 10.4.10.4.1: Using Greens formula, show that F d2f dx2 = 2F() + eix 2 df dx if .
 10.4.10.4.2: For the heat equation, u(x, t) is given by (10.4.1). Show that u 0 ...
 10.4.10.4.3: (a) Solve the diffusion equation with convection: u t = k 2u x2 + c...
 10.4.10.4.4: (a) Solve u t = k 2u x2 u,
 10.4.10.4.5: (a) Solve u t = k 2u x2 u, (b) Determine U. *(c) Solve for u(x, t) ...
 10.4.10.4.6: The Airy function Ai(x) is the unique solution of d2y dx2 xy = 0 th...
 10.4.10.4.7: (a) Solve the linearized KortewegdeVries equation u t = k 3u x3 , 0.
 10.4.10.4.8: Solve 2u x2 + 2u y2 = 0, 0
 10.4.10.4.9: Solve 2u x2 + 2u y2 = 0, y > 0
 10.4.10.4.10: Solve 2u t2 = c2 2u x2 , 0 u t (x, 0) = 0. (Hint: If necessary, see...
 10.4.10.4.11: Derive an expression for the Fourier transform of the product f(x)g(x)
 10.4.10.4.12: Solve the heat equation, 0 subject to the condition u(x, 0) = f(x).
Solutions for Chapter 10.4: 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
ISBN: 9780321797056
Solutions for Chapter 10.4: Infinite Domain Problems: Fourier Transform Solutions of Partial Differential Equations
Get Full SolutionsSince 12 problems in chapter 10.4: Infinite Domain Problems: Fourier Transform Solutions of Partial Differential Equations have been answered, more than 7775 students have viewed full stepbystep solutions from this chapter. This textbook survival guide was created for the textbook: Applied Partial Differential Equations with Fourier Series and Boundary Value Problems, edition: 5. Chapter 10.4: Infinite Domain Problems: Fourier Transform Solutions of Partial Differential Equations includes 12 full stepbystep solutions. Applied Partial Differential Equations with Fourier Series and Boundary Value Problems was written by and is associated to the ISBN: 9780321797056. This expansive textbook survival guide covers the following chapters and their solutions.

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.

Companion 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 nl  An).

Condition number
cond(A) = c(A) = IIAIlIIAIII = 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.

Ellipse (or ellipsoid) x T Ax = 1.
A must be positive definite; the axes of the ellipse are eigenvectors of A, with lengths 1/.JI. (For IIx II = 1 the vectors y = Ax lie on the ellipse IIA1 yll2 = Y T(AAT)1 Y = 1 displayed by eigshow; axis lengths ad

Fibonacci numbers
0,1,1,2,3,5, ... satisfy Fn = Fnl + Fn 2 = (A7 A~)I()q A2). Growth rate Al = (1 + .J5) 12 is the largest eigenvalue of the Fibonacci matrix [ } A].

Length II x II.
Square root of x T x (Pythagoras in n dimensions).

Linear transformation T.
Each vector V in the input space transforms to T (v) in the output space, and linearity requires T(cv + dw) = c T(v) + d T(w). Examples: Matrix multiplication A v, differentiation and integration in function space.

Lucas numbers
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.

Minimal polynomial of A.
The lowest degree polynomial with meA) = zero matrix. This is peA) = det(A  AI) if no eigenvalues are repeated; always meA) divides peA).

Plane (or hyperplane) in Rn.
Vectors x with aT x = O. Plane is perpendicular to a =1= O.

Positive definite matrix A.
Symmetric matrix with positive eigenvalues and positive pivots. Definition: x T Ax > 0 unless x = O. Then A = LDLT with diag(D» O.

Reflection matrix (Householder) Q = I 2uuT.
Unit vector u is reflected to Qu = u. All x intheplanemirroruTx = o have Qx = x. Notice QT = Q1 = Q.

Row picture of Ax = b.
Each equation gives a plane in Rn; the planes intersect at x.

Schwarz inequality
Iv·wl < IIvll IIwll.Then IvTAwl2 < (vT Av)(wT Aw) for pos def A.

Singular Value Decomposition
(SVD) A = U:E VT = (orthogonal) ( diag)( orthogonal) First r columns of U and V are orthonormal bases of C (A) and C (AT), AVi = O'iUi with singular value O'i > O. Last columns are orthonormal bases of nullspaces.

Standard basis for Rn.
Columns of n by n identity matrix (written i ,j ,k in R3).

Transpose matrix AT.
Entries AL = Ajj. AT is n by In, AT A is square, symmetric, positive semidefinite. The transposes of AB and AI are BT AT and (AT)I.

Unitary matrix UH = U T = UI.
Orthonormal columns (complex analog of Q).

Vector v in Rn.
Sequence of n real numbers v = (VI, ... , Vn) = point in Rn.