MA Handout 2: Notation and Background Concepts from Analysis


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1 MA Hndout 2: Nottion nd Bckground Concepts from Anlysis This hndout summrises some nottion we will use nd lso gives recp of some concepts from other units (MA20023: PDEs nd CM, MA20218: Anlysis 2A, nd MA20219: Anlysis 2B) tht will be needed.!!!! If you only lern two things in the course, mke them Green s identities (Theorem 3.6) nd the necessry conditions for locl extrem (Theorem 4.1).!!!! 1 Geometric nottion x R m x = (x 1, x 2,..., x m ), x i R, 1 i m. [x R m x = (x 1, x 2,..., x m ), x i R, 1 i m] x := ( m ) 1/2 [ i=1 x2 i. x := ( m ) ] 1/2 i=1 x2 i B(x, r) := {y R m : y x < r} is n open bll of rdius r centered t x. [B(x, r) := {y R m : y x < r}] will usully denote bounded, open set in R m (lthough we ll lwys sy this explicitly), = boundry of, nd = = closure of. Remrk 1.1 (Why open sets?) Recll tht is open if every point x hs little bll round it tht is lso in (i.e. for every x there exists n ε > 0 such tht B(x, ε) ). When we pose PDE in this course, it will lwys be in n open set. The reson is tht, if u : R, the definition of the derivtive of u t x requires tht there is little bll round x. Thus, since PDEs re reltionships bout derivtives, it is nturl to pose them in open sets. Definition 1.2 (Connected) R m is connected if there do not exist open disjoint sets 1, 2 R m such tht 1, 2 nd 1 2. ( You cn t split into two disjoint open sets.) Exmple 1.3 (Exmple of connected nd not connected sets) Let x 1, x 2 R m nd r 1, r 2 > 0. Define i := B(x i, r i ), i = 1, 2. Clerly, 1 nd 2 re connected. Set 3 = 1 2. Suppose tht x 1 x 2 nd set ρ := x 2 x 1 > 0 (drw picture!). If r 1, r 2 > ρ/2, then 1 2 nd 3 is connected. If r 1, r 2 < ρ/2, then 1 2 = nd 3 is not connected. 2 Function spces C() = {u : R : u is continuous } C() = {u : R : u is continuous nd cn be continuously extended to }. Exmple 2.1 Let m = 1 nd = (0, 1). Then u(x) = 1/x is in C() but not C(). Theorem 2.2 ( A continuous function on compct set is bounded nd ttins its bounds ) Suppose tht R m is compct nd let u C(). Then the function u is bounded nd ttins its mximum nd minimum, i.e. there exist x 1, x 2 such tht u(x 1 ) = sup{u(x) : x }, u(x 2 ) = inf{u(x) : x }. 1
2 C k () = {u : R : u nd its prtil derivtives up to order k re in C()}, Exmple 2.3 If m = 2 then continuous function u : R is in C 2 () if the following prtil derivtives exist nd re continuous on : u x 1, u x 2, x 2, 1 x 2, 2 x 1 x 2, x 2 x 1. C k () is the set of ll u C k () such tht u nd ll its prtil derivtives of order k cn be continuously extended to the closure. C () := k=1 Ck (), nd similrly for C (). For u : R m R we define the support, supp u R m, of u by supp u := {x R m : u(x) 0}. C k c (R m ) = {u C k (R m ) : u hs compct support}. In prticulr, if u C k c (R m ), then there exists ρ 0 such tht u(x) = 0 for ll x R m with x > ρ. Exmple 2.4 ( Exmple of C c (R m ) function ) Define v(t) = { 0 for t 0, e 1/t for t > 0. Then, v C (R) (since ll its derivtives t 0 re 0), but v Cc functions tht belong to Cc (R m ): define u : R m R by (R). Using v, we cn build u(x) := v(1 x 2 ) = { 0 if x 1, e 1/(1 x 2 ) if x < 1. Now v C (R) nd the function x 1 x 2 is in C (R m ). By the chin rule, u C (R m ). Moreover, supp u = B(0, 1), nd therefore u C c (R m ). If F is vector field, i.e. F : R m then we sy F is in C() if ech of its components, F i for i = 1,..., m, re in C(). Remrk 2.5 ( Help, these function spces look scry! ) When we prove results in the course we will be precise bout wht function spces the vrious functions re in. To begin with, you should just focus on understnding the ides of the proofs, becuse once you ve done tht you cn work out the function spces using the following three rules: 1. If u stisfies u = 0 in some set then u must be in C 2 (); this ensures tht the second derivtives exist nd re continuous. (Similr logic pplies for PDEs other thn Lplce s eqution.) 2. If we tlk bout the vlue of u on the boundry of then u must be in C() (so we know tht u hs vlue on the boundry!). 3. If the result uses the divergence theorem, Theorem 3.4 below, then whtever quntity we re using s F needs to be in C 1 () (e.g. if F = u u then u must be in C 2 ()). This lst rule is n exmple of the philosophy: don t lern the conditions in the theorem, lern the method used in the proof, nd then work out the conditions from the method. 2
3 3 Vector clculus, the divergence theorem, nd Green s identities Definition 3.1 (Grdient) If u C 1 () is sclr function (i.e. u : R), the grdient of u is vector defined by ( ) u u u u(x) := (x), (x),..., (x), x 1 x 2 x m i.e. ( u) i = u/ x i. Definition 3.2 (Divergence) If F C 1 () is vectorvlued function, the divergence of F is defined by m F i F(x) = (x) x i (where F i is the ith component of F). Exmple: x = m. Lemm 3.3 (Three importnt vector clculus identities) (i) ( x ) = x x. (3.1) (ii) If f C 1 (R) then i=1 f( x ) = f ( x ) x x. (3.2) (iii) If u nd F re both in C 1 () for some open set then, in this set, Proof. (i) Using the chin rule we hve tht (uf) = u F + u F. (3.3) x = (x x 2 x i x m) 1/2 2x i = i 2(x x2 m) 1/2 which implies tht (ii) By the chin rule so ( x ) = x x. f( x ) = f ( x ) x, x i x i f( x ) = f ( x ) ( x ) = f ( x ) x x using prt (i). (iii) Using the definition of the divergence we hve tht The product rule implies tht (uf) = n j=1 x j (uf j ). (uf j ) = u F j + u F j, x j x j x j nd thus, using the definitions of the grdient nd the divergence, we hve tht (uf) = u F + u F. 3
4 Theorem 3.4 (Divergence theorem or Guss s integrl theorem) Let R m be bounded, open set with sufficiently smooth boundry. Let n(x) denote the outwrdpointing unit norml vector t x. Let F : R m be in C 1 (). Then F(x) dx = F(x) n(x) ds(x). (3.4) We mention tht the bove theorem holds under weker smoothness conditions F C() C 1 () nd F(x) dx <. In 3d the lefthnd side of (3.4) is volume integrl, nd the righthnd side is surfce integrl. In 2d the lefthnd side is n integrl over 2d re, nd the righthnd side is line integrl over curve. The elementl volume dx cn lso be denoted dv or dv(x) (or even d m x if x = (x 1,..., x n )). Remrk 3.5 ( How smooth is sufficiently smooth for the divergence theorem to hold?) Short version: domins with corners (in 2d) nd edges (in 3d) re ok. Longer version: Theorem 3.4 is vlid if is Lipschitz domin. This mens tht loclly is the grph of Lipschitz function. A function f is Lipschitz if, for some constnt L > 0, f(x) f(y) L x y for ll x nd y where f is defined. One wy to think of this is f is Lipschitz if it is slightly better thn continuous, but not quite differentible. A Lipschitz domin does not possess norml vector t every x (think of squre) but t lmost every x. For u C 1 (), where R m is bounded, open set with smooth boundry, we define the norml derivtive of u on by For u C 2 (), the Lplcin of u, u, is defined by u (x) := u(x) n(x) for x. (3.5) n u(x) := u(x) = 2 u(x) = m i=1 x 2 (x) for x. (3.6) i Theorem 3.6 (!!!! Green s identities!!!! ) Let R m be bounded, open set tht is sufficiently smooth enough for the divergence theorem to hold. If u C 1 () nd v C 2 () then (u v + u v) dx = u v ds(x) (Green s 1st identity), (3.7) n nd if both u nd v re in C 2 () then ( (u v v u) dx = u v n v u ) ds(x) n (Green s 2nd identity) (3.7b) Proof. Set F = u v. Then, by (3.3) we hve tht F = u v + u v nd (3.7) follows from the divergence theorem (3.4) pplied to F. Moreover, interchnging u nd v in (3.7) gives (v u + v u) dx = v u n ds(x). Subtrcting (3.7) from this eqution yields (3.7b). 4
5 Alterntively, (3.7b) cn be thought of s the divergence theorem pplied to u v v u = (u v v u). Function spces: for F = u v to be in C 1 () we need u C 1 () nd v C 2 (); this gives the conditions for (3.7) to hold. Similrly, for F = v u to be in C 1 () we need v C 1 () nd u C 2 (). Thus, for (3.7b) to hold we need both u nd v to be in C 2 (). Remrk 3.7 Comprison of 1dimensionl cse nd multidimensionl cse 1dimension Fundmentl theorem of clculus: Product rule: b v dx = v(b) v() (uv) = u v + uv. So hve integrtion by prts b uv dx = [uv] b b u v dx. Multidimensions Divergence theorem: F dx = Importnt identity (3.3) F n ds. (uf) = u F + u F. So hve generl form of integrtion by prts u F dx = u F n ds u F dx. (importnt specil cse: F = v Green s 1st identity (3.7) Corollry 3.8 (Importnt for Chpter 1 of notes) If u C 2 () then pplying the divergence theorm, eqution 3.4, to F = u we get u u dx = n ds(x). This cn lso be understood s setting v 1 in Green s 1st or 2nd identity. Remrk 3.9 ( Wekening the hypotheses on u nd v in Theorem 3.6) The divergence theorem, Theorem 3.4, is still true if F i is in C 1 () C() for ech 1 i m nd F dx < (i.e. the first derivtives need not be continuous on the boundry but the integrl over still needs to mke sense). (This cn be proved by pproximting by sequence of domins from the inside.) Thus Theorem 3.6 is vlid if u, v C 1 () C 2 () nd lso u dx <, v dx <. Theorem 3.6 ws stted with these conditions in previous versions of the course, nd so you will see these less restrictive conditions in some pst exm ppers. 4 Locl extrem Let R m be n open set. A function u : R hs locl mximum t point x 0 if there exists ε > 0 such tht u(x 0 ) u(y) for ll y B(x 0, ε). The function u is sid to hve locl minimum t x 0 if u hs locl mximum t x 0. Theorem 4.1 (!!!! Necessry condition for locl extrem!!!! ) Let R m be n open set. If function u C 2 () hs locl mximum or locl minimum t point x 0, then u(x 0 ) = 0. 5
6 Moreover, if u hs locl mximum t x 0, then x 2 i nd if u hs locl minimum t x 0, then x 2 i (x 0 ) 0, i = 1, 2,..., m, (x 0 ) 0, i = 1, 2,..., m. Lemm 4.2 If b + ε for every ε > 0 then b. Proof. We suppose tht > b nd rech contrdiction. If > b then ( b) > 0 nd thus we cn let ε = ( b)/2. With this vlue of ε, the fct tht b + ε implies tht b, which is contrdiction. 5 Useful results bout integrls For simplicity these re ll stted in one dimension, but we will lso use the corresponding multidimensionl versions. Theorem 5.1 ( The integrl of the modulus is less thn the modulus of the integrl ) If f C[, b] then b b f(x) dx f(x) dx. (5.1) Sketch proof of Theorem 5.1. Integrte the inequlity f(x) f(x) f(x) over [, b]. Theorem 5.2 ( Pulling things out of integrls ) If f nd g re in C[, b] then b ( ) b f(x)g(x) dx mx f(x) g(x) dx. (5.2) x [,b] Corollry 5.3 ( Integrl is less thn length times mx ) If f is in C[, b] then b f(x) dx (b ) mx f(x). (5.3) x [,b] Sketch proof of Theorem 5.2. By Theorem 5.1, b f(x)g(x) dx b f(x)g(x) dx. Then integrte the inequlity ( ) f(x)g(x) mx f(x) g(x). x [,b] At couple of points in the course we will need to differentite under the integrl sign. 6
7 Theorem 5.4 (Differentiting under the integrl sign) Given f(x, y) on [, b] [c, d], define F (y) by F (y) := b f(x, y) dx, c y d. If both f(x, y) nd f y (x, y) re continuous on [, b] [c, d] then F (y) exists nd F (y) := b f (x, y) dx, c y d. y (Brodly speking, if the resulting integrl mkes sense fterwrds, it s ok. ) Following Struss (1992), we cll the next theorem the vnishing theorem. Theorem 5.5 ( Vnishing theorem ) Let f C[, b] nd suppose tht f 0 on [, b]. If b f(x) dx = 0, then f(x) = 0 for ll x [, b]. For proof, see Struss (1992) A.1, p Corollry 5.6 (If the verge of function equls the mx, then the function is constnt (nd equl to the mx)) Let f C[, b] with f(x) M for ll x [, b]. If then f(x) = M for ll x [, b]. 1 b b f(x) dx = M, Proof. We hve tht so b f(x) dx = M(b ) = M b b ( M f(x) ) dx = 0. dx, Now M f(x) 0 for ll x [, b], so by the vnishing theorem (Theorem 5.5) M f(x) = 0 for ll x [, b]. 6 Distributions 6.1 The delt function The multivrite delt function, δ, is defined by the following two properties: (D1) δ(x) = 0 for ll x R m \ {0}. (D2) δ(x)φ(x) dx = φ(0) for ll functions φ tht re continuous t zero R m (in prticulr for the set of test functions φ Cc (R m )). 1 Of course, there does not exist function in the usul sense tht stisfies properties (D1) nd (D2). However δ cn be understood s limit of usul functions, see Lemm 6.1 below (you lso sw this in MA ). 1 In MA30044 this second property ws replced by (D2) : R m δ(x) dx = 1. By letting φ 1 in (D2) you cn see tht (D2) = (D2). 7
8 Why do we need to consider the delt function? The gret thing bout liner PDEs is tht (by definition) if u 1 nd u 2 re both solutions, then so is u 1 + bu 2 for, b constnts (this is liner superposition). The whole ide of seprtion of vribles/trnsform methods (tht you studied in MA20023/MA30044) is to express the solution of liner PDE s superposition (either or n ) of solutions to ODEs. It turns out tht very useful ide in PDEs is to express the solution to liner PDE, sy L y u(y) = 0 2, s integrls (i.e. superposition) involving the fundmentl solution, N x (y), where L y N x (y) = δ(y x). Concrete exmple (see Theorem 1.1 in Lecture notes): if Lu u = 0 in then ( u(x) = u(y) N x n (y) N x(y) u ) n (y) ds(y) for x (this is clled Green s integrl representtion). The delt function cn be interpreted s the limit of sequence of functions in the usul sense. In MA you sw this in 1d with the sequence of functions δ ε (x) = ε/[π(ε 2 + x 2 )] s ε 0. Here we will provide generl recipe for pproximting the delt function in R m. Lemm 6.1 (Delt function pproximtion lemm) Let f : R m R stisfy (H1) supp f is compct ; (H2) f(x) dx < R m nd For ε > 0, define f ε : R m R by Then f ε ( ) δ( ) s ε 0 in the sense tht (1) There exists ρ > 0 such tht for ech ε > 0 R m f(x) dx = 1. f ε (x) = 1 ( ) 1 ε m f ε x. f ε (x) = 0, for ll x R m with x ερ, (AD1) i.e. the support of f ε hs width O(ε) bout 0 (this is the nlogue of (D1). (2) For ll bounded functions φ : R m R tht re continuous t 0 (nd so in prticulr for ll φ Cc (R m )) lim f ε (x)φ(x) dx = φ(0). (AD2) ε 0 R m (this is the nlogue of (D2)). Ide of Lemm 6.1 f hving compct support, condition (H1), mens tht f ε will hve support of width O(ε), i.e. condition (AD1). The 1/ε m scling in the definition of f ε is chosen to keep f R m ε (x)dx = 1. Thus, f ε is becoming more nd more concentrted ner 0, nd its integrl is lwys 1. Intuitively we cn see tht f R m ε (x)φ(x)dx will tend to φ(0). The trick is to chnge vribles so tht the ε ppers in the rgument of φ, nd then use the fct tht φ is continuous t 0. 2 the subscript y in L y is to emphsise the differentition is with respect to y 8
9 Proof of Lemm 6.1. To prove sttement (1) suppose tht f hs compct support. Then there exists ρ > 0 such tht supp f B(0, ρ). Let ε > 0 nd x R m be such tht x ερ. Then x/ε ρ nd so x/ε supp f, which in turn implies tht f(x/ε) = 0 nd so f ε (x) = 1 ( ) 1 ε m f ε x = 0. In order to prove sttement (2), suppose tht f stisfies (H2) nd let φ : R m R be bounded nd continuous t 0. By chnge of vribles we obtin f ε (x)φ(x) dx = 1 ( ) 1 R ε m f m R ε x φ(x) dx = f(y)φ(εy) dy. m R m Using this lst eqution, the continuity of φ t 0, nd the fct tht f(x) dx = 1, we find tht R m lim ε 0 f ε (x)φ(x) dx = R m f(y)(lim φ(εy)) dy = R m ε 0 f(y)φ(0) dy = φ(0) ; R m thus we hve proved (AD2). 6.2 Generlized derivtive For certin functions (distributions) which re not differentible in the usul sense, generlized derivtive cn be defined. It turns out tht in the following exmple the generlized derivtive is the delt function. Exmple: Let m = 1 nd consider the unitstep function (or Heviside function) U : R R given by { 0 for x < 0, U(x) = 1 for x 0. Clerly, U is not differentible t 0. Nevertheless we try to define generlized derivtive U. If we postulte tht the integrtion by prts formul still holds for the generlized derivtive, then we obtin for ll test functions φ Cc (R) U (x)φ(x) dx = [U(x)φ(x)] + Since φ hs compct support we see tht nd thus [U(x)φ(x)] + = 0, 0 U(x)φ (x) dx = [U(x)φ(x)] + φ (x) dx = φ( ) φ(0) = φ(0), U (x)φ(x) dx = φ(0). 0 φ (x) dx. It follows tht U = δ, which is consistent with our intuition tht the derivtive of U should be zero everywhere prt from shrp spike t 0. To define the derivtive of nsty functions: multiply by test function nd integrte by prts. Consider the 1d cse. If φ Cc (R) we will cll φ test function. If f is in C 1 (R) nd φ is test function then, by integrtion by prts, f (x)φ(x) dx = f(x)φ (x) dx. (6.3) The integrls converge, nd the boundry terms from the integrtion by prts vnish, becuse φ equls zero for x ρ, for some ρ (this is precisely the reson why we consider φs with compct support). 9
10 The term on the lefthnd side of (6.3) only mkes sense if f is differentible; however the term on the righthnd side mkes sense under much less restrictions on f. In fct we only need Indeed, ρ ρ ρ f(x)φ (x) dx = f(x)φ (x) dx ρ ρ ρ f(x) dx < for every ρ > 0. (6.4) for some ρ > 0 (since φ hs compct support) f(x) φ (x) dx by Theorem 5.1, mx x [ ρ,ρ] φ (x) ρ ρ f(x) dx by Theorem 5.2, so if (6.4) holds then the righthnd side of (6.3) is finite. Given n f stisfying (6.4), if there exists g (either usul function, or one such s the delt function) such tht g(x)φ(x)dx = f(x)φ (x)dx for ll φ Cc (R), (6.5) then we sy tht g is the distributionl derivtive of f. (Of course, if f is differentible in the stndrd sense, then (6.3) shows tht the distributionl derivtive of f is equl to f the stndrd derivtive.) More generlly, if there exists n h (either usul function, or one such s the delt function) such tht h(x)φ(x)dx = ( 1) n f(x)φ (n) (x)dx for ll φ Cc (R), (6.6) then we sy tht h is the nth distributionl derivtive of f. (We consider test functions, φ, tht re infinitely differentible so tht the righthnd side of (6.6) mkes sense for ny n Z +.) In more thn one dimension the ide is the sme, but we now use Green s identities, (3.7) nd (3.7b), insted of integrtion by prts. Thus y N x (y) φ(y) dy := R m N x (y) φ(y) dy, R m (6.7) where we hve used Green s 2nd identity (3.7b) to move ll the derivtives from N x onto the test function φ (the boundry terms t re zero since φ hs compct support, i.e. φ is zero outside some bll). 10
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