Question

The band matrixAshown below can be used to estimate
the unsteady conduction of heat in a rod when the temperatures at points \({p_{\bf{1}}},...,{p_{\bf{5}}}\) on the rod change with time.

The constantCin the matrix depends on the physical nature
of the rod, the distance \(\Delta x\) between the points on the rod,
and the length of time \(\Delta t\) between successive temperature
measurements. Suppose that for \(k = 0,1,2,...,\) a vector \({{\bf{t}}_k}\)
in \({\mathbb{R}^5}\) lists the temperatures at time \(k\Delta t\). If the two ends of the
rod are maintained at \(0^\circ \), then the temperature vectors satisfy
the equation \(A{t_{k + 1}} = {t_k}\)\(\left( {k = 0,1,...} \right)\), where \[A = \left[ {\begin{array}{*{20}{c}}{\left( {{\bf{1}} + {\bf{2}}C} \right)}&{ - C}&{}&{}&{}\\{ - C}&{\left( {{\bf{1}} + {\bf{2}}C} \right)}&{ - C}&{}&{}\\{}&{ - C}&{\left( {{\bf{1}} + {\bf{2}}C} \right)}&{ - C}&{}\\{}&{}&{ - C}&{\left( {{\bf{1}} + {\bf{2}}C} \right)}&{ - C}\\{}&{}&{}&{ - C}&{\left( {{\bf{1}} + {\bf{2}}C} \right)}\end{array}} \right]\]

1. Find the LU factorization ofAwhen\[C = 1\]. A matrix
   such as Awith three nonzero diagonals is called a tridiagonal matrix. The Land Ufactors are bidiagonal matrices.

2. Suppose \[C = 1\] and\({t_0} = \left( {10,12,12,12,10} \right)\). Use the
   LU factorization of Ato find the temperature distributions \({t_1}\), \({t_2}\), \({t_3}\), and \({t_4}\).

Short answer

1. The LU factorizations are

\(L = \left[ {\begin{array}{*{20}{c}}1&0&0&0&0\\{ - 1/3}&1&0&0&0\\0&{ - 3/8}&1&0&0\\0&0&{ - 8/21}&1&0\\0&0&0&{ - 21/55}&1\end{array}} \right]\),\(U = \left[ {\begin{array}{*{20}{c}}3&{ - 1}&0&0&0\\0&{8/3}&{ - 1}&0&0\\0&0&{21/8}&{ - 1}&0\\0&0&0&{55/21}&{ - 1}\\0&0&0&0&{144/55}\end{array}} \right]\).

2. The temperature distributions are \({{\bf{t}}_1} = \left[ {\begin{array}{*{20}{c}}{6.5556}\\{9.6667}\\{10.4444}\\{9.6667}\\{6.5556}\end{array}} \right]\), \({{\bf{t}}_2} = \left[ {\begin{array}{*{20}{c}}{4.7407}\\{7.6667}\\{8.5926}\\{7.6667}\\{4.7407}\end{array}} \right]\), \({{\bf{t}}_3} = \left[ {\begin{array}{*{20}{c}}{3.5988}\\{6.0556}\\{6.9012}\\{6.0556}\\{3.5988}\end{array}} \right]\), \({{\bf{t}}_4} = \left[ {\begin{array}{*{20}{c}}{2.7922}\\{4.7778}\\{5.4856}\\{4.7778}\\{2.7922}\end{array}} \right]\).

Step-by-step solution

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

Consider the matrix at\(C = 1\)as shown below:

\(A = \left[ {\begin{array}{*{20}{c}}3&{ - 1}&0&0&0\\{ - 1}&3&{ - 1}&0&0\\0&{ - 1}&3&{ - 1}&0\\0&0&{ - 1}&3&{ - 1}\\0&0&0&{ - 1}&3\end{array}} \right]\)

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

\[ > > \left[ {\begin{array}{*{20}{c}}{\rm{L}}&{\rm{U}}\end{array}} \right] = {\rm{lu}}\left( {\rm{A}} \right)\]

The LU factorization is shown below:

\(L = \left[ {\begin{array}{*{20}{c}}1&0&0&0&0\\{ - 1/3}&1&0&0&0\\0&{ - 3/8}&1&0&0\\0&0&{ - 8/21}&1&0\\0&0&0&{ - 21/55}&1\end{array}} \right]\)

\(U = \left[ {\begin{array}{*{20}{c}}3&{ - 1}&0&0&0\\0&{8/3}&{ - 1}&0&0\\0&0&{21/8}&{ - 1}&0\\0&0&0&{55/21}&{ - 1}\\0&0&0&0&{144/55}\end{array}} \right]\)

Use the following MATLAB command to obtain the\(LU - A\)factorization of the matrix:

\[ > > LU - A\]

The output is shown below:

\(\begin{array}{c}LU - A = \left[ {\begin{array}{*{20}{c}}0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{array}} \right]\\ = {\bf{0}}\end{array}\)

Thus, it is verified that \(LU = A\).

(b) Step 2: Solve the temperature distribution

Suppose the solution of\(L{{\bf{s}}_{k + 1}} = {{\bf{t}}_k}\)is\({{\bf{s}}_{k + 1}}\), where \(\left( {k = 0,1,...} \right)\). So, the solution of\(U{{\bf{t}}_{k + 1}} = {{\bf{s}}_{k + 1}}\)is\({{\bf{t}}_{k + 1}}\).

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

\[\begin{array}{l} > > {\rm{b}} = \left[ {\begin{array}{*{20}{c}}{10}&{12}&{12}&{12}&{10}\end{array}} \right]\\ > > {\rm{X}} = {\rm{A}}\backslash {\rm{b}}'\end{array}\]

The solutions are shown below:

\({{\bf{s}}_1} = \left[ {\begin{array}{*{20}{c}}{10.0000}\\{15.3333}\\{17.7500}\\{18.7619}\\{17.1636}\end{array}} \right]\),\({{\bf{t}}_1} = \left[ {\begin{array}{*{20}{c}}{6.5556}\\{9.6667}\\{10.4444}\\{9.6667}\\{6.5556}\end{array}} \right]\), \({{\bf{s}}_2} = \left[ {\begin{array}{*{20}{c}}{6.5556}\\{11.8519}\\{14.8889}\\{15.3386}\\{12.4121}\end{array}} \right]\), \({{\bf{t}}_2} = \left[ {\begin{array}{*{20}{c}}{4.7407}\\{7.6667}\\{8.5926}\\{7.6667}\\{4.7407}\end{array}} \right]\), \({{\bf{s}}_3} = \left[ {\begin{array}{*{20}{c}}{4.7407}\\{9.2469}\\{12.0602}\\{12.2610}\\{9.4222}\end{array}} \right]\), \({{\bf{t}}_3} = \left[ {\begin{array}{*{20}{c}}{3.5988}\\{6.0556}\\{6.9012}\\{6.0556}\\{3.5988}\end{array}} \right]\), \({{\bf{s}}_4} = \left[ {\begin{array}{*{20}{c}}{3.5988}\\{7.2551}\\{9.6219}\\{9.7210}\\{7.3104}\end{array}} \right]\), \({{\bf{t}}_4} = \left[ {\begin{array}{*{20}{c}}{2.7922}\\{4.7778}\\{5.4856}\\{4.7778}\\{2.7922}\end{array}} \right]\).

Thus, the temperature distributions are \({{\bf{t}}_1} = \left[ {\begin{array}{*{20}{c}}{6.5556}\\{9.6667}\\{10.4444}\\{9.6667}\\{6.5556}\end{array}} \right]\), \({{\bf{t}}_2} = \left[ {\begin{array}{*{20}{c}}{4.7407}\\{7.6667}\\{8.5926}\\{7.6667}\\{4.7407}\end{array}} \right]\), \({{\bf{t}}_3} = \left[ {\begin{array}{*{20}{c}}{3.5988}\\{6.0556}\\{6.9012}\\{6.0556}\\{3.5988}\end{array}} \right]\), \({{\bf{t}}_4} = \left[ {\begin{array}{*{20}{c}}{2.7922}\\{4.7778}\\{5.4856}\\{4.7778}\\{2.7922}\end{array}} \right]\).