Math 124A – October 28, 2010
Viktor Grigoryan
10
Heat equation: interpretation of the solution
Last time we considered the IVP for the heat equation on the whole line
u
t

ku
xx
= 0
(
∞
< x <
∞
,
0
< t <
∞
)
,
u
(
x,
0) =
φ
(
x
)
,
(1)
and derived the solution formula
u
(
x, t
) =
Z
∞
∞
S
(
x

y, t
)
φ
(
y
)
dy,
for
t >
0
,
(2)
where
S
(
x, t
) is the heat kernel,
S
(
x, t
) =
1
√
4
πkt
e

x
2
/
4
kt
.
(3)
Substituting this expression into (2), we can rewrite the solution as
u
(
x, t
) =
1
√
4
πkt
Z
∞
∞
e

(
x

y
)
2
/
4
kt
φ
(
y
)
dy,
for
t >
0
.
(4)
Recall that to derive the solution formula we first considered the heat IVP with the following particular
initial data
Q
(
x,
0) =
H
(
x
) =
1
,
x >
0
,
0
,
x <
0
.
(5)
Then using dilation invariance of the Heaviside step function
H
(
x
), and the uniqueness of solutions to
the heat IVP on the whole line, we deduced that
Q
depends only on the ratio
x/
√
t
, which lead to a
reduction of the heat equation to an ODE. Solving the ODE and checking the initial condition (5), we
arrived at the following explicit solution
Q
(
x, t
) =
1
2
+
1
√
π
Z
x/
√
4
kt
0
e

p
2
dp,
for
t >
0
.
(6)
The heat kernel
S
(
x, t
) was then defined as the spatial derivative of this particular solution
Q
(
x, t
), i.e.
S
(
x, t
) =
∂Q
∂x
(
x, t
)
,
(7)
and hence it also solves the heat equation by the differentiation property.
The key to understanding the solution formula (2) is to understand the behavior of the heat kernel
S
(
x, t
). To this end some technical machinery is needed, which we develop next.
10.1
Dirac delta function
Notice that, due to the discontinuity in the initial data of
Q
, the derivative
Q
x
(
x, t
), which we used in
the definition of the function
S
in (7), is not defined in the traditional sense when
t
= 0. So how can
one make sense of this derivative, and what is the initial data for
S
(
x, t
)?
It is not difficult to see that the problem is at the point
x
= 0. Indeed, using that
Q
(
x,
0) =
H
(
x
)
is constant for any
x
6
= 0, we will have
S
(
x,
0) = 0 for all
x
different from zero. However,
H
(
x
) has a
jump discontinuity at
x
= 0, as is seen in Figure 1, and one can imagine that at this point the rate of
growth of
H
is infinite. Then the “derivative”
δ
(
x
) =
H
0
(
x
)
(8)
is zero everywhere, except at
x
= 0, where it has a spike of zero width and infinite height. Refer to Figure
2 below for an intuitive sketch of the graph of
δ
. Of course,
δ
is not a function in the traditional sense,
1