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18 CHAPTER 1. INTRODUCTION
In the previous example the initial condition was of a very
specific form. In fact, theinitial condition f(x) is one of the
eigenfunctions sin( nx
) for n= 2. Now suppose that the
initial condition had beenf(x) =x( x). Then the initial
condition is not an eigenfunctionand hence can not be satisfied for
any n. Does this mean that our method doesnt work forthis case?
But, the PDE is linear and homogeneous (as are the boundary
conditions) so theprinciple of superposition can be applied. Doing
so yields
u(x, t) =Nn=1
cnun(x, t),
or, more generally
u(x, t) =n=1
cnun(x, t) =n=1
cnXn(x)Tn(t) =n=1
cnenkt sin(
nx
).
Now applying the initial condition yields
f(x) =n=1
cnXn(x), i.e. x( x) =n=1
cn sin(nx
).
At this point a few questions come to mind:
1. Does the infinite series above converge?
2. What class of functions can be represented by an infinite
series of the type given above?
3. If a functions f can be represented by such a series, how
does one find the constantcns?
1.7 Physical basis of the Heat EquationConsider a region R3 (see
Figure 1.1), consisting of some substance with boundary givenby =S.
Let us introduce the following notation:
Figure 1.1: Region
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1.7. PHYSICAL BASIS OF THE HEAT EQUATION 19
u temperature at the point (
x,y,z )
at timet
;K thermal conductivity of substance at (x,y,z );
density of substance at (x,y,z );
c heat capacity per unit mass at (x,y,z );
J heat flux vector (gives magnitude and direction of heat
flow);
h rate of internal heat generation per unit volume at (x,y,z
).
The above quantities have the following physical dimensions:
[u] T; T temperature;
[K] H
LT t; H heat;
[] m
L3; t time;
[c] H
mT; L length;
[h] H
L3t; m mass.
Suppose we let
dV an element of volume;
dS an element of surface area;
n outward pointing unit vector ofdS.
The vector
upoints in the direction of the most rapid increase ofu, that
is
uis orthogonalto the surfaceu constant. We useFouriers law,
which states that heat flows from regions
of high temperature to regions of low temperature, where the
flux
J satisfies
J = K
u. (1.19)
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20 CHAPTER 1. INTRODUCTION
Figure 1.2: Flux J through a surface element dS
We have
Amt. of heat
in vol. element
= mass
heat capacity
temp. =cu dV
;
Total amt. ofheat in
=
cudV;
Rate at which heatenters through dS
=
J
n dS;
Total rate at which heat entersthrough the boundary S
= S
J
n dS;
Total rate at which heatis generated internally
=
h dV.
Conservation of energy implies thatRate of change
of heat in
=
Rate at which heat
enters through boundary S
+
Rate at which heat
is generated internally
,
in other words
t
cudV=
S
J
n dS+
h dV.
Using the divergence theorem from advanced calculus, we can
rewrite this as
cu
tdV=
J dV +
h dV,
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1.8. PHYSICAL BASIS OF THE WAVE EQUATION 21
or
c
u
t+
J h
dV= 0.
This must hold for an arbitrary region so the integrand must be
identically zero. Usingthe definition for the flux vector in Eq.
(1.19), we get
u
t =
1
c
(K
u) + h
c . (heat equation)
It is often the case that , c andKare constant. In that case we
define
k:= K
c, F :=
h
c
The constant k is called the thermal diffusivityand F, which is
not necessarily constant,is called the forcing. The heat equation
then becomes:
u
t =k 2u +F . (heat equation in standard form) (1.20)
Written explicitly in terms of Cartesian coordinates:
u
t =k
2u
x2+
2u
y2+
2u
z2
+F, (3-d heat equation) (1.21)
u
t =k
2u
x2+F. (1-d heat equation) (1.22)
1.8 Physical basis of the Wave Equation
Here we will only give a derivation for the onedimensional wave
equation. Consider a stringstretched over an interval [a, b]. Let
us introduce the following notation:
u displacement above the x-axis,
surface tension;
density per unit length;
t time;
h vertical force per unit length;
, angles indicated in the figure.
The above quantities have the following physical dimensions:[u]
L; L length
[] F; F force;
[] m
L; m mass;
[h] F
L.
