Differential Analysis II Lectures 22-24

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1 Lecture : 3 April Differential Analysis II Lectures - Instructor: Larry Guth Trans.: Kevin Sacel.1 Precursor to Schrödinger Equation Suppose we are trying to find u(, t) with u(, 0) = f() satisfying one of the following three PDEs: A: t u = u B: t u = u C: t u = i u We can tae then tae the Fourier transform of each of these, corresponding to A : t û(w, t) = (πiw) u(w, t) B : t u(w, t) = (πiw) u(w, t) C : tû(w, t) = i(πiw) u (w, t) Each of these can now be formally solved in this frequency domain, since these are just separable differential equations (and the initial data f corresponds to initial data f). However, such solutions might or might not correspond to legitimate solutions u, since if the formal solution for û is not well enough behaved in w, then one cannot tae the Fourier transform to get a corresponding solution u. We analyze each case separately. A: Solving yields û(w, t) = ep(cw t)f (w). For almost all initial data, this solution goes bananas for t > 0, because ep(cw t) grows too fast. Thus, we can t possibly tae a Fourier transform to find a legitimate solution u(, t). For eample, if f = e π, which decays quite nicely, then f = e πw, which although it still decays nicely, is taen over by ep(cw t) for any t > 0. Of course, if f C c, then we get a solution. B: Solving yields u(w, t) = ep( Cw t)f(w). For t > 0, for most f, u will become immediately quite smooth. If f L 1, then f is bounded, and so û(w, t) decays rapidly, which means that taing the Fourier transform, u(, t) is C. In fact, we can say a bit more. Proposition 1. u(, 1) L C f L 1 Proof. We have u(, 1) u(w, 1) L 1 w = (πiw) e iw f(w) L 1 (πiw) e iw L 1 f L C f L 1 1

2 Even further, one can use either a scaling argument (since if u(, t) is a solution to the differential equation, so is u(λ, λ t)) or a direct argument (presented below) to prove the following proposition: Proposition. u(, t) L t 1/ f L 1 Proof. u(, t) u(w, t) L 1 w Cw = e t f (w) L 1 e Cwt L 1 f L 1/ t C f L 1 C: Solving in the frequency domain yields (up to rescaling) u(w, t) = e iw t f(w). Note that this means that if f is Schwartz, then so is û, so in particular, if f is itself Schwartz, then so is f, hence u, and so we can tae the Fourier transform and get a Schwartz function u satisfying the PDE. So at the very least, we can solve for f Schwartz, which is a decently large class. Proposition 3. u(, t) L is independent of t for each 0. Proof. By an application of Plancherel s Theorem, this is equal to (πiw) e iw t f(w) L, and so the w eponential term has constant modulus 1 and does not affect this value. What this means is that studying C is much more subtle than studying A or B, since the previous proposition tells us that the total mass in the solution of this equation is independent of t, so things will not just blow up or become more regular as t increases. If we want to study decay, we need to be careful. By scaling, if solutions to this PDE do decay, then it would need to do so in a way resembling the following: u(, t) L t 1/ f L 1. The main question is, does this hold?. The Schrödinger Equation Instead of PDE C from the previous subsection, we will focus on the Schrödinger equation, which as we will see immediately behaves quite similarly to the above. The Schrödinger Equation: Find u(, t), R d, t R, satisfying the initial condition u(, 0) = f() such that t u = i u. In the frequency domain, this amounts to solving the equation û(w, t) = e i w t f (w), and issues of convergence are relatively similar to case C of the previous subsection. In fact, the same eact proves carry over to show: Proposition. If f S, then we get a solution u S. Proposition 5. For any multi-inde I, I u(, t) L is independent of t. We wish to understand the decay of the Schrödinger equation. In particular, we will prove the following proposition: Proposition 6. u(, t) L t d/ f L 1. Remar 7. There is in fact a nice formula for the solution, given by u(, t) = t d/ e i /t f(), so that the proposition becomes trivial. However, we want to have a more generalizable gadget for completing this proof.

3 Proof. We will complete the proof in the net lecture. For now, let us reduce the problem to a lemma. First of all, note that u(, t) = u(w, t)e iw dw = lim e δw e i w t f (w)e iw dw δ 0 Now if we set S δ,t () equal to the inverse Fourier transform of e δ w e i w t with respect to w, then we see that u(, t) = lim S δ,t () f(). δ 0 In particular, we included the term e δ w so that we could tae this Fourier transform in the first place. Hence, we find that u(, t) L lim inf S δ,t L f L 1. It suffices therefore to prove that S δ,t L t d/ uniformly as δ 0. By scaling, we need only prove in fact that S δ,1 L is uinformly bounded by a constant. But this integral breas up into a product as follows: S δ,1 () = e δ w e i w e iw dw R d d = e δw j e iw j e iw j j dw j j=1 R So it suffices to prove the one-dimensional case, the second of the following two lemmas: Lemma 1. e δw e iw dw 1 uniformly in δ. R Lemma. e δw e iw e iw dw 1 uniformly in δ,. R We will prove these net time. Again, the second of these lemmas is what we actually need, but proving the first gives us a good indication for how we might prove the second. Lecture 3: 6 April Completion of proof Proof Setch of Lemma 1. Set η = w. Then changing variables accordingly, we see e δw e iw dw = e δη η 1/ e iη dη, 0 0 and this ind of loos similar to the integration estimates in the Gauss circle problem. A bounded distance away from 0, we can use the triangle inequality, while away from 0, we can integrate by parts. The main term of integration will be η 3/ e iη, which is integrable. Proof Setch of Lemma. This is similar, but now e iw is replaced with e i(w+w ). But w + w is just a quadratic centered around w 0 := /, and the same proof carries over where we change coordinates to integrate around this center instead. Remar 8. There is an alternative proof of these two. The oscillatory term gets faster and faster away from w 0, and so if we brea the integral up into corresponding full periods in the real and imaginary parts, we will get alternating sums which are at most as large as the first term, which can be bounded by the triangle inequality. 3

4 3. Statement of Strichartz Theorem In fact, we see that by the same method, we can study S δ,t () still for all δ,, but also for all t, not just t 0. Say we fi δ = 1. Of course, we now that the mass given by S 1 (t, ) L remains constant in time. At time t = 0, S 1 (t, ) has about height 1 and width 1, while at time t >> 1, it is defocused with height about t d/ and width about t. For negative t, we get focusing data, while for positive t, we get defocusing data, because if we solve the Schrödinger equation with this initial data, we just flow with t getting larger, and for starting at negative t, this collimates around t = 0 and then defocuses again. Qualitatively, the focusing and defocusing data are quite similar, so one epects it to be difficult to use purely qualitative estimates to learn about what happens as we flow in time. Nonetheless, we might be interested in what inds of estimates we can still obtain. To motivate this, consider the following question: Question 1. For which p do we have the estimate u(, t) p L,t f L = mass 1/? Intimately related is the question of how large is the set {(, t): u(, t) > t}? (d+1) d Remar 9. For the function S 1 we have been considering, we get such an estimate if and only if p >, which we will see later. So what about solutions to the Schrödinger equation? Note that if u solves the Schrödinger equation, then u λ (, t) := u(/λ, t/λ ) also solves the Schrödinger equation with initial data f λ () = f(/λ), and we have d+) u p λ L λ ( /p u L p, f λ L λ d/ f L.,t,t Hence, the only possible p which could possibly answer our question is when p = (d + )/d. In fact, this equation is satisfied uniformly for all S δ uniformly in δ if and only if p = (d + )/d. Of course, it is worth noting that S δ,t () is by definition the solution to Schrödinger equation with initial data given by a Gaussian (in particular the inverse Fourier transform of e δw ). So one might epect the following theorem: Theorem 10 (Strichartz, 1970 s). Set p = (d + )/d. Then the estimate u(, t) L p,t u(, 0) L holds for all solutions to the Schrödinger equation. The proof of this theorem will come later, but even here, we stumble upon a general sort of analytic question: Question. Let T be a given linear operator. Then find all p, q such that there is an inequality of the form T f L p f L q. Eample 11. Just something to thin about: Consider g() := e i (1 + ) 1/, and let T f := f g. Then for what (p, q) do we get an inequality as above. This is of course a difficult question, and currently, there is no universal way to solve this question. However, there are many tools which can be used to attac it. We will see how interpolation is one etremely useful tool, allowing us to tae two given estimates of this form and yield an interval-worth of these estimates. 3.3 Interpolation Interpolation of L p spaces comes in two flavors, both from the first half of the 0th century. The first is due to Riesz-Thorin, while the second is due to Marciniewicz. We will follow the Marciniewicz approach instead, although see Remar 13. We begin with a special case of interpolation. Theorem 1 (Interpolation, Special Case). If T is a linear operator, 1 p < r < q, and T f L p p q r f L p for all f L and T f L q f L q for all f L, then for all f L, we have T f L r C(p, q, r) f L r.

5 Remar 13. In this special case, Riesz-Thorin actually gives C(p, q, r) = 1, which is optimal. What the Riesz-Thorin approach gives in terms of better bounds, the Marciniewicz approach maes up for in slightly more general hypotheses, to be stated later, ecept for some fairly uninteresting cases which Marciniewicz doesn t cover but Riesz-Thorin does. We shall not concern ourselves with these matters from here on. We won t quite finish the proof today, but we ll still do a little bit of wor. The idea of the proof of interpolation is to loo at f and T f at different height scales. Define Sf := {: f() < +1 }, f = f χ S f. p Then we have f = f, and we get f L p S (f) p. The idea now is to use estimates on S l (T f ) and glue them together carefully to arrive at a proof. We shall wor through this over the remainder of this lecture and into net lecture. We begin with the estimate on S l (T f ). Proposition 1. If T f p f p for all f L p, then S l (T f ) p ( l)p S (f). Proof. We have p p p S l (T f ) lp T f p p f p S (f). Hence, the proposition above gives us bounds on S l (T f ) with respect to both p and q. Say p < q. Then we see that if l >, then the q-bound is better, and vice versa if l <. This allows us to prove the following corollary: Corollary 15. If T f p f p for all f L p and T f q f q for all f L q with p < q, then there is a constant ɛ(p, q, r) such that for p < r < q, (T f ) ɛ l L l r f L r. Proof. Suppose first that l. Then using the q-bounds, we find and so (T f ) l r r rl S l (T f ) rl q (l )q S (f) = f r r rl r q (l )q (T f ) (l )(1 q/r) q/r f l r r (l )(1 q/r) q/p and 1 q/r < 0. We have a similar bound when l, but instead we get our ɛ value is p/r 1 (and the etra q/p term disappears, though this doesn t matter). Lecture : 8 April Finishing the special case Let s add some etra notation. Set V λ (f) := {: f() > λ}. Of course, V λ (f) λ p T f p p. Now, we will actually show that the theorem we stated last time is true in a slightly weaer setting (and in fact, we ve already down some of the wor). Theorem 16 (Interpolation, Tae II). Suppose T is sublinear (meaning T (f + g)() T f() + T g() ), and we have the so-called wea type estimates V λ (T f) λ p f p p for all f L p q and Vλ(T f) f q for all f L q, say with 1 p < r < q. Then T f L r C(p, q, r) f L r. Remar 17. If we had the wea type estimates, but with constants A p and A q, then our concluding estimate would have an etra factor of A α p A 1 α q, with 1 = α 1 + (1 α) 1 r p q. This is a simple scaling argument, so it suffices to prove for A p = A q = 1. We have Lemma 3. V λ(t f ) ( /λ) p S (f) (and also for q). 5

6 Proof. V λ (T f ) λ p f p p S (f) p. Lemma. V l(t f ) lr S (f) r ɛ l Proof. Essentially done last time. Corollary 18. T f L r f L r Proof. We have T f l l χv l (T f ), so taing L r norms and using the previous lemma gives the result. Our goal is now to add the contributions from all of these height scales. The naïve approach would be to simply use the sublinearity of T to give T f L r T f L r f L r At this point, we d have trouble bounding by f L r. Instead, we want to use the info from our bounds on S (f) to change the first inequality. The correct way to do this is with the following lemma, which is the tricy part of the argument. Lemma 5. V T f ( l ) S r lr ɛ l (f) Proof. This would be easy if T f() l implied that there eists a with T f () l. Unfortunately, we don t have this, but we can be a little bit more clever. For each l, choose weights w with w = 1, which we will choose later. Then the sublinearity of T means that if T f() l, then there is some with T f () w l. So we find that V l(t f) V (T f ) S (f) r lr w c ɛ l w l So if we choose w 1 with constant chosen so that (l ) w = 1, then the w term is dominated by the eponential decay of the ɛ l term, and so we can replace ɛ by an arbitrarily smaller value and obtain the same result. The rest of the proof is now easy. T f r r V l(t f) l l lr S (f) r ɛ l Summing in l, we see this is on the order of S (f) r, which is itself about f r r.. Statement of the General Case The following is essentially the content of a problem on the th PSet, and is again discussed in the April 15 notes. Theorem 19 (Marciniewicz Interpolation). Suppose 1 p i q i for i = 0, 1 with q 0 = q 1. If T sublinear and for all f in the appropriate spaces we have the wea type estimates that for all λ, c. (V λ (T f) λ qi ) 1/qi A i f pi then for each 0 < θ < 1, setting 1 = θ 1 + (1 θ) 1 p θ p 0 p 1 and similarly for q θ, we have strong type estimates of the form θ T f q θ θ A 1 0 A θ 1 f pθ. In other words, the space of estimates of the form T f q f p is conve in the (1/p, 1/q)-domain, and in order to interpolate, one only needs wea estimates at the boundary. 6

7 Question 3. Here s a challenge. Suppose we are given (p 0, q 0 ), (p 1, q 1 ) as in the hypotheses of the theorem. Is there a sublinear operator T such that T f q f p if and only if (p, q) lie in the interpolating interval between these two endpoints? How about if we choose any conve subset of the (1/p, 1/q)-domain? Remar 0. If T is the Fourier transform on R d, then we have strong estimates on L by Plancherel (in fact equality) and L by the triangle inequality. Hence, we get strong estimates by interpolation of the form f L q f L p with p q = 1. In fact, these are the only such estimates for the Fourier transform. 7

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