Differential Equations with Boundary-Value Problems (MindTap Course List)
Differential Equations with Boundary-Value Problems (MindTap Course List)
9th Edition
ISBN: 9781305965799
Author: Dennis G. Zill
Publisher: Cengage Learning
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Chapter 15, Problem 1RE

Consider the boundary-value problem

2 u x 2 + 2 u y 2 = 0 , 0 < x < 2 , 0 < y < 1 u ( 0 , y ) = 0 , u ( 2 , y ) = 50 , 0 < y < 1 u ( x , 0 ) = 0 , u ( x , 1 ) = 0 , 0 < x < 2

Approximate the solution of the differential equation at the interior points of the region with mesh size h = 1 2 . Use Gaussian elimination or Gauss-Seidel iteration.

Expert Solution & Answer
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To determine

The approximate solution of the given differential equation 2ux2+2uy2=0 at the interior points of the region with mesh size h=12 using Gaussian elimination or Gauss-Seidel iteration.

Answer to Problem 1RE

The approximate solution of the differential equation at the interior points of the region is u11=0.8929,u21=3.5714,u31=13.3928_.

Explanation of Solution

Formula used:

uij=14[ui+1,j+ui,j+1+ui1,j+ui,j1]

Calculation:

Consider the given boundary value problem.

2ux2+2uy2=0,0<x<2,0<y<1u(0,y)=0,u(2,y)=50,0<y<1u(x,0)=0,u(x,1)=0,0<x<2

For constructing a mesh with size with h=12, the following values are required.

P41=P(4h,h)P41=P(2,12)

Since x=2,y=12, the value of u(2,y) is evaluated as follows.

u(2,y)=50

Similarly, P12=P(12,1) and u(x,1) gives u(12,1)=0. All other values are similarly obtained and the figure below shows the values of u(x,y) along the boundaries.

Differential Equations with Boundary-Value Problems (MindTap Course List), Chapter 15, Problem 1RE

For P11 apply the above formula when i=1 and j=1.

4u11=[u21+u12+u01+u10]

From the given boundary conditions u01=u(0,1)=0, u12=u(12,1)=0 and u10=u(1,0)=0. Thus, the above equation becomes u214u11=0.

Repeat this for P21,P31, which in turn gives three equations with three unknowns as follows.

u21+0+0+04u11=0u31+0+u11+04u21=050+0+u21+04u31=0

Since there are less number of unknowns, use Gauss-elimination method to solve the system of equations.

[410141014][u11u21u31]=[0050]

Here, use row column transformations to obtain the values of u11,u21,u31.

[10.250141014|0050]R1R14[10.25003.751014|0050]R2R2R1[10.250010.267014|0050]R2R23.75[100.067010.267013.733|0050]R1R1+0.25R2

Further reduce the matrix as follows.

[100.067010.267011|0013.3928]R3R33.733[100010011|0.89293.571413.3928]R1R1+0.067R3

Therefore, the approximate solution of the differential equation at the interior points of the region is u11=0.8929,u21=3.5714,u31=13.3928_.

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