Before we get into actually solving partial differential
equations and before we even start discussing the method of separation of
variables we want to spend a little bit of time talking about the two main
partial differential equations that we’ll be solving later on in the chapter. We’ll look at the first one in this section
and the second one in the next section.
The first partial differential equation that we’ll be
looking at once we get started with solving will be the heat equation, which
governs the temperature distribution in an object. We are going to give several forms of the
heat equation for reference purposes, but we will only be really solving one of
them.
We will start out by considering the temperature in a 1D
bar of length L. What this means is that we are going to
assume that the bar starts off at and ends when we reach . We are also going to so assume that at any
location, x the temperature will be
constant at every point in the cross section at that x. In other words,
temperature will only vary in x and
we can hence consider the bar to be a 1D bar.
Note that with this assumption the actual shape of the cross section (i.e. circular, rectangular, etc.) doesn’t matter.
Note that the 1D assumption is actually not all that bad of
an assumption as it might seem at first glance.
If we assume that the lateral surface of the bar is perfectly insulated
(i.e. no heat can flow through the
lateral surface) then the only way heat can enter or leave the bar as at either
end. This means that heat can only flow
from left to right or right to left and thus creating a 1D temperature
distribution.
The assumption of the lateral surfaces being perfectly
insulated is of course impossible, but it is possible to put enough insulation
on the lateral surfaces that there will be very little heat flow through them
and so, at least for a time, we can consider the lateral surfaces to be
perfectly insulated.
Okay, let’s now get some definitions out of the way before
we write down the first form of the heat equation.
We should probably make a couple of comments about some of
these quantities before proceeding.
The specific heat, ,
of a material is the amount of heat energy that it takes to raise one unit of
mass of the material by one unit of temperature. As indicated we are going to assume, at least
initially, that the specific heat may not be uniform throughout the bar. Note as well that in practice the specific
heat depends upon the temperature.
However, this will generally only be an issue for large temperature
differences (which in turn depends on the material the bar is made out of) and
so we’re going to assume for the purposes of this discussion that the
temperature differences are not large enough to affect our solution.
The mass density, ,
is the mass per unit volume of the material.
As with the specific heat we’re going to initially assume that the mass
density may not be uniform throughout the bar.
The heat flux, ,
is the amount of thermal energy that flows to the right per unit surface area
per unit time. The “flows to the right”
bit simply tells us that if for some x
and t then the heat is flowing to the
right at that point and time. Likewise
if then the heat will be flowing to the left at
that point and time.
The final quantity we defined above is
and this is used to represent any external
sources or sinks (i.e. heat energy
taken out of the system) of heat energy.
If then heat energy is being added to the system
at that location and time and if then heat energy is being removed from the
system at that location and time.
With these quantities the heat equation is,
While this is a nice form of the heat equation it is not
actually something we can solve. In this
form there are two unknown functions, u
and ,
and so we need to get rid of one of them.
With Fourier’s law we can
easily remove the heat flux from this equation.
Fourier’s law states that,
where is the thermal
conductivity of the material and measures the ability of a given material to
conduct heat. The better a material can
conduct heat the larger will be.
As noted the thermal conductivity can vary with the location in the
bar. Also, much like the specific heat
the thermal conductivity can vary with temperature, but we will assume that the
total temperature change is not so great that this will be an issue and so we
will assume for the purposes here that the thermal conductivity will not vary
with temperature.
Fourier’s law does a very good job of modeling what we know
to be true about heat flow. First, we
know that if the temperature in a region is constant, i.e. ,
then there is no heat flow.
Next, we know that if there is a temperature difference in a
region we know the heat will flow from the hot portion to the cold portion of
the region. For example, if it is hotter
to the right then we know that the heat should flow to the left. When it is hotter to the right then we also
know that (i.e.
the temperature increases as we move to the right) and so we’ll have and so the heat will flow to the left as it
should. Likewise, if (i.e.
it is hotter to the left) then we’ll have and heat will flow to the right as it should.
Finally, the greater the temperature difference in a region
(i.e. the larger is) then the greater the heat flow.
So, if we plug Fourier’s law into (1), we
get the following form of the heat equation,
Note that we factored the minus sign out of the derivative
to cancel against the minus sign that was already there. We cannot however, factor the thermal
conductivity out of the derivative since it is a function of x and the derivative is with respect to x.
Solving (2) is quite difficult due to
the non uniform nature of the thermal properties and the mass density. So, let’s now assume that these properties
are all constant, i.e.,
where c, and are now all fixed quantities. In this case we generally say that the
material in the bar is uniform. Under these assumptions the heat equation
becomes,


(3)

For a final simplification to the heat equation let’s divide
both sides by and define the thermal diffusivity to be,
The heat equation is then,
To most people this is what they mean when they talk about
the heat equation and in fact it will be the equation that we’ll be
solving. Well, actually we’ll be solving
(4)
with no external sources, i.e. ,
but we’ll be considering this form when we start discussing separation of
variables in a couple of sections. We’ll
only drop the sources term when we actually start solving the heat equation.
Now that we’ve got the 1D heat equation taken care of we
need to move into the initial and boundary conditions we’ll also need in order
to solve the problem. If you go back to
any of our solutions of ordinary differential equations that we’ve done in
previous sections you can see that the number of conditions required always
matched the highest order of the derivative in the equation.
In partial differential equations the same idea holds except
now we have to pay attention to the variable we’re differentiating with respect
to as well. So, for the heat equation
we’ve got a first order time derivative and so we’ll need one initial condition
and a second order spatial derivative and so we’ll need two boundary
conditions.
The initial condition that we’ll use here is,
and we don’t really need to say much about it here other
than to note that this just tells us what the initial temperature distribution
in the bar is.
The boundary conditions will tell us something about what
the temperature and/or heat flow is doing at the boundaries of the bar. There are four of them that are fairly common
boundary conditions.
The first type of boundary conditions that we can have would
be the prescribed temperature
boundary conditions, also called Dirichlet
conditions. The prescribed
temperature boundary conditions are,
The next type of boundary conditions are prescribed heat flux, also called Neumann conditions. Using Fourier’s law these can be written as,
If either of the
boundaries are perfectly insulated, i.e. there is no heat flow out of them
then these boundary conditions reduce to,
and note that we will often just call these particular
boundary conditions insulated
boundaries and drop the “perfectly” part.
The third type of boundary conditions use Newton’s law of cooling and are
sometimes called Robins conditions. These are usually used when the bar is in a
moving fluid and note we can consider air to be a fluid for this purpose.
Here are the equations for this kind of boundary condition.
where H is a
positive quantity that is experimentally determined and and give the temperature of the surrounding fluid
at the respective boundaries.
Note that the two conditions do vary slightly depending on
which boundary we are at. At we have a minus sign on the right side while
we don’t at . To see why this is let’s first assume that at
we have . In other words the bar is hotter than the
surrounding fluid and so at the heat flow (as given by the left side of
the equation) must be to the left, or negative since the heat will flow from
the hotter bar into the cooler surrounding liquid. If the heat flow is negative then we need to have
a minus sign on the right side of the equation to make sure that it has the
proper sign.
If the bar is cooler than the surrounding fluid at ,
i.e. we can make a similar argument to justify the
minus sign. We’ll leave it to you to
verify this.
If we now look at the other end, ,
and again assume that the bar is hotter than the surrounding fluid or, . In this case the heat flow must be to the
right, or be positive, and so in this case we can’t have a minus sign. Finally, we’ll again leave it to you to
verify that we can’t have the minus sign at is the bar is cooler than the surrounding
fluid as well.
Note that we are not actually going to be looking at any of
these kinds of boundary conditions here.
These types of boundary conditions tend to lead to boundary value
problems such as Example 5 in
the Eigenvalues and Eigenfunctions section of the previous chapter. As we saw in that example it is often very
difficult to get our hands on the eigenvalues and as we’ll eventually see we
will need them.
It is important to note at this point that we can also mix
and match these boundary conditions so to speak. There is nothing wrong with having a
prescribed temperature at one boundary and a prescribed flux at the other boundary
for example so don’t always expect the same boundary condition to show up at
both ends. This warning is more
important that it might seem at this point because once we get into solving the
heat equation we are going to have the same kind of condition on each end to
simplify the problem somewhat.
The final type of boundary conditions that we’ll need here
are periodic boundary
conditions. Periodic boundary conditions
are,
Note that for these kinds of boundary conditions the left
boundary tends to be instead of as we were using in the previous types of
boundary conditions. The periodic
boundary conditions will arise very naturally from a couple of particular
geometries that we’ll be looking at down the road.
We will now close out this section with a quick look at the
2D and 3D version of the heat equation.
However, before we jump into that we need to introduce a little bit of
notation first.
The del operator
is defined to be,
depending on whether we are in 2 or 3 dimensions. Think of the del operator as a function that
takes functions as arguments (instead of numbers as we’re used to). Whatever function we “plug” into the operator
gets put into the partial derivatives.
So, for example in 3D we would have,
This of course is also the gradient of the function .
The del operator also allows us to quickly write down the
divergence of a function. So, again
using 3D as an example the divergence of can be written as the dot product of the del
operator and the function. Or,
Finally, we will also see the following show up in our
work,
This is usually denoted as,
and is called the Laplacian. The 2D version of course simply doesn’t have
the third term.
Okay, we can now look into the 2D and 3D version of the heat
equation and where ever the del operator and or Laplacian appears assume that
it is the appropriate dimensional version.
The higher dimensional version of (1) is,
and note that the specific heat, c, and mass density, ,
are may not be uniform and so may be functions of the spatial variables. Likewise, the external sources term, Q, may also be a function of both the
spatial variables and time.
Next, the higher dimensional version of Fourier’s law is,
where the thermal conductivity, ,
is again assumed to be a function of the spatial variables.
If we plug this into (5) we
get the heat equation for a non uniform bar (i.e. the thermal properties may be functions of the spatial
variables) with external sources/sinks,


(6)

If we now assume that the
specific heat, mass density and thermal conductivity are constant (i.e. the bar is uniform) the heat
equation becomes,
where we divided both sides by to get the thermal diffusivity, k in front of the Laplacian.
The initial condition for the 2D or 3D heat equation is,
depending upon the dimension we’re in.
The prescribed temperature boundary condition becomes,
where or ,
depending upon the dimension we’re in, will range over the portion of the
boundary in which we are prescribing the temperature.
The prescribed heat flux condition becomes,
where the left side is only being evaluated at points along
the boundary and is the outward unit normal on the surface.
Newton’s law of cooling will become,
where H is a
positive quantity that is experimentally determine, is the temperature of the fluid at the
boundary and again it is assumed that this is only being evaluated at points
along the boundary.
We don’t have periodic boundary conditions here as they will
only arise from specific 1D geometries.
We should probably also acknowledge at this point that we’ll
not actually be solving (7) at any point, but we will
be solving a special case of it in the Laplace’s
Equation section.