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Consider the following boundary value problem (if necessary, see Section 2.4.1): u t = k

Applied Partial Differential Equations with Fourier Series and Boundary Value Problems | 5th Edition | ISBN: 9780321797056 | Authors: Richard Haberman ISBN: 9780321797056 284

Solution for problem 2.3.7 Chapter 2.3

Applied Partial Differential Equations with Fourier Series and Boundary Value Problems | 5th Edition

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Applied Partial Differential Equations with Fourier Series and Boundary Value Problems | 5th Edition | ISBN: 9780321797056 | Authors: Richard Haberman

Applied Partial Differential Equations with Fourier Series and Boundary Value Problems | 5th Edition

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Problem 2.3.7

Consider the following boundary value problem (if necessary, see Section 2.4.1): u t = k 2u x2 with u x(0, t)=0, u x(L, t)=0, and u(x, 0) = f(x). (a) Give a one-sentence physical interpretation of this problem. (b) Solve by the method of separation of variables. First show that there are no separated solutions which exponentially grow in time. [Hint: The answer is u(x, t) = A0 + n=1 Ane nkt cos nx L . What is n?(c) Show that the initial condition, u(x, 0) = f(x), is satisfied if f(x) = A0 + n=1 An cos nx L . (d) Using Exercise 2.3.6, solve for A0 and An(n 1). (e) What happens to the temperature distribution as t ? Show that it approaches the steady-state temperature distribution (see Section 1.4).

Step-by-Step Solution:
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K-2 CCSS Strategies Add To: Result Unknown Level 1: Count All 5 + 4 = IIIII IIII Level 2: Counting On Say 5 and then, using your fingers to model, 6, 7, 8, 9 Level 3: Doubles Plus/Minus 5 + 4= 5+ 4= 1 + 4 +4 = 1 + 8 = 9 Add To: Change Unknown Level 1: N/A Level 2: Counting On 5 + = 9 Say 5, and then, using your fingers to model, 6, 7, 8, 9 Level 3: Rearrange 5 + = 9  9...

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Chapter 2.3, Problem 2.3.7 is Solved
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Textbook: Applied Partial Differential Equations with Fourier Series and Boundary Value Problems
Edition: 5
Author: Richard Haberman
ISBN: 9780321797056

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Consider the following boundary value problem (if necessary, see Section 2.4.1): u t = k

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