Explanation Given: The boundary-value problem is ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 , 0 < y < 1 , x > 0 , u ( 0 , y ) = u 0 , lim x → ∞ u ( x , y ) = 0 , ∂ u ∂ y | y = 0 = 0 and ∂ u ∂ y | y = 1 = − h u ( x , 1 ) , h > 0 , x > 0 . Calculation: Consider a function u as a function of ( x , y ) as follows: u ( x , y ) = X ( x ) Y ( y ) Separate variables X and Y in above expression. X ″ Y + X Y ″ = 0 The boundary conditions is u ( 0 , y ) = u 0 , lim x → ∞ u ( x , y ) = 0 , ∂ u ∂ y | y = 0 = 0 and ∂ u ∂ y | y = 1 = − h u ( x , 1 ) , h > 0 , x > 0 that corresponds to X ( x ) Y ′ ( 0 ) = 0 and X ( x ) Y ′ ( 1 ) = − h X ( x ) Y ( 1 ) . The value of Y ′ ( 0 ) is 0 and Y ′ ( 1 ) = − h Y ( 1 ) . As the homogeneous boundary conditions are in terms of Y , separate the differential equation as follows: Consider X ″ X = − Y ″ Y as α 2 . The equation X ″ X = α 2 is reduced as follows: X ″ X = α 2 X ″ X − α 2 = 0 X ″ − α 2 X X = 0 X ″ − α 2 X = 0 The characteristic equation of above expression is given below. m 2 − α 2 = 0 m 2 = α 2 m = ± α Therefore, the solution is X ( x ) = c 3 e − a x + c 4 e a x . Consider the equation − Y ″ ( y ) Y ( y ) as α 2 . Therefore, the equation is reduced as follows: − Y ″ ( y ) Y ( y ) = α 2 Y ″ ( y ) Y ( y ) + α 2 = 0 Y ″ ( y ) + α 2 Y ( y ) Y ( y ) = 0 Y ″ ( y ) + α 2 Y ( y ) = 0 The characteristic equation of above expression is given below. m 2 + α 2 = 0 m = − α 2 Therefore, the solution is given below. Y ( y ) = c 1 cos α y + c 2 sin α y Now, differentiate the expression Y ( y ) = c 1 cos α y + c 2 sin α y as follows: Y ′ ( y ) = − c 1 α sin α y Substitute the given boundary condition Y ′ ( 1 ) = − h Y ( 2 ) in the above expression. − c 1 α sin α ( 1 ) = − h c 1 cos α α sin α = h cos α cos α sin α = α h cot α = α h The graph of y = cot α and y = α h is shown below in Figure 1. From Figure 1, it is observed that roots of above graph are infinite. So, the Eigen value of the given problem is consecutive positive roots α n for n = 1 , 2 , 3 , … . The Eigen functions are Y n ( y ) = c 1 cos α n y . The condition lim x → ∞ u ( x , y ) = 0 is equivalent to lim x → ∞ X ( x ) = 0

6th Edition

ISBN: 9781284127003

Author: ZILL

Publisher: JONES+BARTLETT LEARNING,LLC (CC)

See similar textbooks

Concept explainers

Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

Videos

Question

Chapter 13.7, Problem 4E

To determine

The steady-state temperature u(x,t) for the boundary-value problem ∂2u∂x2+∂2u∂y2, 0<y<1, x>0, u(0,y)=u0, limx→∞u(x,y)=0, ∂u∂y|y=0=0 and ∂u∂y|y=1=−hu(x,1), h>0,x>0.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 13.7, Problem 3E

Chapter 13.7, Problem 5E

Students have asked these similar questions

Can you help me with this problem using linear recurrance: Find an explicit formula for the recurrence relation an = 2can−1 + 3c2an−2 where c not equal to 0 with initial conditions a0=4c and a1 = 0

Can you help me solved this problem using generalized combination:How many combinations are there to pick r objects from 2n objects numbered from 1to 2n when repetitions are allowed and at least one object of odd type does not appear?

Chapter 13 Solutions

EBK ADVANCED ENGINEERING MATHEMATICS

Ch. 13.1 - Prob. 1E Ch. 13.1 - Prob. 2E Ch. 13.1 - Prob. 3E Ch. 13.1 - Prob. 4E Ch. 13.1 - Prob. 5E Ch. 13.1 - Prob. 6E Ch. 13.1 - Prob. 7E Ch. 13.1 - Prob. 8E Ch. 13.1 - Prob. 9E Ch. 13.1 - Prob. 10E

Knowledge Booster

$Advanced Math$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

The steady-state temperature u ( x , t ) for the boundary-value problem ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 , 0 < y < 1 , x > 0 , u ( 0 , y ) = u 0 , lim x → ∞ u ( x , y ) = 0 , ∂ u ∂ y | y = 0 = 0 and ∂ u ∂ y | y = 1 = − h u ( x , 1 ) , h > 0 , x > 0 .

Solution Summary: The author explains the steady-state temperature u(x,t) for the boundary-value problem.

Concept explainers

Videos

The steady-state temperature u(x,t) for the boundary-value problem ∂2u∂x2+∂2u∂y2, 0<y<1, x>0, u(0,y)=u0, limx→∞u(x,y)=0, ∂u∂y|y=0=0 and ∂u∂y|y=1=−hu(x,1), h>0,x>0.

Want to see the full answer?

Chapter 13 Solutions

The steady-state temperature u ( x , t ) for the boundary-value problem ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 , 0 &lt; y &lt; 1 , x &gt; 0 , u ( 0 , y ) = u 0 , lim x → ∞ u ( x , y ) = 0 , ∂ u ∂ y | y = 0 = 0 and ∂ u ∂ y | y = 1 = − h u ( x , 1 ) , h &gt; 0 , x &gt; 0 .

Solution Summary: The author explains the steady-state temperature u(x,t) for the boundary-value problem.

Concept explainers

Videos

The steady-state temperature u(x,t) for the boundary-value problem ∂2u∂x2+∂2u∂y2, 0<y<1, x>0, u(0,y)=u0, limx→∞u(x,y)=0, ∂u∂y|y=0=0 and ∂u∂y|y=1=−hu(x,1), h>0,x>0.

Want to see the full answer?

Chapter 13 Solutions

The steady-state temperature u ( x , t ) for the boundary-value problem ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 , 0 < y < 1 , x > 0 , u ( 0 , y ) = u 0 , lim x → ∞ u ( x , y ) = 0 , ∂ u ∂ y | y = 0 = 0 and ∂ u ∂ y | y = 1 = − h u ( x , 1 ) , h > 0 , x > 0 .