18.155 LECTURE 21, 1 DECEMBER 2015
RICHARD MELROSE
Abstract. Notes before and after lecture – if you have questions, ask!
Before lecture
• Last time I showed that the flat wave operator
= Dt2 − ∆
(1)
has a (necessarily unique) forward and tempered fundamental solution
E+ (t, x) = δ(t)δ(x), E+ ∈ S(Rn+1 ), supp(E+ ) ⊂ {t ≥ |x|} = K
(2)
which has support in the ‘wave cone’ (this looks the same as the characteristic cone, τ ≥ |ξ| only because of our choice of coordinates, they are really
dual to each other).
• Now, consider f ∈ Cc∞ (R × Rn ) then u = E+ ∗ f ∈ S 0 (Rn+1 ) sastifies
u = f, supp(u) ⊂ {t ≥ 0}.
(3)
In fact the support of u is contained in the union of the cones t−t0 ≥ |x−x0 |
over (t0 , x0 ) ∈ supp(f ). This follows from the estimate on the supports of
convolutions that we proved earlier.
• We know that u is the unique tempered solution to (3). If we multiply it
by e−at for a ≥ 0 then it is still tempered and satisfies
e−at u = ((Dt − ia)2 − ∆)(v) = g, v = e−at u, g = e−at f.
(4)
Taking the Fourier transform in all variables we find that
(5)
((τ − ia)2 − |ξ|2 )v̂(τ, ξ) = ĝ(τ, ξ) =⇒ |v̂| ≤
1 −1
a (a + |τ | + |ξ|)−1 |ĝ|, (τ, ξ) ∈ Rn+1 .
2
Here I have used the estimate
|(τ − ia)2 − |ξ|2 | = |(τ − ia − |ξ|)||(τ − ia + |ξ|) ≥ |τ | + a + |ξ|
(6)
which follows since when τ ≤ 0 we can estimate the first term from below
by a and the second from below by
1
(|τ | + |ξ|)2 + a2 ) 2
and on the other side the other way.
•
(7)
Lemma 1. The operator g = e−at f −→ v = e−at (E+ ∗ f ) extends by
continuity to a bounded operator
E(a) ∈ B {g ∈ L2 (Rn+1 ); supp(g) ⊂ {t ≥ 0}} , kE(a)k ≤ a−2 .
1
2
RICHARD MELROSE
• Now, suppose that V ∈ Cc∞ (Rnx ) is a ‘potential’ (you can see from the
argument below that we can get away with far less regularity than this) –
meaning we would like to solve
u − V u = f
(8)
where f is given with support in t ≥ 0 and we want u to have support there
too. Multiply through by e−at , observe that this commutes with V, and
conclude that we can try to solve
((Dt − ia)2 − ∆)v − V v = g,
(9)
where v and g are as above (but we don’t know that v exists). If we write
this in the form of an ‘integral equation’ with unknown v we get
(10)
((Dt − ia)2 v − ∆)v = V v + g −→ v = E(a)g + E(a)V v =⇒ (Id −E(a)V )v = E(a)g.
•
Proposition 1. If a is large enough (just related to the supremum norm
of V ) then Id −E(a)V is invertible as an operator as in (7) and v =
(Id −E(a)V )−1 E(a)g is the unique tempered (in fact arbitrary distributionals) solution of (9) with support in t ≥ 0.
Department of Mathematics, Massachusetts Institute of Technology
E-mail address: rbm@math.mit.edu