Download SOLUTION OF PROBLEM 1/PAGE 158 Problem: Show that the

SOLUTION OF PROBLEM 1/PAGE 158
Problem: Show that the distributional solution of
ut = ∆u + f (x, t) in U ⊂ Rn+1
is smooth, whenever f ∈ C ∞ (U ).
Proof. It suffices to show that for every fixed point ξ = (x, t) ∈ U , we have ε > 0, so
that on B(ξ, ε) the distribution u coincides with a C ∞ (U ) function.
Let, as in the proof of the Theorem on p. 155, φ ∈ C0∞ (B(ξ, 4ε)), so that B(ξ, 4ε) ⊂
U and φ = 1 on B(ξ, 3ε). Introduce w = φu, which is of course a distribution too.
Compute (in a distributional sense)
(∂t − ∆)w = [(∂t − ∆)u]φ + [(∂t − ∆)φ]u − ∇φ · ∇u = f φ + [(∂t − ∆)φ]u − ∇φ · ∇u.
Observe that the first term on the right hand side is a C ∞ (Rn+1 ) function, while
H = [(∂t − ∆)φ]u − ∇φ · ∇u is a distribution, supported on B(ξ, 4ε), which vanishes
on B(ξ, 3ε) (since φ is a constant there). We need the following
Lemma 1. Let z1 , z2 be two distributions, so that z1 = (∂t − ∆)z2 and z2 = 0 on
B(ξ, 3ε), suppz2 ⊂ B(ξ, 4ε). Then, there is the representation formula
z2 = (ψ K̃) ∗ z1
on B(ξ, ε)
where the convolution is both in t and x and K̃(t, x) is defined on p.150 (equation
(22)) and ψ is a function, so that ψ = 1 on B c (x, 2ε), ψ = 0 on B(x, ε).
This lemma is proved similar to (36) on page 155. Essentially, it avoids integrating
K̃(y, s) close to s = 0, so one can convert the formula
Z ∞Z
Z tZ
K̃(y, s)z1 (x − y, t − s)dsdy
K(y, s)z1 (x − y, t − s)dsdy =
0
Rn
−∞
Rn
We can now represent w = w1 + w2 , where
(∂t − ∆)w1 = f φ
(∂t − ∆)w2 = H
Clearly w1 is smooth everywhere, since the right hand side is C0∞ (Rn+1 ) function, so
it remains to show that w2 is smooth on B(ξ, ε) as well (recall u coincides with w on
B(ξ, 3ε)). We now use Lemma 1 to get
w2 = (ψ K̃) ∗ H
and this is of course a smooth function, since it is written as a convolution of Schwartz
function with a distribution H.
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