- 10.6.1: In each of 1 through 8, find the steady state solution of the heat ...
- 10.6.2: In each of 1 through 8, find the steady state solution of the heat ...
- 10.6.3: In each of 1 through 8, find the steady state solution of the heat ...
- 10.6.4: In each of 1 through 8, find the steady state solution of the heat ...
- 10.6.5: In each of 1 through 8, find the steady state solution of the heat ...
- 10.6.6: In each of 1 through 8, find the steady state solution of the heat ...
- 10.6.7: In each of 1 through 8, find the steady state solution of the heat ...
- 10.6.8: In each of 1 through 8, find the steady state solution of the heat ...
- 10.6.9: Let an aluminum rod of length 20 cm be initially at the uniform tem...
- 10.6.10: (a) Let the ends of a copper rod 100 cm long be maintained at 0C. S...
- 10.6.11: Consider a rod of length 30 for which 2 = 1. Suppose the initial te...
- 10.6.12: Consider a uniform rod of length L with an initial temperature give...
- 10.6.13: Consider a bar of length 40 cm whose initial temperature is given b...
- 10.6.14: Consider a bar 30 cm long that is made of a material for which 2 = ...
- 10.6.15: Consider a uniform bar of length L having an initial temperature di...
- 10.6.16: In the bar of 15, suppose that L = 30, 2 = 1, and the initial tempe...
- 10.6.17: Suppose that the conditions are as in 15 and 16 except that the bou...
- 10.6.18: Consider the problemX+ X = 0, X(0) = 0, X(L) = 0. (i)Let = 2, where...
- 10.6.19: The right end of a bar of length a with thermal conductivity 1 and ...
- 10.6.20: Consider the problem2uxx = ut, 0 < x < L, t > 0;u(0, t) = 0, ux(L, ...
- 10.6.21: 21 through 23 deal with this kind of problem.Write u(x, t) = v(x) +...
- 10.6.22: 21 through 23 deal with this kind of problem.(a) Suppose that 2 = 1...
- 10.6.23: 21 through 23 deal with this kind of problem.(a) Let 2 = 1 and s(x)...
Solutions for Chapter 10.6: Other Heat Conduction Problems
Full solutions for Elementary Differential Equations and Boundary Value Problems | 10th Edition
peA) = det(A - AI) has peA) = zero matrix.
Circulant matrix C.
Constant diagonals wrap around as in cyclic shift S. Every circulant is Col + CIS + ... + Cn_lSn - l . Cx = convolution c * x. Eigenvectors in F.
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.
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!).
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.
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.
lA-II = l/lAI and IATI = IAI.
The big formula for det(A) has a sum of n! terms, the cofactor formula uses determinants of size n - 1, volume of box = I det( A) I.
= 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.
Normal equation AT Ax = ATb.
Gives the least squares solution to Ax = b if A has full rank n (independent columns). The equation says that (columns of A)·(b - Ax) = o.
Particular solution x p.
Any solution to Ax = b; often x p has free variables = o.
The diagonal entry (first nonzero) at the time when a row is used in elimination.
Rank one matrix A = uvT f=. O.
Column and row spaces = lines cu and cv.
Semidefinite matrix A.
(Positive) semidefinite: all x T Ax > 0, all A > 0; A = any RT R.
Symmetric matrix A.
The transpose is AT = A, and aU = a ji. A-I is also symmetric.
Constant down each diagonal = time-invariant (shift-invariant) filter.
Tridiagonal matrix T: tij = 0 if Ii - j I > 1.
T- 1 has rank 1 above and below diagonal.
v + w = (VI + WI, ... , Vn + Wn ) = diagonal of parallelogram.