section 111.4).The differential equation (111.620) follows by appropriate substitutions in (111.5l),which delivers (111.54).
( l p @ T e d A )e =l(eTe)pdA,
Using subscript 1 and 2 to refer to quantities associated to the two plane, parallel faces, if nl = e we must take n2 as  e, and so (111.714) yields
111.7.2
Equilibrium of forces requires that F1 = F2 ; equilibrium of moments, that p o ( d l )  p 0 ( d 2 ) be parallel to e. Taking Y as p @ p in (111.73)yields at once
Forming the combination indicated by (111.710)1and then using Cauchy’s Second Law yields
111.8.1
Let the constant g denote the gravitational acceleration, let p denote the density of the heavy liquid, and let z denote the distance downward from the surface of the liquid. Then p = pgz on the surface of
366
APPENDIX
m.SOLUTIONS OF EXERCISES
the submerged part of the body, say Y . The resultant surface force and surface torque upon x(g, t ) are, respectively,
F$ =  p g L ( x
 xg) AzndA.
The formulae used to denote the two integrands serve to extend them smoothly to the interior of the part of x ( S , t ) below the plane z := 0. (This fact expresses STEVIN’S Principle of Solidification: The load exerted by one part of a heavy fluid body upon another is unchanged if either is replaced by a rigid solid.) Thus we can apply Green’s transformation to express the two surface integrals as volume integrals over the submerged part Y . Since gradz = k , a unit vector pointing downwards,
LzndA = ( L d V ) k = V(flk. Likewise
L(xw)AzndA =
s,
(xxO)dVhk,
p, being the position vector of the center of buoyancy.’
111.9.1
To consider a heavy body, invoke the result of Exercise 111.1.1 to conclude that the load on that body is equipollent to two parallel forces: the weight of the body, acting downward at its center of mass, and the weight of the displaced fluid, acting upward at the center of buoyancy. Consideration of a simple vector diagram suffices to conclude the exercise. Since d x ( B , t) has a differentiable unit normal field n, that field can be extended smoothly into a small region containing dx(9?, t ) in its interior. A standard theorem of differential geometry asserts then that the mean curvature
k = div n.
(4
‘To define the centroid of a region, in (1.828) replace M by Vand D by the region considered.
CHAPTER
367
Iv
If n is any differentiable field of unit vectors, and if c is a constant vector field, then in a 3dimensional space n.curl(n x c) = c.(divn)n, n.curl{n x [c x (x
IV.4.1
%)I} = c [(x  xg) x (div n)n].
By Kelvin's transformation the integral of n. div(p A q) over a surface Y is equal to the value of a line integral around d Y . If Y is a surface without boundary, that value is 0. Thus the integrals of the righthand sides of (B) over dx(33, t) both equal 0. Use of (A) completes the proof. By hypothesis, QK = KQ VQ. For Q take the reflection & in the plane normal to e. Then &v = v if and only if v is proportional to e. But &Ke = K&e = Ke, so Ke is proportional to e, no matter what be e. Suppose now that Ke = a e , Kf = Pf, K(e+f) = y(e+f). Then a e (3f = y(e f). Choosing e and f as linearly independent shows that a = (3 = y. Note. We may ask if the condition RK = KR for all rotations R implies (IV.423). The answer is no if the dimension of the vector space is 2, for then all rotations commute. If the dimension of the vector space is odd, the answer is obviously yes, since the tensors k R exhaust the orthogonal tensors. The answer is yes also for vector spaces of even dimension greater than 2 but is not so obvious. It is easy to prove that a symmetric tensor which commutes with every rotation is proportional to 1. Follow the procedure given in the proof of (IV.42). If g in (IV.41) is an affine function of F,
+
IV.4.2 IV.4.3

(B)
+
T = A +B[F], A being a constant tensor and B being a tensorvalued linear function of tensors. In order for this constitutive equation to satisfy the Principle of Material FrameIndifference, it is necessary and sufficient that Q(A
+ B[F])QT = A + B[QF]
( *) L
for all invertible F and all orthogonal Q. Put F = C1, C obtain QAQ'  A
+ Cf(B, Q) = 0
+ 0, to
368
APPENDIX
111. SOLUTIONS OF EXERCISES
Since the real number C is arbitrary, both terms in this affine function of C vanish. Because A commutes with all orthogonal tensors, A = A l . The functional equation (*) reduces to
+
IV.6.1.
IV.7.1
Taking Q = 1 shows that B[F] = 0. (Note. T = a1 /3V does satisfy the principle, but the righthand side is not an affine function of F.) (IV.62) and (IV.63) are straightforward. A frameindifferent, elastic constraint equivalent to (IV.61) is of the form p ( C ) = 0, where p vanishes if and only if y vanishes. This statement is logically equivalent to the last sentence of the exercise. (Note that p and y are not claimed to be functionally dependent, though of course they may be.) Only (IV.714) requires care, because CAUCHY’S Theorem in Section IV.4 refers to a function whose domain is the space of all symmetric tensors. If the domain of g is the subspace of traceless, symmetric tensors, we define as follows a function f on all symmetric tensors: f(D) := g(D  i(tr D)1). If g is affine and isotropic, so is f. Thus CAUCHY’S Theorem applies to
f. Specializing the conclusion to traceless tensors D yields (IV.714). IV.7.2
Since t = pn, the last statement follows at once from the formula proved in Exercise 111.6.5. More generally, (111.618) becomes
If p = const. on d x ( 9 , t), the righthand side reduces to p
J
divxdV,
X(S,O
IV.9.1 IV.9.2
which vanishes since the flow is isochoric. A glance at (11.115) and (IV.97)1 gives the assertion.
+ tF1)] = det Fo det( 1 + tFl), det(1 + tF1) = 1 + t(trF1) + ;t2[(trF1)’  trF:] + t3 detF1 , det F ( t ) = det[Fo( 1
CHAPTER
IV.9.3
Iv
369
because the second principal invariant of a tensor S equals [(tr S)2tr s2]. Generalizing the spectral decomposition (IV.913), we allow not only the latent roots of U but also the orthonormal basis vectors to depend differentiably upon t. Then
and hence
Because e k . e,
+ e k . e, = 0, k,q = 1 , 2, 3, a calculation yields
Consequently a necessary and sufficient condition that UU = UU is
If the orthonormal basis is constant in time, the statement of the exercise follows. Another sufficient condition, obviously, is that U shall have only one proper number. The student will distinguish and assemble other solutions of the above differential equation. For the first instance, write
and conclude the first statement following (IV.913). Taking the ei as the axes, let the block be the region included by the planes Xk = f a k . Show that it is deformed into a similar block. Since it is already proved that u k = C?k + b k t if C?k > 0 and b k > 0, the motion
370
APPENDIX
m.SOLUTIONS OF EXERCISES
will be isochoric if and only if 3
k=l
IV.12.1
This condition holds if and only if 010203 = 1, bl = bt = b3 = 0. For a pure stretch k = 0, R = 1. Use (11.1126)~and (11.1142). For an unconstrained simple body T = 8(F'), and for the corresponding incompressible simple body T = p ( a  h ) l 8 ( F f ) . F(t) is given by (IV.910). In the unconstrained body, every component of T is determined uniquely. In the incompressible body, the function h is arbitrary. For example, if a = 0, then by choice of h we may let any one of the tractions T,, , T,, , and T,, be any function we please, e.g. 0. If H I , Ht EY., then
IV.12.2
where the first step follows because H:! E %. ' and the second beThus H1Hz EP.. Similar arguments verify the cause HI other axioms of a group. Since 8. satisfies Axiom N3,
IV.10.2 IV.10.3
+
.
IV.15.5
This statement when combined with (IV.126) implies that Q' E %g * The change of reference placement is described as follows by use of the Cartesian coordinates X,9 , Z :
z =FP +62 + R ; IV.16.1
A , B,. . .R are constants, and A(CG  bF)= 1. By NOLL'S rule Q* = PQP', which can be written as Q*RoUo = RoQQ'uoQ. Use the uniqueness of the polar decomposition.
CHAPTER
IV.16.2
Iv
371
If
x ; = UI,
x; =A x 2 ,
x ; = px3,
x + p,
f x , then NOLL'Srule gives
then [PI = diag(X, A, p ) , and so if H E
If cos cp
sin cp
0
H = sin cp cos cp 0
0
0
1 [HI = 0 0
1
0
0
cos cp
sin cp
sin cp
cos cp
,
then
1 [H*] = 0 0
IV.16.3 JV.16.4 IV.17.2 IV.17.3
Since in generalyx
+
0
0 cos cp P 
sin cp
x
0
 sin cp P
.
cos cp
,yKwill not be an invariant subgroup of a .
Thusf:f(= RY~R')will not be equal to Pfi * Put UO= K1 in Theorem 2. Note that the righthand side of (IV.173)depends upon through px / J . Use (11.1153).
K
only
372
IV.18.1
APPENDIX
111. SOLUTIONS OF EXERCISES
For (IV.181) [D] = diag((a + b / r 2 ) , b / r 2 ,a). For (IV.182)
.
[D] =
h/r2 $l/r
.
0
IV.18.3
Clearly h / r 2 is a principal stretching, and it is constant if and only if h = 0. In that case the other principal stretchings are f I/r. In (IV.181) the constants c and g may be removed by superposing a rigid motion. Then
while for (6)
A =1,
B =2hs,
E=ls, IV.21.1
IV.21.2 IV.21.3
D=O,
C = 1,
K = h ,
L=rns+ilks2.
F=l,
Substitute (IV.213) into (11.88) to get (IV.21.13). The other relations follow easily from the definitions (11.112) and (11.1131). The condition tr G = 0 is necessary and sufficient that the motion be isochoric, so (IV.2116) follows. If A and B commute, then eAeB= eA+B. The most general form of A2 is
“421 =
u
a
b
a
u
c
b
c
w
Remembering that (IV.2122) does not hold, show that this A2 commutes with M  M as given by (IV.2127) if and only if x = 0. When A1 = a l , by the lemma A1 commutes with every skew tensor. Therefore (M  MT)A1 = A I ( M  MT). Using (IV.2115)3, conclude that MMT = MTM, and then arrive at (IV.2126).
Iv
CHAPTER
IV.21.4
313
If N3 = 0, the HamiltonCayley equation reduces to (trN)N2 = $[(trN)2  trN2]N. Thus N+O&N2=0
trN2=O&trN=0,
=+
while N2
+0
+
(tr N)N3 = 0 = i[(tr N)2  tr N2]N2
=+ [(trNI2 = trN2]
IV.21.6 IV.21.7
trN = 0
=+
=+
trN2 = 0.
Use (IV.2115)4,7 to get (IV.2130). Then (IV.2131) follows by use of (IV.2114)2 and (IV.2115)7. Use (11.83) and (11.84) to obtain the relative description of the motion whose spatial velocity field is (IV.2133): 42 = (7  O P X l
41 = X I ,
4 3 = (7  t ) ( X x l
+
vx2)
+x2,
+ ;(t  7 ) 2 p v x 1 + x 3 .
Hence Ft(7)assumes the form (IV.2113) with the special values Q = 1 and 0 0 0
[/A"] =
/A
0 0 =[GI,
A
V
O
components being taken with respect to the Cartesian coordinate basis. Since ( K N ~=) ~0 and ( K N o ) ~ = 0 + pv = 0, the first two assertions of the exercise follow. The relative description of the motion whose spatial velocity field is (IV.2134) is
374
APPENDIX
m.SOLUTIONS OF EXERCISES
Thus
[F] =
IV.21.8
calf
0
0
0
e‘zt
0
0
0
east
= [U];
(IV.915) is satisfied; and the last sentence of the exercise follows by the conclusion of Exercise IV. 10.2. Use (IV.2115)2, (11.1122), and the statement in Exercise IV.21.4. A nonvanishing isochoric dilatation (IV.21.34) superposed on a rigid motion of spin W, provides a monotonous motion of NOLL’S third class. For it @ = 1BrI/d2(a: + a $ (21a2), which for a fixed W, by choice of a1 and a2
[email protected] an arbitrary value in 30, GO[. (11.83), (11.8S), and (IV.2139) show that
+
IV.21.9
cf.(Iv.2129)1. IV.21.10 Using (11.83) and (11.84), integrate (IV.2144). So as to calculate physical components of Ft(7) with respect to {ei(()} and {ei(x)}, evaluate the quantities ei(t).Ft(E)ej(x).Since the bases {ei([)} and {ej(x)} are orthonormal, there is an orthogonal tensor function Q such that ej(E(7)) = Q(T)ej(x). Thus
Now show that Fo(7) = Q(7)(1
+ 7KNO), where
0 0 0 [No] =
CY
0 0
P
O
with respect to the basis {ei(x)}.
0
Writing (IV.2147) as ik = Rek , calculate the matrix of No with respect to the basis {ik}. IV.21.11 Inspect (IV.1532) and (IV.1561) to show that (IV.2150) and (IV.2151) are universal. Proceed as in Exercise IV.21.7 to show by use of (11.87) that the flows are monotonous and that N2 = 0. IV.21.12 Inspect the solution of Exercise IV.18.3.
CHAPTER
Iv
375
IV.21.13 Use (IV.2154)l to perceive (IV.2154)2,3. Follow the method of Exercise IV.21.7 and use its notations to obtain
which delivers (IV.2155). IV.21.14 From (IV.2154)2,3 conclude that
and then use (IV.2154)1 and (IV.2155). Since G3 = R2G, comparison of (IV.2154) with (IV.213) shows that No is not nilpotent unless R = 0. Use (IV.2154) to calculate the components of A l , A2, A3, and & and so establish (IV.2158) when n = 1. Next, suppose that if n 2 4
If so, (11.1133) shows that
IV.22.1
whence by induction (*) is proved to hold if n 2 4. The conclusions of Exercise IV.21.14 show that A3 = R2A1, and so the third argument of (IV.221) may be replaced by R. Calculation of Al and A2 shows that they together determine R unless f’ = g’ = 0. In that case the motion is rigid, and so the assertion of the exercise becomes trivial.