References [1] Adams R.A., Sobolev Spaces. Academic Press (New York) 1975. [2] Apostol T.M., Mathematical Analysis: A Modern Approach to Ad- vanced Calculus. Addison-Wesley (Reading, Mass.) 1957. [3] Babuska 1. and Aziz A.K., Survey Lectures on the Mathematical Foun- dations of the Finite Element Method, in The Mathematical Founda- tions of the Finite Element Method with Applications to Partial Dif- ferential Equations (ed. A.K. Aziz). Academic Press (New York) 1972. [4] Baiocchi C. and Capelo A., Variational and Quasi- Variational Inequal- ities. Wiley (New York) 1984. [5] Becker E.B., Carey G.F. and Oden J.T., Finite Elements, Volume 1: An Introduction. Prentice-Hall (Englewood Cliffs, N.J.) 1981. [6] Binmore K.G., Mathematical Analysis: A Straightforward Approach. Cambridge University Press (Cambridge) 1977. [7] Binmore K.G., The Foundations of Analysis: A Straightforward Intro- duction. Book 2: Topologicalldeas. Cambridge University Press (Cam- bridge) 1981. [8] Brenner S. and Scott L.R., The Mathematical Theory of Finite Ele- ment Methods. Springer-Verlag (New York) 1994. [9] Burnett D.S., Finite Element Analysis. Addison-Wesley (Reading, Mass.) 1987.
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References
[1] Adams R.A., Sobolev Spaces. Academic Press (New York) 1975.
[2] Apostol T.M., Mathematical Analysis: A Modern Approach to Advanced Calculus. Addison-Wesley (Reading, Mass.) 1957.
[3] Babuska 1. and Aziz A.K., Survey Lectures on the Mathematical Foundations of the Finite Element Method, in The Mathematical Foundations of the Finite Element Method with Applications to Partial Differential Equations (ed. A.K. Aziz). Academic Press (New York) 1972.
[4] Baiocchi C. and Capelo A., Variational and Quasi- Variational Inequalities. Wiley (New York) 1984.
[5] Becker E.B., Carey G.F. and Oden J.T., Finite Elements, Volume 1: An Introduction. Prentice-Hall (Englewood Cliffs, N.J.) 1981.
[6] Binmore K.G., Mathematical Analysis: A Straightforward Approach. Cambridge University Press (Cambridge) 1977.
[7] Binmore K.G., The Foundations of Analysis: A Straightforward Introduction. Book 2: Topologicalldeas. Cambridge University Press (Cambridge) 1981.
[8] Brenner S. and Scott L.R., The Mathematical Theory of Finite Element Methods. Springer-Verlag (New York) 1994.
[9] Burnett D.S., Finite Element Analysis. Addison-Wesley (Reading, Mass.) 1987.
436 References
[10] Carey G.F. and Oden J.T., Finite Elements, Vol. 2; A Seeond Course. Prentice-Hall (Englewood Cliffs, N.J.) 1983.
[11] Ciarlet P.G., The Finite Element Method for Elliptie Problems. NorthHolland (Amsterdam) 1978.
[12] Ciarlet P.G and Raviart P.-A., Interpolation theory over curved elements with applications to finite element methods. Computer Methods in Applied Meehanies and Engineering 1 (1972) 217-249.
[13] Dautray R. and Lions, J.-L., Mathematieal Analysis and Numerieal Methods for Seienee and Teehnology, Vol. 2; Functional and Variational Methods. Springer-Verlag (Berlin) 1988.
[14] Dhatt G. and Touzot G., The Finite Element Method Displayed. Wiley (New York) 1984.
[15] Duvaut G. and Lions J.L., Inequalities in Meehanies and Physies. Springer-Verlag (Berlin) 1976.
[16] Glowinski R., Numerieal Methods for Nonlinear Variational Problems. Springer-Verlag (Berlin) 1984.
[18] Halmos P., Finite Dimensional Veetor Spaees. Van Nostrand Reinhold (New York) 1958.
[19] Hewitt E. and Stromberg K.R., Real and Abstmct Analysis; A Modern Treatment of the Theory of Funetions of a Real Variable. SpringerVerlag (New York) 1965.
[20] Hoffman K. and Künze R., Linear Algebm. Addison-Wesley (Reading, Mass.) 1973.
[21] Horgan C.O., Kom's inequalities and their applications in continuum mechanics, SIAM Review 37 (1995) 491-511.
[22] Hughes T.J.R., The Finite Element Method; Linear Statie and Dynamie Analysis. Prentice-Hall (Englewood Cliffs, N.J.) 1987.
[23] Johnson C., Numerical Solutions of Partial Differential Equations by the Finite Element Method. Cambridge University "Press (Cambridge) 1987.
[24] Kardestuncer H. (ed.), Finite Element Handbook. McGraw-Hill (New York) 1987.
References 437
[25] Kolmogorov A.N. and Fomin S.V., Elements of the Theory of Funetions and Funetional Analysis. Volume 1: Metrie and Normed Spaees. Graylock Press (Rochester, N.Y.) 1957.
[26] Kolmogorov A.N. and Fomin S.V., Elements of the Theory of Funetions and Functional Analysis. Volume 2: Measure, Lebesgue Integrals and Hilbert Spaee. Academic Press (New York) 1961.
[27] Kreyszig E., Introduetory Functional Analysis with Applications. Wiley (New York) 1978.
[28] Lang S., Introduetion to Linear Algebra. 2nd edition. Springer-Verlag (New York) 1986.
[29] Lang S., Undergraduate Analysis. Springer-Verlag (Ncw York) 1983.
[30] Lions J.L. and Magenes E., Non-Homogeneous Boundary- Value Problems and Applieations, Volume 1. Springer-Verlag (New York) 1972.
[31] Lipschutz S., Set Theory and Related Topies. Schaum Outline Series. McGraw-Hill (New York) 1964.
[32] Loula A.F.D., Hughes T.J.R. and Franca L.P., Petrov-Galerkin formulations of the Timoshenko beam problem. Computer Methods in Applied Meehanies and Engineering 63 (1987) 115-132.
[33] Naylor A.W. and Seil G.R., Linear Operator Theory in Engineering and Seienee. Springer-Verlag (Berlin) 1982.
[34] Necas .I., Les Methodes Direetes en Theorie des Equations Elliptiques. Masson (Paris) 1967.
[36] Oden J.T., Applied Funetional Analysis: An Introduetory Treatment for Students of Meehanics and Engineering Seience. Prentice-Hall (Englewood Cliffs, N.J.) 1979.
[38] Oden J.T. and Reddy J.N., An Introduetion to the Mathematical Theory of Finite Elements. Wiley (Ncw York) 1976.
[39] Raviart P.-A. and Thomas J.M., Introduetion a l'Analyse Numerique des Equations aux Derivees Partielles. Masson (Paris) 1983.
[40] Reed M. and Simon B., Methods 0/ Modern Mathematical Physics I: Functional Analysis. Acadcmic Press (New York) 1980.
438 References
[41) Rektorys K., Variational Methods in Mathematics, Science and Engineering. 2nd edition. D. Reidel (Dordrecht) 1980.
[42) Roman P., Some Modern Mathematics for Physicists and Other Outsiders, Volume 1: Algebra, Topology and Measure Theory. Pergamon (Oxford) 1975.
[43) Roman P., Some Modern Mathematics for Physicists and Other Outsiders, Volume 2: Functional Analysis with Applications. Pergamon (Oxford) 1975.
[44) Royden H.L., Real Analysis. 3rd edition. Collier-Macmillan (London) (1988).
[45) Rudin W., Real and Complex Analysis. 2nd edition. McGraw-Hill (New York) 1974.
[46) Schwartz L., Theorie des Distributions. Hermann (Paris) 1950.
[47) Schwartz L., Mathematics for the Physical Sciences. Hermann (Paris) 1966.
[48) Showalter R.E., Hilbert Space Methods for Partial Differential Equations. Pitman (Boston) 1977.
[49) Smirnov V.L, A Course of High er Mathematics, Volume 5: Integration and Functional Analysis. Pergamon (Oxford) 1964.
[50) Strang G., Linear Algebra and its Applications. Academic Press (New York) 1976.
[51) Strang G. and Fix G.J., An Analysis of the Finite Element Method. Prentice-Hall (Englewood Cliffs, N.J.) 1973.
[52) Zauderer E., Partial Differential Equations of Applied Mathematics. 2nd edition. Wiley (New York) 1989.
[53) Zeidler E., Nonlinear Functional Analysis and Its Applications. Volume IIA: Linear Monotone Operators. Springer-Verlag (Berlin) 1990.
[54) Zeidler E., Applied Functional Analysis: Applications of Mathematical Physics. Springer-Verlag (Berlin) 1995.
[55) Zeidler E., Applied Functional Analysis: Main Principles and Their Applications. Springer-Verlag (Berlin) 1995.
[56) Zienkiewicz O.C. and Taylor R.L., The Finite Element Method. Volume 1: Basic Formulation and Linear Problems. McGraw-HiIl (London) 1989.
References 439
[57] Zienkiewicz O.C. and Taylor R.L., The Finite Element Method. Volume 2: Solid and Fluid Mechanics, Dynamics and Nonlinearity. McGraw-Hill (London) 1991.
Solutions to Exercises
Chapter 1
1.1. A = {-2,3}, B = {-3,-2,-I,O,I,2,3}. AUB = B; AnB = A; AnZ+=3, A-Z+={-2}.
1.2. Au C = {I, 2, 9} so B x (A U C) = {(7, 1), (7,2), (7,9), (8, 1), (8, 2), (8,9)}; An C = {I} so (A n C) x B = {(I, 7), (1, 8)}.
1.3. Let x E An (B U C). Then x E A and x E B or C; Le., x E A and xE B, or xE A and x E C. Hence x E (AnB) U (AnC). The second identity is proved in a similar way.
1.4. n(A U B U C) = n(A) + n(B) + n(C) - n(A n B n C) - n(A nBC) - n(B n C - A) - n(C nA - B).
The rationals can be listed by writing down the numbers in the preceding table in the order shown, omitting those already listed (e.g.,
442 Solutions
omit 2/2 = 1). This then gives a listing of all rationals whose numerator and denominator add up to 2, then 3, and so on. In this way all positive rationals are covered. Multiply by -1 to get negative rationals.
1.7. (i) [a, b]; (ii) lR; (iii) [0,1].
1.8. (i) Not open: A = {±I/n7r, n = I, ... } and for every x E A there is a nhd N(x) such that N(x) - {x} ri A. Not closed: 0 ri Ais a point of accumulation. (ii) Neither open nor closed. (iii) Open, not closed since {±I/n7r} are points of accumulation, but are not in A.
1.9. Assurne I is closed. Let x E 1'; since x ri I, the distance from x to I is finite. Hence we can set up a neighborhood of radius E < d about x that lies entirely in 1'. Hence l' is open. Conversely, assurne l' is open. We always have I c 1, so we want to show that 1 c I. Let x E 1 and assurne x ri I. Then x is in 1'. Since l' is open, there is a neighborhood N of x with N nI = 0, which is a contradiction. Thus x E I and so 1 EI.
1.10. Points of accumulation: A = {z: x 2 - y2 = I}. A is open.
1.12. (a) Converges to -3/2; (b) not convergent; (c) converges to 1.
1.13. 1(3n + 2)/(n - 1) - 31 = 15/(n - 1)1 < 0.001. Assurne n > 1, so that 5 < O.OOI(n - 1) => n > 5001. Take n = 5001.
1.14. Suppose U n is monotone increasing, with sup = m. For any E > 0 there exists N such that IUn - ml < E far n > N, so Un -> m. The same reasoning applies if U n is monotone decreasing.
1.15. (i) maxA = 1 = supA,minA is undefined, inf A = O. (ii) maxA, minA undefined; sup A = 1, inf A = -1. (iii) min A, max A do not exist, inf A = -00; supA = c. (iv) Iz2 + 11 :<::: Iz2 1 + 1 = Izl 2 + 1 :<::: 2. Maximum achieved at z = ±1. Minimum is achieved at z = ±i.
1.16. Y = inf A => a :<::: y :<::: x for all x E A and lower bounds a. Thus -x :<::: -y :<::: -a so that -y is the least upper bound of -A.
1.17. Take A = (-1,0) and B = (-2,0); then a = b = O. But C = (0,2) so that sup C = 2 =I ab.
1.18. (i) Lct p = supI; then x:<::: p for any x E I. Let J = {ax: xE I}; since a > 0, ax :<::: ap. Hence J is bounded above by ap. Let the supremum of J be q (we must prove that q = ap). Since ap is an upper bound for J and q is the least upper bound, q :<::: ap. But for
Solutions 443
any y E J we have y ::::; q =} a-1y ::::; a-1q. But 1= {a-1y : y E J}, hence a-1q is an upper bound for I. Thus p ::::; a-1q or ap ::::; q. Since q:::: ap also, we have ap = q.
1.19. (i) Closed. (ii) Open. Set of limit points is nU {x : x2 + y2 + z2 = a2, z > O} U {x: x2 + y2 < a2, z = O}.
1.20. (a) V2j (b) 2a.
1.21. (a) Not an equivalence relation, but a partial ordering; (b) equivalence relation, not a partial orderingj (c) neither an equivalence relation nor a partial orderingj (d) not an equivalence relation, but a partial ordering.
1.23. Take c E A a n Ab. Then c'" a implies that a '" c. Also, c'" b. Thus a '" b by transitivity, and b '" a by reflexivityj hence a E Ab and b E A a . Take any x E A a : x '" aj hence x '" b, so x E Ab. Thus A a C Ab· Similarly show that Ab C A a ·
1.24. A a is the set of ordered pairs of integers lying on the "diamond" {z: lxi + lyl = const} on which a is located.
Chapter 2
2.1. (a) not continuous at x = ±1; (b) continuous on (-oo,Ol.
2.2. (a) Supposethat Ix-yl < o. Then Ip(x)-p(y)1 = la1(x-y)+a2(x2 -
y2) + ... + ak(xk - yk)1 ::::; Ix - yl[la11 + la211x +yl + ... + lakllxk-1 + xk - 2 y + ... + yk- 1ll < oC since term in square brackets is bounded above. Set 0 = E/C. (b) For ° :::: y ::::; :z: we have..jY::::; ,fi =} 2y ::::; 2yXfj =} x-2yXfj+y ::::; x - y or (,fi - Vy)2 :::: x -y. If Ix - yl < 0, then Iu(x) - u(y)1 < 01/2 .
For given E set 0 = E2 •
2.3. (b) Ij(x) - j(y)1 = Ix-1 - y-11 = Iy - xl/lxyl. But x > a, y > a, so xy > x2 or 1/xy < 1/a2. Hence Ij(x) - j(y)1 < a-2lx - yl.
2.5. Set je) = d(-, E). Then Ij(x) - j(y)1 = I infzEA Ix - zl- infzEA Iyzil :::: Ilx - Yl + inf Iy - zi - inf Iy - zll = Ix - yl· Given E > 0, choose 0 = E.
444 Solutions
2.6. If(xo) - f(x)1 < E whenever Ixo - xl < 8, Le., for xE (xo - 8, Xo + 8). Pick any such x: either 0 < f(xo) - fex) < E in which case fex) > f(xo) - E or 0< fex) - f(xo) < Ein which case fex) < f(xo) + E. For the first case choose E smaller than I(xo) so that fex) is positive. For the seeond ease fex) > f(xo) > o.
2.7. Assume that f(a) < 0, f(b) > o. Sinee f(a) < 0, there is an interval [a, c] in which fex) < o. Let the l.u.b. of such points c be e; then fee) ~ o. We cannot have fee) < 0 sinee we would then be able to find an interval about e for whieh fex) < 0, which would imply that e is not a l.u.b. Hence fee) = o. A similar argument applies if f(a) > 0 and I(b) < O.
2.8. (a) U E G(-I, 1); (b) U E Goo([O, 71"] X [0,1]); (e) U E Gl[O, I].
2.9. Iu(x) - u(y)1 = Ilxl - lyll ~ Ix - Yl sinee lxi = Ix - Y + Yl ~ Ix - Yl + lyl·
2.10. Choose 8 = E/ L in the definition of continuity.
2.11. I = IQ U I', where IQ and I' are the subsets of rationals and irrationals. J.L(I') = J.L(I) - J.L(IQ) = J.L(I).
2.12. Let M be an arbitrary measurable set in IR. If 1 E M, 0 f/. M, then XE/(M) = E; 1 f/. M, 0 E M => XE/(M) = E'; 1 f/. M, 0 f/. M => XE/(M) = 0; 1 E M, 0 E M => XE/(M) = dom XE. Thus XIi/(M) is a measurable set. Conversely, if E is not measurable, then XE cannot be measurable.
2.13. Put 8n = 2-n . For eaeh n and for every x there is an integer kn
such that kn8n ~ x < (kn + 1)8n . Set ifJn(x) = kn(x)8n if 0 ~ x < n, ifJn(x) = 0 for n ~ x. Then x - 8n < ifJn(x) ~ x if 0 ~ x ~ n; furthermore 0 ~ ifJl ~ ifJ2 ~ ... ~ x and ifJn(x) ----> x as n ----> 00, for x E [0,00]. Set Sn = ifJn 0 f.
2n 1 2.14. First caleulate JlR Sk(X) dx = I:k=~ (k/2n )(I/2n) = (1/22n ) I: k.
Then use the formula I:~l k = m(m - 1)/2.
215 f+( ) = {I, 0 ~ x ~ 1 r() = {I, -1 ~ x < 0 •• X 0 otherwise,' x 0 otherwise.
,{jRf+ dx = JlRf- dx = 1, so JlRf dx = O. JlRg+ dx = +00, JlRg- dx = 1, so JlRg dx = +00.
2.16. Use the fact that III = f+ + 1-, and that integrability of f implies that of f+ and f-. For the converse use f = f+ - f-· Show that - J r - J f- ~ J f+ - J f- ~ J r + r·
2.17. (a) ap> -1; (b) ap< -1.
Solutions 445
2.18. All real a exeept a = -~, -~.
2.19. Consider 0< In lu(x) - av(x)1 2 dx for any a E R Expand and then ehoose a = In uv dx/ In Ivl2 dx.
Chapter 3
3.1. (a) Veetor spaee; (b) not a veetor spaee; (e) not a veetor spaee; (d) veetor space; (e) not a veetor space.
3.2. (a) Subspaee; (h) not a subspace: 0 ft V.
3.3. (a) Subspace; (b) not a subspace; (e) subspace; (d) subspaee.
3.4. Suppose that U = V EB W, and let u = VI + WI = V2 + W2 for VI, V2 E V and WI, W2 E W. Then VI -V2 = WI -W2. But VI -V2 E V and WI -W2 E W, so that VI -V2 = WI -W2 = 0, or VI = V2, Wl = W2.
Conversely, suppose that u = V + W for V E V, W E W with V and W uniquely defined. If V n W f {O}, then there exists z E V n W with z f o. Henee we ean write u = (V + z) + (w - z) so that the deeomposition of u is not unique, a eontradietion.
3.5. For any u E e[O,l],u(x) = v(x) + w(x), where v(x) = Hu(x) + u(-x)) and w(x) = Hu(x) - u(-x)). Thus V E V and W E W. Also, vn W = {v: V is even and odd} = {O}.
3.6. aß -::; areaA + areaB, henee aß -::; aP /p+ ßq /q sinee A = Ioo. xp- I dx = aP /p, ete. The proof now follows easily from the hints given.
3.7. (u,w) = (v,w) => (u - v,w) = 0 for aB w. Set w = u - v.
3.8. (u,v)o = 0; (u,vh = (u,v)o + (u',v')o f O.
3.9. Ilull = Ilu - V + vII :s: Ilu - vii + IIvll. Repeat with u.
3.10. Ilu + vl12 + lIu - vll 2 = (u + v,u + v) + (u - v,u - v). Expand and rearrange.
3.11. If v = au, then lIu + vii = (u + au,u + au)1/2 = (1 + a)lIull. But Ilull+llvll = (1+a)llull· Conversely, assumethat Ilu+vll = Ilull+llvll· Then Ilu + vll 2 = IIull2 + IIvl1 2 + 2(u, v) = (Ilull + IIvl1)2 = lIull2 + IIvll2 + 211ullllvll. Hence lIullllvll = (u,v) or (u,v) = 1, where u = uillull, fJ = v/llvll. Suppose v f u; then fJ = u + w, and 1 = (u,u + w) = 1 + (u, w) => (u, w) = O. Also, IIfJII 2 = 1 = 1 + IIwll 2 + 2(u, w); Le., IIwll = 0 => w = O. Hence fJ = u or v = au for some a.
3.12. Assurne that IIx - ylllly - zll = Ilx - zll. Square and rearrange to get (a, b) = 1, where a = a/llall, a = x - y, b = Y - z. Thus b = a which gives Y = ax + (1- a)z, where a = lIy - zli/(llx - yll + IIY - zll). The converse is straightforward.
446 Solutions
3.13. Verify that (. , .) defined by (u, v) = J: u' Vi dx is an inner product on X.
3.22. J IUTVT I dx ::; [f luIT(p/Tlr/p[f Ivlr(q/rlt/q· Take rth roots of both sides.
3.23. Follow the argument of Example 20.
Chapter 4
4.1. 2.
4.2. I(un,vn) - (u,v)1 = I(un - U,Vn - v) + (u,vn - v) + (v,un - u)l ::; Ilun - ullllvn - vii + llullllvn - vii + llvlillun - ull -t 0 as n ---> 00.
Set Vn = v (Le. the sequence v, v, ... ) to get (un , v) -t (u, v). Finally, l(un,v) - (u,v)l ::; l(un - u,vll ::; llun - ullllvll, hence (un,v) -t
(u, V l. Set V n = U n to get the final result.
4.3. llu - wll = llu - U n + U n - wll ::; Ilu - unll + Ilun - wll < E + a. The inequality follows from the arbitrariness of a.
4.4. (a) (-1,1]; (b) (-00,00).
4.5. (a) un(x) -t 0 pointwise. But Ilun - ulli2 = J12/:: n2 dx = n -t 00
as n ---> 00; (b) un(x) -t pointwise since un(x) = n3 / 2x/ exp(n2x2 ) =
Solutions 447
n 3/ 2x/[1 + n2x 2 + ~n4x4 + ... ] -> 0 as n -> 00. But Ilun - ulli,2 = f'n y2 exp( _2y2) dy (setting y = nx) = -H[yexp( -2y2)]~n
- J:'n exp(-2y2)dy = -~(O + )(7f/2)) as n -> 00.
4.6. sup lun(x)1 = 1/2at x = l/n. Thus in [0,1], un(x) - .. 0 pointwise but Ilun - ulloo = 1/2, so convergence is not uniform. But convergence is uniform in (a,l] (a > 0) : sup lun(x)1 = na/(l + n 2 a2 ) at x = a for n > l/a (check this by sketching un(x)) and sup lun(x)1 -> 0 as n -> 00.
b 4.7. sup Iun(x) - u(x)1 < E for n > N. Hence Ja Iun(x) - u(x)IP dx :s;
(sup Iun(x) - u(x)I)P' (b - a) < (b - a)EP.
4.8. Ilull = 0 does not imply that u = 0; 111 . 111 is also not a norm.
4.9. Ilu - U 11 2 = ~ - 2mn + ~ = 2 (m_n)2 Nu-n m L2 n+2 mn+m+n m+2 (m+2)(n+2)(mn+m+n)' merator (m - n)2 :s; (m + n)2. Now show that Ilun - um lli2 -> 0 as n,m -> 00.
4.lO. Ilun - umllu = Jo1 Ixn - xml dx = n~1 - m~1 (taking m > n)
= (n+0C';-:+I) :s; (n+1)(m+l) -> 0 as n,m -> 00. Hence {un } is a Cauchy sequence.
4.11. {un} is Cauchy, so suplun(x) - um(x)1 < E for m,n > N. For any Xo, Iun(xo) - um(xo)1 < E, so {un(xo)} is a Cauchy sequence of real numbers. IR is complete, so un(xo) -> u(xo), say, which defines a function u( x). Thc rest of the proof follows easily from the hints given.
4.12. Let {x k } be a Cauchy sequence in IRn : Ilxk - xIII< E for k, I > N; i.e., 2:; IXki - xlilP < EP • Hence IXki - Xlil P < EP for each i. But IR is complete so xki -> Xi, say. Hence X -> X in IRn.
4.13. Assume {un } convergent: Ilun - ull < E for n > N. Also Ilum - ull < E'
for m > Nt Hence Ilun - Um 11 = II(un - U) + (um - u)11 :s; Ilun - ull + Ilum - ull < E + E' for n, m > N (assume N > N').
11 11 2 Jl/2+1/n[ ( 1) ( 1 )]2 d J 1/2+1/m[ 4.14. Un - Um = 1/2 n X - '2 - m X - '2 :r + 1/2+1/n 1-m(x - ~)J2 dx. Show that this -> 0 as m, n -> (Xl, SO that {Un } is
Cauchy. Also, Ilun _u11 2 = Jllg+l/n[n(x-~) -1]2 dx -> 0 as n -> 00.
So u n -> u in L 2 .
4.15. Take V n E Y with Vn -> v. It is required to show that v E Y. From Exercises 3.9 and 3.22, Illvn Ilu -Ilvii u I :s; Ilvn - vllu :s; cllvn - vllp· Thus 11-llvllul < E so that v E Y.
448 Solutions
4.16. Let v(x) E C[~I, 1] be defined by v(x) = I/E', ~E ::; x::; E, { ~1 ~1::;x<~E,
+1, E<X::;1.
We have Ilu ~ vlli2 = J~« ~1 ~ C I)2 dx + Jo«1 ~ C I)2 dx = E3 /3 ~ E2 + E. Hence v can be made arbitrarily elose to u by choosing E small enough.
4.17. Ilu ~ vll oo = supll ~ v(x)l, where Iv(x)1 < 1 and v(O) = O. Hence Ilu~vlloo = 1; neighborhoods ofu ofradius less than 1 do not contain members of V, so u is not a point of accumulation.
4.18. v E B(Uo,T) ~ Iluo ~ vll oo ::; T; Le., suplsin21rx ~ COS21rTI ::; T. sup lUD ~ vi = v'2 (at x = 3/8) so we require T ~ 3/8.
4.19. Cf. solution to Exercise 1.9.
4.20. Assume that Y is complete, and let v be a point of accumulation of Y. Then each open ball B(v, l/n), n = 1,2, ... , contains a point vn , say, in Y. The sequence {un } is convergent, hence Cauchy, in Y. Since Y is complete, v E V. Hence Y contains all its points of accumulation, and is elosed. Conversely, assume that Y is elosed, and let {vn } be a Cauchy sequence in Y. Then {vn } is a Cauchy sequence in X, and so converges to v in X. From Theorem 3 of Chapter 4, v is in Y also, so Y is complete.
4.21. W dense in X ~ for any v EX there is a w E W such that Ilw~vll < E. Similarly, for any u E Y there is v E X such that Ilv ~ ull < E.
Hence Ilu ~ wll ::; Ilu ~ vii + Ilv ~ wll < 2t:, so W is dense in Y.
4.22. Take i E LP. For given E > 0 choose a bounded function 9 in LP, where 9 has compact support, for example, Igl ::; M in [a, b] and 9 = 0 otherwise. Select 9 so that Ili ~ gllp < E. Bounded functions with compact support are dense in LP, so we can find ihn} in Co such that 9 = limhn a.e. Assume that Ihni::; M in [a,b] and 0 otherwise. Then Ig ~ hnlP ::; (2M)P on [a, b] and Iig ~ hnll p --> 0 from the Dominated Convergence Theorem. Choose n so that Ilg~hnll ::; E
and use the Minkowski inequality.
4.23. Suppose that there are two points vo, vb such that Ilu - Vo II = d. Then w = (vo + vb)/2 is in M hence, using the parallelogram law, it can be shown that d2 ::; IluD ~ wl1 2 < ~llu ~ vol1 2 + ~llu ~ v'II 2 =~, a contradiction.
4.24. Consider {un } C y.l with limit UD in X. We must show that UD E y.l also. By definition (un, v) = 0 for any v E Y; thus 0 = limn->oo(un, v) = (limn~ooun,v)=(uo,v)=O~uo E V.l.
Solutions 449
4.25. Theorem 7(b), which requires completeness of H, is used in Lemma 1.
4.26. Let u E X and w E Y 1.. Then u E Y also, so (u, w) == 0. u is arbitrary; hence w E Xl. =} yl. C Xl..
Chapter 5
5.1. (i) R(M) = points on the upper unit semicircle, N(M) = 0; (ii) R(K) = [0, 00), N(K) = {al; (iii) R(f) = (0, 00), N(f) = 0.
5.2. N(S) = {al; N(T) = {a( -8,4, In.
5.3. (i) One-to-one, not surjective; (ii) one-to-one, surjective (T is a refiection about a line at 45° through the origin).
5.10. No; e.g., d(x, B) + d(y, B) =1= d(x + y, B) in general. Null space is the set B.
5.11. For u =1= 0, IITII = sup(IITull/llull) = sup IIT(u/llulDIl (T is linear) = sup 11 Tu 11 , lIull = 1. To prove the second result, consider IITull ::; IITlillull. For every E > 0, there is a Uo such that IITuoli > (IITII -E)lIuoll· If lIull ::; 1, then IIAull ::; IIAlillull ::; 11 All =} sup 11 Au 11 ::;
IIAII, lIull ::; 1. Eut ifwe put U1 = uo/iluoll, then IIAutil = lIuo 1I-11IAuo 11
> IIAII-E, so for lIull ::; 1, sup IIAul1 2- IIAutli > IIAII-E or sup 11 Au 11 ::;
maxl:S:i:S:n 2::7=1 lAi] I maxl:S:j:S:n IXjl = maxl:S:i:S:n 'L7=1 IAijlllxll=· Hence IIAII = sup(IIAxll=/lIxll=) ::; maXi:S:i:S:n 'L7=1IAij l. Suppose maximum occurs for i = k. Then for x such that Xj = +1 if Akj 2-0, Xj = -1 if A kj < ° we have IIAxll=/llxll oo = 2::7=1 IAijl·
450 Solutions
5.13. For x =f. 0, (IIAxll/llxllf = (a + b)2 - 2ab(x - y)2/(x2 + y2). Take the supremum (at y = x) to find IIAI12.
5.14. Illull = Ilull; I is bounded. Consider u(x) = sin nx: lIullv = 1 but Illullw = 1 + n which cannot be bounded.
5.16. Let {un} C N(T) with limit u in U. Then TUn = O. Thus 0 = limn -+oo TUn = T(limn -+oo un) = Tu =} u E N(T).
5.17. T is one-to-one since, if TUI = TU2 = v, then IITul - TU211 = 0 ~ Kllul - u211. IIT-1vll = Ilull ~ K-1 11Tull = K- 1 I1vll·
5.18. u(x) = I; u'(s) dx ~ sUPoSxsllu'(x)1 = IIDull. Take sup of both sides.
5.19. (I - P)(l - P) = 12 - PI - IP + p 2 = 1- P.Range : R(l - P) = N(P), R(P) = N(I - P).
5.20. From Theorem 8, IlPuli ~ Ilull. Thus IIPII ~ 1. But for u E R(P) we have Pu = u, so IIPul1 = Ilull. Hence IIPII = 1.
5.21. Take, for example, the map on lR2 that takes a point x to the point in B(O, 1) dosest to x. This is a projection, but the map is not homogeneous.
5.22. Let u E N(P). By definition (u,v) = 0 for v E R(P). Hence N(P) C R(P).L. Let u E R(P).L. Then (u, z) = 0 for z E R(P). By Theorem 9, u = v+w for v E R(P), w E N(P), so Pu = Pv + Pw = Pv = v. Also, 0 = (u,z) = (v,z) + (w,z) = (v,z); hence v = o. Thus Pu = o =} u E N(P).
5.23. T is a projection since T is linear and T 2u = Tv (where v = u(x) if lxi< 1 and 0 otherwise) = v = Tu. R(T) = {u E L2(lR): u(x) = 0 for lxi ~ I}, N(T) = {u E L2 (lR): u(x) = 0 for lxi< I}.
5.24. v(y) = Pu(y) = I~1 exp(i(y - z))u(z) dz; show that Pv(x) == p 2u(x) = Pu(x). Pis an orthogonal projection.
5.25. (i) x satisfies Ax = 1 where 1 = (1, ... ,1); (ii) x satisfies Ax = 0 = (1,0, ... ,0).
e3~1 (~e - e3 ) = 101 g(x)2x dx; so 9 satisfies Jo1(gj - u)dx = O.
5.28. Let {Pn} be a Cauchy sequence in X'. Then for any u E X, I(Pn,u)(Pm, u)1 :S IIPn - Pmlillull -> 0 as m, n -> 0 so {(Cn, u)} is a Cauchy sequence in lR, with limit (C, u), say. Complete the proof by showing that P is bounded and linear, and Cn -> C in X'.
Solutions 451
5.29. In the use of the projection theorem.
5.30. If there are two elements Ul, U2 such that (Ul,V) = (U2'V) = (P,u), then (Ul - U2,V) = o. Set v = Ul - U2: IIUI - u211 2 = Oor Ul = U2· II Pli = sup(I(P,v)I/lIvID (for v =I- 0) = sup((u,v)/llvID :'S sup(llullllvll/llvll) = Ilull· Also, I(P,u)1 = (u,u) = IIul1 2 :'S IIPllllul1 so IIPII ~ lIull· Hence IIPII = Ilull·
5.31. Take I = Iglq-l sgng; then I/IP = Iglq, so I E LP, and II/lb = IlgIl1:;-1. Then show that (Pg , I) = II/lbl!gl!Lq.
5.32. P = 0 <=} (P, v) = 0 far all v E X. Given v E X there exists a sequence {vn } in Y such that Vn -+ v. Thus (P, v) == (P, limn--->oo vn ) = limn--->oo(P, vn ) = o.
6.2. L;=1 akeikx = 0 =? L;=1 ak cos kx = 0 and L;=1 ak sin kx = 0 which halds only for all ak = O. Hence {eik"'} is linearly independent.
6.3. If u, v E X, then (au + ßv)" - 2(au + ßv)' + (au + ßv) = a(u" -2u' + u) + ß(v" - 2v' + v) = 0, hence au + ßv EX. dim X = 2. Basis for Xis {Ul(X) = e"', U2(X) = xe"'}.
6.4. dimM = 9, dimK = 4.
6.5. Let dimV = m with basis {Vl, ... ,Vm } and dimW = n with basis {Wl' ... , wn }. Every u E V ffi W is of the form u ,= v + W for some v E V, W E W. But v = Li aiVi and W = Lj ßjWj so u = Li aivi+ Lj ßjWj. Hence B = {VI, ... , Vm ,Wl,·· .,Wn } spans VffiW. It remains to show that B is linearly independent.
6.9. All = ~(e2-1), A l2 = A 2l = ~(I-e-2), A22 = i(1-e-6 ). detA =I 0.
6.10. Consider I : Xl ---> X 2 : IIIul12 = IIUl12 S kllulh (show this using Lemma 1; see also Theorem 4). Similarly, Ilulll S Kllul12 if we consider I: X 2 ---> Xl.
6.11. Tl2 = 2, T23 = 6, others zero.
6.12. Tll = 27r, T22 = cosx, others zero.
6.13. (b, c) = (Ta, c) = (a, TT c) = ° if c E N(TT). Let d E R(T)l... Then (d, Tu) = ° = (TT d, u) => d E N(TT). Conversely, if d E
N(TT), then if Tu = v we have (TT d,u) = ° = (d,v) => d E R(T)l... Hence N(TT) = R(T)l.. => N(TT)l.. = R(T). N(TT) =
{(1,1,-1)}, b=(a,ß,a+ß).
6.14. (0'.2, -0'.1,0), (0'.3,0, -ad·
6.15. Let BI = {el, ... ,en } and B2 = {h, ... ,fn} be orthonormal bases of X and ffi:n, respectively. For any u E X we have u = L uiei, Ui = (u, eil. Define the map T: X ---> IRn by T(u) = (Ul,"" un ). Then T is an isomorphism (show this) and Ilulli = (u,u) = (Luiei, LUjej) =
LU; = IITullffi.n.
6.16. 111:'11 = max lail·
6.17. (i) u(x) = y'2;(I/y'2;); (ii) u(x) = L;;'=1(2/k)(I- (-I)k)sinkx.
6.18. Uo = -V2/4, Ul = 5V3/6V2, U2 = ,;5/8V2.
6.19. Ck = ~(U2k - iU2k-l) for k = 1,2, ... , Ck = ~(U2k + iU2k-l) for k = -1, -2, ... , Co = uo/V2.
6.20. ° S Ilu- L;':1 (u, cPi)cPi 11 2 = lIu11 2 - L;':1 (u, cPi)2, hence L;':1 (u, cPi)2 S Ilu11 2 . Since sum is bounded, we can let N ---> 00.
6.21. Use the property PcPk = cPk to show that p 2u = Pu. Clearly R(P) c V. Conversely, if v E V, show that Pv = v so that R(P) = V. Orthogonality: take v E R(P) and W E N(P); then (w, v) = (w, Pu). Use this to show that (w,v) = 0.
6.22. See Exercise 6.8. Pu = L!=o(U,4>k)cPk = ..J275cPo + (8/35)-/572cP2'
Solutions 453
6.23. (a) Set u(r, e) = R(r)8(e) to get (8' sine)' + '\8sine = O. Set E = cos e to get Legendre's equation. General solution is u( r, e) = 2:~=o[anrn + bnr-(n+1)]Pn(cose). (b) an = (2n + 1)/2 Jo71: f(e)Pn(cose) deo
6.25. Use integration by parts and the boundary conditions to show that (Lu, u) 2: O. Nonnegativity ofthe eigenvalues follows from 0 :S (Lu, u) :S A(U,U). Since L2 is separable there is at most a countable number of nonzero mutually orthogonal vectors.
6.26. Let the minimizer be u, and set w = u + EV; then consider R(w) = R(E) over all w that satisfy (w,el) = (w,e2) = ... = (w,en-l) = O. Set [dR/dE]<=o = 0; expand and differentiate to find that A = R(u) and u = en .
6.27. (Lsn, rn ) = (L 2:~=1 ukq;k, rn) = (2:~=1 ukAkq;k, rn ) = 0 since (rn, q;k) = o (Proof of Theorem 6.12).
6.28. Return to (6.34): for symmetry of L, (a) p(x) ----> I) as x ----> ±oo; (b) p( -L) = p(L).
6.29. (c) Show that H~(x) = 2xHn(x)-Hn+l (x). Set f(x) = exp( _x2 ) and show that f(n+1)+2xf(n)+2nf(n-l) = 0; multiply by (_l)n+l exp(x2 )
xlyO D(l,O) f(O) + XOyl D(O,l) f(O) = x Bf I +- Y Bf I etc 1!0! 0!1! 8x 0 8y 0 .
7.2. J~a 8(x)q;(x) dx :S Cl J~a 8(x) dx since supq;a(x) = e- l . If 8 were
locally integrable, then lima--->o J~l 8(x) dx = O. But left-hand side = q;(0) = e- l .
7.3. f(x)q;(x) E C(O). Assume f # 0, but J N dx = O. In particular, if f(xo) # 0, then f(x) # 0 for all x E (xo - h, Xo + h) for some h. Choose arbitrary rp with compact support inside (:r:o - h, Xo + h); can always find q; such that J fq; dx # 0, a contradiction.
7.4. Consider 0 C lR?, for example; for 10:1 = m, JoJDau)v dx = In(8rnu/ 8xk8yrn-k)v dx, where 0 :S k :S m. Use Green's theorem repeatedly.
7.lO. (a) u E H 2(0, 3); (b) u E HI((O, 1) x (0,2)).
7.1l. u..L v in HI(O, 2).
7.13. D"'u E L 2 (fJ) far lai = 2; so m = 2 > n/2 = l.
7.14. Consider {un }, {vn } C Cl (f!) such that Un -> u and Vn -> u in the Hl-norm with u, v E HI(fJ) (H I is the closure of Cl). Then DO:Un ->
D"'u, DO:vn -> DO:v in L2, for lai ::::: l. Also, Vn -> v and Un -> U in L2(r). Thus, for example, (ßUn/ßxi,vnh2(o.) = (un,vnvih2(r) -(un , aVn/ßXi)P(o.). Take limn~oo.
7.15. Assume 0, c lR?; then left-hand side is 10. (~:~ + ~:~) (~:~ + ~:~) dx.
f ö2u ö2v f (ö2u ÖV ö3u) f ö4u d Now Jo. öx2 öx2 dx = Jr öx2 öx - öx3V Vx ds+ Jo. öx4V dx. Procee
in this manner; use ß/av = Vla/ßX + V2a/ßy.
7.16. Let {vn } be a sequence in 'D(fJ) with limit v E H{j(fJ). We have Ilvnllp ::::: clvnlHl; V n -> v in H I implies that IIvn llL2 --+ IIvllL2 and IvnlHJ -> IvlHl. I·IHJ is positive-definite since lviI = 0 implies that J l'Vvl 2 dx = 0, so that v = const = 0, given the boundary value of v.
Solutions 455
7.17. Show that (u,v) == InLlal=mDauDav dx is an inner product. In particular, (u, u) = 0 =} In(Da u)2 dx = 0 for Inl = m, hence Dau = o for Inl = m. But u E HO'(n); so u = O.
But in {)x {)y dx = - in {)x2 {)y {)x dx = in {)y2 {)x2 dx.
7.19. Require sup I (8, v) I to be defined, Le., v continuous. Hence m > n/2. For example, 8: HJ(n) --+ R is not defined for 0. C R2 .
7.20. u E HJ(n).L =} (U,V)Hl = 0 for all v E HJ(n); Le., 0 = In(uv + Llal=l DOtuDOtv) dx. Set v = </J E D(n): 0 = I(u. - V 2 u)</J dx using Green's theorem =} V 2u = u. Since D(n) is dense in HJ, we can extend this result in the usual way. u E HJ(n).L for 0. = (0,1) =}
u" - u = o. Basis for HJ(n).L is {eX,e-X}.
7.21. I(lnx?dx = x(lnx)2 - 2xlnx + 2x. Then use Theorem 9.
Chapter 8
8.1. (a) Second order, nonlinear, 0. = upper unit semicircle. (b) Fourthorder, linear, 0. = triangle with vertices at (0,0), (1,0), (0, I).
8.2. (a) 1t In' pv dx = In' Q dx + fr" t ds. Use Cauchy's law t = ern and the divergence theorem to rewrite the surface integral as In' diver dx. The left-hand side equals In' p{)2u /{)t2 dx. Regroup and invoke the arbitrariness of 0.' to obtain (8.5). (b) (J"ij = 'x(divu)Iij + 2J-tEij(U). Substitute in (8.5).
8.3. The argument is as in Example 2 of the Introduction: simply replace f by f - ku, ku being the force of the foundation.
8.4. (a) Lj {)(J"aj/{)Xj = Lß {)(J"Otß/{)xß + {)(J"Ot3/{)Z, where Z = X3. Integrate with respect to z and use the definitions of Sa. and Maß. (b) Follows as part (a). (c) Differentiate (8.14h, with respect to x"' sum on n, and use (8.14h to eliminate Sa:; this gives La,ß {)2 M aß/ {)xa{)xß = -q. Next, use (8.13). This gives La,ß{)2Ma:ß/{)xa{)xß = -D[VLa [j2(V2w)/{)x; + (1 - v) La:,ß {)4w/{)x;{)x~1·
8.5. (a) w' = 0, Will = 0; (b) w = 0, w" = O.
8.6. Elliptic in A = {x: x > l,y > I} U {x: X < l,y < I}; strongly elliptic in any open subset of A.
8.7. LIOtI, IßI=l aa:ßE,Ot+ß = -(1 + x2)e + 3.,,2 + 2(1 + X2)(2 = 0 at any Xo for any E, such that e = [3.,,2 + 2(1 + z~)(2l1(1 + x6).
456 Solutions
8.8. Li,j,k,l Cijkl~i"'j~k"'l = JLle121771 2 + (>. + JL)(e· 77)2 = JLle1 21771.12 + (>. + 2JL) (e . 77)2, where 771. is the component of 77 orthogonal to e. The result follows from the independence of.,,1. and (e· 77). Pointwise stability: f == Li,j,k,l CijklMijMkl = (3)'+2JL)IMs I2+2JLIMD I2 [Ms = ~(tr M)I and MD = M -MsJ. Show that IMI2 = IMs I2 + IMDI2: then f ~ cIMI2 Hf 3>' + 2{t > ko and {t > {to.
[)2U [)2U [)2U 8.9. -[) 2 V~ + 2-8 [) VI V2 + -8 2 V~ = g. Have to check Llal=2 baaa =
Xl Xl X2 X 2
v?a~+2VIV2ala2+V2a2 = (VIai +V2a2)2 = (v?+vi)2 f= 0 if a = v.
8.10. Use (8.13) and (8.14). The BC can be rewritten as ~ +V a~~~2 = o. With 10:1 = 3, Lbava = b(3,0)V? + b(I,2)VIVi = VI [V? + vviJ f= 0 along X = L, for which VI = 1, V2 = o.
8.11. Irl + Iß - 31 f= o.
8.12. ku·v-t·v = O. n = 2j bn = k, b22 = -Cn = Ij all other components are zero. So (8.33) is satisfied.
8.13. u· v = 0, t· s = {tt· v with t = UV, and u = Ce(v).
8.16. Set 8vif8xj = eijj then since u is symmetrie, Li,j O'ijeij = Li,j O'jieij (i)j also, by swapping indices, Li,j O'ijeij = Li,~ O'jieji (ii)j add (i) and (ii) to get desired result. To obtain (8.49), use the fact that Jn Lk,l O'klfkl(U) dx = Jn Lk,l O'kl(äuk/8xt} dx = Jr Lk,l O'klVIUk dsJn Lk,I([)O'k!/[)XI)Uk dx. Set u = Ce(v).
Jol xf(x) dx = O. Solution is unique if J; u(x) dx = Jol xu(x) dx = O.
8.20. N(A) = {u: u const.}j unique solution if Jn u(x) dx = O. N(A*) = {u: u(x) = 0:1 + 0:2(X - Y)}j solution exists if Jn f dx = Jn(xy) f dx = O. If n = (-1, 1) x (-1, 1), then Jn f dx = 0 if f is odd in X or Yj Jn(x - y)f(x) dx = 0 if f(x,y) = f(y,x).
8.22. (b) From (a), A : N(A)1. --> R(A) is bounded. Hence, using the Banach theorem, A-I : R(A) --> N(A)1. is linear, bounded '* IIA-IVIl s Kllvll for all v E R(A), so setting v = Au we have lIull s KIlAull for u E N(A)1.. If {vn} is a Cauchy sequence in R(A) with limit v, then with Un = A-1vn we have lIum - unll s Kllvm - vnll --> 0 as m, n --> 00; so {um} is a Cauchy sequence in N(A)1.. N(A)1. is closed;
Solutions 457
SO Um ----> u in N(A)1-. Since A is continuous, Vn 0= AUn =* v = Au. Hence v E R(A) =* R(A) is closed.
8.23. Flexible foundation: N(A) = {O} and a unique solution exists. Coulomb friction: N(A) = {clel + c2ed, where Cl and C2 are constants. A unique solution exists if and only if h =, 12 = O. If friction is not limiting, then N(A) = {O}.
Chapter 9
9.1. V = {v E H 2(O,I): v(O) = 0, v'(O) = O}. I~ [kUli v" + du'v' + cuJ dx = I; Iv dx + ßv(l) + Qv'(l).
9.2. Let angle between T and v be ß. Boundary term in VBVP is Ir vVu· v ds, v E HI(n). But T = vcosß + ssinß (s = tangent = (-V2,VI)), or T = (VICOSß-V2sinß, V2COSß+VIsinß) =* v = (Tl cos ß + T2 sin ß, - Tl sin ß + T2 COS ß). Boundary term is thus Ir v(g cos ß - Vu· IL sin ß) ds, where IL is normal to T.
9.3. a(w,v) follows by direct substitution of (8.13).
9.4. For continuity of a, use the Sobolev Embedding Theorem to obtain Iv(l)1 ~ CllvliHl ~ CllvllH2, etc.
9.5. a( v, v) ~ 10. J-l L.i t;, t;, dx using strong ellipticity. Complete by using the Poincare-Friedrichs inequality.
9.6. Use (8.13) to obtain the first part. For the second part return to Exercise 9.3: the remaining boundary term is Ir M,,(w)8v/öv ds = O. Use the identity a2 + 2vab + b2 ::>: (1- v)(a2 + b2 ) and (7.18) to show that a(v, v) ::>: (1 - v) 10. L.1<>1=2 D<>u1 2 dx. Here v is Poisson's ratio.
Use the Poincare inequality (7.18) to get Io.(~';?dx ~ c Io.[(~,;)2 +
(a~::X2?J dx, etc. Then apply the Poincare-Friedrichs inequality to obtain a similar bound on Io.v2 dx. This leads to a(v,v) ~ C(l-v)llvll~2'
9.7. a( u, v) = 101 (pu'v' +TUV) dx+p(l)u(l)v(l). V-ellipticity: use Theorem I, Chapter 7, to get a(v,v) ::>: Qllvll~" Q = min(po,To). Continuity:
9.8. (b) VBVP is: Jol U"V" dx + [h)v(l) - gIV'(l) - hov(O) + gov'(O)J = 101 Iv dx, v E H 2 (O, 1); so P = PI (0,1). Hence Q = {v E H 2 (O, 1) :
101 v dx = Jo1 xv dx = O.} Q-ellipticity is tricky, but see Rektorys [39], Chapter 35. A unique solution exists if and only if 0 = (P.,p) =
I; fp dx + [glP'(l) - hlP(1) + hop(O) - gOp'(O)J for all P E PI(O,I).
458 Solutions
9.9. See Exercise 8.8: use Korn's inequality.
9.11. a(u + p, v + p) = a(u, v) for p = Prer + p()e() , where Pr,P() E Po(!1). Q = {v E V: In Vr dx = In v() dx = O}; a unique solution exists if f satisfies In fr dx = In f() dx = O.
9.12. (DJ(u),v) = lim()->oe-1[J(u+ev) - J(u)] by definition. Set f(e) = J(u+ev) for any given u,v. Then (DJ(u),v) = lim()->oe-1[f(e)f(O)] = 1'(0) = (d/de)(DJ(u)v)I()=o.
9.13. äJ / äXi = lim()->o[J(x+e(O, ... , Yi, . .. ,0) )-J(x )]/(eYi) (Yi in ith slot). Multiply by Yi and sum over i to get rcsult.
9.14. J(eu + (1- e)v) = He2a(u, u) + (1- e)2a(v, v) - 2e(1 - e)a(u, v)}e(C,u) - (1 - e)(C,v). a(u - v,u - v) > 0 since a is V-elliptic, so 2a(1L, v) < a(u, u) + a(v, v). Use this to obtain strict convexity of J.
9.15. J(eu+(I-e)v) = eJ(u)+(I-e)J(v)-~e(l-e)a(u-v,u-v). Thelast term on the right is nonnegative. To show that u is a minimizer: from convexity, J(v) - J(u) :::: e- 1 [J(u + e(v - u)) - J(u)] = (DJ(u), v) when e ---7 0 (see Example 15).
9.17. Since u is a minimizer, J(u) :::; J((1 - e)u + ev) for 0 < e < 1 (since V must be convex). Expand and rearrange to get a(u, v - u) - (C, vu) + ~ea(v - u,v - u) :::: o. Let e ---7 O.
Chapter 10
10.3. Un satisfies a( u, v) = (f., v) or (un , CPk)a = (C, CPk/. Also, Un =
L~=1 (u n , CPk)aCPk = L~=l (C, CPk)CPk. Now J(u) = -~ Ilull; (show this); but J(un ) = ~llun - ull; - ~llull;; hence Ilun - ull a ---7 O. The result Ilun - uliH ---7 0 follows from continuity of a(·, .).
10.5. Ilu - uhll; = a(u - Uh, U - Uh) = Ilull; - Iluh II~ - 2a(uh, u - Uh)' The last term is zero.
10.9. Replace v by AVh in Green's formula (G(u,v) = 0); (10.37) gives (AVh,f) = (AVh,Auh) =?- (vh,A*f) = (vh,A*Auh)' If A = _\72 ,
then A* = _\7 2 .
10.10. (a) I;(-u~ +Uh -- sinX)Vh dx = 0, Vh E vh C L 2 (0, 1), Uh E Uh C H 2 (0, 1) n HJ(O, 1). (b) Least squares: solve MT a = F, where Mij = 10\ -CP:' +CPi)( -'ljJ~' +
Solutions 459
'l/Jj) dx and Pj = J; (sinx)( -'l/Jj' +'l/Jj) dx. Collocation: solve L:=1 (-cp% (Xi) + CPk(xd)ak = f(xi), i = 1, ... , N. (c) Solve MT a = F, where Mij = J; CPi( -'l/Jj' + 'l/Jj) dx and Pj = Jo1 f'I/Jj dx.
Chapter 11
11.1. Must show that a function v, say, exists such that Jn ViCP dx = - Jn VOCP/OXi dx. For each Oe, Jne (OV/OXi)CP dx = Jre V!/iCP ds -Jne V OcpjOXi dx since vbe E H 1(O).
11.2. Optimal B = 5.
11.4. emax exists at point x, where e' = 0; then e(:l:i) = 0 = e(x) + ~e"(z)(xi - x? => le(x)1 = ~!e"(Z)I(Xi - x? If i is node nearer to x, then lXi - xl ::::: ~h. Hence le(x)1 ::::: ~h2Ie"(x)l. Maximize ovcr all elements to get result.
11.5. f"(x) = 27rcos7rx-7r2xsin7rx.IMax.valuel = 12r.1 at X = 0,1. Hence lIell= ::::: ~h2 ·271" = 7rh2/4. See whether log lIell= ::= 2h + const.
11.7. Retain 1,~, TJ, e, ~TJ, TJ2, eTJ, ~TJ2. Then, for example, if node 5 is located at ce, TJ) = (0, -1), N5 (e, TJ) = ~Cl - e)(:l- TJ)·
11.10. One needs to solve a system of 21 equations uniquely, for any given right-hand side. Equivalently, show that any polynomial for which D"p = 0 for lai::::: 2 at the vertices, and p" = 0 at the midpoints is identically zero. See Ciarlet [11] (Theorem 2.2.11) for full details.
4' , 11.11. x = LA=1 xANA(~,TJ), whcrc NA are given by (11.27). Substitute
and use the geometry of the parallelogram to verify that x = A~ + b for suitable A and b.
11.12. j = H2d - (e + TJ)(I- d)] > 0 for all ~ E n if d > ~.
11.15. (a) a = [1 1 1 IjT, b = 2[-1 1 1 - IjT, c = 2[1 1 - 1 - I]T, d = [1 -11 _1]T. (b) C'ilN)T = [-~b+TJd ~c+~d].
11.16. Show by direct integration over reference element; for examplc, Jne (a + b1x + b2 y) dxdy = Ae(a + bTx) where b = [bI b2 jT and x = (1/6)[1 1jT, for a polynomial of degrec one.
Chapter 12
12.1. IIT;;-lll = sup IIT;:-I Y II/lIylI, Y f O. Set z = Pey/llyll; thcn IIT;111 = supllp;IT;;-lzll· Pick x,y in Oe such that IIx - ylI = Pe: IIT;111 = p;l sup IIT;;-l(x - b + b - y)11 = p;1 sup IIx - yll = h/ Pe.
460 Solutions
12.2. //lv//m,1l = /lv/lm,1l ~ /lv/lk+l,n since m ~ k + 1. /lrrv/lm,n = /lI:iV(Xi)~i/lmll ~ I:i/iJ(Xi)"/~i/lmll ~ CsUP/iJ(Xi)/ (C is inde-pendent of iJ). ' ,
12.3. Let the triangle have angles o:,ß,"'( with (Je = 0: ~ ß ~ "'(. Let the sides opposite 0:, ß, "'( be a, b, c, respectively. Then a ~ b ~ c and he = c. The largest cirele inscribed in the tri angle touches all sides. Draw a sketch and show that he = (Pe/2) (cot 0:/2 + cotß/2). Now 0: < 7r/2,ß < 7r/2; so cotß/2 ~ coto:/2. Hence he/Pe ~ coto:/2 ~ a if we prescribe 0: ~ Bo, so that a = cot Bo/2.
e k+l - e e k+l e = max(ao, .. . ,a-2m )). Given K > 0 we can always find E > 0 such that the term in square brackets < 1 + K provided he < E.
12.6. 'OiJ(a) = I:i :;, ai = I:i,j t:, ~ai = I:i,j t:] Tjiai = 'Ov(Ta). Proceed in the same way for higher derivatives. Then for k = 2, for example, ID"'iJ(x) I ~ /l'02iJ/I = sup 1'02 iJ(a, b)/ (/la/l ~ I, /lbll ~ 1)
12.7. Any v E X h also belongs to L2 (n), so it is required to find Vi E L 2 (n) such that Jn Vicj; dx = - Jn vß/ßx; dx Vcj; E '0(0.). Use Green's theorem applied to the function ßW/ßXi' where w = vln e ; then sum over
all elements to get In ViCP dx = - In Ußcj;/ßXi dx = I::=1 Jarle Wcj;Vi ds; the boundary integrals vanish.
12.8. a(w,e) = /lell1,2' where e = u - Uh. Also, a(wh,e) = 0; so a(w-wh,e) = lIe/l1,2' Hence /le 11 1,2 ~ Kllw-wh//I,n/lel/I,n ~ KCh/Lllwllp,nM3I1ullr,n for w E HP(n), u E Hr(n), where J-l = min(k,p - 1) and ß =
min(k,r - 1). Since Aw = e, we have w E H 2 (n) and I/w/l2m,n ~ ellellv; so /lel/L2 ~ C1hv llull r ,n.
12.9. /lu - Uh/lU ~ C1hvllu/lr,rl, where v = min(2,r) for linear or bilinear elements.
12.10. Using the Hermite basis functions and making appropriate changes (e.g., replace C(n) by CI(n)), the estimate (12.24) remains valid.
The VBVP is: find u E Hg(O, 1) such that I; (u"v" + k(x)uv) dx = 101 fv dx for all v E Hg(O,I). We obtain an error estimate from
lIu - uhl/2,n ~ Kllu - uhll2,rl = K (I:e /Iu - uhll~,nY/2 ~ Kh;-1
Solutions 461
(Ee lul~+l,oY/2 = Khk- 1 Iu lk+1,O, provided that Pk(O) c X c H 2 (0) and Hk+l(O) C C 1 (0) and u E Hk+l(n).
12.12. This follows as in Theorem 9. In particular k"ilv . v ~ kol"ilvl2 so
that a~(vh,vh) ~ koE~=lWtl"ilVh(~l)12 = kolvhltoe' since "ilvh E [P1(ne )J2 and a rule of order three is exact for quadratic functions.
normal boundary conditions, 274 normal derivative, 5 normed spaee, 18, 92, 95 norms
equivalent, 97 on IRn , 93, 103
null space, 135 numerical integration, 402
on square, 404 on triangle, 404 order, 402
one-to-one, 186 one-to-one operator, 138 onto, 135 open ball, 117 open mapping, 151 Open Mapping Theorem, 151 open neighborhood, 116 open set, 30, 116, 117