NINTH ASSIGNMENT, DUE NOVEMBER 20 IN CLASS18.155 FALL 2001RICHARD MELROSEProblem 1. Work out the elementary behavior of the heat equation.i) Show that the function on R × Rn, for n ≥ 1,F (t, x) =(t−n2exp−|x|24tt > 00 t ≤ 0is measurable, bounded on the any set {|(t, x)| ≥ R} and is integrable on{|(t, x)| ≤ R} for any R > 0.ii) Conclude that F defines a tempered distibution on Rn+1.iii) Show that F is C∞outside the origin.iv) Show that F satisfies the heat equation(∂t−nXj=1∂2xj)F (t, x) = 0 in (t, x) 6= 0.v) Show that F satisfies(1) F (s2t, sx) = s−nF (t, x) in S0(Rn+1)where the left hand side is defined by duality “F (s2t, sx) = Fs” whereFs(φ) = s−n−2F (φ1/s), φ1/s(t, x) = φ(ts2,xs).vi) Conclude that(∂t−nXj=1∂2xj)F (t, x) = G(t, x)where G(t, x) satisfies(2) G(s2t, sx) = s−n−2G(t, x) in S0(Rn+1)in the same sense as above and has support at most {0}.vii) Hence deduce that(3) (∂t−nXj=1∂2xj)F (t, x) = cδ(t)δ(x)for some real constant c.Hint: Check which distributions with support at (0, 0) satisfy (2).viii) If ψ ∈ C∞c(Rn+1) show that u = F ? ψ satisfies(4)u ∈ C∞(Rn+1) and supx∈Rn, t∈[−S,S](1 + |x|)N|Dαu(t, x)| < ∞ ∀ S > 0, α ∈ Nn+1, N.12 RICHARD MELROSEix) Supposing that u satisfies (4) and is a real-valued solution of(∂t−nXj=1∂2xj)u(t, x) = 0in Rn+1, show thatv(t) =ZRnu2(t, x)is a non-increasing function of t.Hint: Multiply the equation by u and integrate over a slab [t1, t2] × Rn.x) Show that c in (3) is non-zero by arriving at a contradiction from theassumption that it is zero. Namely, show that if c = 0 then u in viii) satisfiesthe conditions of ix) and also vanishes in t < T for some T (depending onψ). Conclude that u = 0 for all ψ. Using prop erties of convolution showthat this in turn implies that F = 0 which is a contradiction.xi) So, finally, we know that E =1cF is a fundamental solution of the heatoperator which vanishes in t < 0. Explain why this allows us to show thatfor any ψ ∈ C∞c(R × Rn) there is a solution of(5) (∂t−nXj=1∂2xj)u = ψ, u = 0 in t < T for some T.What is the largest value of T for which this holds?xii) Can you give a heuristic, or indeed a rigorous, explanation of whyc =ZRnexp(−|x|24)dx?xiii) Explain why the argument we used for the wave equation to show that thereis only one solution, u ∈ C∞(Rn+1), of (5) does not apply here. (Indeedsuch uniqueness does not hold without some growth assumption on u.)Department of Mathematics, Massachusetts Institute of TechnologyE-mail address:
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