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Note: 1 lecture, §9.7 in [EP], §10.8 in [BD]

Suppose we have an insulated wire, a plate, or a 3-dimensional object. We apply certain ﬁxed temperatures on the ends of the wire, the edges of the plate, or on all sides of the 3-dimensional object. We wish to ﬁnd out what is the steady state temperature distribution. That is, we wish to know what will be the temperature after long enough period of time.

We are really looking for a solution to the heat equation that is not dependent on time. Let us ﬁrst solve the problem in one space variable. We are looking for a function that satisﬁes

but such that for all and . Hence, we are looking for a function of alone that satisﬁes . It is easy to solve this equation by integration and we see that for some constants and .

Suppose we have an insulated wire, and we apply constant temperature at one end (say where ) and on the other end (at where is the length of the wire). Then our steady state solution is

This solution agrees with our common sense intuition with how the heat should be distributed in the wire. So in one dimension, the steady state solutions are basically just straight lines.

Things are more complicated in two or more space dimensions. Let us restrict to two space dimensions for simplicity. The heat equation in two space variables is

(4.18) |

or more commonly written as or . Here the and symbols mean . We will use from now on. The reason for using such a notation is that you can deﬁne to be the right thing for any number of space dimensions and then the heat equation is always . The operator is called the Laplacian.

OK, now that we have notation out of the way, let us see what does an equation for the steady state solution look like. We are looking for a solution to (4.18) that does not depend on , or in other words . Hence we are looking for a function such that

This equation is called the Laplace equation^{5}. Solutions to the Laplace equation are called harmonic functions and have many nice properties and applications far beyond the steady state heat problem.

Harmonic functions in two variables are no longer just linear (plane graphs). For example, you can check that the functions and are harmonic. However, if you remember your multi-variable calculus we note that if is positive, is concave up in the direction, then must be negative and must be concave down in the direction. Therefore, a harmonic function can never have any “hilltop” or “valley” on the graph. This observation is consistent with our intuitive idea of steady state heat distribution; the hottest or coldest spot will not be inside.

Commonly the Laplace equation is part of a so-called Dirichlet problem^{6}. That is, we have a region in the -plane and we specify certain values along the boundaries of the region. We then try to ﬁnd a solution deﬁned on this region such that agrees with the values we speciﬁed on the boundary.

For simplicity, we consider a rectangular region. Also for simplicity we specify boundary values to be zero at 3 of the four edges and only specify an arbitrary function at one edge. As we still have the principle of superposition, we can use this simpler solution to derive the general solution for arbitrary boundary values by solving 4 diﬀerent problems, one for each edge, and adding those solutions together. This setup is left as an exercise.

We wish to solve the following problem. Let and be the height and width of our rectangle, with one corner at the origin and lying in the ﬁrst quadrant.

The method we apply is separation of variables. Again, we will come up with enough building-block solutions satisfying all the homogeneous boundary conditions (all conditions except (4.23)). We notice that superposition still works for the equation and all the homogeneous conditions. Therefore, we can use the Fourier series for to solve the problem as before.

We try . We plug into the equation to get

We put the s on one side and the s on the other to get

The left hand side only depends on and the right hand side only depends on . Therefore, there is some constant such that . And we get two equations

Furthermore, the homogeneous boundary conditions imply that and . Taking the equation for we have already seen that we have a nontrivial solution if and only if and the solution is a multiple ofFor these given , the general solution for (one for each ) is

(4.24) |

We only have one condition on and hence we can pick one of or to be something convenient. It will be useful to have , so we let . Setting and solving for we get that

After we plug the and we into (4.24) and simplify, we ﬁnd

We deﬁne . And note that satisﬁes (4.19)–(4.22).

Observe that

Suppose

Then we get a solution of (4.19)–(4.23) of the following form.

As satisﬁes (4.19)–(4.22) and any linear combination (ﬁnite or inﬁnite) of must also satisfy (4.19)–(4.22), we see that must satisfy (4.19)–(4.22). By plugging in it is easy to see that satisﬁes (4.23) as well.

Example 4.9.1: Suppose that we take and we let . We compute the sine series for the function (we will get the square wave). We ﬁnd that for we have

Therefore the solution , see Figure 4.22, to the corresponding Dirichlet problem is given as

This scenario corresponds to the steady state temperature on a square plate of width with 3 sides held at 0 degrees and one side held at degrees. If we have arbitrary initial data on all sides, then we solve four problems, each using one piece of nonhomogeneous data. Then we use the principle of superposition to add up all four solutions to have a solution to the original problem.

A diﬀerent way to visualize solutions of the Laplace equation is to take a wire and bend it so that it corresponds to the graph of the temperature above the boundary of your region. Cut a rubber sheet in the shape of your region—a square in our case—and stretch it ﬁxing the edges of the sheet to the wire. The rubber sheet is a good approximation of the graph of the solution to the Laplace equation with the given boundary data.

Exercise 4.9.3: Let be the region described by and . Solve the problem

for some constant . Hint: Guess, then check your intuition.Exercise 4.9.4: Let be the region described by and . Solve

Hint: Try a solution of the form (diﬀerent separation of variables).

Exercise 4.9.6: Let be the region described by and . Solve the problem

The solution should be in series form using the Fourier series coeﬃcients of .Exercise 4.9.7: Let be the region described by and . Solve the problem

The solution should be in series form using the Fourier series coeﬃcients of .Exercise 4.9.8: Let be the region described by and . Solve the problem

The solution should be in series form using the Fourier series coeﬃcients of .Exercise 4.9.11 (challenging): Using only your intuition ﬁnd , for the problem , where for , and for . Explain.

^{5}Named after the French mathematician Pierre-Simon, marquis de Laplace (1749–1827).

^{6}Named after the German mathematician Johann Peter Gustav Lejeune Dirichlet (1805–1859).