Explanation Calculation: Consider the given telegraph problem, ∂ 2 u ∂ t 2 + ∂ u ∂ t + u = α 2 ∂ 2 u ∂ x 2 , 0 < x < L , t > 0 , u ( 0 , t ) = u ( L , t ) = 0 , t > 0 , u ( x , 0 ) = f ( x ) , 0 < x < L , ∂ u ∂ t ( x , 0 ) = 0 , 0 < x < L . Let, u be the function of x and t . u ( x , t ) = X ( x ) T ( t ) Partial differentiate on the both side of the equation u ( x , t ) = X ( x ) T ( t ) with respect to t , ∂ u ( x , t ) ∂ t = ∂ ∂ t ( X ( x ) T ( t ) ) = X ( x ) T ' ( t ) Again partial differentiate with respect to t , ∂ 2 u ( x , t ) ∂ t 2 = X ( x ) T " ( t ) Double Partial differentiate the above equation with respect to x ∂ 2 u ( x , t ) ∂ x 2 = X " ( x ) T " ( t ) Substitute the value of ∂ 2 u ∂ t 2 , ∂ u ∂ t , u and ∂ 2 u ∂ x 2 in the given equation ∂ 2 u ∂ t 2 + ∂ u ∂ t + u = α 2 ∂ 2 u ∂ x 2 . ∂ 2 u ∂ t 2 + ∂ u ∂ t + u = α 2 ∂ 2 u ∂ x 2 X ( x ) T " ( t ) + X ( x ) T ' ( t ) + X ( x ) T ( t ) = α 2 X " ( x ) T ( t ) X ( x ) [ T " ( t ) + T ' ( t ) + T ( t ) ] = α 2 X " ( x ) T ( t ) T " ( t ) + T ' ( t ) + T ( t ) α 2 T ( t ) = X " ( x ) X ( x ) The ratio of double derivative of the function to the function itself gives a constant and let the constant be − λ therefore, T " ( t ) + T ' ( t ) + T ( t ) α 2 T ( t ) = − λ and X " ( x ) X ( x ) = − λ Consider the differential equation in x . X " ( x ) X ( x ) = − λ X " ( x ) = − λ X ( x ) X " ( x ) + λ X ( x ) = 0 The auxiliary equation of the differential equation in x is given by, r 2 + λ = 0 The roots of the equation is given by r 2 = − λ r = ± i λ The solution of the function X ( x ) is given by, X ( x ) = c 1 e i λ x + c 2 e − i λ x = c 1 ( cos λ x + i sin λ x ) + c 2 ( cos λ x − i sin λ x ) = ( c 1 e 1 m + c 2 e − 1 m ) cos λ x + i ( c 1 − c 2 ) sin λ x = C 1 cos λ x + C 2 sin λ x Therefore the solution of the function is X ( x ) = C 1 cos λ x + C 2 sin λ x Use the property u ( 0 , t ) = u ( L , t ) = 0 and substitute 0 for x in the solution of the function. X ( x ) = C 1 cos λ x + C 2 sin λ x X ( 0 ) = C 1 cos λ ( 0 ) + C 2 sin λ ( 0 ) 0 = C 1 + 0 C 1 = 0 Substitute L for x in the solution of the function, X ( L ) = C 1 cos λ L + C 2 sin λ L 0 = ( 0 ) cos λ L + C 2 sin λ L C 2 sin λ L = 0 Since, C 2 ≠ 0 so, sin λ L = 0 λ L = n π λ = n π L Therefore the general solution of X ( x ) is given by X n ( x ) = C n sin ( n π x L ) C n is an arbitrary nonzero constant. The arbitrary nonzero constants a N is given by, a n = 2 L = ∫ 0 x f ( x ) sin ( n π x L ) d x Consider the second differential equation in t . T " ( t ) + T ' ( t ) + T ( t ) α 2 T ( t ) = − λ T " ( t ) + T ' ( t ) + T ( t ) = − α 2 T ( t ) λ T " ( t ) + T ' ( t ) + T ( t ) ( 1 + λ α 2 ) = 0 The auxiliary equation of the differential equation in x is given by, r 2 + r + 1 + λ α 2 = 0 r = − 1 ± − 3 − 4 λ α 2 2 = − 1 ± i 3 + 4 λ α 2 2 = − 1 2 + i 3 + 4 λ α 2 2 , − 1 2 − i 3 + 4 λ α 2 2 The roots of the equation is given by r 2 + r + 1 + λ α 2 = 0 r = − 1 ± − 3 − 4 λ α 2 2 = − 1 ± i 3 + 4 λ α 2 2 = − 1 2 + i 3 + 4 λ α 2 2 , − 1 2 − i 3 + 4 λ α 2 2 The solution of the function T ( t ) is given by, T ( t ) = c 1 e ( − 1 2 + i 3 + 4 λ α 2 2 ) t + c 2 e ( − 1 2 − i 3 + 4 λ α 2 2 ) t = c 1 e − 1 2 t ( cos ( 3 + 4 λ α 2 2 ) t + i sin ( 3 + 4 λ α 2 2 ) t ) + c 2 e − 1 2 t ( cos ( 3 + 4 λ α 2 2 ) t − i sin ( 3 + 4 λ α 2 2 ) t ) = ( c 1 + c 2 ) e − 1 2 t cos ( 3 + 4 λ α 2 2 ) + i ( c 1 − c 2 ) e − 1 2 t sin ( 3 + 4 λ α 2 2 ) = C 1 e − 1 2 t cos ( 3 + 4 λ α 2 2 ) + C 2 e − 1 2 t sin ( 3 + 4 λ α 2 2 ) u be the function of x and t . u ( x , t ) = X ( x ) T ( t ) Partial differentiate on the both side of the equation u ( x , t ) = X ( x ) T ( t ) with respect to t , ∂ u ( x , t ) ∂ t = X ( x ) ∂ T ( t ) ∂ t Use the property ∂ u ∂ t ( x , 0 ) = 0

Fundamentals of Differential Equations and Boundary Value Problems

7th Edition

ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Question

Chapter 10.6, Problem 11E

To determine

A formal solution of the telegraph problem if

∂2u∂t2+∂u∂t+u=α2∂2u∂x2, 0<x<L, t>0,u(0,t)=u(L,t)=0, t>0,u(x,0)=f(x), 0<x<L,∂u∂t(x,0)=0, 0<x<L.

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Chapter 10.6, Problem 10E

Chapter 10.6, Problem 12E

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Chapter 10 Solutions

Fundamentals of Differential Equations and Boundary Value Problems

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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A formal solution of the telegraph problem if ∂ 2 u ∂ t 2 + ∂ u ∂ t + u = α 2 ∂ 2 u ∂ x 2 , 0 < x < L , t > 0 , u ( 0 , t ) = u ( L , t ) = 0 , t > 0 , u ( x , 0 ) = f ( x ) , 0 < x < L , ∂ u ∂ t ( x , 0 ) = 0 , 0 < x < L .

Solution Summary: The author explains that the formal solution of the telegraph problem is u(x,t)=displaystyle undersetn=1

A formal solution of the telegraph problem if

∂2u∂t2+∂u∂t+u=α2∂2u∂x2, 0<x<L, t>0,u(0,t)=u(L,t)=0, t>0,u(x,0)=f(x), 0<x<L,∂u∂t(x,0)=0, 0<x<L.

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Chapter 10 Solutions

A formal solution of the telegraph problem if ∂ 2 u ∂ t 2 + ∂ u ∂ t + u = α 2 ∂ 2 u ∂ x 2 , 0 &lt; x &lt; L , t &gt; 0 , u ( 0 , t ) = u ( L , t ) = 0 , t &gt; 0 , u ( x , 0 ) = f ( x ) , 0 &lt; x &lt; L , ∂ u ∂ t ( x , 0 ) = 0 , 0 &lt; x &lt; L .

Solution Summary: The author explains that the formal solution of the telegraph problem is u(x,t)=displaystyle undersetn=1

A formal solution of the telegraph problem if ∂2u∂t2+∂u∂t+u=α2∂2u∂x2, 0<x<L, t>0,u(0,t)=u(L,t)=0, t>0,u(x,0)=f(x), 0<x<L,∂u∂t(x,0)=0, 0<x<L.

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Chapter 10 Solutions

A formal solution of the telegraph problem if ∂ 2 u ∂ t 2 + ∂ u ∂ t + u = α 2 ∂ 2 u ∂ x 2 , 0 < x < L , t > 0 , u ( 0 , t ) = u ( L , t ) = 0 , t > 0 , u ( x , 0 ) = f ( x ) , 0 < x < L , ∂ u ∂ t ( x , 0 ) = 0 , 0 < x < L .

A formal solution of the telegraph problem if

∂2u∂t2+∂u∂t+u=α2∂2u∂x2, 0<x<L, t>0,u(0,t)=u(L,t)=0, t>0,u(x,0)=f(x), 0<x<L,∂u∂t(x,0)=0, 0<x<L.