Explanation Given: The boundary value problem is ∂ u ∂ t − ∂ 2 u ∂ x 2 = e − | x | , − ∞ < x < ∞ , t > 0 and the boundary condition is u ( x , 0 ) = 0 . Calculation: The given boundary value problem is, ∂ u ∂ t − ∂ 2 u ∂ x 2 = e − | x | . (1) Take Fourier transform of both sides of the above equation, F { ∂ u ∂ t } − F { ∂ 2 u ∂ x 2 } = F { e − | x | } The above equation becomes, d U d t + α 2 U ( α , t ) = F { e − | x | } . (2) The value of F { e − | x | } is, F { e − | x | } = ∫ − ∞ ∞ e − | x | e i α x d x By Euler’s formulae e i α = cos α + i sin α and substitute in the above equation, F { e − | x | } = ∫ − ∞ ∞ e − | x | ( cos α x + i sin α x ) d x = ∫ − ∞ ∞ e − | x | cos α x d x + i ∫ − ∞ ∞ e − | x | sin α x d x = ∫ − ∞ ∞ e − | x | cos α x d x + 0 = ∫ − ∞ ∞ e − | x | cos α x d x The integration function of the above equation is odd function of x , F { e − | x | } = 2 ∫ 0 ∞ e − | x | cos α x d x So, the equation is expressed as, F { e − | x | } = 1 1 + α 2 Substitute the value of F { e − | x | } in equation (2), d U d t + α 2 U ( α , t ) = 1 1 + α 2 Apply Fourier cosine transform then the particular solution of the above equation is, U ( α , t ) = 1 α 2 ( 1 + α 2 ) − c e − α 2 t

Differential Equations with Boundary-Value Problems

9th Edition

ISBN: 9781337632515

Author: Dennis G. Zill

Publisher: Cengage Learning US

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Chapter 14 Solutions

Differential Equations with Boundary-Value Problems

Ch. 14.1 - (a) Show that erf(t)=10ted. (b) Use part (a), the...Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Prob. 5E Ch. 14.1 - Prob. 6E Ch. 14.1 - Prob. 7E Ch. 14.1 - Prob. 8E Ch. 14.1 - Prob. 9E Ch. 14.1 - Use the third and fifth entries in Table 14.1.1 to...

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Chapter 14 Solutions