Explanation Given: The boundary conditions are as follows: ∂ 2 u ∂ x 2 − u = ∂ u ∂ t 0 < x < 1 , t > 0 u ( 1 , t ) = 0 ∂ u ∂ x | x = 0 = 0 u ( x , 0 ) = 1 − x 2 Calculation: The boundary conditions are given as follows: ∂ 2 u ∂ x 2 − u = ∂ u ∂ t 0 < x < 1 , t > 0 u ( 1 , t ) = 0 ∂ u ∂ x | x = 0 = 0 u ( x , 0 ) = 1 − x 2 Consider the product solution as follows: u ( x , t ) = X ( x ) T ( t ) … … ( 1 ) Substitute u = X T in ∂ 2 u ∂ x 2 − u = ∂ u ∂ t . ∂ 2 ( X T ) ∂ x 2 − ( X T ) = ∂ ( X T ) ∂ t X ″ T − X T = X T ′ X ″ T = X T ′ + X T X ″ T = X ( T ′ + T ) Divide the above equation by X T . X ″ T X T = X ( T ′ + T ) X T X ″ X = T ′ + T T Separate the variables with the help of separation constant and take the separation constant λ as follows: X ″ X = T ′ + T T = − λ The separated equations are written as follows: X ″ + λ X = 0 … … ( 2 ) T ′ + ( 1 + λ ) T = 0 … … ( 3 ) Substitute 1 for x in equation ( 1 ) . u ( 1 , t ) = X ( 1 ) T ( t ) X ( 1 ) T ( t ) = 0 ( u ( 1 , t ) = 0 ) X ( 1 ) = 0 Differentiate equation ( 1 ) with respect to x . u ′ x ( x , t ) = X ′ ( x ) T ( t ) Substitute 0 for x in the above equation. u ′ x ( 0 , t ) = X ′ ( 0 ) T ( t ) X ′ ( 0 ) T ( t ) = 0 ( u ′ x ( 0 , t ) = 0 ) X ′ ( 0 ) = 0 There are three cases for the value of λ . First case is λ = 0 . Substitute 0 for λ in equation ( 2 ) . X ″ = 0 The general solution of the equation is as follows: X ( x ) = c 1 + c 2 x … … ( 4 ) Substitute 1 for x in equation ( 4 ) . X ( 1 ) = c 1 + c 2 c 1 + c 2 = 0 ( X ( 1 ) = 0 ) c 1 = − c 2 Substitute c 1 = − c 2 in equation ( 4 ) . X ( x ) = − c 2 + c 2 x = c 2 ( x − 1 ) Differentiate the above equation with respect to x . X ′ ( x ) = c 2 Substitute 0 for x in the above equation. X ′ ( 0 ) = c 2 c 2 = 0 ( X ′ ( 0 ) = 0 ) Since, the value of c 1 and c 2 is 0 so, for λ = 0 the non-trivial solution does not exist. Second case is λ = − α 2 ( λ < 0 ) . Substitute λ = − α 2 in equation ( 2 ) . X ″ − α 2 X = 0 The general solution of the equation is as follows: X ( x ) = c 3 cos h ( α x ) + c 4 sinh ( α x ) … … ( 5 ) Substitute 1 for x in equation ( 5 ) . X ( 1 ) = c 3 cos h ( α ) + c 4 sinh ( α ) c 3 cos h ( α ) + c 4 sinh ( α ) = 0 ( X ( 1 ) = 0 ) c 3 = − c 4 sinh ( α ) cos h ( α ) Substitute the value of c 3 in equation ( 5 ) . X ( x ) = − ( c 4 sinh ( α ) cos h ( α ) ) cos h ( α x ) + c 4 sinh ( α x ) = c 4 ( sinh ( α x ) − sinh ( α ) cos h ( α x ) cos h ( α ) ) Differentiate the above equation with respect to x . X ′ ( x ) = c 4 ( α cosh ( α x ) − α sinh ( α ) sinh ( α x ) cos h ( α ) ) Substitute 1 for x in the above equation. X ′ ( 1 ) = c 4 ( α cosh ( α ) − α sinh ( α ) sinh ( α ) cos h ( α ) ) α c 4 ( ( cosh ( α ) ) 2 − ( sinh ( α ) ) 2 cos h ( α ) ) = 0 ( X ′ ( 1 ) = 0 ) α c 4 ( 1 cos h ( α ) ) = 0 c 4 = 0 Substitute the value of c 4 in c 3 = − c 4 sinh ( α ) cos h ( α ) . c 3 = − ( 0 ) ⋅ sinh ( α ) cos h ( α ) c 3 = 0 Since, the value of c 3 and c 4 is 0 so for λ = − α 2 the non-trivial solution does not exist. The third case is λ = α 2 ( λ > 0 ) . Substitute λ = α 2 in equation ( 2 ) . X ″ + α 2 X = 0 The general solution of the equation is as follows: X ( x ) = c 5 cos ( α x ) + c 6 sin ( α x ) … … ( 6 ) Substitute 1 for x in equation ( 6 ) . X ( 1 ) = c 5 cos ( α ) + c 6 sin ( α ) c 5 cos ( α ) + c 6 sin ( α ) = 0 ( X ( 1 ) = 0 ) c 5 = − c 6 sin ( α ) cos ( α ) Substitute the value of c 5 in equation ( 6 ) . X ( x ) = − c 6 sin ( α ) cos ( α ) cos ( α x ) + c 6 sin ( α x ) = c 6 ( sin ( α x ) − c 6 sin ( α ) cos ( α ) cos ( α x ) ) Differentiate the above equation with respect to x

Differential Equations with Boundary-Value Problems (MindTap Course List)

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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 12.7, Problem 8E

To determine

The solution u(x,t) of the differential equation ∂2u∂x2−u=∂u∂t such that u(1,t)=0, ∂u∂x|x=0=0 and u(x,0)=1−x2.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 12.7, Problem 7E

Chapter 12.7, Problem 9E

Students have asked these similar questions

H.W: Find f,, f,and f; if f(x,y,z) = x- \y² + z²

3. Find all functions from X = {a,b} to Y = {p, r, t}.

Example: Calculate the value of (aw/@x) at the point (x, y, z) = (2,-1,1) if:- w = x2 + y2 + z? z3 – xy + yz +y3 = 0 Assuming that x and y are the independent variables.

Chapter 12 Solutions

Differential Equations with Boundary-Value Problems (MindTap Course List)

Knowledge Booster

Learn more about

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

The solution u ( x , t ) of the differential equation ∂ 2 u ∂ x 2 − u = ∂ u ∂ t such that u ( 1 , t ) = 0 , ∂ u ∂ x | x = 0 = 0 and u ( x , 0 ) = 1 − x 2 .

Solution Summary: The author calculates the boundary conditions of the differential equation partial.

Concept explainers

Videos

The solution u(x,t) of the differential equation ∂2u∂x2−u=∂u∂t such that u(1,t)=0, ∂u∂x|x=0=0 and u(x,0)=1−x2.

Want to see the full answer?

Chapter 12 Solutions