By Matthew P. Coleman

ISBN-10: 1439898472

ISBN-13: 9781439898475

Advent What are Partial Differential Equations? PDEs we will be able to Already remedy preliminary and Boundary stipulations Linear PDEs-Definitions Linear PDEs-The precept of Superposition Separation of Variables for Linear, Homogeneous PDEs Eigenvalue difficulties the massive 3 PDEsSecond-Order, Linear, Homogeneous PDEs with consistent CoefficientsThe warmth Equation and Diffusion The Wave Equation and the Vibrating String InitialRead more...

summary: advent What are Partial Differential Equations? PDEs we will Already clear up preliminary and Boundary stipulations Linear PDEs-Definitions Linear PDEs-The precept of Superposition Separation of Variables for Linear, Homogeneous PDEs Eigenvalue difficulties the massive 3 PDEsSecond-Order, Linear, Homogeneous PDEs with consistent CoefficientsThe warmth Equation and Diffusion The Wave Equation and the Vibrating String preliminary and Boundary stipulations for the warmth and Wave EquationsLaplace's Equation-The power Equation utilizing Separation of Variables to unravel the massive 3 PDEs Fourier sequence creation

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**Extra resources for An Introduction to Partial Differential Equations with MATLAB, Second Edition**

**Sample text**

Find all product solutions of the wave equation utt = uxx which also satisfy the boundary conditions u(0, t) = u(1, t) = 0. ) 19. a) Show that the ODE y + λy = 0 has nontrivial periodic solutions of period L, that is, which satisfy y(x + L) = y(x) if λ = λn = 2πn 2 L for all x , n = 0, 1, 2, . . b) Show that the statement actually is an if and only if. ) Hence, show that the only solutions of y + λy = 0 of period 2π are the functions y0 = 1, yn = c1 cos nx + c2 sin nx, n = 1, 2, . . c) Solve, instead, the eigenvalue problem y + λy = 0 y(−π) = y(π) y (−π) = y (π) and show that we get the same eigenvalues and eigenfunction as above.

Case 2: λ = 0 The general solution here is y = c1 x + c2 and, since y = c1 , the boundary conditions give y(0) = 0 = c2 y(1) + y (1) = 0 = 2c1 + c2 . Therefore, c1 = c2 = 0, so λ0 = 0 is not an eigenvalue. Case 3: λ > 0, λ = k 2 , k > 0. Here, as usual, y = c1 cos kx + c2 sin kx, so y = −c1 k sin kx + c2 k cos x. Then, y(0) = 0 = c1 y(1) = 0 = c2 (sin k + k cos k). This system has only the solution c1 = c2 = 0 unless k is such that sin k + k cos k = 0. Therefore, the eigenvalues correspond to those values of k satisfying −k = tan k.

Case 2: λ = 0 In this case, we have the repeated root r = 0, giving us the linearly independent solutions x0 and x0 ln x. So the general solution is y = c1 + c2 ln x. Then, y(1) = 0 = c1 y(e) = 0 = c1 + c2 so, again, c1 = c2 = 0, and λ = 0 is not an eigenvalue. Case 3: λ < 0, λ = −k 2 , k > 0 Here we have the roots r = ±ik and corresponding linearly independent solutions cos(k ln x) and sin(k ln x). The general solution is y = c1 cos(k ln x) + c2 sin(k ln x). Introduction 31 Applying the boundary conditions, we have y(1) = 0 = c1 ⇒ c1 = 0 y(e) = 0 = c2 sin k and the latter equation forces c2 = 0, except for those values of k satisfying sin k = 0; that is, c2 is arbitrary when k = π, 2π, 3π, .

### An Introduction to Partial Differential Equations with MATLAB, Second Edition by Matthew P. Coleman

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