Nonlinear Wave Equations

Nonlinear Wave Equations Notes taken from lectures of Professor Monica Visan Yunfeng Zhang 04 Contents Fundamental Solutions Symmetries and Conservation Laws 7 3 The Energy-Flux Identity 9 4 Morawetz Identity 5 Strichartz Estimates 5 6 Local Well-posedness I 7 7 Local Well-posedness II 9 8 Local Well-posedness III 3 9 An Ill-posedness Result 5 0 Global Well-posedness 8

Fundamental Solutions We would love to find explicit solutions to the wave equation tt u(t, x) x u(t, x) + F (t, x) = 0 with initial conditions u(0, x) = u 0 (x), t u(0, x) = u (x). In matrix notation, the wave equation can be rewritten as ( ) ( ) ( ) ( ) u 0 u 0 t = +. u t 0 F Our strategy is to first find fundamental solution R(t, x) to the homogenous equation (that is when F = 0) with initial data u(0, x) = 0 and t u(0, x) = δ 0 (x). Then t R should be the fundamental solution to one with initial data u(0, ) = δ 0 and t u(0, ) = 0. Combined with Duhamel s principle, we have that the solution to our original equation is u(t) = t R(t) u 0 + R(t) u u t t 0 R(t s) F (s)ds. Let s consider the one space dimension case first, to find the fundamental solution R as indicated above. The wave equation becomes u tt u xx = 0. Let v(ξ, η) = u( ξ+η, ξ η ), we have v ξ,η = 0. This gives v = f(ξ) + g(η). So u = f(x+t)+g(x t). Using initial data, we have f(x)+g(x) = 0 and f(x) ġ(x) = δ 0. So f = ġ = δ 0 then f = g = H, H denoting the Heaviside function. Thus the fundamental solution R(t, x) = H(x + t) H(x t). Then the full solution is u(t) = t R(t) u 0 +R(t) u t 0 R(t s) F (s)ds = (u 0(x + t) + u 0 (x t)) + t t u (x y)dy t s F (s, x y)dyds. t 0 (t s) Remark.. We have Huggen s principle: solutions for -dim wave equation depend on u 0 at x ± t, u on [x t, x + t], F on the solid-triangle {(s, y) : 0 s t, x y t s}. Let s move to three space dimensions. In order to find the fundamental solution, we introduce the method of taking spherical average. Fix x. Let ũ(x, t, r) = 4π u(t, x + rw)dw. In order to find the PDE of ũ, let s rewrite the Laplacian in spherical coordinates. Let s do the computation in R n. xj = r r x j + i θi θ i x j,

xj x j = rr ( r x j ) + i θi r r θi r + r x j x j x + j i θi r r x j θi x j + i,k θi θk θ k x j θ i x j + i θi θ i x j. Summing over j, noting that r θ i = 0, we get = rr + n r + θ k θ i θi θk + r x j x j i,k,j i,j In fact, we can show that i,k,j θi θk θ k x j θ i x j + i,j θi θ i x j. θi θ i x j = r S n u, where S n is the Laplace-Beltrami operator on the sphere. So using spherical coordinates, u solves tt u rr u r ru r Su = 0. Integrate it on the sphere y x = r, noting that S S u = 0, we get tt ũ rr ũ r rũ = 0. Now let v(t, r) = rũ(t, r), then tt v rr v = 0. v(0, r) = rũ(0, r) = 0, t v(0, r) = r 4π S δ 0 (x + rw)dw. Now let r be any real number and x 0. We want to figure out what is r S δ 0 (x + rw)dw as a distribution on R. Pick a test function φ(r), φ(r)r R S B ɛ (0) χ B ɛ(0)(x + rw)dwdr φ(r) = R S r χ B ɛ(0)(x + rw)r dwdr φ(r) = 0 S r χ B ɛ(0)(x + rw)r φ( r) dwdr + 0 S r χ B ɛ(0)(x rw)r dwdr, which is the average of φ( y ) φ( y ) y y on the ball y x ɛ minus the average of on the same ball. Letting ɛ 0, we will have the limit of the above 3

to be φ( x ) x φ( x ) x. So r S δ 0 (x + rw)dw = x (δ x δ x ). Then by the solution to one space dimensional wave equation, Finally, v(t, r) = (H(r + t) H(r t)) r (δ x δ x ). 4π x v(t, r) u(t, x) = ũ(x, t, 0) = lim = r v(t, 0) r 0 r = (δ(r + t) δ(r t)) r (δ(r x ) δ(r + x )) 4π x = 8π x (δ(r + t x ) δ(r t x ) δ(r + t + x ) + δ(r t + x )) r=0 = (δ(t x ) + δ(t + x )) 4π x = { 4πt 4πt δ(t x ), t > 0 δ(t + x ), t < 0 So u(t, x) = t R(t) u 0 + R(t) u t 0 R(t s) F (s)ds, where t R(t) u 0 = t ( u 0 (x tw)t dw) 4πt S = 4πt u 0 (x tw)t dw w u 0 (x tw)t dw S 4πt S = 4πi u 0 (y)ds t (y) w u 0 (y)ds t (y), 4πt t 0 x y =t x y =t R(t) u = u (y)ds t (y), 4πt x y =t t R(t s) F (s)ds = F (y)ds t (y)ds. 4π(t s) 0 x y =t s For two space dimensions, we use Hadamard method of descent: the two dimension problem is a particular case of three dimension problem where the initial data does not depend on the last coordinate. Thus we should have, u(t, x) denoting the solution to the two dimension problem, ũ denoting the three dimension one, with initial data u(0) = 0 and t u(0) = u, u(t, x) = ũ(t, x) = 4πt 4 x ỹ =t u (y)ds t (ỹ).

Here x = (x, 0), ỹ = (y, y 3 ). We parameterize the sphere x ỹ = t by y (y, ± t x y t ). Then ds t (ỹ) =. So t x y u(t, x) = πt x y t t u (y) t x y dy. From this we get the fundamental solution to be R(t, x) = π t x χ B t (x). Then we have Huygen s principle: the solution depends on u 0, u on { x y t} and F on the solid cone {(s, y) : 0 s t, x y t s}. We last state the results of fundamental solution to general dimensions. We have for odd dimensions, for even dimensions, R(t, x) = C d ( t t) d 3 R(t, x) = C d ( t t) d δ(t x ); t t x χ B t (x). Now we move to the Fourier analytic approach. If we do a Fourier transform on the space variable for the homogeneous equation, we will get { tt û(t, ξ) + ξ û(t, ξ) = 0, Solving this linear ODE, we get Then u(t, x) = u 0 (cos t ξ ) + u ( cos t := (cos t ξ ), û(0) = û 0, t û(0) = û. sin(t ξ ) û(t, ξ) = û 0 (ξ) cos(t ξ ) + û. ξ sin t := ( sin t ξ ξ ). Let s introduce the notation sin t ξ ξ ). Thus u(t, x) = cos tu 0 + sin t u. For the inhomogeneous solution, we could apply Duhamel s principle as before to get the solution: u = cos(t)u 0 + sin t t u 0 sin(t s) F (s)ds. We can also use variation of parameters, that is, to find solution of the form cos(t)v 0 (t) + 5 sin t v (t).

With additional assumption we get cos(t) v 0 (t) + sin t v (t) = 0, sin(t) v 0 (t) + cos(t) v (t) = F. From these two equations we can solve for v 0 and v, from which we get v 0 and v thus u. 6

Symmetries and Conservation Laws Let s study symmetries of solutions to the wave equation. Noether s theorem indicates, in a concise sentence, continuous symmetries corresponds to conservation laws. Consider { u + F (u) = 0, u(0) = u 0, t u(0) = u. Let G (u) = F (u). We have time-translation invariance: if u(t, x) is a solution, so is u(t t 0, x) for any t 0. This corresponds to the conservation of energy, E(t) = R d u + u t + G(u)dx. We have space-translation invariance: if u(t, x) is a solution, so is u(t, x x 0 ) for any x 0. This gives rise to the conservation of momentum, P (t) = u t udx. R d Indeed, t P (t) = u tt u + u t u t = R d where u kk u j = u u F (u) u + u t, u k u jk = j(u k ), F (u) u = G(u), from which we get t P (t) = 0. We have time-reversal invariance: u( t, x) is a solution. But this does not correspond to conservation law. We have space rotation and reflection symmetry: u(t, Ox) is a solution for any O in the orthogonal group of R d. This corresponds to conservation of angular momentum, L jk (t) = u t (x k u j x j u k ), j k. R d 7

We have Lorentz invariance: u( t vx, x vt, v v x ) is a solution for any v <. It suffices to check that t t x x = tt xx for t = x = x vt v. We have thus ( t x ( t ) = x ( v v v ) = ( 0 t t x x = ( t, x ) 0 = v ( t, x ) = tt xx. ( v v v ) ( t ( v v x ) ) ( 0 0 ) ( t x ) ( t x ), ), ) ( v v t vx v ) ( t If we examine the above computation closely, we shall then define the Lorentz group to be the matrices that preserve the Minkowski metric g = dt dx dx n, then the solutions to the wave equation is symmetric under this group. Note that this group contains the rotation and reflection on the space as well as the above special Lorentz transformation. What conservation law does the Lorentz group correspond to? x ) and 8

3 The Energy-Flux Identity We introduce the energy-flux identity, which says that if a time-dependent region (think about light cone time sections) is contracting with or faster than the speed of light, then the energy trapped on the region is nonincreasing. Or say, the energy flux through the boundary is non-negative. In fact we may define the energy flux as the energy at the bottom minus energy at the top. Let s demonstrate with the wave equation as usual. Here we assume that G(u) = F (u)du 0. The time dependent region is sections of the light cone, x t. Consider Let s compute t E(t) = E(t) = + = = = = x =t x t x t x =t x =t x =t x t u t + u + G(u)dx. u t + u + G(u)dS t (x) u t u tt + u u t + F (u)u t dx ( uu t )dx ( uu t ) x x ds t(x) Here u r = u x x x, and u = u x u r. u t + u + G(u) + u r u t ds t (x) (u t + u r ) + u + G(u)dS t (x) 0. Remark 3.. In order to make energy flux zero, we need the derivative of the solution to vanish in the direction of t + r and all the angular direction, so the solution should depend only on x t. This solution has to be a fundamental solution, since with no energy flux, the energy at the bottom would be concentrated at the vertex point of the light cone, which could be a positive number. In one dimension, we shall consider a variant E(t) = x t u t + u dx. Any solution of the form u(x t) would make E 9

constant, or say it has zero energy flux across the boundary x = t. Similarly, solution of the form u(x+t) has zero energy flux across the boundary x = t. Now we prove the equipartition of energy for Klein-Gordon equation: { u + m u = 0, u(0) = u 0, t u(0) = u, u 0, u S(R d ) We have the conservation of energy E(u(t)) = R d u t + u + m u dx. We have u t ( u + m u ) = u( u m u) + u t = uu tt + u t = t uu t. From this we have for m 0 t t u t ( u + m u )dxdt = R d (uu t )(t ) (uu t )(t ) which gives t t t 0 t t u t ( u + m u )dxdt R d When m = 0, same computation gives u L t L x u t L t L x E/m E = 4E m, 4E m(t t ) 0, as t. R d (u t u )dxdt u L t L x ([0,t] Rd ) u t L t L x ([0,t] Rd ) E u L t L x ([0,t] Rd ). sin t Now u(t) = cos(t)u 0 + u. cos(t)u 0 L x = cos(t ξ )û 0 L. ξ Then sin(t) u L = sin (t ξ ) û x ξ (ξ) dξ. When d 3, the above integral is integrable and uniformly bounded in t. For d = and d =, we have sin t ξ ξ û (ξ) dξ, ξ 0

since u is Schwartz. sin t ξ ξ û (ξ) dξ ξ t sin t ξ t ξ ξ û (ξ) dξ So at the end of the day, we have sin t L t L x([0,t] R d ) This suffices to give t t 0 ξ t t ξ ξ t dξ dξ { t, d =, d = { t, d = log t, d = (u t u )dxdt 0, as t. R d { t, d = log t, d = The energy flux identity gives the uniqueness of solutions to the wave equation. { u + F = 0 Proposition 3.. Let u solve and let v solve u(0) = u 0, t u(0) = u { v + F = 0 Here F = F (t, x). If u v(0) = v 0, t v(0) = v 0, = v 0, on B r (x 0 ), then u = v on {(t, x) : 0 t r, x x 0 r t}. Proof. Consider w = u v. E(t) = x x 0 r t w + w t dx. Then E(t) E(0) for 0 t r from energy flux. But E(0) = 0 since w vanishes on B r (x 0 ). So w = 0 on the specified light cone. Question* 3.3. We expect finite speed of propagation for general nonlinear wave equations. { u + F = 0 Corollary 3.4. Let u solve Assume u u(0) = u 0, t u(0) = u 0, = 0 on { x x 0 r} and F {():0 t T, x x0 r+t} = 0. Then u {():0 t T, x x0 r+t} = 0.

4 Morawetz Identity Now we introduce the Morawetz identity. The intuition comes from classical mechanics, where t ( x x p) 0. Here H = p, ẋ = p, ṗ = 0. For the wave equation, the momentum density is u t u. The analogue for the wave equation is R u d t [ x x u + ( x x u)]dx. We perform the computation for a general weight A(x) = a(x), in our case a(x) = x so A(x) = x x. Consider the Morawetz action M(t) = u t [A(x) u + (A(x)u)]dx = u t [ a u + a jju]dx. Let s compute t M(t) = (u kk F (u))(a j u j + a jju) + u t (a j u tj + a jju t )dx = u kk a j u j + a j (G(u)) j a jju kk u a jjf (u)u + a j(u t ) j a jju t = a jk u k u j + a j u k u jk + a jj ( F (u)u G(u)) + a jjku k u + a jju k = a jk u k u j + a j u k u jk + a jj ( F (u)u G(u)) + a jjku k u + a jju k = a jk u k u j + a j(u k ) j + a jju k + a jj( F (u)u G(u)) + 4 a jjk(u ) k = a jk u k u j + a jj ( F (u)u G(u)) 4 a jjkku. For d 3. Let a(x) = x. Then a j = x j x, a jk = δ jk { 4πδ, d = 3 ( x ) = d 3 x 3, d 4 p+ u p+ = p (p+) u p+. We get for d = 3, x x j x k x x, a = d x, Take F (u) = u p u, then u p+ F (u)u G(u) = R3 u t M(t) = x x x x u + x p p + u p+ dx + π u(t, 0), which is positive. Similarly we have the formula for d > 3 and t M(t) is positive. Note that u x x x x u = x u. Now we have M(t ) M(t ) = t t x u + x p p + u p+ dxdt + π t t u(t, 0).

Now we would love to control M(t). We have M(t) = u t ( x x u + d x u)dx u t u + u t u(t) x. Lemma 4.. For 0 s < d, we have f S(Rd ) Using this lemma, we get In particular, we get for d 3, f x s s f. M(t) u t u E(t). t t Rd u(t, x) p+ dxdt E. x This says that the solution cannot live near the origin for large times. To prove the lemma, we introduce the Littlewood-Paley theory of decomposing a function into portions with localized frequencies. Lemma 4.. For 0 s < d, we have f S(Rd ) Proof. Let 0 < s < d. physical space), we have f(x) x s R Z f x s s f. Divide space into dyadic shells (localization on x R f(x) x s dx R s f(x) dx, R Z x R which is R s χ x R f l R. Now, localizing on the frequency space, χ x R f N Z χ x R f N. On one hand, χ x R f N f N N s s f N ; 3

On the other hand, χ x R f N f N R d/ N d/ f N R d/ N s (NR) d/ s f N. Thus f x s R s min{n s, N s (NR) d/ } s f N l R N Z = M Z min{m s, M d/ s } s f M/R l R M Z min{m s, M d/ s } s f M/R l R s R Z ξ s ψ M/R (ξ) ˆf(ξ) dξ ξ s ψ M/R (ξ) ˆf(ξ) dξ R Z = s f. 4

5 Strichartz Estimates We say that (q, r) is wave-admissible if q + d r (q, r, d) (,, 3). d 4, q, r, d [, ) and Question* 5.. Where does the wave admissible condition come from? Theorem 5. (Strichartz estimate). Let d and (q, r), q, r are waveadmissible pairs with r, r < and q + d r = d γ = q + d r for some { u + F = 0 γ > 0. Assume u solves on I R u(0) = u 0, t u(0) = u d with 0 I. Then u L t Ḣ γ x + tu L t Ḣ γ x with spacetime norms on I R d. + u L q u t Lr 0 Ḣγ + u x Ḣγ + F L q t L r x To figure out the relation equation for the exponents, we consider scaling u λ (t, x) = u(λt, λx). Then u λ satisfies u λ + F λ = 0 with F λ (t, x) = λ F (λt, λx). u λ L t Ḣ γ x = λγ d u L t Ḣ γ x, uλ L q t Lr x F λ L q t L r x = λ q d r F L q t L r x λ γ d ( u 0 Ḣγ + u Ḣγ )+λ q d r F L q γ d = q d r. Then q d r = γ d. = λ q d r u L q t Lr x,. So λ γ d ( u L t Ḣx γ + u t L t Ḣx γ )+λ q d r u L q t Lr x t L r x { Consider the initial value problem: (NLW). Take F = 0, we would have u ± u p u = 0 u(0) = u 0, t u(0) = u, p > 0. We would love to solve this for (u 0, u ) Ḣs x Ḣs x for some s 0. In addition to the symmetries discussed before, (NLW) enjoys scaling property: u(t, x) u λ (t, x) = λ p u(λt, λx), u 0 (x) u λ 0 (x) = λ p u 0 (λx), u (x) u λ (x) = λ p + u (λx) leaves the class of solutions invariant. Combining the Lorentz invariance u(t, x) u( t x v, x tv, v v x ) for v with the scaling symmetry with λ = v, we get the normalized Lorentz invariance u(t, x) ( v ) p u(t x v, x tv, ( v ) x ). For p = 4 d, we have the conformal geometry u(t, x) (t x ) d t u( inside the cone { t < x }. Question* 5.3. Why is (t x ) d t u( What s about the conformal geometry? t x, t x, x t x ) x t x ) still a solution? 5

Let s find the homogeneous Sobolev space Ḣs Ḣs that is left invariant by scaling. u λ 0 Ḣ = u x s 0 Ḣs ( u λ Ḣx s = u Ḣs ), since u λ 0 Ḣ = x s λ p +s d u 0 Ḣs, s = d p =: s c. If s > s c, we call the problem subcritical (rather well understood); if s = s c, we call it critical (has seen a lot of progress in the past 5-0 years); if s < s c, we call it supercritical (not well understood). Assume that the initial data is of size A, which corresponds to a solution up to time T. Then the scaled solution is of initial date of size λ s sc A, and with solution up to time λ T. For the supercritical case, this suggests that we can construct arbitrarily small data that blows up arbitrarily quickly. We expect well-posedness for s s c (delicate for s = s c ) and ill-posedness for s < s c. The normalized Lorentz invariance suggests s l = s c + = d+ 4 p is critical for well-posedness, thus we expect well-posedness if s max{s c, s l } and ill-posedness otherwise. Question* 5.4. Remark 5.5. It turns out that for d 4 we needs to also impose the (apparently artificial) condition: p( d 4 s) ( d+3 s). 6

6 Local Well-posedness I Now we need to discuss what is a solution. Classical solution: smooth and decaying enough to make sense of all computation: Run into trouble when u p u is not smooth. Distribution solution: 0 R u φ+f (u)φdxdt+ d R u d 0 t φ(0) R u d φ(0) = 0, φ Cc (R R d ). Problems: Even if u is a function, F (u) need not be a distribution. For example, u(x) = x d p+ χ x L for p >, but F (u) = x d χ x / L loc. And we have trouble justifying the conservation laws. Strong solution: u : I R d R s.t. (u, u t ) C t (I, H s x H s x satisfies ( u u t ) (t) = ( cos(t) sin(t) sin(t) cos(t) ) ( u0 u ) t 0 ) and u ( sin((t ζ)) cos((t ζ)) and u L q t Lr x(i R d ) for some wave admissible (q, r) with q + d r = d s. Definition 6.. We say (NLW) is well-posed (locally) if it admits a unique local-in-time strong solution and the map from the initial data to the solution is continuous. Theorem 6.. Let d 3, p >. s = max{s c, s l } = max{ 3 p, p } = { 3 p, p p, < p < Let (u 0, u ) Ḣs Ḣs. Then there exists T > 0 and a unique strong solution u : [0, T ] R 3 R s.t. (u, u t ) C t ([0, T ], Ḣs s x Ḣ s ) with u L s t L ([0, T ] R 3 ) if p < and u L p ([0, T ] R3 ) if p. Moreover, if p we have the following. () η = η(p) > 0, s.t. if (u 0, u ) Ḣs η, then we may take T =. ()Let T be the maximal time existence. Ḣs Then either T = or u L p. Proof. First consider the case < p < (in the case s = s l > s c ). Let φ(u)(t) = cos(t)u 0 + sin(t) u t sin(t ζ) 0 F (u(ζ))dζ. To use fixed point theorem, we would love to find a complete metric space (B, d) such that d(φ(u), φ(v)) d(u, v) if u, v B. We want to use the Strichartz estimate s φ(u), + φ(u) q,r s u(0) + F (u) q, r, ) F (u(ζ))dζ L4 t Ḣ4,s x ([0,T ) R 3 ) = 7

q + 3 r = 3 s = q + 3 r. We then ask that q + r = (wave admissible). Also r = p+ r, which is suggested from the following Hölder inequality: u p u q, r ([0,T ] R 3 ) T q p+ q u p+ q,r. From these conditions we get the exponents, r = s, q = s, r = (p+)( s), q = 4 3p+s(3p+). Then we need to check that the wave admissible condition r, q, r, q. Then we have u p u q, r ([0,T ] R 3 ) T p(s sc) u p+ q,r. Let B = {(u, u t ) C t ([0, T ], Ḣs Ḣs ), u L s s t Lx ([0, T ] R 3 ), u L t Ḣ s C(p) u(0) Ḣs, u L s s t Lx ([0,T ] R 3 ) under the metric d(u, v) = u v By Strichartz, φ(u) s, s L s s t Lx C(p){ u(0) Ḣs + u p u q, r } C(p){ u(0) Ḣs + T p(s sc) u p+ x ([0,T ] R 3 ) C(p) u(0) Ḣs }, T = c(p) u(0) s sc Ḣ s. We can show that B is complete. s, s C(p){ u(0) Ḣs + T p(s sc) [C(p) u(0) p+ Ḣ s ]} C(p) u(0) Ḣs, if C(p) = C(p) and T p(s sc) [C(p) u(0) Ḣs ] p, T c(p) u(0) s sc Ḣ s. Now d(φ(u), φ(v)) p u p u v p v q, r p u v ( u p + v p ) q, r p p+ p u v p+ p+ q, r ( u p+ q, r + v p+ } p p+ q, r ) p u v s, { u p s s, + v p s s, }T p(s sc) s T p(s sc) [C(p) u(0) Ḣs ] p d(u, v) d(u, v), if c(p) is small enough. So φ is a contraction on B, thus it has a unique fixed point, which is our strong solution. 8

7 Local Well-posedness II Theorem 7. (LWP). Let d = 3, p, (u 0, u ) Ḣsc Ḣsc where s c = 3 p [, 3 ). Then there exists T > 0 and a unique strong solution (u, u t) C([0, T ], Ḣsc, Ḣsc ) such that u L p ([0, T ] R3 ). Moreover, η η(p) s.t. if (u 0, u ) Ḣsc < η, then we may take T =. Finally, we have Ḣsc the blowup criterion: if u L p ([0,T 0] R 3 ) + sc u L 4 ([0,T 0 ] R 3 ) <, then there exists δ = δ(u 0, u ) s.t. the solution extends to a strong solution on [0, T 0 + δ]. Proof. Let φ(u)(t) = cos(t)u 0 + sin(t) u t sin((t ζ)) 0 (± u p u)(ζ)dζ. We will use the Strichartz inequality on [0, T ] R 3 : φ(u) L t Ḣx sc sc F (u) L 4. 3 + sc φ(u) L 4 + φ(u) L p u(0) Ḣsc + Theorem 7. (Fractional chain rule, Christ-Weinstein). Let F C (C), 0 < s <, < q, q, q < such that q = q + q. Then Using the estimate s F (u) L q s u L q F (u) L q. sc F (u) L 3 4 sc u L 4 u p L sc u L 4 u p, L p the Strichartz inequality becomes φ(u) L t Ḣx sc + sc φ(u) L 4 + φ(u) L p u(0) Ḣsc + sc u L 4 u p. L p Strichartz inequality for the linear equation gives cos(t)u 0 + sin(t) u L p (R R3 ) + sc (cos(t)u 0 + sin(t) u ) L 4 (R R 3 ) (u 0, u ) Ḣsc. Pick T such that cos(t)u Ḣsc 0 + sin(t) u L p ([0,T ] R3 ) + sc (cos(t)u 0 + sin(t) u ) L 4 ([0,T ] R 3 ) η. Let B = {(u, u t) C([0, T ], Ḣsc Ḣsc ), u L t,ḣsc x u(0) Ḣsc, u L p η, sc u L 4 η}. Let d(u, v) = u v L p + sc (u v) L 4. Let s see φ : B B. 9

Let u B. φ L t Ḣ sc x u(0) Ḣsc + C p sc u L 4 u p L p u(0) Ḣsc + C p (η) p+ u(0) Ḣsc (C p p+ η p ) for η < C(p) min{, u(0) Ḣsc }. By Strichartz, φ(u) L p t 0 cos(t)u 0 + sin(t) u L p + sc u L 4 u p, L p sin((t ζ)) F (u(ζ))dζ L p sc F (u) L 4, 3 φ(u) L p η + C(p)(η) p+ η for η C(p). Let s prove φ is a contraction. Let u, v B. d(φ(u), φ(v)) sc [F (u) F (v)] L 4 3, F (u) F (v) = (p+)(u v) Theorem 7.3 (Christ-Weinstein, Fractional product rule). s (uv) q s u q v q + u q3 s v q4, with s > 0, q = q + q = q 3 + q 4. 0 v+θ(u v) p dθ. d(φ(u), φ(v)) sc (u v) L 4 { u L p + v L p } p + u v L p { sc u L 4 + sc v L 4 } { u L p + v L p } p η p d(u, v) + η p d(u, v) d(u, v). Let s see the continuity of the solution map from Ḣsc Ḣsc to L t (Ḣsc Ḣ sc ). By Strichartz inequality, (u v) L t Ḣ sc + d(u, v) [u(0) v(0)] Ḣsc + sc [F (u) F (v)] L 4 3 [u(0) v(0)] Ḣsc + η p d(u, v). For small η this gives d(u, v) [u(0) v(0)] Ḣsc and so (u v) L t Ḣ sc [u(0) v(0)] Ḣsc. To prove the local well-posedness, we still need to see uniqueness in L p. Let (v, v t ) C t ([0, T ], Ḣsc Ḣsc ) be another solution to (NLW) such that v 0

L p. Want to show u = v. Since uniqueness is a local statement, it suffices to show u = v on C t0 := {(t, x) : t 0 t t 0 + τ, x x 0 r 0 (t t 0 )}, for some (t 0, x 0 ), r 0, and sufficiently small τ, such that u L p ([t 0,t 0 +τ] R 3 ) + v L p ([t 0,t 0 +τ] R 3 ) η. By Hölder, u L 4 (C t0 ) + v L 4 (Ct 0 ) τ 4 p u L p (Ct 0 ) + τ 4 p u L p. By (Ct 0 ) Strichartz, u v L 4 (C t0 ) F (u) F (v) L 4 3 (C t0 ) u v L 4 (C t0 ) { u L p (Ct 0 ) + v p L p (Ct 0 )} η p u v L 4 (C t0 ) u v L 4 (Ct 0 ). for η small enough sin(t) For small data GWP use cos(t)u 0 + sin(t) u L p (R R3 ) + sc (cos(t)u 0 + u ) L 4 (R R 3 ) (u 0, u ) Ḣsc η. Then run the same argu- Ḣsc ment with u L t Ḣ sc ([0,T ] R 3 ) u(0) Ḣsc replaced by u L t Ḣ sc (R R 3 ) η. Theorem 7.4 (LWP in H x L x). Let d = 3, 3 p < 4 and (u 0, u ) H L. Then there exists T = T ( u 0, u Ḣ L ) > 0 and a unique solution (p+) (u, u t ) C t ([0, T ], H L p ) and u Lt Lx (p+) ([0, T ] R 3 ) for p. Also the data to solution map is continuous in Hx L x. Proof. Let p < 4. Let φ(u)(t) = cos(t)u 0 + sin(t) u t 0 sin((t ζ)) F (u(ζ))dζ. The Strichartz inequality on [0, T ] R 3 : φ(u), + u (p+) p,(p+) u(0) + F (u), u(0) + T p u p+ (p+) p,(p+). Let B = {(u, u t) C t ([0, T ], H L ) : u L t L x u(0), u (p+) p,(p+) C(p) u(0) } with T = c(p) u(0) d(u, v) = u v (p+) p,(p+),[0,t. For uniqueness, use Strichartz again, ] R3 (u v), + u v (p+) p,(p+) (u v)(0) + F (u) F (v), (u v)(0) + T p u v (p+) p,(p+) (u v)(0) + T p { u(0) + v(0) } p u v (p+),(p+). p p 4 p.

Take T sufficiently small, we get uniform continuous dependence of data to solution map on bounded sets of H L : t u(t) dx = u(t) t u(t)dx u(t) t u(t), t u(t) t u(t), thus u(t) L t L x ([0,T ] R3 ) u(0) L x + T t u L t L x ([0,T ). ] R3 For the defocusing case (F (u) = u p u), the conservation of energy E = u + tu + p + u p+ gives the uniform bound on (u, u t ) Ḣ L, thus we can iterate the existence theorem to extend solution from T = T ( u 0, u Ḣ L ) to T + T ( u 0, u Ḣ L ), then to a global solution.

8 Local Well-posedness III Theorem 8. (LWP in Hx L x). Let d = 3, 0 < p <, and (u 0, u ) Hx L x. Then there exists T = T ( (u 0, u ) H L) > 0 and a unique strong solution (u, u t ) C t ([0, T ], H L ). Moreover, the data to solution map is continuous on bounded sets of Hx L x. Corollary 8.. For d = 3, 0 < p < 4, and (u 0, u ) Hx L x, solutions to (NLW) conserve the energy E(u(t)) = u(t, x) + tu(t, x) + ± p+ u(t, x) p+ dx. Proof. Sobolev embedding tells us that u,6 u, <, with u, <, u,p+ <, 0 < p < 4. Proof. Consider 3 p <. We will prove φ is a contraction on B = {(u, u t ) C t ([0, T ], Hx L x) : u L t L x u(0) } with respect to d(u, v) = (u v) L t L for T to be determined shortly. Use the x Strichartz inequality φ(u), u(0) + C F (u),. F (u), T u,6 u p,3 T u, u p,3p (Sobolev embedding) T u, u pθ, u p( θ),6 T u, u p, u 3p Using u, u(0) + T t u,, we have Thus for (u, u t ) B,, = T u p, F (u), T u 3p, { u 0 + T t u, } p. u 3p,. φ(u), u(0) +C(p)T u(0) 3p, { u 0 +T u(0) } p. Choose T u(0) 3p u(0) p, T 4 p u(0) p, so T min{ u(0) 3p u(0) p, u(0) p 4 p }. Let s see φ is a contraction. Let u, v B. d(φ(u), φ(v)) = (φ(u) φ(v)), F (u) F (v), T u v,6 { u,3p + v,3p } p T d(u, v){ u p 3p, u, p + v, 3p v, }, 3

where u v,6 d(u, v) is by Sobolev embedding. Then the above is smaller than T d(u, v){[ u(0) p + T p + [ v(0) p + T p u(0) p u(0) p ] u(0) 3p ] v(0) 3p } d(u, v). Global well-posedness. Consider the defocusing initial value problem (with F (u) = u p u, (u 0, u ) H L ). We have conservation of energy E(u) = u(t) + tu(t) + p+ u(t) p+ dx. Recall (NLW) is invariant under scaling u(t, x) u λ (t, x) = λ p u( t λ, x λ ). E(u λ 4 ) = λ p +d [ u( t λ ) + t u( t λ ) ] + p + λ p (p+)+d u( t λ ) p+ dx = λ ( d p )E(u) = λ (sc ) E(u). When s c = ( p = 4 d ), the energy is left invariant by the scaling and we call the problem energy critical. When s c < ( p < 4 d ), we call the problem energy-subcritical. In this case E(u λ ) as λ 0. Thus concentration costs an infinite amount of energy. We expect conservation of energy to preclude concentration. When s c > ( p > 4 d ), we call the problem energy-supercritical. In this case, E(u λ ) 0 as λ 0. So energy conservation favours concentration. Expect ill-posedness. Theorem 8.3 (Energy subcritical GWP). Let d = 3, 0 < p < 4, (u 0, u ) H L. Then there exists a unique global solution (u, u t ) C t (R, H L ) to defocusing (NLW) such that u u L Proof. Iterate the LWP result. (p+) p t,loc L (p+) x for p. 4

9 An Ill-posedness Result Ill-posedness for u u p u = 0, d = 3, p > 3 for data (u 0, u ) Ḣs Ḣs when 3 < s < s c. Let s first show there exists Schwartz initial data which lead to solutions that blow up in finite time. Let u 0 Cc with suppu 0 B (0). Let u = 0. Then E(u(0)) = u 0 p+ u 0 p+ <. Multiplying u 0 by a large constant we may assume E(u(0)) < 0. This energy is conserved. Indeed for p < 4 this follows from LWP in the energy space. For p = 4, this follows from LWP in H L. For p > 4, this follows from persistance of regularity. where u, + u 4,4 u(0) + ( u p u) 4 3, 4, 3 ( u p u) 4 3, 4 3 u 4,4 u p p,p. Split [0, T ] into subintervals I j = [t j, t j+ ] where u p L (I j η. For η R 3 ) small enough, we conclude by the above Strichartz inequality that u L t u(t j ) L. By Hölder, u p+ u θ 6 u θ 3p u θ u θ Ḣ sc. You can also see that E(u(t)) < for all times of existence by finite speed of propagation. Indeed, suppu(t) B +t (0). u(t) u(t) 3p ( + t) 3 ( p+ 3p ) u(t) Ḣsc ( + t) p 4 p <. p+ t u(t) t u(t) 3p ( + t) 3 ( p+ 3p ) t u(t) Ḣsc ( + t) p 4 p <. p+ So the energy is conserved and negative. Now consider V (t) = u(t, x) dx < for all time of existence. V t (t) = uu t dx. V t (t) = u t + u( u + u p u)dx = u t u + u p+ = u t u + (p + )( u + u t ) (p + )E = (p + 4) u t + p u (p + )E > 0. L x (I j R 3 ) 5

Thus V t is increasing, V t (0) = 0 V t > 0, t > 0. So V is increasing, with V (0) > 0, V > 0, t > 0. Now Integrating against t, we find V t 4 u u t < 4 p + V V tt, V t V < 4 V tt, t > 0. p + V t log V (t) V (ɛ) < 4 p + 4 log V t(t), t < ɛ. V t (ɛ) V t (ɛ)v (ɛ) p+4 4 < V t (t)v (t) p+4 4, integrating it again, (t ɛ)v t (ɛ)v (ɛ) p+4 4 < 4 p (V (ɛ) p 4 V (t) p 4 ) < 4 p V (ɛ) p 4. Let t, we get a contradiction. λ Consider u λ 0 (x) = λ p u 0 (λx). Then u λ 0 Ḣ = λ s s sc u 0 Ḣs 0. If u blows up at T, then u λ blows up at T λ. Thus we find arbitrarily small data that blows up arbitrarily quickly. This means the data to solution map is not continuous. In fact, there exists initial data in Ḣs Ḣs for which there is no solution on any time interval. Let u j 0 (x) = (j ) p u 0 ( j x), v 0 (x) = j>c(t ) uj 0 (x j e ). Then v 0 j C(T ) ( j ) s sc u 0 Ḣs <. There is no solution however with initial data (v 0, 0) for any time interval. Indeed, if there exists such a solution on [0, t]. Then by finite speed of propagation, it would have to agree with u j (x j e ) which blows up at T. j For a second method, look for a solution that does not depend on x. It must solve u tt = u p u. The solution is u(t, x) = c(p) t p with c(p) = { (+p) } p p. This blows up at 0, but does not lie in any Sobolev space. { x Let χ(x) = 0 x 3. χ(x)dx = 0. ut 0 = χ( x T u( T, x)), ut (x) = χ( x T ) tu( T, x). The corresponding solution u T agrees with u inside the cone {(t, x) [ T, 0] R 3 : x T t}. So it blows up on an open set at t = 0 if not before. u T 0 Ḣs T p χ( x T ) Ḣ s T sc s T 0 0, 6

u T Ḣs T p χ( x T ) Ḣ T sc s T 0 0. s Then similarly, initial data v 0 (x) = j u j 0 (x j e ), v (x) = u j (x j e ) j would do the blow up at t = 0 trick. Recall that when < p <, we have s c < s l = p and in this case we proved LWP in Ḣs l Ḣs l. Will show ill-posedness for max 0, s c s < s l. Recall that we have a space-independent solution to NLW, u(t, x) = c(p)t p. Applying a Lorentz boost, we obtain the solution u v (t, x) = c(p)( v ) p t x v p, t < x v. Take the initial data w0 T (x) = uv ( T, x)χ v ( { x T ), wt (x) = u v t ( T, x)χ v ( x 0 r T ), where χ(r) = 8 r, χ v (x) = χ( v v [( + x ) x ]), 6 argχ x = (x, x 3 ). B (0) {argχ 0}. By finite speed of propagation, the corresponding solution w T agrees with u v on an open neighborhood of the light cone {(t, x) [ T, 0) R 3, x t }. Thus w T cannot be continued beyond t = 0 at least in the classical sense. We will show () w T ( ɛ) Ḣs ɛ sc s, w T L p ([ T,0) R3 ) =. () w0 T Ḣ + w T s Ḣ ( v)s l s. s T s sc Then this means we can construct arbitrarily small initial data (v ) that lead to solutions that blow up arbitrarily quickly (T 0). This proves that the data to solution map is not continuous. 7

0 Global Well-posedness Defocusing { energy critical NLW on R 3. u + u 4 u = 0 (NLW) (u(0), t u(0)) = (u 0, u ) Ḣ L (R 3 ). E(u(t)) = R 3 u + 6 u 6 dx = E(u(0)), which is invariant under scaling u λ (t, x) = λ u(λ t, λ x). Local theory: small data to global well-posedness, arbitrary data to local well-posedness. In construction to solutions, the time of existence T is chosen such that which comes from cos tu 0 + sin(t) u 4,([0,T ] R 3 ) ɛ 0, cos tu 0 + sin(t) u 4,(R R 3 ) (u 0, u ) H L. Our T in general depends on the profile of initial data not only on its norm, so the method of using energy conservation to iterate the local well-posedness to get global well-posedness does not work in general. Theorem 0.. (NLW) is GWP. Proof. Prove by contradiction. Suppose u : [0, T ) R 3 R is a maximal solution with T <. The result follows from the following two contradictory propositions. Proposition 0.. x 0 R, ɛ 0 > 0, s.t. lim sup t0 T B(x 0,T t 0 ) u 6 dx ɛ 0. This means if we have blow up, we have energy concentration. Proposition 0.3. There is no energy concentration. Lemma 0.4 (Exterior energy decay). Define E Ω (u) = Ω [ ]dx. inf σ>0 lim sup t0 T E B(x0,T t 0 +σ)\b(x 0,T t 0 )(u(t 0 )) = 0, x 0 R 3. Proof. let Flux[t 0, t ] = E B(x0,T t 0 )(u(t 0 )) E B(x0,T t )(u(t ))(u(t )). Flux[0, t], t [0, T ) is increasing and bounded by E B(x0,T )(u(0)). Fix ɛ > 0, choose t 0, such that Flux[t 0, t] < ɛ for t 0 < t < T. Then choose σ > 0, s.t. E B(x0,T t 0 +σ)\b(x 0,T t 0 ) < ɛ. Then a triangular inequality tells us that E B(x0,T t+σ)\b(x 0,T t) < ɛ for all t 0 < t < T. 8

Proposition 0.5. Let u t0 (t) = cos(t t 0 )u(t 0 ) + sin(t t 0) t u(t 0 ). x 0 R, ɛ > 0, s.t. lim sup t0 T u t 0 4,(D+B(x0,4(T t 0 ))) ɛ. Here D + B(x 0, 4(T t 0 )) is the domain of dependence that depends on B(x 0, 4(T t 0 )). Proof. Suppose x 0, ɛ > 0, lim sup t0 T u t0 4,(D+B(x0,T t 0 )) < ɛ. Fix x 0. Fix η > 0. Pick σ, t 0 s.t. σ > 00(T t 0 ) and E B(x0,T t 0 +σ)\b(x 0,T t 0 ) < η and u t0 L 4 t L < η. Pick a bump function χ on R3 that is on B(x x 0, 4(T t 0 )) and 0 outside of B(x 0, 5(T t 0 )). Let E be the truncated light cone with base B(x 0, 4(T t 0 )), E be the infinite truncated light cone with basis B(x 0, 5(T t 0 )) c, E 3 is the complement. Then (χu) t0 L 4 t L x ([t 0,T ] R 3 ) E + E + E3 η, where by Strichartz and finite speed of propagation, for some χ being on (T t 0 ) x 7(T t 0 ) and 0 outside, (χu) t0 E3 (χ u) t0 E3 E(χ u(t 0 )) E B(x0,T t 0 +σ)\b(x 0,T t 0 ) < η. 9