Explanation Calculation: Consider the following given differential equation: ∂ 2 u ∂ t 2 = α 2 ∂ 2 u ∂ x 2 , − ∞ < r < ∞ , t > 0 , ...... (1) For the given function f ( x ) = 0 and g ( x ) = cos x . Boundary condition: u ( x , 0 ) = f ( x ) , − ∞ < x < ∞ , ∂ u ∂ t ( x , 0 ) = g ( x ) , − ∞ < x < ∞ , Apply the change of variable as: ψ = x + α t and η = x − α t Here α is parameter, Differentiate the function ψ = x + α t and η = x − α t with respect to x , ∂ ψ ∂ x = 1 and ∂ η ∂ x = 1 ...... (2) Differentiate the function ψ = x + α t and η = x − α t with respect to t , ∂ ψ ∂ t = α and ∂ η ∂ t = − α ...... (3) Consider the function as function of ( ψ , η ) , u ( x ( ψ , η ) , t ( ψ , η ) ) Differentiate the function u ( x ( ψ , η ) , t ( ψ , η ) ) with respect to x and t , ∂ u ∂ x = ∂ u ∂ ψ ∂ ψ ∂ x + ∂ u ∂ η ∂ η ∂ x And ∂ u ∂ t = ∂ u ∂ ψ ∂ ψ ∂ t − ∂ u ∂ η ∂ η ∂ t ...... (4) Substitute 1 for ∂ ψ ∂ x , 1 for ∂ η ∂ x , α for ∂ ψ ∂ t , and − α for ∂ η ∂ t in the equation (4). ∂ u ∂ x = ∂ u ∂ ψ + ∂ u ∂ η and ∂ u ∂ t = α ( ∂ u ∂ ψ − ∂ u ∂ η ) ...... (5) Differentiate the equation (5) with respect to x and t , ∂ 2 u ∂ x 2 = ( ∂ ∂ ψ + ∂ ∂ η ) ( ∂ u ∂ ψ + ∂ u ∂ η ) = ∂ 2 u ∂ ψ 2 + ∂ 2 u ∂ η ∂ ψ + ∂ 2 u ∂ ψ ∂ η + ∂ 2 u ∂ η 2 = ∂ 2 u ∂ ψ 2 + 2 ∂ 2 u ∂ η ∂ ψ + ∂ 2 u ∂ η 2 and ∂ 2 u ∂ t 2 = α ( ∂ ∂ ψ − ∂ ∂ η ) ( ∂ u ∂ ψ − ∂ u ∂ η ) = ∂ 2 u ∂ ψ 2 − ∂ 2 u ∂ η ∂ ψ − ∂ 2 u ∂ ψ ∂ η + ∂ 2 u ∂ η 2 = ∂ 2 u ∂ ψ 2 − 2 ∂ 2 u ∂ η ∂ ψ + ∂ 2 u ∂ η 2 Substitute ∂ 2 u ∂ ψ 2 + 2 ∂ 2 u ∂ η ∂ ψ + ∂ 2 u ∂ η 2 for ∂ 2 u ∂ x 2 , α ( ∂ 2 u ∂ ψ 2 − 2 ∂ u ∂ η ∂ ψ + ∂ 2 u ∂ η 2 ) for ∂ 2 u ∂ t 2 in the equation (1). α 2 ( ∂ 2 u ∂ ψ 2 + 2 ∂ 2 u ∂ η ∂ ψ + ∂ 2 u ∂ η 2 ) = α 2 ( ∂ 2 u ∂ ψ 2 − 2 ∂ u ∂ η ∂ ψ + ∂ 2 u ∂ η 2 ) ( ∂ 2 u ∂ ψ 2 + 2 ∂ 2 u ∂ η ∂ ψ + ∂ 2 u ∂ η 2 ) = ( ∂ 2 u ∂ ψ 2 − 2 ∂ u ∂ η ∂ ψ + ∂ 2 u ∂ η 2 ) 2 ∂ 2 u ∂ η ∂ ψ = − 2 ∂ u ∂ η ∂ ψ ∂ u ∂ η ∂ ψ = 0 Integrate the differential equation ∂ u ∂ η ∂ ψ = 0 with respect to ψ , ∂ u ∂ η = b ( η ) Here, b ( η ) is arbitrary function of η , Again integrate the equation ∂ u ∂ η = b ( η ) with respect to η , u ( ψ , η ) = A ( ψ ) + B ( η ) Here, A ( ψ ) and B ( η ) = ∫ b ( η ) d η are arbitrary functions. Substitute x + α t for ψ , x − α t for η in the solution u ( ψ , η ) = A ( ψ ) + B ( η ) , u ( x , t ) = A ( x + α t ) + B ( x − α t )

Fundamentals Of Differential Equations And Boundary Value Problems Plus Mylab Math With Pearson Etext -- Title-specific Access Card Package (7th ... Fundamentals Of Differential Equations)

7th Edition

ISBN: 9780134768717

Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Publisher: PEARSON

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Chapter 10.6, Problem 14E

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To solve: The initial value problem ∂2u∂t2=α2∂2u∂x2,−∞<r<∞,t>0, with boundary condition u(x,0)=f(x),−∞<x<∞ and ∂u∂t(x,0)=g(x),−∞<x<∞ for the given function f(x)=x2 and g(x)=0.

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Chapter 10.6, Problem 13E

Chapter 10.6, Problem 15E

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Chapter 10 Solutions

Fundamentals Of Differential Equations And Boundary Value Problems Plus Mylab Math With Pearson Etext -- Title-specific Access Card Package (7th ... Fundamentals Of Differential Equations)

Ch. 10.2 - In Problems 1-8, determine all the solutions, if...Ch. 10.2 - Prob. 2E Ch. 10.2 - Prob. 3E Ch. 10.2 - Prob. 4E Ch. 10.2 - In Problems 1-8, determine all the solutions, if...Ch. 10.2 - Prob. 6E Ch. 10.2 - In Problems 1-8, determine all the solutions, if...Ch. 10.2 - In Problems 1-8, determine all the solutions, if...Ch. 10.2 - In Problems 9-14, find the values of eigenvalues...Ch. 10.2 - In Problems 9-14, find the values of eigenvalues...

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The initial value problem ∂ 2 u ∂ t 2 = α 2 ∂ 2 u ∂ x 2 , − ∞ < r < ∞ , t > 0 , with boundary condition u ( x , 0 ) = f ( x ) , − ∞ < x < ∞ and ∂ u ∂ t ( x , 0 ) = g ( x ) , − ∞ < x < ∞ for the given function f ( x ) = x 2 and g ( x ) = 0 .

Solution Summary: The author explains how to solve the initial value problem, partial 't2', with boundary condition, and the given function.

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To solve: The initial value problem ∂2u∂t2=α2∂2u∂x2,−∞<r<∞,t>0, with boundary condition u(x,0)=f(x),−∞<x<∞ and ∂u∂t(x,0)=g(x),−∞<x<∞ for the given function f(x)=x2 and g(x)=0.

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Chapter 10 Solutions

The initial value problem ∂ 2 u ∂ t 2 = α 2 ∂ 2 u ∂ x 2 , − ∞ &lt; r &lt; ∞ , t &gt; 0 , with boundary condition u ( x , 0 ) = f ( x ) , − ∞ &lt; x &lt; ∞ and ∂ u ∂ t ( x , 0 ) = g ( x ) , − ∞ &lt; x &lt; ∞ for the given function f ( x ) = x 2 and g ( x ) = 0 .

Solution Summary: The author explains how to solve the initial value problem, partial 't2', with boundary condition, and the given function.

Videos

To solve: The initial value problem ∂2u∂t2=α2∂2u∂x2,−∞<r<∞,t>0, with boundary condition u(x,0)=f(x),−∞<x<∞ and ∂u∂t(x,0)=g(x),−∞<x<∞ for the given function f(x)=x2 and g(x)=0.

Want to see the full answer?

Chapter 10 Solutions

The initial value problem ∂ 2 u ∂ t 2 = α 2 ∂ 2 u ∂ x 2 , − ∞ < r < ∞ , t > 0 , with boundary condition u ( x , 0 ) = f ( x ) , − ∞ < x < ∞ and ∂ u ∂ t ( x , 0 ) = g ( x ) , − ∞ < x < ∞ for the given function f ( x ) = x 2 and g ( x ) = 0 .