Partial Differential Equations

u t (x, t) =
∂ 2 u
∂x i ∂x j
(x, t) =
∫ t
0
∫ t
This proves the first claim and allows the calculation
0
1.3 The Heat Equation 23
∫
∫
Φ(y, s)f t (x − y, t − s) dy ds + Φ(y, t)f(x − y, 0) dy,
R n R
∫
n ∂ 2
Φ(y, s) f(x − y, t − s) dy ds.
R ∂x n i ∂x j
∫ t ∫
u t (x, t) − ∆u(x, t) = Φ(y, s) (( ∂
∂t − ∆ )
x)
f(x − y, t − s) dy ds
0 R
∫
n
+ Φ(y, t)f(x − y, 0) dy
R
∫ n t ∫
= Φ(y, s) (( ∂
∂t − ∆ )
x)
f(x − y, t − s) dy ds
ε R
} n {{ }
I ε
∫ ε ∫
+ Φ(y, s) (( ∂
∂t − ∆ )
x)
f(x − y, t − s) dy ds
0 R
} n {{ }
J ε
∫
+ Φ(y, t)f(x − y, 0) dy .
R
} n {{ }
K
A suitable estimate for J ε follows from Lemma 1.3.3:
|J ε | ≤ (‖f t ‖ ∞ + ‖D 2 f‖ ∞ )
∫ ε
0
∫
R n Φ(y, s) dy ds ≤ εC.
Integration by parts and Green’s formula yield an We obtain an estimate for I ε since Φ solves the heat
equation:
I ε =
=
=
∫R n ∫ t
ε
∫ t ∫R n
ε
Φ(y, s) (( − ∂ ∂s − ∆ y)
f(x − y, t − s)
)
ds dy
(( ∂
∂s − ∆ ) )
y Φ(y, s) f(x − y, t − s) ds dy
∫
∫
+ Φ(y, ε)f(x − y, t − ε) dy −
R
∫
n Φ(y, ε)f(x − y, t − ε) dy − K.
R n
For every (x, t) ∈ R n × ]0, ∞[, we together obtain
R n Φ(y, t)f(x − y, 0) dy
∫
u t (x, t) − ∆u(x, t) = lim Φ(y, ε)f(x − y, t − ε) dy = f(x, t),
ε→0
R n
where the limit is computed by the procedure of the proof of Theorem 1.3.4 (decomposition of the
integration domain into B(x; δ) and R n \ B(x; δ)). Therefore, the second claim is proven.
The last claim follows since ‖u(·, t)‖ ∞ ≤ t‖f‖ ∞ → 0 for t → 0.
Exercise 1.3.7. Let I be an interval and F : I ×I → R a continuous function which is differentiable with
respect to the first variable. Show that g : I → R defined by g(t) := ∫ t
F (t, s) ds is differentiable with
0
derivative
g ′ (t) = F (t, t) +
∫ t
0
F t (s, t) ds.