- 188.8.131.52.1: *(a) Assume that u(r, , t) = a(t)(r, ), where (r, ) are the eigenfu...
- 184.108.40.206.2: The solution of (8.6.6), d2bn dy2 n L 2 bn = qn(y), subject to bn(0...
- 220.127.116.11.3: Solve (using two-dimensional eigenfunctions) 2u = Q(r, ) inside a c...
- 18.104.22.168.4: Solve Exercise 8.6.3 using one-dimensional eigenfunctions.
- 22.214.171.124.5: Consider 2u = Q(x, y) inside an unspecified region with u = 0 on th...
- 126.96.36.199.6: Solve the following example of Poissons equation: 2u = e 2y sin x s...
- 188.8.131.52.7: Solve 2u = Q(x, y, z) inside a rectangular box (0 < x < L, 0 < y < ...
- 184.108.40.206.8: Solve 2u = Q(r, , z) inside a circular cylinder (0 < r < a, 0 << 2,...
- 220.127.116.11.9: On a rectangle (0 < x < L, 0 <y (a) Show that a solution exists onl...
- 18.104.22.168.10: Reconsider Exercise 8.6.9 for an arbitrary two-dimensional region.
Solutions for Chapter 8.6: Nonhomogeneous Problems
Full solutions for Applied Partial Differential Equations with Fourier Series and Boundary Value Problems | 5th Edition
Characteristic equation det(A - AI) = O.
The n roots are the eigenvalues of A.
cond(A) = c(A) = IIAIlIIA-III = amaxlamin. In Ax = b, the relative change Ilox III Ilx II is less than cond(A) times the relative change Ilob III lib II· Condition numbers measure the sensitivity of the output to change in the input.
Conjugate Gradient Method.
A sequence of steps (end of Chapter 9) to solve positive definite Ax = b by minimizing !x T Ax - x Tb over growing Krylov subspaces.
When random variables Xi have mean = average value = 0, their covariances "'£ ij are the averages of XiX j. With means Xi, the matrix :E = mean of (x - x) (x - x) T is positive (semi)definite; :E is diagonal if the Xi are independent.
S. Permutation with S21 = 1, S32 = 1, ... , finally SIn = 1. Its eigenvalues are the nth roots e2lrik/n of 1; eigenvectors are columns of the Fourier matrix F.
Eigenvalue A and eigenvector x.
Ax = AX with x#-O so det(A - AI) = o.
Fourier matrix F.
Entries Fjk = e21Cijk/n give orthogonal columns FT F = nI. Then y = Fe is the (inverse) Discrete Fourier Transform Y j = L cke21Cijk/n.
Independent vectors VI, .. " vk.
No combination cl VI + ... + qVk = zero vector unless all ci = O. If the v's are the columns of A, the only solution to Ax = 0 is x = o.
Current Law: net current (in minus out) is zero at each node. Voltage Law: Potential differences (voltage drops) add to zero around any closed loop.
Linearly dependent VI, ... , Vn.
A combination other than all Ci = 0 gives L Ci Vi = O.
Ln = 2,J, 3, 4, ... satisfy Ln = L n- l +Ln- 2 = A1 +A~, with AI, A2 = (1 ± -/5)/2 from the Fibonacci matrix U~]' Compare Lo = 2 with Fo = O.
Markov matrix M.
All mij > 0 and each column sum is 1. Largest eigenvalue A = 1. If mij > 0, the columns of Mk approach the steady state eigenvector M s = s > O.
= Xl (column 1) + ... + xn(column n) = combination of columns.
IIA II. The ".e 2 norm" of A is the maximum ratio II Ax II/l1x II = O"max· Then II Ax II < IIAllllxll and IIABII < IIAIIIIBII and IIA + BII < IIAII + IIBII. Frobenius norm IIAII} = L La~. The.e 1 and.e oo norms are largest column and row sums of laij I.
Orthonormal vectors q 1 , ... , q n·
Dot products are q T q j = 0 if i =1= j and q T q i = 1. The matrix Q with these orthonormal columns has Q T Q = I. If m = n then Q T = Q -1 and q 1 ' ... , q n is an orthonormal basis for Rn : every v = L (v T q j )q j •
Ps = pascal(n) = the symmetric matrix with binomial entries (i1~;2). Ps = PL Pu all contain Pascal's triangle with det = 1 (see Pascal in the index).
Permutation matrix P.
There are n! orders of 1, ... , n. The n! P 's have the rows of I in those orders. P A puts the rows of A in the same order. P is even or odd (det P = 1 or -1) based on the number of row exchanges to reach I.
Iv·wl < IIvll IIwll.Then IvTAwl2 < (vT Av)(wT Aw) for pos def A.
Standard basis for Rn.
Columns of n by n identity matrix (written i ,j ,k in R3).
Sum V + W of subs paces.
Space of all (v in V) + (w in W). Direct sum: V n W = to}.