- 10.5.10.5.1: Consider F() = e, > 0 ( 0). (a) Derive the inverse Fourier sine tra...
- 10.5.10.5.2: Consider f(x) = ex,> 0 (x 0). (a) Derive the Fourier sine transform...
- 10.5.10.5.3: Derive either the Fourier cosine transform of ex2 or the Fourier si...
- 10.5.10.5.4: (a) Derive (10.5.26) using Greens formula. (b) Do the same for (10....
- 10.5.10.5.5: (a) Show that the Fourier sine transform of f(x) is an odd function...
- 10.5.10.5.6: There is an interesting convolution-type theorem for Fourier sine t...
- 10.5.10.5.7: Derive the following; If a Fourier cosine transform in x, H(), is t...
- 10.5.10.5.8: Solve (10.5.1)(10.5.3) using the convolution theorem of Exercise 10...
- 10.5.10.5.9: Solve (10.5.1)(10.5.3) using the convolution theorem of Exercise 10...
- 10.5.10.5.10: Determine the inverse cosine transform of e. (Hint: Use differentia...
- 10.5.10.5.11: Consider u t = k 2u x2 , x > 0 t > 0 u(0, t)=1 u(x, 0) = f(x). (a) ...
- 10.5.10.5.12: Solve u t = k 2u x2 (x > 0) u x(0, t)=0 u(x, 0) = f(x).
- 10.5.10.5.13: Solve (10.5.28)(10.5.30) by solving (10.5.32).
- 10.5.10.5.14: Consider u t = k 2u x2 v0 u x (x > 0) u(0, t)=0 u(x, 0) = f(x). (a)...
- 10.5.10.5.15: Solve 2u x2 + 2u y2 = 0, 0
- 10.5.10.5.16: Solve 2u x2 + 2u y2 = 0, 0
- 10.5.10.5.17: The effect of periodic surface heating (either daily or seasonal) o...
- 10.5.10.5.18: Reconsider Exercise 10.5.17. Determine u(x, t) exactly. (Hint: See ...
- 10.5.10.5.19: (a) Determine a particular solution of Exercise 10.5.17, satisfying...
- 10.5.10.5.20: Solve the heat equation, 0
Solutions for Chapter 10.5: Infinite Domain Problems: Fourier Transform Solutions of Partial Differential Equations
Full solutions for Applied Partial Differential Equations with Fourier Series and Boundary Value Problems | 5th Edition
Solutions for Chapter 10.5: Infinite Domain Problems: Fourier Transform Solutions of Partial Differential EquationsGet Full Solutions
peA) = det(A - AI) has peA) = zero matrix.
Put CI, ... ,Cn in row n and put n - 1 ones just above the main diagonal. Then det(A - AI) = ±(CI + c2A + C3A 2 + .•. + cnA n-l - An).
Diagonal matrix D.
dij = 0 if i #- j. Block-diagonal: zero outside square blocks Du.
0,1,1,2,3,5, ... satisfy Fn = Fn-l + Fn- 2 = (A7 -A~)I()q -A2). Growth rate Al = (1 + .J5) 12 is the largest eigenvalue of the Fibonacci matrix [ } A].
Free variable Xi.
Column i has no pivot in elimination. We can give the n - r free variables any values, then Ax = b determines the r pivot variables (if solvable!).
The nullspace N (A) and row space C (AT) are orthogonal complements in Rn(perpendicular from Ax = 0 with dimensions rand n - r). Applied to AT, the column space C(A) is the orthogonal complement of N(AT) in Rm.
Invert A by row operations on [A I] to reach [I A-I].
Gram-Schmidt orthogonalization A = QR.
Independent columns in A, orthonormal columns in Q. Each column q j of Q is a combination of the first j columns of A (and conversely, so R is upper triangular). Convention: diag(R) > o.
Hankel matrix H.
Constant along each antidiagonal; hij depends on i + j.
Hypercube matrix pl.
Row n + 1 counts corners, edges, faces, ... of a cube in Rn.
Kronecker product (tensor product) A ® B.
Blocks aij B, eigenvalues Ap(A)Aq(B).
Least squares solution X.
The vector x that minimizes the error lie 112 solves AT Ax = ATb. Then e = b - Ax is orthogonal to all columns of A.
The pivot row j is multiplied by eij and subtracted from row i to eliminate the i, j entry: eij = (entry to eliminate) / (jth pivot).
Pivot columns of A.
Columns that contain pivots after row reduction. These are not combinations of earlier columns. The pivot columns are a basis for the column space.
Projection matrix P onto subspace S.
Projection p = P b is the closest point to b in S, error e = b - Pb is perpendicularto S. p 2 = P = pT, eigenvalues are 1 or 0, eigenvectors are in S or S...L. If columns of A = basis for S then P = A (AT A) -1 AT.
Rank r (A)
= number of pivots = dimension of column space = dimension of row space.
Rayleigh quotient q (x) = X T Ax I x T x for symmetric A: Amin < q (x) < Amax.
Those extremes are reached at the eigenvectors x for Amin(A) and Amax(A).
Skew-symmetric matrix K.
The transpose is -K, since Kij = -Kji. Eigenvalues are pure imaginary, eigenvectors are orthogonal, eKt is an orthogonal matrix.
Solvable system Ax = b.
The right side b is in the column space of A.
Transpose matrix AT.
Entries AL = Ajj. AT is n by In, AT A is square, symmetric, positive semidefinite. The transposes of AB and A-I are BT AT and (AT)-I.