Question

(M) Study how changes in boundary temperatures on a steel plate affect temperatures on a steel plate affect the temperatures at interior points on the plate.

1. Begin by estimating the temperatures \({T_1}\), \({T_2}\), \({T_3}\), \({T_4}\) at each of the sets of four points on the steel plate shown in the figure. In each case, the value of \({T_k}\) is approximated by the average of the temperature at the four closest points. See Exercises 33 and 34 in section 1.1, where the values (in degrees) turn out to be \(\left( {20,27.5,30,22.5} \right)\). How is this list of values related to your results for the points in the set (a) and (b)?
2. Without making any computations, guess the interior temperatures in (a) when the boundary temperatures are all multiplied by 3. Check your guess.
3. Finally, make a general conjecture about the corresponds from the list of eight boundary temperatures to the list of four.

Short answer

a. The temperature in section 1.1 is the sum of the temperatures of interior points in fig (a) and fig (b).

b. \(\left( {30,30,30,30} \right)\)

c. Using the equation \(A{\bf{x}} = {\bf{b}}\), the interior temperatures of the loop can be approximated. Matrix \(A\) depends on the position of the interior points, and \({\bf{b}}\) vector in \({\mathbb{R}^4}\) is the matrix for boundary temperature.

Step-by-step solution

Form the equation for temperatures

By fig. (a), the equations of temperatures are as shown below.

\(4{T_1} = 0 + 20 + {T_2} + {T_4}\)

\(4{T_2} = {T_1} + 20 + 0 + {T_3}\)

\(4{T_3} = {T_4} + {T_2} + 0 + 20\)

\(4{T_4} = 0 + {T_1} + {T_3} + 20\)

The augmented matrix can be written as follows:

\(\left[ {\begin{array}{*{20}{c}}4&{ - 1}&0&{ - 1}&{20}\\{ - 1}&4&{ - 1}&0&{20}\\0&{ - 1}&4&{ - 1}&{20}\\{ - 1}&0&{ - 1}&4&{20}\end{array}} \right]\)

Convert the matrix into the row-reduced echelon form

Consider the matrix \(\left[ {\begin{array}{*{20}{c}}4&{ - 1}&0&{ - 1}&{20}\\{ - 1}&4&{ - 1}&0&{20}\\0&{ - 1}&4&{ - 1}&{20}\\{ - 1}&0&{ - 1}&4&{20}\end{array}} \right]\).

Use the code in MATLAB to obtain the row-reduced echelon form, as shown below:

\[\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left[ {\begin{array}{*{20}{c}}4&{ - 1}&0&{ - 1}&{20}\end{array};{\rm{ }}\begin{array}{*{20}{c}}{ - 1}&4&{ - 1}&0&{20}\end{array};{\rm{ }}\begin{array}{*{20}{c}}0&{ - 1}&4&{ - 1}&{20}\end{array};{\rm{ }}\begin{array}{*{20}{c}}{ - 1}&0&{ - 1}&{ - 4}&{20}\end{array}{\rm{ }}} \right];\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\]

\[\left[ {\begin{array}{*{20}{c}}4&{ - 1}&0&{ - 1}&{20}\\{ - 1}&4&{ - 1}&0&{20}\\0&{ - 1}&4&{ - 1}&{20}\\{ - 1}&0&{ - 1}&4&{20}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&0&0&{10}\\0&1&0&0&{10}\\0&0&1&0&{10}\\0&0&0&1&{10}\end{array}} \right]\]

Form the equation for temperatures

For fig. (b), the equation of temperatures are as shown below:

\(4{T_1} = 10 + 0 + {T_2} + {T_4}\)

\(4{T_2} = {T_1} + 0 + 40 + {T_3}\)

\(4{T_3} = {T_4} + {T_2} + 40 + 10\)

\(4{T_4} = 10 + {T_1} + {T_3} + 10\)

The augmented matrix for fig. (b) is

\(\left[ {\begin{array}{*{20}{c}}4&{ - 1}&0&{ - 1}&{10}\\{ - 1}&4&{ - 1}&0&{40}\\0&{ - 1}&4&{ - 1}&{50}\\{ - 1}&0&{ - 1}&4&{20}\end{array}} \right]\).

Convert the matrix into the row-reduced echelon form

Consider the matrix \(\left[ {\begin{array}{*{20}{c}}4&{ - 1}&0&{ - 1}&{10}\\{ - 1}&4&{ - 1}&0&{40}\\0&{ - 1}&4&{ - 1}&{50}\\{ - 1}&0&{ - 1}&4&{20}\end{array}} \right]\).

Use the code in MATLAB to obtain the row-reducedechelon form, as shown below:

\[\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left[ {\begin{array}{*{20}{c}}4&{ - 1}&0&{ - 1}&{10}\end{array};{\rm{ }}\begin{array}{*{20}{c}}{ - 1}&4&{ - 1}&0&{40}\end{array};{\rm{ }}\begin{array}{*{20}{c}}0&{ - 1}&4&{ - 1}&{50}\end{array};{\rm{ }}\begin{array}{*{20}{c}}{ - 1}&0&{ - 1}&{ - 4}&{20}\end{array}{\rm{ }}} \right];\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\]

\[\left[ {\begin{array}{*{20}{c}}4&{ - 1}&0&{ - 1}&{10}\\{ - 1}&4&{ - 1}&0&{40}\\0&{ - 1}&4&{ - 1}&{50}\\{ - 1}&0&{ - 1}&4&{20}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&0&0&{10}\\0&1&0&0&{17.5}\\0&0&1&0&{20}\\0&0&0&1&{12.5}\end{array}} \right]\]

Interpret the results

The solution temperature of fig. (a) is \(\left( {10,10,10,10} \right)\). The solution temperature of fig. (b) is \(\left( {10,17.5,20,12.5} \right)\), and the solution temperature in section 1.1 is \(\left( {20,27.5,30,22.5} \right)\).

It can be observed from the results that the temperature in the problem of section 1.1 is the sum of the temperatures in fig. (a) and fig. (b).

Guess the interior temperature

When the boundary temperatures are multiplied by 3, the interior temperature also becomes three times.

Therefore, the interior temperature in fig. (a) is \(\left( {30,30,30,30} \right)\).

Estimate the eight boundary temperatures

The linear transformation of eight boundary temperatures gives the four interior temperatures.

Using the equation \(A{\bf{x}} = {\bf{b}}\), the interior temperatures of the loop can be approximated. Matrix \(A\) depends on the position of interior points, and \({\bf{b}}\) vector in \({\mathbb{R}^4}\) is the matrix for boundary temperature.