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The band matrixAshown below can be used to estimate
the unsteady conduction of heat in a rod when the temperatures at points p1,...,p5 on the rod change with time.

The constantCin the matrix depends on the physical nature
of the rod, the distance Δx between the points on the rod,
and the length of time Δt between successive temperature
measurements. Suppose that for k=0,1,2,..., a vector tk
in R5 lists the temperatures at time kΔt. If the two ends of the
rod are maintained at 0, then the temperature vectors satisfy
the equation Atk+1=tk(k=0,1,...), where A=[(1+2C)CC(1+2C)CC(1+2C)CC(1+2C)CC(1+2C)]

  1. Find the LU factorization ofAwhenC=1. A matrix
    such as Awith three nonzero diagonals is called a tridiagonal matrix. The Land Ufactors are bidiagonal matrices.
  1. Suppose C=1 andt0=(10,12,12,12,10). Use the
    LU factorization of Ato find the temperature distributions t1, t2, t3, and t4.

Short Answer

Expert verified
  1. The LU factorizations are

L=[100001/3100003/8100008/211000021/551],U=[3100008/31000021/81000055/2110000144/55].

2. The temperature distributions are t1=[6.55569.666710.44449.66676.5556], t2=[4.74077.66678.59267.66674.7407], t3=[3.59886.05566.90126.05563.5988], t4=[2.79224.77785.48564.77782.7922].

Step by step solution

01

(a) Step 1: State the LU factorization of matrix A

Consider the matrix atC=1as shown below:

A=[3100013100013100013100013]

Use the following MATLAB command to obtain theLU factorization of the matrix:

>>[LU]=lu(A)

The LU factorization is shown below:

L=[100001/3100003/8100008/211000021/551]

U=[3100008/31000021/81000055/2110000144/55]

Use the following MATLAB command to obtain theLUAfactorization of the matrix:

>>LUA

The output is shown below:

LUA=[0000000000000000000000000]=0

Thus, it is verified that LU=A.

02

(b) Step 2: Solve the temperature distribution

Suppose the solution ofLsk+1=tkissk+1, where (k=0,1,...). So, the solution ofUtk+1=sk+1istk+1.

Use the following MATLAB command to solve the system using LU factorization:

>>b=[1012121210]>>X=Ab

The solutions are shown below:

s1=[10.000015.333317.750018.761917.1636],t1=[6.55569.666710.44449.66676.5556], s2=[6.555611.851914.888915.338612.4121], t2=[4.74077.66678.59267.66674.7407], s3=[4.74079.246912.060212.26109.4222], t3=[3.59886.05566.90126.05563.5988], s4=[3.59887.25519.62199.72107.3104], t4=[2.79224.77785.48564.77782.7922].

Thus, the temperature distributions are t1=[6.55569.666710.44449.66676.5556], t2=[4.74077.66678.59267.66674.7407], t3=[3.59886.05566.90126.05563.5988], t4=[2.79224.77785.48564.77782.7922].

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