The Navier Stokes Equations for Incompressible Flows: Solution Properties at Potential Blow Up Times

The Navier Stokes Equations for Incompressible Flows: Solution Properties at Potential Blow Up Times Jens Lorenz Department of Mathematics and Statistics, UNM, Albuquerque, NM 873 Paulo Zingano Dept. De Matematica Pura E Aplicada, Universidade Federal Do Rio Grande Do Sul, Porto Alegre, RS 959-9, Brasil Abstract In this paper we consider the Cauchy problem for the 3D Navier Stokes equations for incompressible flows. The initial data are assumed to be smooth and rapidly decaying. It is not known if a classical solution can develop singularities in finite time. Assuming the maximal interval of existence to be finite, we review various properties of the solution as time approaches the blow up time. Key words: Navier Stokes equations, incompressible flows, blow up AMS subject classification: 35G2, 35Q3, 76D3, 76D5 Abbreviated Title: Incompressible Navier Stokes Introduction In this paper we consider the Cauchy problem for the 3D Navier Stokes equations, u t + u u + p = u, u =, u(x, ) = f(x) for x IR 3. (.) The current mathematical theory of the problem (.) remains fundamentally incomplete: It is known that a weak solution exists for all time t if f L 2, f =, but it is not known if weak solutions are unique. This is recognized as a major open problem

since the fundamental paper of Leray [5]. On the other hand, if f is more regular, then a unique classical solution exists in some maximal time interval, t < T f, but it is not known if T f can be finite or is always infinite. In other words, we do not know if classical solutions can break down in finite time. The Navier Stokes equations are of fundamental importance in continuum mechanics. When one derives the equations from the balance laws of mass and momentum and from principle assumptions relating the stresses to velocity gradients, then smoothness of the solution is assumed. To make the model of the Navier Stokes equations self consistent, one would like to prove that singularities in the solution do not develop in finite time, from smooth initial data with finite energy. Thus far, however, this aim has not been achieved. It remains one of the fundamental open problems in nonlinear analysis. In this paper we consider only classical solutions of the problem (.), and under our assumptions on f these will be C functions. If one normalizes the pressure so that p(x, t) as x, then the solution ( u(x, t), p(x, t) ) is unique. Its maximal interval of existence is denoted by t < T f. Assuming T f to be finite, certain norms of u(, t) will tend to infinity as t T f while other norms will remain bounded. This issue is well studied in the literature. We will review some results that we consider to be most important and will also derive lower bounds on some blow up rates. Another issue is to compare two functionals of u(, t) which blow up as t T f. Which one will blow up faster? We believe that a better understanding of this issue is important for further progress on the blow up question and give some results in Section 5. The intent of this paper is to give rather complete proofs of some solution properties for (.) that must hold, as t approaches T f, if T f is finite. These results and their proofs may be helpful if one wants to construct a solution that actually does blow up. They may also be helpful to show that blow up is ultimately impossible. For a treatment of many recent developments regarding the Navier Stokes equations, using methods of modern analysis, the reader is referred to [6]. For simplicity of presentation (and to avoid inessential complications near t = ) we put strong smoothness and decay assumptions on the initial function f and require f to be a divergence free C function for which all derivatives lie in L 2. We will use the following notations (all integrals are over IR 3 ): u 2 = u 2 + u 2 2 + u 2 3 for u IR 3 α = α + α 2 + α 3 for a multi index α = (α, α 2, α 3 ) D α = D α D α 2 2 D α 3 3, D j = / x j u(, t) L q = ( u(x, t) q dx ) /q, q < u(, t) = u(, t) L 2 2

u(, t) = sup u(x, t) x D j u(, t) L q = ( D α u(x, t) q dx ) /q, q < α =j D j u(, t) = D j u(, t) L 2 D j u(, t) = sup D j u(x, t) x (u, v) = α =j 3 j= u j (x)v j (x) dx When convenient, we will also write u(t) L q or simply u L q instead of u(, t) L q. An outline of the paper is as follows: In Section 2 we show that a bound on the maximum norm u(, t) in some interval t < T implies bounds for all derivatives of u in the same interval. This result is well known (see, for example, [4]), but since it is the basis for all our blow up results we will prove it here. An implication is the following: If u(, t) is bounded in t < T and if T is finite, then u can be continued as a C solution beyond T. This follows from well known local constructions of solutions. (See, for example, [4] for an elementary proof. See also [2, 3] for the development of a local theory under much weaker assumptions on f.) In other words, we can state the following first blow up result: Theorem. If T f < then sup u(, t) =. t<t f In Section 3 we show that boundedness of Du(, t) L q in some interval t < T implies boundedness of u(, t) in the same interval, as long as 3 < q 2. Therefore, Theorem. yields the first part of the following result: 2 Theorem.2 a) Let 3 2 < q 2. If T f < then sup Du(, t) L q =. t<t f b) There is a constant c >, independent of t and f, with the following property: If T f < then Du(, t) 2 c(t f t) /2, t < T f. (.2) 3

As we will show in Section 3, the algebraic blow up rate (.2) follows from the differential inequality ψ Kψ 3 satisfied by the function ψ(t) = Du(, t) 2. We note that the blow up rate of Du(, t), as expressed by the estimate (.2), is of course consistent with the fundamental upper bound Tf Du(, t) 2 dt 2 f 2 (.3) that we show in Theorem 2.. In Section 4 we consider the norms u(, t) L q, 3 < q <. If any of these norms stays bounded in some interval t < T, then Du(, t) is also bounded in this interval. Consequently, by the previous theorem, sup t<t f u(, t) L q =, 3 < q <, if T f <. In fact, one can again derive an algebraic lower bound for the blow up rate: Theorem.3 For 3 < q < there is a constant c q >, independent of t and f, so that the following holds: If T f < then with u(, t) L q c q (T f t) κ, t < T f, (.4) κ = q 3 2q The corresponding result also holds for u(, t) with κ = 2, u(, t) c (T f t) /2, t < T f. Remark: A recent result [] covers the limit case q = 3 showing that if T f is finite. We will prove Theorem.3 by writing u(, t) = e t f( ) +. sup u(, t) L 3 = (.5) t<t f t e (t s) Q(, s) ds, Q = u u p (.6) 4

and then using standard estimates for the heat semi group, e t, and a Gronwall type lemma. These auxiliary results, together with properties of the Helmholtz projector, will be reviewed in an Appendix. If there is blow up, one can ask: Do certain norms blow up faster than others? The answer is yes. For example, if 3 < q < r then we will prove that the L r norm blows up faster than the L q norm. A further result related to (.5) will also be discussed in Section 5. Acknowledgment. This paper is based on lectures given by the first author during his visit to the Universidade Federal do Rio Grande do Sul in 22. We are grateful to CNPq for the financial support provided for this visit. 2 A Bound for u(, t) Implies Bounds for All Derivatives Under our assumptions on f stated in the introduction, the Cauchy problem (.) has a unique C solution ( u(x, t), p(x, t) ), defined in some interval t < T, with D α u(, t) L 2 for all α and t < T. (We always require p(x, t) as x to make p unique.) We set J 2 n(t) = α =n D α u(, t) 2, n =,,... The most basic estimate for the solution of the Navier Stokes equations is the following energy estimate. Theorem 2. We have thus 2 d dt J 2 (t) = J 2 (t), t < T, J (t) f for t < T and T J 2 (s) ds 2 f 2. (2.7) Proof: Using integration by parts, 2 d dt J 2 (t) = (u, u t ) = i,j u j u i D i u j dx J 2 (t) 5

with i u j u i D i u j dx = i = i u j D i (u i u j ) dx (D i u j )u i u j dx Therefore, the sum is zero. This proves the theorem. Note that the integral bound in (2.7) proves (.3). To prove the next result, we will use the Sobolev inequality which implies v C v H 2, D j u C(J m + J ) for m j + 2. (2.8) (The bound J 2 n C(J 2 m + J 2 ) for m > n follows by Fourier transformation.) Theorem 2.2 Assume that sup u(, t) =: M <. (2.9) t<t Then all functions J n (t), n =,, 2,... are bounded in t < T. Proof: Using (2.9) we will first prove that J 2 (t) is bounded. Then boundedness of J simply follows from J 2 C(J2 2 + J 2 ). Once boundedness of J 2 is shown, we make an induction argument in n. We have, for any multi index α, and D α u t + D α (u u) + D α p = D α u 2 d dt J 2 n(t) = α =n (D α u, D α u t ) = (D α u, D α (u u)) Jn+ 2 α =n =: NT n J 2 n+ It is convenient to use the short notation 6

for any integral (D i u, D j ud k u) D α u ν D β u ν2 D γ u ν3 dx with ν, ν 2, ν 3 {, 2, 3} and α = i, β = j, γ = k. Let n = 2 and consider the nonlinear terms appearing in NT 2, Thus NT 2 is a sum of terms of the form (D α u, D α (u u)), α = 2. (D 2 u, ud 3 u) and (D 2 u, DuD 2 u). Using integration by parts, a term (D 2 u, DuD 2 u) can also be written as a sum of terms (D 2 u, ud 3 u). Since we obtain that (D 2 u, ud 3 u) MJ 2 J 3 d dt J 2 2 (t) CMJ 2 J 3 2J3 2 C 2 M 2 J2 2 Boundedness of J 2 in t < T follows. Let n 2 and assume J n (t) to be bounded in t < T. We have d 2 dt J n+(t) 2 = NT n+ Jn+2(t) 2 where NT n+ is a sum of terms a) Let j n 2. We have T j = (D n+2 u, D j ud n+ j u), j n. T j C n D j u J n+2 (J n+ + J ) C n J n+2 (J n+ + J ) 7

In the latter estimate we have used (2.8) and the induction hypothesis. b) Let j = n. We have T n C n D n u J n+2 (J 2 + J ) C n D n u J n+2 C n (J n+ + J )J n+2 In the second estimate we have used that bounds for J and J 2 are already shown. c) Let j = n. We have T n C n J n J n+2 Du C n J n+2 (J 3 + J ) In the last estimate we have used that n 2. These bounds prove that C n J n+2 (J n+ + J ) d dt J 2 n+(t) C n J n+ (J n+2 + J ) 2J 2 n+2(t), and boundedness of J n+ in t < T follows. If u(, t) is bounded in some interval t < T, then all functions J n (t) are also bounded in t < T and, using (2.8), all space derivatives of u are bounded in maximum norm. Estimates for the pressure and its derivatives follow from the Poisson equation p = i,j D i D j (u i u j ). Time derivatives and mixed derivatives of u can be expressed by space derivatives, using the differential equation (.). Therefore, if u(, t) is bounded in some interval t < T, then all derivatives of u are bounded in the same interval, and therefore the solution (u, p) can be continued as a C solution beyond T. This proves Theorem.. 3 Blow Up of Du(, t) A physically important quantity is the vorticity, ω = u, and the so called total enstrophy of the flow, ω(x, t) 2 dx. 8

If ˆv(k) = (2π) 3/2 e ik x v(x) dx denotes the Fourier transform of a field v(x), then we have and, therefore, ˆω(k, t) = ik û(k, t), k u(k, t) =, Using Parseval s relation one finds that ˆω(k, t) = k û(k, t). ω(x, t) 2 dx = ˆω(k, t) 2 dk = ki 2 û j (k, t) 2 dk i,j = Du(, t) 2. In the section we study blow up of Du(, t) in the L q norm, 3 2 < q 2. 3. Boundedness of Du(, t) Implies Boundedness of u(, t) Theorem 3. If is finite, then sup Du(, t) =: C (3.) t<t sup u(, t) (3.) t<t is also finite. Therefore, if T f <, then sup t<tf Du(, t) =. Proof: We Fourier transform the Navier Stokes equations: û t + (u u)ˆ + ( p)ˆ = k 2 û where k 2 = k 2. If we set Q = u u p then the Navier Stokes equations take the form u t = u + Q 9

or, on the Fourier side, Since û t = k 2 û + ˆQ. (u u)ˆ = ˆQ ( p)ˆ is the orthogonal decomposition of the vector (u u)ˆ into a vector orthogonal to k and a vector parallel to k, it follows that ˆQ (u u)ˆ. One obtains, for each Fourier mode, thus t û(k, t) = e k2 t ˆf(k) + e k2 (t s) ˆQ(k, s) ds û(k, t) e k2t ˆf(k) + Integrating in k one finds that t e k2 (t s) (u u)ˆ(k, s) ds. u(, t) û(k, t) dk ˆf t L + e k2 (t s) (u u)ˆ(k, s) dkds. Apply the Cauchy Schwarz inequality to bound the inner integral by with I = I /2 I /2 2 e 2k2 (t s) dk = C(t s) 3/2. (3.2) (Note that e τy2 dy = π/τ for τ >, and (3.2) follows form Fubini s theorem.) Also, by Parseval s relation, I 2 = (u u)(x, s) 2 dx u(, s) 2 Du(, s) 2 C 2 u(, s) 2

where we have used (3.). Thus we have shown the estimate u(, t) ˆf t L + CC (t s) 3/4 u(, s) ds for t < T. By Gronwall s lemma, see Lemma 6.2, boundedness of u(, t) in the interval t < T follows. 3.2 Blow Up of Du(, t) L q for 3 2 < q 2 We now sharpen the above arguments slightly by using Hölder s inequality instead of the Cauchy Schwarz inequality and by using the Hausdorff Young inequality instead of Parseval s relation. Theorem 3.2 Let 3 2 < q 2. If is finite, then is also finite. Therefore, if T f <, then sup Du(, t) L q t<t sup u(, t) t<t sup t<t f Du(, t) L q = for 3 2 < q 2. Proof: We use the same notations as in the previous subsection. Applying Hölder s inequality to the integral I = e k2 (t s) (u u)ˆ(k, s) dk we obtain the bound with I I /q I /q 2, q + q =, I = e qk2 (t s) dk = C(t s) 3/2 and

I /q 2 = (u u)ˆ L q C u u L q C u Du L q. In the first estimate we have used the Hausdorff Young inequality. We obtain that u(, t) C + C t Here the exponent κ of t s satisfies (t s) 3/2q u(, s) Du(, s) L q ds. κ = 3 2q < since q > 3. As above, the proof is completed by Lemma 6.2. 2 Remarks:. The Navier Stokes equations on all space IR 3 enjoy the following scaling invariance: If ( u(x, t), p(x, t) ) solves the Navier Stokes equations, then, for every scaling parameter λ >, ( λu(λx, λ 2 t), λ 2 p(λx, λ 2 t) ) also solves the equations. The norms v L 3 and Dv L 3/2, which occur for the limiting values q = 3 in Theorem.3 and q = 3/2 in Theorem 3.2, are invariant under λ scaling. Better understanding of the scale invariant norms, u(, t) L 3 and Du(, t) L 3/2, as t T f is likely to be important for further progress on the blow up question. 2. Theorem 3.2 can also be deduced from the results of Section 4 using a Sobolev inequality. 3.3 A Differential Inequality for Du(, t) Lemma 3. There is a constant K with Proof: We have d dt Du(, t) 2 K Du(, t) 6, t < T f. (3.3) 2 d dt Du 2 = j = j (D j u, D j u t ) =: NT D 2 u 2 (D j u, D j (u u)) D 2 u 2 2

Since (D j u, u D j u) = (by an argument as in the proof of Theorem 2.) the nonlinear term can be estimated by C Du 3 L 3. Using the Gagliardo Nirenberg inequality for v = D j u and then Young s inequality, v L 3 C v /2 Dv /2 ab aα α + bβ β for α + β =, with α = 4, β = 4/3, one obtains (3.3). In the remaining part of this section K will denote the constant in (3.3). Set ψ(t) = Du(, t) 2, t < T f, thus ψ Kψ 3. Assuming that T f <, we have sup t<tf ψ(t) = by Theorem 3.. We will now show that the local growth control of ψ, expressed by the inequality ψ Kψ 3, implies a lower bound for ψ(t). Theorem 3.3 Assume that T f < and let K denote the constant in (3.3). Then we have with K = (2/K) /2. Du(, t) 2 K (T f t) /2, t < T f, Proof: Fix t < T f and let φ(t) be the solution of It is elementary to show that φ = Kφ 3, φ(t ) = ψ(t ) =: φ. The blow up time of φ(t) is φ 2 (t) = φ 2 K 2 φ2 (t t ). t = t + 2 Kφ 2 Because ψ Kψ 3 and because ψ(t) blows up at T f, one obtains that Therefore, T f t. 3.

T f t 2, φ 2 Kφ 2 2/K. T f t Replacing t with t and φ with ψ(t), the theorem is proved. 4 Blow Up of u(, t) L q for 3 < q < Using the Helmholtz projector P H, which we discuss in Section 6.3, one can write the Navier Stokes equations as u t = u P H (u u), and if e t denotes the heat semi group, then one obtains that u(, t) = e t f( ) t e (t s) P H (u u)(, s) ds. We will use boundedness of P H in L q spaces, see Theorem 6.2, as well as estimates for e t, which we review in Section 6.2. It is not difficult to show that the linear operators P H and e t commute, and these operators also commute with differentiation, D j = / x j. 4. Boundedness of u(, t) L q Implies Boundedness Du(, t) In the following estimate, let 3 < q < and let 6 5 < r < 2 be defined by Setting D = D j we have q + 2 = r. Therefore, with Du(t) = e t Df t De (t s) P H (u u)(s) ds. the following estimates hold: κ = 3 2 ( r ) + 2 2, Du(t) Df + C Df + C t t (t s) κ u u(, s) L r ds (t s) κ u(, s) L q Du(, s) ds 4

In the first estimate we have applied (6.23); the second estimate follows from Hölder s inequality. Let us note that since q > 3. Therefore, by Lemma 6.2, if κ = 3 2q + 2 < then sup u(, t) L q t<t < sup Du(, t) <. t<t Applying Theorem 3., we have proved the following result: Theorem 4. Let 3 < q <. If T f < then sup t<t f u(, t) L q =. 4.2 An Integral Inequality for u(, t) L q We show a simple nonlinear integral inequality for the scalar function u(, t) L q, which implies local control of the growth of u(, t) L q. This local control of growth together with Theorem 4. imply a lower bound for u(, t) L q if T f <. Lemma 4. Let 3 < q < and set There is a constant C = C q with κ = 3 2q + 2 <. u(, t) L q f L q + C t (t s) κ u(, s) 2 L q ds. Proof: In the following, let r = q 2 We also have (since 2r = q), and note that κ = 3 2 ( r ) + q 2. u i u j L r C u 2 L q. 5

From u(, t) = e t f( ) j t e (t s) P H D j (u j u)(, s) ds we obtain that u(, t) L q f L q + C j f L q + C t t (t s) κ u j u(, s) L r ds (t s) κ u(, s) 2 L q ds The estimate of the previous lemma can be started at any t < T f ; this yields u(, t) L q u(, t ) L q + C q t t (t s) κ u(, s) 2 L q ds, t t < T f, where 3 < q <, κ = 3 +. Using Lemma 6.3 we obtain that u(, t) 2q 2 L q for t t t + τ with τ = c q u(t ) µ, µ = κ = 2q q 3. Suppose that T f <. Then, applying Theorem 4., we find that T f > t + τ, thus T f t > c q / u(t ) µ Lq. This yields the lower bound is bounded u(t ) L q c q ( Tf t ) (q 3)/2q, t < T f, and we have proved Theorem.3 for 3 < q <. similarly. The case q = can be treated 5 Comparison of Blow Up Functions Let 3 < q < r and assume that T f <. Theorem.3 yields the lower bounds, u(t) L q c q (T f t) κ(q), u(t) L r c r (T f t) κ(r) with positive constants c q, c r and 6

< κ(q) = q 3 2q < κ(r) = r 3 2r Thus, the lower bound for the L r norm blows up faster than the lower bound for the L q norm. This suggests that u(t) L r u(t) L q as t T f if 3 < q < r. (5.4) A precise result can be obtained by using boundedness of the L 2 norm and interpolation: Define < λ < by and recall the interpolation estimate q = λ 2 + λ r. which yields u(t) L q C u(t) λ u(t) λ L r u(t) λ L r C u(t) L r u(t) L q. (5.5) Using the lower bound on the blow up of u(t) L r provided by Theorem.3, one obtains (5.4) with the following algebraic lower bound c(t f t) λκ(r), κ(r) = r 3 2r, λ = for the quotient in (5.4). We have shown: q r 2 r, Theorem 5. Let 3 < q < r and assume that T f <. There is a constant c = c(q, r) >, with where u(t) L r u(t) L q c(t f t) γ, t < T f, γ = r 3 2r q r 2 r >. 7

In the remaining part of this section we compare the growth of the L q norm, u(t) L q, for 3 < q < with the growth of the maximum norm, u(t), as t T f. Setting r = and λ = 2/q in (5.5), we have u(t) 2/q C u(t), t t < T f. u(t) L q Together with (.5) one obtains the following theorem: Theorem 5.2 Let 3 < q < and assume that T f <. Then we have sup t<t f u(t) q u(t) q L q u(t) L3 u(t) 2 =. (5.6) Proof: Since the proof of (.5) in [] is very involved, we give here an elementary argument for (5.6). We use the notation u, v = i u u v i for the Euclidean inner product in IR 3. We have q d dt u q L = q u q u t dx = u q 2 u, u t dx = u q 2 u, p dx =: T p + T c + T v u q 2 u, u u dx + u q 2 u, u dx Using integration by parts one obtains that T c =. Also, T v = i,j u q 2 u i D 2 j u i dx = i,j u q 2 Du 2 dx (q 2) j u q 4 u, D j u 2 dx. The pressure term is T p = u q 2 u j D j p dx j = (q 2) u q 4 u, D j u u j p dx j C q u q 2 Du p 8

For p we use the bound p C i,j u i u j C u 2 L 4 C u L 3 Du. Thus we have shown the estimate Set d dt u q L q C u q 2 u L 3 Du 2. (5.7) h(t) = u(t) q u(t) q L q u(t) L3 u(t) 2. (5.8) If h(t) would be bounded by h max in t < T f, then the estimate (5.7) yields d dt u q L q Ch max Du 2 u 2 L q. Since T f Du 2 ds is finite by Theorem 2., one then obtains boundedness of u(t) L q in t < T f. This contradiction proves (5.6). 6 Appendix: Auxiliary Results 6. Gronwall s Lemma and Extensions A simple and well known form of Gronwall s lemma is the following. Lemma 6. Let A, B >. Let φ C[, T ) satisfy φ(t) A + B Then φ(t) satisfies the bound t φ(s)ds for t < T. φ(t) Ae Bt, t < T. The following extension is often useful in applications to partial differential equations. 9

Lemma 6.2 Let A, B >, < κ <. Let φ C[, T ) satisfy φ(t) A + B Then φ(t) is bounded in t < T. Proof: Choose ε > with t (t s) κ φ(s)ds for t < T. and set ε ξ κ dξ = ε κ κ 2B M := ε κ = max ε ξ T ξ κ. Let t < T be arbitrary. There is t t with Case ε t t : Then we have φ(t) = max s t φ(s). thus φ(t) A + B A + BM t ε t (t s) κ φ(s) ds + B t φ(s) ds + B 2B φ(t), t ε (t s) κ φ(s) ds Case 2 t ε: We have φ(t ) φ(t) 2A + 2BM t φ(s) ds. φ(t) A + B t A + Bφ(t) A + 2 φ(t) (t s) κ φ(s) ds ε ξ κ dξ thus 2

We have shown that φ(t ) φ(t) 2A. φ(t ) 2A + 2BM The previous lemma implies the bound t φ(s)ds for t < T. sup φ(t) 2Ae 2BMT t<t and the lemma is proved. Next we assume that φ(t) satisfies a nonlinear estimate. This will imply some local control of φ(t). Lemma 6.3 Let A >, B >, < κ <. Let φ C[, T ) satisfy φ(t) A + B t If < τ T is chosen so small that (t s) κ φ 2 (s) ds for t < T. then 4ABτ κ < κ φ(t) < 2A for t < τ. (6.9) Proof: Suppose (6.9) does not hold. Then let t < τ denote the smallest time with One obtains Therefore, 2A = φ(t ) A + B φ(t ) = 2A. t (t s) κ 4A 2 ds. 4AB t (t s) κ ds = 4ABt κ /( κ) Since t < τ, this contradicts the choice of τ. 2

6.2 Heat Equation Estimates Consider the Cauchy problem for the heat equation, u t = u, u = f at t =, x IR N. We assume that f L r for some r. The solution is u(x, t) = Φ(x y, t)f(y) dy (6.2) where Φ is the heat kernel, We have Φ(x, t) = (4πt) N/2 e x 2 /4t. < Φ(x, t) (4πt) N/2, Φ(x, t)dx = for t >. (6.2) It is elementary to show that the function u(x, t) defined by (6.2) is a C function for t >. Furthermore, for t >, all derivatives of u(x, t) can be obtained by differentiating the convolution integral (6.2) under the integral sign. In particular, we have for all space derivatives D α u, D α u(x, t) = Dx α Φ(x y, t)f(y) dy. We show the following solution estimates: Theorem 6. Let r q. Then we have u(, t) L q (4πt) λ f L r (6.22) and, for every space derivative D α, with D α u(, t) L q Ct λ α /2 f L r (6.23) λ = N 2 ( r ). q The constant C depends on r, q, α, and N, but is independent of t and f. To prove (6.22) we first show that the L r norm of u(, t) cannot grow in time: 22

Lemma 6.4 For r, u(, t) L r f L r, t. Proof: For r = and for r = the estimate follows directly from Φ(x, t) dx =. For < r < define r by Using Hölder s inequality, r + r =. u(x, t) Φ(x y, t) /r Φ(x y, t) /r f(y) dy ( Φ(x y, t)dy ) /r ( Φ(x y, t) f(y) r dy ) /r, thus u(x, t) r Φ(x y, t) f(y) r dy. Integration in x proves the lemma. In the next lemma we estimate the maximum norm of u(, t) by f L r. Lemma 6.5 For r < we have u(, t) ( 4πt) N/2r f L r. Proof: For r = the estimate follows from the bound Φ(x, t) (4πt) N/2. If < r < define r by + = and obtain r r u(x, t) Φ(x y, t) /r Φ(x y, t) /r f(y) dy (4πt) N/2r Φ(x y, t) /r f(y) dy (4πt) N/2r f L r. In the last estimate we have used Hölder s inequality, and the lemma is proved. Using the bounds of the two previous lemmas, the estimate (6.22) follows by a simple argument: For r q < we have 23

u(, t) q L = u(x, t) q dx q = u(x, t) q r u(x, t) r dx u(t) q r u(t) r L r (4πt) (q r)n/2r f q L r Taking the q th root we obtain (6.22). Estimates of Derivatives: Consider the function It is easy to see that g(z) = e z 2 /4, z 2 = z 2 +... + z 2 N. D α z g(z) = p α (z)g(z) where p α (z) is a polynomial. Therefore, D α z g(z) is a bounded function and D α z g(z) dz =: c(α, N) <. Since we have Φ(x, t) = (4πt) N/2 g(xt /2 ) D α x Φ(x, t) = (4πt) N/2 t α /2 (D α z g)(xt /2 ). Integrating over x IR N and using the substitution one obtains that xt /2 = z, dx = t N/2 dz, D α x Φ(x, t) dx = (4πt) N/2 t (N α )/2 c(α, N) = c(α, N)t α /2. One can now repeat the arguments given above for the case α = and derive estimates of D α u(, t) L q in terms of f L r. Instead of Φ(x, t) dx = 24

one uses the equation D α x Φ(x, t) dx = c t α /2. In this way the estimate (6.23) follows. Remark: Consider a convolution integral v(x) = k(x y)f(y) dy where k L, f L r. Arguing as in the proof of Lemma 6.4 one obtains that v L r k L f L r, which is Young s inequality. As above, define r by + r r proof of Lemma 6.5 one obtains the bound =. If one argues as in the v k L r f L r. Therefore, the estimates (6.22) and (6.23) can also be proved by computing the L q norms of Φ(, t) and D α Φ(, t). 6.3 The Helmholtz Projector P H Let v : IR 3 IR 3 denote a given vector field. If v satisfies some restrictions, there is a unique decomposition of v into a divergence free field, w, and a gradient field, φ, v = w φ, w =. (6.24) Then, in suitable function spaces, the assignment v w =: P H v defines a projection operator P H, called the Helmholtz projector. An important and nontrivial result is the boundedness of P H in L q for every q with < q <. We will not prove this result here, but only discuss how it is related to the Calderon Zygmund theory of singular integrals. To begin with, let v C, i.e., v is smooth and compactly supported. First assume that a decomposition (6.24) holds with C functions w and φ. Then, taking the divergence of the equation v = w φ, one obtains that Assuming φ(x) as x, we have φ = v. φ(x) = x y v(y) dy. (6.25) 4π One can now reverse the process, define φ by (6.25) and set 25

w = v + φ =: P H v. Then the Helmholtz decomposition (6.24) of v is obtained. This leads to the following definition. Definition 6.: If v C then define w = P H v by w j (x) = v j (x) + 4π = v j (x) + 4π i i D xj x y D i v i (y) dy (6.26) x y D j D i v i (y) dy, j 3. (6.27) The following result about P H is easy to prove: Lemma 6.6 Let v C. a) Define w = P H v by (6.27) and define φ by (6.25). Then w C, φ C and (6.24) holds. Furthermore, w(x) + φ(x) as x. (6.28) b) Conversely, if w(x), φ(x) are C functions satisfying (6.24) and (6.28), then w and φ agree with the functions defined in (6.27) and (6.25). So far we have assumed v C. To obtain estimates of P H v in terms of v, which allow to extend P H to less regular functions, the theory of singular integrals can be applied. To discuss this, let Then we have K ij (z) := D i D j ( z ) for z IR 3, z. K ij (z) = 3z i z j z 5 for i j and K jj (z) = (3z 2 j z 2 ) z 5. Thus the kernels K ij (z) are homogeneous of degree 3, i.e., K ij (z) = z 3 K ij (z ), z = z/ z. If one integrates by parts in (6.25) and formally differentiates under the integrals sign, then one obtains D j φ(x) = 4π K ij (x y)v i (y) dy. (6.29) i 26

However, since K ij (z) z 3, the integrals in (6.29) do not exist as Lebesgue integrals, but must be interpreted as principle values, D j φ(x) = 4π i lim K ij (x y)v i (y) dy. (6.3) ε x y ε Together with the equation w = v + φ one is lead to the following definition of the Helmholtz projection P H v. Definition 6.2: If v C then define w = P H v by w j (x) = v j (x) + 4π i lim K ij (x y)v i (y) dy, j 3. ε x y ε Remark: It is not difficult to show that Definitions 6. and 6.2 are equivalent for functions v C. Definition 6.2 has the advantage that, formally, the function v is not required to be differentiable. Also, it is the form w = P H v = v + φ with φ determined by the singular integral in (6.3), to which the Calderon Zygmund theory of singular integrals can be applied. Using the estimate and large x it is easy to see that K ij (x y) C + x 3 for y C w(x) C( + x 3 ) with C = C(v), thus w L q for < q. Here, as before, we have assumed that w = P H v and v C. Furthermore, we have w C since the integrals in (6.3) can be differentiated under the integral sign if all derivatives are moved through integration by parts from the kernels to v i (y). We have shown the following: If < q, then the Helmholtz projector P H maps C into C L q. We state the following theorem without proof: Theorem 6.2 For < q < there is a constant C q with P H v L q C q v L q, v C. Since C is dense in L q, the operator P H can be extended uniquely as a bounded linear operator from L q into itself. 27

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