LECTURE 6
The Heat Equation
We now turn our attention to the Heat (or Di¤usion) Equation:
2 ut k uxx = 0 : This PDE is used to model systems in which heat or some other property (e.g. the concentration of a solvent in a solution) distributes itself throughout a body.
1. Derivation of the Heat Equation
Definition 6.1. Heat is the energy transferred from one body to another due to a di¤erence in temperature. (Better: heat is the kinetic energy of the molecules that compose the material.)
There are two basic principles governing the physical concept of heat.
(i) The total heat energy H contained in a uniform, homogeneous body is related to its temperature T and mass in the following simple way
H = sMT
where s is the speci…c heat capacity of the material (a measurable constant speci…c to the material from with the body is made). More generally, in a situation for which neither the temperature nor the density of the material is constant we have
(1) H (t) = s (x) T (x; t) dx ZV (ii) The rate of heat transfer across a portion S of the boundary of a region R of the body is proportional directional derivative of T across the boundary and the area of contact
(2) Heat ‡ux across S = T ndS r ZS where n = n (x) is the direction normal to the surface of contact at the point x, and is another constant speci…c to the material from with the body is constructed. is called heat conductivity constant.
Applying Gauss’sdivergence theorem to (2) we have
(3) Heat ‡ux entering/leaving a region = T ndS = T dx r r r Z@R ZR
This should be the (total) rate at which heat enters or leaves the region R, which in turn should correspond to the rate of change of the total amount of heat energy contained in the region: dH dT (4) = x dt s dt ZR 23 2.PROPERTIESOFSOLUTIONSOFTHEHEATEQUATION 24
Equating (3) and (4) we thus obtain
@T T dx =s dx r r @t ZR ZR or, after putting both integrals on one side,
@T 0 = 2T s dx r @t ZR Since the region R can be chosen arbitrarily, the two terms in the integrand must coincide at every point of the body. We thus obtain @T (The Heat Equation) 2T s = 0 r @t
1.1. The 1-dimensional Heat Equation. Above we derived the 3-dimensional heat equation. Let me now reduce the underlying PDE to a simpler subcase. Consider a long uniform tube surround by an insulating material like styroform along its length, so that heat can ‡ow in and out only from its two ends:
In such a situation we can assume that the temperature really only depends on the position x along the length of the heat pipe. Then @2T @2T @2T @T 2 2T + + r @x2 @y2 @z2 @x2 and the heat equation reduces to a 2-dimensional PDE of the form
@T @2T (5) 2 = 0 @t @x2 where = r s (Replacing the ratio =(s) by a perfect square will prove convenient later on when we actually solve the heat equation.)
2. Properties of Solutions of the Heat Equation
We’llnow restrict our attention to the 1-dimensional heat equation, writing it in the form
2 (6) ut a uxx = 0 : Before actually solving the heat equation, it is worthwhile to …rst observe some simple properties of its solutions. 2.PROPERTIESOFSOLUTIONSOFTHEHEATEQUATION 25
2.1. The Maximal Principle. Consider again the 1-dimensional Heat Equation. 2 (1) ut a uxx = 0 : Proposition 6.2 (The Maximum Principal). If u (x; t) satis…es the heat equation (1) in a rectangle, say, R = (x; t) 0 x L; 0 t T f j g in space-time, then the maximum value of u (x; t) in R is assumed either initially (t = 0) or on the lateral sides (x = 0 or x = L).
Proof. From vector calculus, we know
Theorem 6.3. Suppose u (x; t) is a function on a region D R2 with continuous partial derivatives up to order 2. Then
u = (ux; ut) = [0; 0] r 2 (x0; t0) is a local maximum = uxxutt 4 (utx) 0 ) 8 uxx 0 < u = (ux; ut) = [0; 0] : r 2 (x0; t0) is a local minimum = uxxutt 4 (utx) 0 ) 8 uxx 0 < : Now in the situation at hand u (x; t) must satisfy the heat equation, and so 1 (2) u = u xx a2 t
At an interior critical point (x0; t0), we must have
ut (x0; t0) = 0 1 (3) 0 = u (x0; t0) = = uxx (x0; t0) = ut (x0; t0) = 0 r ) ux (x0; t0) = 0 ) a2 But then 2 2 uxxutt 4 (utx) = 0 4 (utx) 0 which contradicts the requirements of Theorem 8.3 unless utx (x0; t0) = 0.
We are thus reduced to analyzing the possibility that u has an interior local maximum at a point (x0; t0) such that ux (x0; t0) = 0 ; ut (x0; t0) = 0 ; uxx (x0; t0) = 0 ; utx (x0; t0) = 0 : These conditions may seem rather onerous, but because they only have to hold at one point it is not yet clear that there is no interior local maximum is possible. So we’ll have to work a little harder.
Recall that a continuous function on any bounded, closed set obtains a maximum value on that set. Let M be the maximum value of u on the union S of left, bottom, and right sides of R (4) S = [x; 0] 0 x < L [0; t] 0 t T [L; t] 0 t T f j g [ f j g [ f j g M = max u(x; t [x; t] S f g j 2 As a union of closed, bounded intervals, S is closed and bounded and so M is a well-de…ned number.
Because we could still have a local maximum if uxx = 0 at an interior critical point, or if we have a maximum along the top boundary, we’ll need have to eliminate these possibilities in order to prove the theorem. We’ll elimate these possibilities by considering a small deformation v (x; t) of our solution u (x; t).
Let " > 0 be a positive constant and consider the function v (x; t) = u (x; t) + "x2. 2.PROPERTIESOFSOLUTIONSOFTHEHEATEQUATION 26
What I shall show is that (5) v (x; t) M + "L2 throughout R. Once this is accomplished, we will have u (x; t) = v (x; t) "x2 M + "L2 "x2 M + " L2 x2 Since this would be true for every " > 0, we will be able to conclude, by taking the limit " 0, that ! u (x; t) M; [x; t] R 8 2 which is what we’retrying to prove.
The …rst step in proving (5) is to eliminate the possibility that v (x; t) has a local maximum in the interior of R. Now while v (x; t) does not satisfy the Heat Equation, it does satisfy a similar PDE; for
2 2 2 2 2 2 2 2 2 (6) @t a @ v (x; t) = @t a @ u (x; t) + "x = 0 + " a @ x = 2a "; [x; t] R: x x x 2 This PDE then implies that at a critical point (where v (x; t) = 0) that r 2 2 a vxx = 2a " vxx = 2" > 0 : ) Since we need vxx 0 for a local maximum, we conclude that v (x; t) has no local maximum and so no maximum in the interior of R.
We also need to eliminate the possibility of a maximum of v (x; t) occuring along the top boundary of R where t = T . Suppose we did have a maximum of v (x; t) along the top boundary, say at the point (x0;T ). We’dthen have
v (x0;T ) v (x0;T ) v (x0;T ) v (x0;T ) 0 ) for all su¢ ciently small . But then
v (x0;T ) v (x0;T ) (7) vt (x0;T ) = lim 0 0 ! (because both the numerator and denominator inside the limit are always non-positive). On the other hand, by (6) we have
2 2 (8) vt (x0;T ) = 2a " + a vxx (x0;T ) If (x0;T ) is to be a maximum, it must in particular be a local maximum as we vary x about x0. This requires
(9) ux (x0;T ) = 0 and uxx (x0;T ) 0 But then (8) and (9) together imply
2 vt (x0;T ) = 2a " + (some number 0) < 0 which contradicts (7). We conclude that there can be no maximum of v (x; t) along the top edge.
But the region R is still a bounded domain, and so v (x; t) has attains a maximum somewhere in R. We’ve eliminated the possibilities of maxima in the interior and upper boundary; and so v (x; t) must attain its maximal value along the left, right or bottom edges of R. Thus, max v (x; t) [x; t] R = max v (x; t) [x; t] S f j 2 g f j 2 g = max u (x; t) + "x2 [x; t] S j 2 = max u (x; t) [x; t] S + max "x2 [x; t] S f j 2 g j 2 M + "L2 This proves (5) and so by the argument immediately following (5), proves the theorem. 3.UNIQUENESS 27
3. Uniqueness
Theorem 6.4. Consider the inhomogeneous Heat Equation with Dirichlet boundary conditions: 2 ut k uxx = f (x; t) ; 0 x L ; t > 0 (10a) u (x; 0) = (x) ; 0 x L (10b) u (0; t) = g (t) ; 0 < t (10c) u (L; t) = h (t) ; 0 < t (10d) Then there is at most one solution of (10a)–(10d).
Proof. Let u1 (x; t) and u2 (x; t) be two solutions of (10a)–(10d) and set
w (x; t) = u1 (x; t) u2 (x; t) Then w satis…es the homogeneous heat equation with homogeneous boundary conditions 2 ut k uxx = 0 ; 0 x L ; t > 0 (11a) u (x; 0) = 0 ; 0 x L (11b) u (0; t) = 0 ; 0 < t (11c) u (L; t) = 0 ; 0 < t (11d) But now, as a solution of the homogeneous heat equation, w (x; t) must satisfy the Maximum Principal. Thus, on any rectangle R = [x; t] 0 x L; 0 t T f j g w (x; t) must be less than or equal to its maximum value on the three boundary lines
`1 = [0; t] 0 t T f j g `2 = [x; 0] 0 x L f j g `3 = [x; t] 0 t T f j g But, by (11b)–(11c) w (x; t) must vanish on these lines. So w (x; t) 0 inside any rectangle R. Applying the same argument to
u2 (x; t) u1 (x; t) = w (x; t) we can similarly conclude w (x; t) 0 and so 0 = w (x; t) = u1 (x; t) u2 (x; t) = u1 (x; t) = u2 (x; t) ) Hence, any solution of (10a)–(10d) is unique.