Application to Temperature Distribution  introduction Consider the cross section of a long rectangular dam on a river. As you can imagine, the boundaries of the dam are subject to three factors: the temperature of the air, the temperature of the water, and the temperature of the ground at its base. The following diagram represents this situation: where the number represent the temperatures (in degree Celsius)  of the boundaries.  Engineers are interested in knowing the temperature distribution inside the dam in a specific period of time so they can determine the thermal stress to which the dam is subjected. Assuming the boundary temperatures are held constant during that specific period of time, the temperature inside the dam will reach certain equilibrium after some time has passed. Finding this equilibrium temperature distribution at different points on the plate (the dam) is desirable, but extremely difficult. However, one can consider a few points on the plate and approximate the temperature of these points. This approximation is based on a very important physical property called the Mean-Value Property:

# If a plate has reached a thermal equilibrium, and P is a point on the plate, C is a circle centered at P and fully contained in the plate, then the temperature at P is the average value of the temperature function over C.

The following diagram illustrates this property: To see how the property works, place a grid over the plate (the cross section of the bridge in our case) and consider the points with the lines of the grid meet. We will be interested in the temperatures at these points only in the plate. Design the grid in such a way that some of the points considered lie on the boundary of the plate. Studying the temperature at these grid points requires the following practical version of the Mean-Value Property:

If a plate has reached a thermal equilibrium and P is a grid point not on the boundary of the plate, then the temperature at P is the average of the temperatures of the four closest grid points to P.

Let us start with a grid with four interior points, and let x1, x2, x3, x4 be the temperatures at these four points. The situation is illustrated in the following diagram: By the second version of the Mean-Value Property, we have the following system of linear equations:  which after simplification gives: The matrix form of the system is AX=b, where X is called the vector of equilibrium temperatures. The solution for the above system is then provided, of course, that A is invertible.

Using the methods you learned in your first linear algebra course, you can compute the inverse of a square matrix. In our case, we find: and the vector of equilibrium temperatures is These answers are in degrees Celsius.

Suppose now that the boundary temperatures change from 25, 20 and 30 to 15, 10 and 20 respectively, then we get a new linear system having the same coefficient matrix, but with a new right hand side and the vector of equilibrium temperatures in this case is: .

The equilibrium temperature approximations found above could be more accurate if we consider a finer grid (more interior points).  Consider the following grid obtained from the first one by reducing the spacing in grid 1 to half: The new grid has 25 interior points. Repeating the same process as in the case of 4 interior points above gives a linear system in 25 equations and 25 unknowns. The coefficient matrix of this new system is (try it yourself): and the right hand side is Solving the above system manually would take a long time, but using an algebra software, like Maple or Matlab can compute the inverse of the above 25x25 matrix and perform the multiplication A-1 b quite rapidly. This gives the following vector of equilibrium temperatures: Again, all the components of this vector are in degrees Celsius. Since the grid is finer than before enough, the above vector gives a better indication of the temperature distribution inside the bridge.

Of course, one can be more accurate by considering even lager grids (more interior points), but the problem of solving the corresponding linear system becomes much harder. In this case, a numerical approach to solve the system is preferable.

Note that the four interior points of the first grid are in particular interior points for the second grid (with the 25 interior points). The following table compares the temperatures of these common points in both cases.

 Temperature in grid 1 Temperature in grid 2 Point 1 16.04 23.10 Point 2 12.70 21.81 Point 3 16.45 25.68 Point 4 14.79 24.39

Since grid 2 is finer than grid 1, the temperatures found in the last column of the above table are closer to the exact equilibrium temperatures than those in the first column.