Question

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

The constantCin the matrix depends on the physical nature
of the rod, the distance $\mathrm{\Delta }x$ between the points on the rod,
and the length of time $\mathrm{\Delta }t$ between successive temperature
measurements. Suppose that for $k=0,1,2,...,$ a vector ${\mathbf{t}}_k$
in ${\mathbb{R}}^5$ lists the temperatures at time $k\mathrm{\Delta }t$. 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}{ccccc}\left(\mathbf{1}+\mathbf{2}C\right)& -C& & & \\ -C& \left(\mathbf{1}+\mathbf{2}C\right)& -C& & \\ & -C& \left(\mathbf{1}+\mathbf{2}C\right)& -C& \\ & & -C& \left(\mathbf{1}+\mathbf{2}C\right)& -C\\ & & & -C& \left(\mathbf{1}+\mathbf{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

$L=\left[\begin{array}{ccccc}1& 0& 0& 0& 0\\ -1/3& 1& 0& 0& 0\\ 0& -3/8& 1& 0& 0\\ 0& 0& -8/21& 1& 0\\ 0& 0& 0& -21/55& 1\end{array}\right]$,$U=\left[\begin{array}{ccccc}3& -1& 0& 0& 0\\ 0& 8/3& -1& 0& 0\\ 0& 0& 21/8& -1& 0\\ 0& 0& 0& 55/21& -1\\ 0& 0& 0& 0& 144/55\end{array}\right]$.

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

Step-by-step solution

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

Consider the matrix at$C=1$as shown below:

$A=\left[\begin{array}{ccccc}3& -1& 0& 0& 0\\ -1& 3& -1& 0& 0\\ 0& -1& 3& -1& 0\\ 0& 0& -1& 3& -1\\ 0& 0& 0& -1& 3\end{array}\right]$

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

$$>>\left[\begin{array}{cc}\mathrm{L}& \mathrm{U}\end{array}\right]=\mathrm{l}\mathrm{u}\left(\mathrm{A}\right)$$

The LU factorization is shown below:

$L=\left[\begin{array}{ccccc}1& 0& 0& 0& 0\\ -1/3& 1& 0& 0& 0\\ 0& -3/8& 1& 0& 0\\ 0& 0& -8/21& 1& 0\\ 0& 0& 0& -21/55& 1\end{array}\right]$

$U=\left[\begin{array}{ccccc}3& -1& 0& 0& 0\\ 0& 8/3& -1& 0& 0\\ 0& 0& 21/8& -1& 0\\ 0& 0& 0& 55/21& -1\\ 0& 0& 0& 0& 144/55\end{array}\right]$

Use the following MATLAB command to obtain the$LU-A$factorization of the matrix:

$$>>LU-A$$

The output is shown below:

$\begin{array}{c}LU-A=\left[\begin{array}{ccccc}0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0\end{array}\right]\\ =\mathbf{0}\end{array}$

Thus, it is verified that $LU=A$.

(b) Step 2: Solve the temperature distribution

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

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

$$\begin{array}{l}>>\mathrm{b}=\left[\begin{array}{ccccc}10& 12& 12& 12& 10\end{array}\right]\\ >>\mathrm{X}=\mathrm{A}\mathrm{\setminus }{\mathrm{b}}^{\prime }\end{array}$$

The solutions are shown below:

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

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