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√
5) If u(x, t) is a solution of (7.1), so is the dilated function u( ax, at), for any a > 0.
√
Prove this by the chain rule: Let v(x, t) = u( ax, at). Then v t = [∂(at)/∂t]u t =
√ √ √ √
au t and v x = [∂( ax)/∂x]u x = au x and v xx = a · au xx = au xx .
Our goal is to find a particular solution of (7.1) and then to construct all the other
solutions using property 4. The particular solution we will look for is the one, denoted
Q(x, t), which satisfies the special initial condition

Q(x, 0) = l for x > 0, Q(x, 0) = 0 for x < 0. (7.3)

The reason for this choice is that this initial condition does not change under dilation.
We’ll find Q in three steps.
Step 1. We’ll look for Q(x, t) of the special form
x
Q(x, t) = g(p) where p = (7.4)
4kt
√
and g is a function of only one variable (to be determined). (The 4k factor is included
only to simplify a later formula.)
Why do we expect Q to have this special form? Because property 5 says that the
√
equation (7.1) doesn’t ”see” the dilation x → ax, t → at. Clearly, (7.3) does not
change at all under the dilation. So Q(x, t), which is defined by the conditions (7.1) and
(7.3), ought not see the dilation either. How could that happen? In only one way: if Q
√ √
depends on x and t solely through the combination x/ t. For the dilation takes x/ t
√ √ √ √
into ax/ at = x/ t. Thus let p = x/ 4kt and look for Q which satisfies (7.1) and
(7.3) and has the form (7.4).
Step 2. Using (7.4), we convert (7.1) into an ODE for g by use of the chain role:
dg ∂p 1 x
0
Q t = = − √ g (p),
dp ∂t 2t 4kt
dg ∂p 1
0
Q x = = √ g (p),
dp ∂x 4kt
dQ x ∂p 1
00
Q xx = = g (p),
dp ∂x 4kt

1 1 1
0
00
0 = Q t − kQ xx = − pg (p) − g (p) .
t 2 4
Thus,
0
00
g + 2pg = 0.
R 2
This ODE is easily solved using the integrating factor exp 2pdp = exp(p ). We get
2
0
g (p) = c 1 exp(−p ) and
Z 2
Q(x, t) = g(p) = c 1 e −p dp + c 2 .
Step 3 We find a completely explicit formula for Q. We’ve just shown that
√
Z
x/ 4kt 2
Q(x, t) = c 1 e −p dp + c 2 .
0

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