Image:Heatequation exampleB.gif
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Contents |
[edit] General Audience Description
Here the horizontal axis represents the location along a bar of metal and the graph records the temperature at that location. It begins with an initial temperature which is hot at one side and cool at the other, and then shows how the temperature of the bar approaches an equilibrium. It is assumed that no heat is lost from the bar and that there are no heat sources. This demonstrates two key properties of the heat equation: approaching an equilibrium, and the maximum principle. The maximum principle says that the temperature will always have a maximum either earlier in time or at the ends of the bar.
[edit] Summary
Graphical representation of the solution to the heat equation for an "infinite slab" of width 1 given by:
where k = .061644 subject to the boundary conditions:
and with the initial heat distribution given by:
In this case, the left face (x=0) and the right face (x=1) are perfectly insulated. This image shows how the heat redistributes, flowing from the warmer left edge to the cooler right edge, then equalizing to a constant temperature throughout. This temperature happens to be the average value of cos(2x) over [0,1], as one might expect.
The solution:
where:
[edit] Solution Details
This solution was obtained using separation of variables.
[edit] Source Code
Mathematica Source:
j = 10;
k = .061644;
A[n_] := (-4 (-1)^n* Sin[2])/(-4 + n^2*Pi^2);
u[x_, t_] := Sin[2]/2 + Sum[A[n]*Cos[n*Pi*x]*Exp[-k(n*Pi)^2*t], {n, 1, j}];
For[i = 0, i <= 12, i += .1,
Plot[u[x, i], {x, 0, 1},
Prolog -> {Line[{{0, 1}, {1, 1}, {1, 1}, {1, 0}}]},
PlotRange -> {0, 1},
AxesLabel -> {"x", "u=temp"},
PlotLabel -> {i}
]
]
[edit] Licensing
File history
Click on a date/time to view the file as it appeared at that time.
| Date/Time | Dimensions | User | Comment | |
|---|---|---|---|---|
| current | 22:01, 17 April 2007 | 288×177 (30 KB) | Wtt | |
| 21:35, 13 April 2007 | 288×177 (29 KB) | Wtt | (== Summary == Graphical representation of the solution to the heat equation for a "slab" of width 1 given by: :<math>\ u_t = ku_{xx}</math> subject to the boundary conditions: :<math>u_x(0,t) = 0,\ \ u_x(1,t)=0</math> and with the initial heat distribu) |
