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LECTURE 12
The Heat Equation
- Derivation of the Heat Equation
Definition 12.1. Heat is the energy transferred from one body to another due to a difference in temperature. (Better: heat is the kinetic energy of the molecules that compose the material.)
There are two basic physical principles governing the notion of heat.
(i) The total heat energy contained in a uniform, homogeneous body is related to its temperature and mass in the following simple way = where is the specific heat capacity of the material (a measurable constant specific 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) () =
Z
(x) (x ) x
(ii) The rate of heat transfer across a portion of the boundary of a region of the body is proportional directional derivative of across the boundary and the area of contact
(2) Heat flux across =
Z
∇ · n
where n = n (x) is the direction normal to the surface of contact at the point x, and is another constant specific to the material from with the body is constructed. is called heat conductivity constant.
Applying Gauss’s divergence theorem to (2) we have
(3) Heat flux entering/leaving a region =
Z
∇ · n =
Z
∇ · ∇ x
This should be the (total) rate at which heat enters or leaves the region , which in turn should correspond to the rate of change of the total amount of heat energy contained in the region:
(4)
Z
x
Equating (3) and (4) we thus obtain
Z
∇ · ∇ x =
Z
x
Since the region can be chosen arbitrarily, the two integrands must coincide at every point of the body. We thus obtain
(The Heat Equation) ∇^2 −
1
- SEPARATION OF VARIABLES 2
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 flow in and out only from its two ends:
In such a situation we can assume that the temperature really only depends on the position along the length of the heat pipe. Then
∇^2 ≡
^2
^2
^2
^2
^2
^2
^2
and the heat equation reduces to a 2 -dimensional PDE of the form
(5)
− ^2
^2
^2
where
=
r
(Replacing the ratio () by ^2 will prove convenient later on.)
- Separation of Variables
We begin by looking for solutions of (5), (6) and (6) of the form
(8) ( ) = () ()
In other words, we look first for a solution which is the product of a function depending only on the spatial variable and a function that depends only the the time variable . Plugging (8) into (5) yields
μ
= ^2
μ ^2 ^2
Dividing both sides by ^2 () () we obtain
(9)
^2
^2
^2
Noting that the left hand side depends only on , that the right hand side depends only on , and that this equation must hold for all and we can conclude that both sides must be constants. (Since the right hand side does not depend on the left hand side must be independent of , and since the left hand side does not dependent on the right hand side must be independent of ). If we denote by the common constant equal to the left and right hand sides of (9) we have
1 ^2
^2
^2
(10) = ^2
^2
^2
Thus, assuming (8) we can replace the PDE (5) with a pair of ordinary differential equations.
- EXAMPLE: HOMOGENEOUS DIRICHLET BOUNDARY CONDITIONS 4
= tan−^1
1 2
q ^21 + ^22 ⇐⇒ 1 = sin 2 = cos
whence
() = sin () cos () + cos sin () = sin ( + )
The easiest way to get nontrivial solutions to satisfy the boundary conditions (13) is to stipulate = 0 and = , with ∈ N. But note also that if we do not choose and in this fashion, then the only way to satisfy the boundary conditions (0) = 0 = () is to take = 0 =⇒ () = 0 for all . In particular,
- the separation constant −^2 first appearing in (9) and (10) can not be completely arbitrary if the solution ( ) = () () is to satisfy the boundary conditions ( 0) = 0 = ( 0).
Let us then set
−^2 = −
^2 ^2
^2
and proceed with the solution of (10).
= −^2 ^2 = −
^2 ^2 ^2
^2
This is just the differential equation of an exponential function, the general solution of which being
() = −(^
)^2
Thus,
(14) ( ) = −(^
)^2
sin
will be a solution of the 1 -dimensional heat equation satisfying the boundary conditions ( 0) = 0 = ( 0).
Unfortunately, the above solution is unlikely to satisfy the boundary condition at = 0:
() = ( 0)
What saves the day here is that fact that (14) actually gives an infinite number of solutions of (5), (12b) and (12c); for we get a distinct solution for each = 0 1 2 3 . Morever, since the PDE (5) is a linear
PDE, given a set of solutions
n ( ) = −(^
)^2
sin
o , we can create even more
solutions by taking linear combinations of the solutions ( ). We thus set
X^ ∞
=
−(^
)^2
sin
Let’s now impose the boundary condition at = 0:
X^ ∞
=
sin
Apparently, the coefficients should be chosen so that they correspond to the Fourier-sine expansion of the function ():
=
Z
0
() sin
- EXAM PLE: NON-HOM OGENEOUS BOUNDARY CONDITIONS 5
We conclude that the solution of
− ^2
^2
^2
is given by
( ) =
X^ ∞
=
−(^
)^2
sin
where
=
Z
0
() sin
- Example: Non-homogeneous boundary conditions
Let us now consider the solution of the 1 -dimensional heat equation
(15)
− ^2
^2
^2
subject to non-homogenous boundary conditions
(16a) (0 ) = 1
(16b) ( ) = 2
(16c) ( 0) = ()
which might correspond to a situation where a long rod with an initial temperature distribution () has its two ends inserted into two different heat baths that are maintained at different temperatures.
Since we expect that eventually as → ∞ the rod will eventually reach a steady state temperature distrib- ution that is independent of time, we shall suppose that if
for sufficiently large ( ) ≈ ()
where () is the (as yet undetermined) final steady state temperature distribution. Since even for large , ( ) must still satisfy (15), (16a) and (16b), we have for sufficiently large
(17) 0 =
− ^2
^2
^2
^2
^2
and
(18) (0) = 1 () = 2
The differential equation
(^2) ^2 = 0^ implies^ ^ is a linear fuction of^ , () = +
and the boundary conditions (18) require the constants and to be
= 1 and =
Thus,
(19) () =
Let us now define an auxiliary function ( ) by
(20) ( ) = () + ( )
- EXAM PLE: HOMOGENEOUS NEUMANN BOUNDARY CONDITIONS 7
We thus look for solutions of
− ^2
^2
^2
(21a) = 0
(21b) (0 ) = 0
(21c) ( ) = 0
(21d) ( 0) = ()
Our method will procede as in Section 3, up to the point where we imposed the boundary conditions (13). Thus, we begin by looking for solutions of the form
( ) = () ()
which reduces the solution of equations (21a) to the solution of
(22a) ^00 = −^2
(22b) ^ ˙ = −^2 ^2
This time, however, we shall write the general solution of the ODE for as
() = cos ( + )
The reason for using this (instead of the other choices () = 1 cos () + 2 sin () or () = sin ( + )) is that the boundary conditions at 0 and
0 = ^0 (0) = − sin (0 + ) 0 = ^0 () = − sin ( + )
are quickly reduced to the requirements that
(23) = 0 ; =
∈ N
whence () = cos
∈ N
With restricted as in (23) the general solution of (22b) is
() = −(^
)^2
and so the functions (^) n ≡ −(^
)^2
cos
| ∈ N
o
will provide us with an infinite set of independent solutions of (21a) - (21c). From these we now construct an ansatz for a function that will also satisfy (21d):
X^ ∞
=
−(^
)^2
cos
Applying the boundary condition at = 0
() = ( 0) =
X^ ∞
=
cos
we see that (24) will indeed satisfy (21a) - (21d) so long as the coefficients are chosen to correspond with the Fourier-cosine coefficients of () :
=
Z
0
cos
