Sep. 1 0, 008 Distributions Distributions are generalized functions. Some familiarity with the theory of distributions helps understanding of various function spaces which play important roles in the study of PDEs. In this lecture we will introduce the theory of distributions, and illustrate its power show via examples. 1. Distributions. 1. 1. Introduction. Definition 1. Distributions A distribution T on Ω is a linear functional on C 0 Ω such that T φ j 0 for every sequence { φ j } C 0 Ω converging to 0, that is satisfying 1 i. the supports of all φ j s are in a common compact set K Ω. Here K is independent of j. ii. sup x K D α φ j 0 as j for all α with α 0. We denote the space of distributions on Ω by D Ω. Example. The following linear functionals are distributions. 1. Let f be a locally integrable function on d. Define linear functional T f by T f φ = f φ. n. Let a d. Consider the Dirac delta function: 3. Let a d, and α a multi-index. Define δ a φ φ a. 3 δ a α φ = 1 α D α φ a. 4 If { T j } D Ω satisfies T j φ 0 for any φ C 0 Ω, we say T j converges to 0 in D Ω, and write T j 0. We say T j T if T j T 0. It turns out that D Ω is complete under this topology 3. For T D Ω, we say that T is zero on an open set V Ω if T φ = 0 for every φ C 0 V. The support of T D Ω is defined as follows: supp T = { x Ω T 0 on any neighborhood of x }. 6 1. This topology is induced from the natural topology on the space C defined by the semi-norms p, k u sup D α u. 1 α k x. The reason for such notation is that L. Schwartz used D to denote C 0 in his original paper. 3. We need to show that if T n φ converges for every φ C 0, the limit T defined by T φ = lim T n φ 5 n is also a distribution. This is a direct consequence of the Banach-Steinhaus Theorem in functional analysis note that C 0 is metrizable. B ut one can also prove it directly via the following argument. 1. If T D, then there are φ n 0 but T φ n c > 0. eplacing φ n by c n φ n for appropriate c n we can make T φ n. We want to construct ψ C 0 such that T n ψ 0, thus getting a contradiction. The idea is to find subsequences ψ i and S i, such that S i ψ i is large possible because T ψ i is large, and S j ψ i small for all j i. Then set ψ = ψ i.. Now take φ i such that T φ i > 1 and then we can find T j such that T j φ i > 1. Set ψ 1 = φ i, S 1 = T j. 3. Choose ψ n such that S i ψ n < 1 / n j for j = 1,, n 1, and T ψ n > n 1 j = 1 T ψ j + n. Then choose S n satisfying S n ψ n > n 1 j = 1 T ψj + n. 4. Set ψ = 1 ψ n. We have S n ψ > n 1. One can further show that if T n 0 in D and φ n 0 in C 0, then T n φ n 0.
Example 3. The support of δ a α is { a }. There is an alternative definition which is much more useful in practice. We present it as a lemma. Lemma 4. Let T D Ω. Then for any compact subset K of Ω, there is n = n K N and C = C K such that T φ C max D α φ x 7 x K α n for all φ supported in K. Conversely, if the above is satisfied for a linear functional T on C 0, then T D Ω. Proof. The converse direction is trivial. For the other direction, we prove by contradiction. Assume that there is T D Ω such that there is a compact subset K of Ω, such that there are φ n C 0 Ω supported in K and T φ n > n max x K Dα φ n x. 8 α n By replacing φ n by c n φ n, we can set T φ n = 1. Thus we have max x K Dα φ n x < 1 n. 9 α n But this implies φ n 0, consequently T φ n 0 and we obtain a contradiction. emark 5. The constant n K may not be uniform. For example, let T D be defined by T φ = φ n n. 1 0 n= 1 If n can be taken to be independent of K, then the smallest uniform n is called the order of the distribution T. For example, the Dirac δ function is a distribution of order 0, the distribution δ a α has order α. emark 6. It turns out that C 0 is dense in D, that is, any distribution T is the limit of a sequence of C 0 functions f n, in the sense that for any φ C 0, T φ = lim f n x φ x dx. 1 1 n d 1.. Operations on distributions. Differentiation. Definition 7. Derivatives of a distribution Let T D Ω, and let α be any multi-index of nonnegative integers. The derivative D α T is defined by Example 8. Consider the Heaviside function D α T φ = 1 α T D α φ. 1 H x = { 0 x < 0 1 x > 0. 1 3 H is locally integrable, and we identify it with the distribution T H. We can show that H = δ. If we define K by { 0 x < 0 K x = x x > 0. 1 4 then K = H.
Example 9. Let { 0 x 0 x λ + x λ x > 0 with 1 < λ < 0. Differentiating naively, we would have λ x { 0 + = λ x λ 1 distribution. The correct computation is as follows. 1 5 but it is not locally integrable and is not a λ x+ φ = λ x+ φ = x λ φ x dx 0 = lim x λ φ x dx ε 0 ε = lim λ x λ 1 φ x dx + ε λ φ ε ε 0 ε = lim λ x λ 1 φ x dx + ε λ φ 0 + ε λ [ φ ε φ 0 ] ε 0 ε = lim ε 0 λ x λ 1 [ φ x φ 0 ]. 1 6 ε Therefore the distributional derivative of x λ + is defined by λ x+ φ = lim λ x λ 1 [ φ x φ 0 ]. 1 7 ε 0 Example 1 0. Poisson summation formula We consider the sum ε cos x + cos x +. 1 8 It turns out the limit of this diverging sequence exists as a distribution. To see this, let T n = n 1 cos k x and S n = n 1 1 sin k x. From basic Fourier analysis we know that S k n π x locally uniformly on 0, π. Since S n is uniformly bounded, one can show that S n converges to the periodic expansion of π x in D. On the other hand we see that T n = S n as distributions, thus T n S where S = lim S n. Therefore we obtain the summation formula cos k x = 1 + π δ x n π. 1 9 1 n Z Example 1 1. Let T D. If T = 0 then T is a constant. To see this, recall that T = 0 implies T φ = 0 0 for all φ C 0. Now take an arbitrary function h C 0 with h = 1. For any ψ C 0 we have ψ x = h x ψ s ds + ψ 1 x 1 Since ψ 1 x = 0, its primitive φ x = x In other words, T ψ = T h x ψ 1 s ds C 0 and consequently ψ s ds + T ψ 1 = T h x ψ s ds. T = T h x 3 is a constant as a distribution. One can show using similar ideas that any T D d with T = 0 is a constant.
Multiplication by smooth functions. Let T D Ω and f C Ω. Then we can define a linear functional f T by We see that f T D Ω by noticing that 1. f φ C 0 Ω when φ C 0 Ω ;. When φ j 0 in C 0 Ω, f φ j 0 too. f T φ = T f φ. 4 C onvolution. In general, the product TS for T, S D Ω is not defined. What we can do is calculate their convolution. Definition 1. Convolution of functions Let f, g be integrable. We define a new function f g by f g x = f y g x y dy. 5 d The convolution operation has the following properties: 1. f g is integrable;. f g = g f; f g h = f g h; 3. supp f g supp f + supp g, where the sum of two sets A, B is defined as A + B = { x x = y + z, y A, z B }. 4. If f is differentiable, so is f g even if g is not, and D α f g = D α f g. We can extend the convolution operation to distributions, using the following calculations as a guide: f g x φ x dx = f y g x y dy φ x dx = f y g x y φ x dy dx = f y g x y φ x dx dy = f y g z φ z + y dz dy = f y T g φ + y dy = T f T g φ + y. 6 Now let T, S D Ω and assume S has compact support denoted as S E Ω 4, we define T S φ T y S x φ x + y. 7 where the subscripts y, x means the distributions T and S are acting in the y-space and x-space respectively. We can show that the above properties still hold: 1. T S D Ω 5 ;. T S = S T; If U E Ω, T S U = T S U ; 3. supp T S supp T + supp S; 4. D α T S = D α T S = T D α S. ecall that distributions are infinitely differentiable in the space of distributions. 4. E Ω is the notation for C Ω used by Schwartz, and compacted supported distributions form the dual space of it, thus the notation E Ω. 5. This is not true without the condition on the support of S. More specifically, S y φ x + y may not still be in C 0 when the support of S is not compact. Consider for example the distribution S 1 obtained by the constant function 1.
Example 1 3. Let T D Ω, we compute T δ. Let φ C 0. We have Therefore More generally, for any multi-index α. T δ φ = T y δ x φ + y = T φ. 8 T δ = T 9 T D α δ = D α T. 30 Example 1 4. Fundamental solutions The power of distributions in the study of linear PDEs comes from the following fact: Let P D be a differential operator with constant coefficients. Let G solves Such a G D is called a fundamental solution of the operator P. Then for any f D we have In other words G f solves the equation P D G = δ 31 P D G f = [ P D G] f = δ f = f. 3 P D u = f. 33 Therefore the solvability of any constant-coefficient equation P D u = f reduces to the existence of the fundamental solution G. Fourier transform of distributions. In this part we consider the case Ω = d. A very powerful tool in the study of PDEs is the Fourier transform, which turns differential equations into algebraic equations. Unfortunately not all distributions have Fourier transforms. The appropriate subset that has Fourier transforms is called tempered distributions and denoted S. To do this we need to first define the space of rapidly decreasing functions. Definition 1 5. apidly decreasing functions We first define the space S of rapidly decreasing functions: { S = f C d : lim x α D β f x } = 0 for all multi-indices α and β. 34 x A sequence { φ n } S is said to converge to 0 in S, denoted φ n 0 in S, if for all k, β. Here lim n sup x d Example 1 6. e x S. Note that it does not belong to C 0. x k D β φ n x = 0. 35 x 1 + x 1 + + x d 1 /. 36 One can show that C 0 S and φ n 0 in C 0 implies φ n 0 in S. Definition 1 7. Tempered distributions The space of tempered distributions, denoted S, consists of those functionals on S such that if φ n 0 in S, then T φ n 0. Equivalently, T S if there are m, C such that T φ C sup x m D α φ x φ S. 37 x d α m Example 1 8. δ a α S for all point a and multi-index α. Now we can define the Fourier transform of T S.
Definition 1 9. Fourier transform of tempered distributions Let T S. Its Fourier transform Tˆ is defined by Tˆ φ = T φˆ 38 for any φ S. emark 0. The above definition is motivated by the following calculation for f, g S. 6 T fˆ g = fˆ ξ g ξ = e iξ x f x g ξ = e iξ x g ξ f x = ĝ x f x = T f ĝ. 41 emark 1. For this definition to make sense, it must hold that whenever φ S, φˆ S too. It turns out that ˆ is an isomorphism on S Interested readers can try to verify this. As a consequence, it is also an isomorphism on the dual space S. We notice that when φ n 0 in S, so is φˆ n. Therefore Tˆ S. Example. We compute δˆ. Take any φ S. we have δˆ φ = δ φˆ = φˆ 0 = π d/ φ x dx. 4 Therefore d where 1 is the constant function. δˆ = π d/ 1 43 Example 3. Consider the tempered distribution defined by the locally integrable function e i a x for a. For any φ S, we have e ia x φ = e ia x φˆ = e ia x φˆ x dx = π n/ φ a = π n/ δ a φ. 44 Therefore d The following properties are useful. Proposition 4. Let D j = i x j. Then e i a x = π n/ δ a. 45 a D j T = ξ j Tˆ; This implies P D T = P ξ Tˆ for any partial differential operator P D whose coefficients are constants. b x j T = D j Tˆ; c T S = π n/ Tˆ Ŝ, here we require S E 7. Example 5. Let P D be a differential operator with constant coefficients here we use D to denote i, then for any u, P D u = P ξ û. 46 6. emember that the Fourier transform of an integrable function f is defined as with inverse transform: fˆ ξ = π n/ d e i ξ x f x dx. 39 f x = π n/ d e i ξ x fˆ ξ dξ. 40 7. This actually makes Ŝ C. ecall that the multiplication between a C function and a distrib ution is well-defined.
. Existence for linear, constant-coefficient PDEs. Let P D be a differential operator of order m with constant coefficients Here we define D j = i x j. We try to solve the equation P D u = f 47 for f S. We know that it suffices to find a fundamental solution G which solves Now we take the Fourier transform of both sides, and obtain P D G = δ. 48 P ξ Ĝ = P D G = δˆ = π n/. 49 Dividing both sides by P ξ we have Ĝ = n/ π. 50 P ξ It suffices to show that Ĝ S, or equivalently, P ξ 1 S. The general proof is complicated. Here we will present the idea of an elementary proof for a special case, that is when f has compact support, following Michael Taylor PDE 3. 1 0. The main steps are the following. 1. By choosing a period box large enough, we can extend f periodically. After re-scaling we reduce the problem to the torus T d.. In this case the Fourier transform becomse Fourier series, and D = S. 3. G D T if and only if G k grows polynomially. 4. Observe that the solvability of P D u = f is the same as P D + α v = g where v = e iα x u and g = e iα x f. Thus we only need one α such that P k + α 1 k m < for some m. It suffices to show P k + α δ k m < for some δ > 0, m > 0. 5. For a polynomial P ξ not identically zero, its zeroes are curves. One can show that there is δ 1 > 0 such that P ξ δ <. 51 d for all 0 < δ < δ 1. 6. Let Q ξ = ξ m P 7. We have 8. Therefore ξ ξ 0 α i 1 ξ 1. Then Q is a polynomial and we obtain δ. Take δ = min δ 1, δ. P ξ δ ξ d dξ = = η 1 η 1 η 1 δ η P dη η m η δ Q η δ dη Q η δ dη <. 5 P k + α δ k d P ξ δ ξ d dξ <. 53 k Z d d Consequently we can find α such that the desired bound holds. emark 6. From the proof we see that when f C 0, the solution u also belongs to C 0.
3. Elliptic PDEs with constant coefficients. ecall that P D is elliptic if and only if for large ξ. We consider P ξ C ξ m 54 P D u = f 55 for such P D. It turns out that we can find E S which is smooth everywhere except the origin, rapidly decreasing at, such that P D E = δ + w 56 where w S d. This E is called a parametrix for P D. Now we have u = δ + w u w u = P D E u w u = E P D u w u = E f w u. 57 Definition 7. Singular support The singular support of a general distribution u D d is the smallest set K such that u is smooth on d \K read: u can be identified with a smooth function on d \K. This set is denoted sing supp u. 58 Example 8. We have sing supp δ = { 0}. 59 One can show that sing supp f g sing supp f + sing supp g. In particular, for the parametrix E, we have sing supp E f sing supp f. 60 Since w u C, we have established Proposition 9. For any u D d solving P D u = f, if P D is elliptic, then Further readings. sing supp u sing supp f. 61 Michael Taylor, Partial Differential Equations, Vol. 1, Chap. 3, Springer-Verlag. Joel Smoller, S hock Waves and eaction-diffusion Equations, Chap. 7, Springer-Verlag. Exercises. Exercise 1. Show that in D, but that lim sin n x = 0 6 n lim n sin n x 0. 63 This is the paradigm example in understanding weak convergence and homogenization theory. Hint: the first part follows directly from the so-called iemann-lebesgue Lemma. Exercise. Find the distributional derivatives of the following functions. 1. f x = ln x, x. { f1 x x > 0. f x = f x x < 0, x, f 1, f C. Exercise 3. Show that the general solution of x y = 0 is c 1 + c H x where H is the Heaviside function. Note that x y is the product of a C function and a distribution. Hint: First show that for φ C 0 which vanishes at the origin, φ/ x C 0 too, then show that y = c δ for some constant c. Exercise 4. Let u x, t = f x + t where f is any locally integrable function on. Then u D. Show that u solves the wave equation u t t u x x = 0 in the sense of distributions. Exercise 5. Let φ C 0 satisfy φ = 1. Define P rove that φ n δ in D. φ n = n φ n x. 64 Exercise 6. Prove that if φ vanishes in a neighborhood of the support of T D, then T φ = 0. Would it suffice if φ vanishes on the support of T? Hint: use partition of unity.