PAA by Douglas Heggie 2011 Notes of Paul Boeck

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PAA by Douglas Heggie

2011

Notes of Paul Boeck

Last changes: May 16, 2011

1 The Generalized Riemann Integral

Definition 1.1

Let f :[a,∞)→_Rbe a function that is Riemann–integrable on each[a,b], b>a. We define Z _∞

a f(x)dx= lim

b→∞ Z _b

a f(x)dx provided that the limit exists.

Example 1.2 • f(x) = _x¹₂,x ∈[1,∞). ThenRb

1 f(x)dx=^h^x₋₂₊₁⁻²⁺¹ⁱ^b

1= (b⁻¹−1) =1−b⁻¹−−−→^b→^∞ 1

• ¹_x,x∈[1,∞),Rb 1 1

xdx= [lnx]^b₁=lnb−−−→^b→^∞ _∞ =⇒ the integral does not exist.

Remark Integrals with∞limits sometimes are called "improper" integrals.

Definition 1.3

Let f : (−∞,a] → Rbe a function which is Riemann–integrable on each [b,a] with b < a. We define Ra

−∞f(x)dx = limb→−∞Ra

b f(x)dx provided the limit exists.

Definition 1.4

Let f :R→Rbe a given function. We defineR_∞

−∞f(x)dx=R0

−∞f(x)dx+R_∞

0 f(x)dx provided both limits exist.

Proposition 1.5 The integralR_∞

a f(x)dx exists iffRb⁰

b f(x)dx→0as b,b⁰→_∞.

Proof: Problem 2.

Proposition 1.6 IfR_∞

a |f(x)|dx exists, so doesR_∞

a f(x)dx.

Proof: Assume R_∞

a |f(x)|dx exists. Then Prop. 1.5. implies Rb⁰

b |f(x)|dx −−−−→^b,b⁰^→^∞ 0. But

Rb⁰

b f(x)dx ≤ Rb⁰

b |f(x)|dx. Hence Prop 1.5 impliesR_∞

a f(x)dxexists.

Proposition 1.7 (Comparison)

Let f,g be Riemann–integrable on every[a,b]and suppose that|f(x)| ≤g(x)on[a,∞)andR_∞

a g(x)dx exists. Then so does R_∞

a f(x)dx.

Proof: Rb

a |f(x)|dx ≤ Rb

a g(x)dx −−−→^b→^∞ R_∞

a g(x)dx from below. Obviously Rb

a |f(x)|dx is a non–decreasing function ofband bounded above byR_∞

a g(x)dx. HenceR_∞

a |f(x)|dxexists, henceR_∞

a f(x)dxexists by Prop.

1.6.

Example 1.8 • TestR_∞

0 cosx 1+x²dx.

cosx 1+x²

≤ _x¹₂ andRb 0 dx

1+x² = [arctanx]^b₀ =arctanb −−−→^b→^∞ ^π₂. Hence convergence by Prop. 1.7.

• TestR_∞

0 e^−x²dx. Nowx²≥xifx ≥1 and so ifb>1:Rb

0 e^−x²dx=R1

0 e^−x²dx+Rb

1 e^−x²dxwhere|e^−x²| ≤e^−x (ifx>1).

Rb

1 e^−xdx= [−e^−x]₁^b=e^−b+e⁻¹−−−→^b→^∞ e⁻¹. Hence the whole integral exists.

• TestR_∞

1 x^ke^−xdx. e^−x = _e¹x ≤ _xn¹/n! for anyn≥0 and allx ≥0. Choosen> k+1. Thusx^ke^−x ≤n!x^k−n and Rb

1 x^k−ndx= _k−n+1¹ ^hx^k−n+1ib

1= _k−n+1¹ e^k−n+1−1 _b→_∞

−−−→ −_k−n+1¹ . Hence convergence.

Theorem 1.9 (Leibniz’ rule for improper integrals) Suppose f(x,τ)_and fx(x,τ)are continuous functions fora ≤ x ≤ bandc ≤ τ < _{∞. Define} F[x,t) = Rt

c f(x,τ)dτand supposeF(x,t) → ϕ(x) ≡ R_∞

c f(x,τ)dτandFx(x,t) → ψ(x)uniformly forx∈[a,b]_.

Thenϕis differentiable andψ(x) =ϕ⁰(x) =R_∞

c fx(x,τ)dτ.

Remark Basically _dx^d R_∞

c f(x,τ)dτ=R_∞

c ∂

∂xf(x,τ)dτprovided that integral on the right has uniform convergence.

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2 FOURIER TRANSFORMS

2 Fourier transforms

2.1 Definition and basic properties

Definition 2.1.1

Let f :R→Cbe such thatR_∞

−∞|f(x)|dx exists. We define theFourier transformof f by fˆ:R→C, where fˆ(ξ) =

Z _∞

−∞f(x)e^−iξ^xdx (ξ∈R) Remark (i) The integral exists, becauseR_∞

−∞

f(x)e^iξ^x

dxexists, and using Prop 1.7.

(ii) ˆf is a bounded function, because for allξ∈R

|f^ˆ(ξ)|=

Z _∞

−∞f(x)e^−iξxdx

≤ Z _∞

−∞|f(x)e^−iξ^x|dx= Z _∞

−∞|f(x)|dx Example 2.1.2 Take f(x) =

(1 if|x| ≤1

0 otherwiseIts F.T. is fˆ(ξ) =

Z _∞

−∞f(x)e⁻^ξxdx= Z ₁

−1e^iξxdx=

−¹ iξe^−iξx

1

−1

=−¹ iξ

e^−iξ−e^iξ

=−¹

iξ(−2isinξ) = ²

ξsinξ if ξ6=0 Forξ=0, ˆf(0) =R1

−1 dx=2=lim_ξ→02 sinξ

ξ . Thus ˆf(ξ) = (_{2 sin}_ξ

ξ ifξ6=0 2 ifξ=0

Remark Sometimes we just write ˆf = ^{2 sin}_ξ ^ξ on the understanding that ˆf(0) =lim_ξ→0fˆ(ξ). Definition 2.1.3

TheSchwartz space(of rapidly decreasing functions) denotedS(_R)is defined to be the set of all functions f :R→_Cwhich are infinitely differentiable and have the following property: For all integers k,l ∈ (0, 1, 2, . . .)the function|x|^k|f^(l)(x)|is bounded.

Example 2.1.4 The function f(x) =e^−x² is inS(R). For f is obviously infinitely differentiable.

Also fixk,l∈ {0, 1, 2, . . .}, we show that|x|^k|f^(l)(x)|is bounded, as follows.

First, do

l=_{0 :}|x|^k|f(x)|=|x|^ke^−x² = |x|^k

e^x² = |x|^k

1+x²+¹₂x⁴+· · ·+_(2k)!^x^2k +· · · ≤ |x|^k 1+_(2k)!^x^2k which is bounded.

Now considerl > 0. First, observe that f(x) = e^−x² =⇒ f⁰(x) = −2xe^−x² =⇒ f⁰⁰(x) = (−2+4x²)e^−x². Clearly (or by induction) f^(l)(x) = hl(x)e^−x², wherehl(x)is a polynomial. Hence|x|^k|f^(l)(x)| = |x^khl(x)|e^−x² =

|q_l,k(x)|e^−x², whereq_l,kis another polynomial. Writeq_k,l(x) = _∑^d_n=0cnx^k. Thenq_k,l(x)e^−x² = _∑^d_n=0cnxⁿe^−x², each term is bounded

Proposition 2.1.5 If f ∈ S(R)thenR_∞

−∞|f(x)|dx exists.

Proof: Since|x|^k|f^(l)(x)|is bounded for eachk,l ∈(0, 1, 2, . . .),|f(x)|andx²|f(x)|are bounded. (l =0,k=0 and 2). Hence(1+x²)|f(x)|is bounded, i.e.∃Ms.t.(1+x²)|f(x)| ≤M. Hence|f(x)| ≤ _1+x^M₂.R_∞

−∞ dx

1+x² exists,

hence result, by Prop 1.7.

Corollary 2.1.6

The Fourier transformf exists for any fˆ ∈ S(R).

Proof: See Def 2.1.1 and remarks and Prop 2.1.5.

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2.1 Definition and basic properties 2 FOURIER TRANSFORMS

Example 2.1.7 Let f : R→ Cbe infinitely differentiable and suppose that f vanishes outside some interval[a,b]. Then f ∈ S(R).

For, given anyk,l the function|x^k||f^(l)(x)|vanishes outside[a,b]and is continuous and hence bounded on[a,b]. Hence it is bounded onR.

Example 2.1.8 The functione^−|x|is not inS(R)as it is not differentiable atx=0.

Proposition 2.1.9

S(R)is closed under (i) differentiation and under (ii) multiplication by polynomials.

Proof: (i) Let f ∈ S(R). Then f is infinitely differentiable. Therefore f⁰is infinitely differentiable. Also for any integersk,l≥0 we havex^k(f⁰)^(l)(x) =x^kf^(l+1)(x), which is bounded.

(ii) See Problem 9.

Let f ∈ S(R)and let h∈R. Define g(x) =e^−ixhf(x). Thengˆ(ξ) = f^ˆ(ξ+h). Proof: gˆ(ξ) =R_∞

−∞g(x)e^−iξxdx=R_∞

−∞f(x)e^−i(ξ+h)dx= f^ˆ(ξ+h)

Let f ∈ S(_R), h∈_Rand define g(x) = f(x+h). Thengˆ(ξ) = f^ˆ(ξ)e^iξh Proof: gˆ(ξ) =R_∞

−∞g(xe^−iξxdx=R_∞

−∞f(x+h)e^−iξxdx. Change variable: Letx⁰ =x+h.

Then ˆg(ξ) =R_∞

−∞f(x⁰)e^−iξ(x⁰^−h)dx⁰=R_∞

−∞e^−iξx⁰e^iξhdx⁰ =e^ihξfˆ(ξ).

Remark Props 2.1.10-11 are calledshift–theorems.

Let f ∈ S(R). The(df⁰)(ξ) =iξfˆ(ξ).

Remark analogue ofcn(f⁰) =incn(f)complex Fourier series.

Proof: (df⁰)(ξ) =R_∞

−∞f⁰(x)e^−iξxdx=f(x)e^−iξx∞

−∞−R_∞

−∞f(x)(−iξ)e^−iξxdx. Now f ∈ S(R) =⇒ |x f(x)| ≤ M =⇒ |f(x)| ≤ _|x|^M for allx6=0.Hence limx→±∞|f(x)e^−iξx|=0. Finally(df⁰)(ξ) =iξfˆ.

Let f ∈ S(R). Then(f^ˆ)⁰(ξ) =gˆ(ξ)where g(x) =−ix f(x). Proof: (f^ˆ)⁰(ξ) = _dξ^d R_∞

−∞ f(x)e^−iξxdx=R_∞

−∞ ∂

∂ξ f(x)e^−iξx

dx=R_∞

−∞(−ix f(x))e^−iξxdxby Prop 1.9.

Remark \_{x f}(x)(ξ) =ifˆ⁰(ξ). Corollary 2.1.14

If f ∈ S(_R)thendf^(k)= (iξ)^kf .^ˆ

Proof: Induction onk. Prop 2.1.12 givesk=1. Also, assuming 2.1.14,\_f^(k+1)=(\f^(k))⁰ =iξdf^(k)=iξ(iξ)^kf^ˆ.

Corollary 2.1.15

If f ∈ S(R)thenf is infinitely differentiable.ˆ fˆ^(k)(ξ) =gˆk(ξ)where gk(x) = (−ix)^kf(x). Proof: Induction based on Prop 2.1.13.

Remark _x\^k_f(x)(ξ) =i^kfˆ^(k)(ξ).

Example Let f(x) = e^−x². f⁰(x) = −2xe^−x² = −2x f(x) (∗).The FT of f⁰ isiξfˆ(Prop 2.1.12) and the FT ofx f(x) isifˆ⁰. Hence(∗) =⇒ iξfˆ= −2ifˆ⁰. We are using the fact that the Fourier transform, being an integral, is a linear operation. Hence ^d_dξ^f^ˆ=−^ξ₂f^ˆ. Separable differential eqn:

Z dfˆ fˆ −

Z ξ

2dξ =⇒ ln ˆf =−^ξ

2

4 +const. =⇒ f^ˆ(ξ) =Ce^−ξ²^/4 Cconst.

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2.2 Gaussians 2 FOURIER TRANSFORMS

Z∞

−∞

e^−x²^−iξxdx x²+iξx=x+^iξ₂²+^ξ₄²

= Z∞

−∞

e^−(x+^iξ²⁾²^−ξ²^/4dx=e^−ξ²^/4 Z∞

−∞

e^−(x+^iξ²⁾²dx x⁰=x+^iξ 2

=e^−ξ²^/4

∞+iξ/2

Z

−∞+iξ/2

e^−x⁰²dx⁰

Now ˆf(0) =R_∞

−∞f(x)dxby Def 2.1.1.C=R_∞

−∞e^−x²dx=√

π. Hence ˆf(ξ) =√ πe^ξ²^/4. Proposition 2.1.16

If f ∈ S(R)thenfˆ∈ S(_R). Proof: We need to show

(a) ˆf is infinitely differentiable and

(b) |ξ|^k|f^ˆ^(l)(ξ)|is bounded for all integersk,l≥0

To proof (b) we have (iξ)^kf^ˆ^(l)(ξ) = (iξ)^kgˆ_l(ξ)where g_l(x) = (−ix)^lf(x). by Cor 2.1.15. Now g_l ∈ S(_R) because it is a product off ∈ S(_R)by a polynomial (Prop 2.1.9).

fˆ(k)=

Z _∞

−∞f(x)e^−iξxdx

≤ Z _∞

−∞|f(x)||e^−iξx|dx

Also(iξ)^kgˆ_l(ξ) = ^dg^(k)_l (ξ)by Cor 2.1.14. But since g_l ∈ S(_R)we haveg^(k)_l ∈ S(_R)(Prop 2.1.9 again). Hence

dg^(k)_l

≤R_∞

−∞|g^(k)_l |dx(by Def 2.1, Remark ii), i.e.

ξ^kfˆ(l)(ξ)=(iξ)^kf^ˆ^(l)

=(iξ)^kgˆ_l(ξ)=

gd^(k)_l (ξ)

is bounded

Remark The mapF :S(_R)→ S(_R), defined byF(f) = f^ˆis linear (as integration). We shall see it is 1–1 and onto and (almost) an isometry.

2.2 Gaussians

i.e. functions of the forme^−ax²,a>0.

Forδ>0define K_δ(x) = ^√¹

2πδe^−x²^/2δ. Lemma 2.2.2

Kˆ_δ(ξ) =e⁻^δξ²^/2. Proof:

Z _∞

−∞

√1

2πδe^−x²^/2δ−iξxdx x=√ 2δx⁰

= Z _∞

−∞

√1

πe^−x⁰²^−2ξx⁰

√2δdx=e^−ξ²^2δ/4=e^−ξ²^δ/2

Remark K1 δ

(x)^{Def 2.2.1}= q

δ

2πe⁻^δx²/2 Lemma 2.2.2= q

δ

2πKˆ_δ(x). Also ˆK1 δ

(ξ) =√

2πδK_δ(ξ).

Example What is the function whose FT isξe⁻^δξ²^/2? We use 2.1.12: bf⁰ = iξfˆ. Let f(x) = K_δ(x) = ^√¹

2πδe^−x²^/2δ. Hence bf⁰ =_iξe^−δξ²^/2. Required function is ¹_iK⁰_δ(_x) =−i^√_2πδ¹ (−^2x_2δ)_e^−x²^/2δ= ^ix

δK_δ(_x)_.

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2.3 Inversion Theorem 2 FOURIER TRANSFORMS

The family K_δ,δ>0has the properties (i) ∀δ>0 :R_∞

−∞K_δ(x)dx=1 (ii) ∀η>0 :R

|x|>ηK_δ(x)dx→0asδ→0.

For f,g∈ S(_R)define theirconvolutionby

f∗g:R→_C where (f∗g)(x) = Z∞

−∞

f(x−y)g(y)dy

Theorem 2.2.5 (Convolution Theorem) Let f,g∈ S(_R)then(\f∗g)(ξ) = f^ˆ(ξ)gˆ(ξ). Proposition 2.2.6

If f ∈ S(R)then f∗K_δ → f uniformly onRasδ→0.

2.3 Inversion Theorem

Proposition 2.3.1 (Multiplication Formula) Let f,g∈ S(R). Then

Z∞

−∞

f(x)gˆ(x)dx= Z∞

−∞

fˆ(x)g(x)dx

Theorem 2.3.2 (Inversion Theorem) Let f ∈ S(_R). Then∀x∈_R f(x) = ¹

2π Z _∞

−∞fˆ(ξ)e^ixξdξ

Remark This is the analogue of the following result for a well–behaved 2π–periodic function f. If cn = ¹

2π Z _π

−π f(x)e^−inxdx then f(x) =

∑

∞

−∞cne^inx Proof: First prove it forx=0, i.e. f(0) = _2π¹ R_∞

−∞fˆ(ξ)dξ(∗). Letg(x) =G_δ(x) = _2π¹ e⁻^δx²^/2where Z _∞

−∞ f(x−y)K_δ(y)

| {z }

ˆ g(y)

dy→ f(x) K_δ= √¹

2πδe^−x²^/2δ) F√¹

2πδe^−x²^/2δ is e⁻^δx²^/2 F

r δ

2πe^δx²^/2 is e^−x²^/2δ (δ→ ¹

δ) Gˆ_δ(ξ) =K_δ(ξ)(remark to Lemma 2.2.2). ThusR_∞

−∞f(x)K_δ(x)dx= R_∞

−∞fˆ(x)^e⁻^δx2/2_2π dx. Now letδ→0. RHS→

1 2π

R_∞

−∞fˆ(x)dxand LHS=R_∞

−∞ f(x)K_δ(x)dx.

Z _∞

−∞f(x−y)K_δ(y)dy→ f(x)

= Z _∞

−∞f(−y)K_δ(−y)y (x=−y)

= Z _∞

−∞f(0−y)K_δ(y)dy sinceK_δ(y) =K_δ(−y)

= (f∗K_δ)(0)→ f(0) as δ→0 by Prop 2.2.6 Hence f(0) = _2π¹ R_∞

−∞ fˆ(ξ)dξ.

Fix x0 ∈ _{R. Define} g(x) = f(x+x0). Apply (∗) to g : g(0) = _2π¹ R_∞

−∞gˆ(ξ)dξ. But g(0) = f(x0) and ˆ

g(ξ) =e^iξx⁰fˆ(ξ)(Prop 2.1.11). Hence f(x0) = _2π¹ R_∞

−∞fˆ(ξ)e^iξx⁰dξ.

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2.4 Plancherel’s Theorem 2 FOURIER TRANSFORMS

Corollary 2.3.3

(i) If f ∈ S(R)and fˆ(0) =0(i.e. fˆ(ξ) =0for allξ∈R) then f =0.

(ii) If f₁,f₂∈ S(_R)and fˆ₁= f^ˆ₂then f₁= f₂ Proof: (i) f(x) = _2π¹ R_∞

−∞0e^iξxdξ=0

(ii) apply (i) to f₁−f₂.

For f ∈ S(_R)we have defined its FT by(Ff)(ξ) = f^ˆ(ξ) =R_∞

−∞f(x)e^−ixξdx.

We now also define theinverse Fourier transformof f by(F⁻¹f)(ξ) = _2π¹ R_∞

−∞f(x)e^ixξdx= f^ˇ(ξ). Remark (i) We can show that if f ∈ S(_R), then ˇf ∈ S(_R).

(ii) The inversion theorem (2.3.2) states thatF⁻¹(Ff) = f. Similarly we can show thatF(F⁻¹f) = f. Corollary 2.3.5

The mapF :S(_R)→ S(R), f 7→ Ff = f is 1–1 and onto (i.e. bijection).^ˆ Proof:onto: If f ∈ S(_R)letg=F⁻¹f. ThenFg= f.

1–1: Ff₁=Ff₂ =⇒ f₁= f₂(Cor 2.3.3(ii))

Remark The inversion theorem is true for many f ∈ S(/ R). Example 2.3.6 Let f(x) =

(e^−x ,x>0

0 ,x≤0. Then ˆf(ξ) =_1+iξ¹ (Tutorial Q4b). Hence f(x) = _2π¹ R_∞

−∞ e^iξx

1+iξdξ. f(−x) =

1 2π

R_∞

−∞e⁻^iξxdξ

1+iξ and withx↔ξwe have f(−ξ) = _2π¹ R_∞

−∞e⁻^iξxdx 1+ix

2.4 Plancherel’s Theorem

Recall:For 2π–periodic functions f :R→_Cwe haveParceval’s identity 1

2π Z _π

−π

|f(x)|²dx=

∑

∞

−∞

|cn|² Theorem 2.4.1 Iff ∈ S(_R)then

Z _∞

−∞|f(x)|²dx= ¹ 2π

Z _∞

−∞|f^ˆ(ξ)|²dξ Remark (i) In terms of theL₂–normkfk₂=R_∞

−∞|f(x)|dx1/2

,Plancherel’s Theoremsays thatkfk₂= ^√¹

2πkf^ˆk₂. (ii) Except for the factor ^√¹

2π, which can be removed by slight redefinition ofF, F is an isometry (preserving length).

Proof: (idea)

Z∞

−∞

|f(x)|²dx= Z∞

−∞

f(x)f(x)dx= ¹ 2π

Z∞

−∞

fˆ(ξ)e^ixξdξf(x)dx

But fˆ(ξ) = Z∞

−∞

f(x)e^−ixξdx

fˆ(ξ) = Z∞

−∞

f(x)e^−ixξdx= Z∞

−∞

f(x)e^−ixξdx= Z∞

−∞

f(x)e^ixξdx

Hence Z∞

−∞

|f(x)|²dx= ¹ 2π

Z∞

−∞

fˆ(ξ)f^ˆ(ξ)dξ= ¹ 2π

Z∞

−∞

|f^ˆ(ξ)|²dξ

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2.5 An application: Heisenberg’s Uncertainty Principle 2 FOURIER TRANSFORMS

2.5 An application: Heisenberg’s Uncertainty Principle

Theorem 2.5.1 Let f ∈ S(R)be such that R∞

−∞|f(x)|²dx=1. Then R∞

−∞x²|f(x)|²dx

! _∞ R

−∞ξ²|f(ξ)|²dξ

!

≥ ^π₂

Proof:

1= Z∞

−∞

1· |f(x)|²dx= Z∞

−∞

dx

dx|f(x)|²dx

=− Z∞

−∞

x

|f(x)|²⁰ dx integration by parts

=− Z∞

−∞

x

f(x)f(x)⁰ dx=− Z∞

−∞

x

f⁰(x)f(x) +f(x)f⁰(x)dx

Hence 1=

Z∞

−∞

x

f⁰(x)f(x) +f(x)f⁰(x)dx

≤ Z∞

−∞

|x||f⁰(x)||f(x)|+|f(x)||f⁰(x)| dx=2 Z∞

−∞

|x||f(x)||f⁰(x)|dx

Cauchy Schwarz

≤2



 Z∞

−∞

|x|²|f(x)|²dx





1/2

 Z∞

−∞

|f⁰(x)|²dx





1/2

(∗)

=2



 Z∞

−∞

|x|²|f(x)|²dx





1/2

 1 2π

Z∞

−∞

|bf⁰(ξ)|²dξ





1/2

by Theorem 2.4.1

=2



 Z∞

−∞

|x|²|f(x)|²dx





1/2

 1 2π

Z∞

−∞

|iξfˆ(ξ)|²dξ





1/2

by Theorem 2.1.12

Squaring: π 2 ≤



 Z∞

−∞

|x|²|f(x)|²dx







 Z∞

−∞

|ξ|²|f^ˆ(ξ)|²dξ





Remark In quantum mechanics, the position of a particle moving on thex–axis is known only probabilistically, in terms of awave functionψ:R→_C,|ψ(x)|²dxis the probability of finding the particle in[x,x+ dx]. Thus|ψ(x)|²is the probability density function and soR_∞

−∞|ψ(x)|²dx=1. Its width is characterized by∆xwhere (_∆x)²=

Z∞

−∞

x²|ψ(x)|²dx

Similarly its velocity (strictly, momentum) has a probability distribution whose width is∆p, where (∆p)²=

Z∞

−∞

| −i¯h∂ψ

∂x|²dx where ¯h= _2π^h wherehisPlanck’s constant. Hence(∗)

=⇒ 1<4(_∆x)² ∆p

¯ h

2

, i.e. ∆x,∆x≥ ^¯^h 2

Hence, if the uncertainty in position is small, the uncertainty in velocity is large and conversely.

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3 THE HEAT EQUATION AND LAPLACES EQUATION

Example 2.5.2 Let f(x) =^pK_δ(x) =^q^√¹

2πδe^−x²^/2δ= _(2πδ)¹_1/4e^−x²^/4δ(Def. 2.2.1). Then Z∞

−∞

|f(x)|²dx= Z∞

−∞

K_δ(x)x=1 Prop 2.2.3

Also Z∞

−∞

x²|f(x)|²dx= Z∞

−∞

x²

√2πδe^−x²^/2δdx= √¹ 2πδ

Z∞

−∞

d

dx(e^−x²^/2δ)(−δ)x dx

= r

δ 2π







he^−x²^/2δxi∞

−∞− Z∞

−∞

e^−x²^/2δdx







=???

Next, f(x) = ¹ (2πδ)^1/4

√

4πδK2δ(x) = (8πδ)^1/4K2δ(x) and so fˆ(ξ) = (8πδ)^1/4e^−δξ² Lemma 2.2.2 By a similar calculationR_∞

−∞|ξ|²|f^ˆ(ξ)|²dξ= _2δ^π. Hence Heisenbergs inequality becomes equality.

3 The Heat Equation and Laplaces Equation

ut=uxx uxx+uyy=0

Interpretation LetT(x,t)be the temperature at timet, at position x along a rod which conducts heat. Then for common substancesk^∂_∂x²^T₂ = ^∂T_∂t. By scalingt= ^t_k^∗,^∂_∂x²^T₂ = _∂t^∂T∗ (heat equation).

NextT(x,y,t)be the temperature at timetat position(x,y)on a flat object which conducts heat. Then

∂T

∂t =k ∂²T

∂x² +^∂

2T

∂y²

If the temperature settles into steady, ^∂T_∂t =0 and so^∂_∂x²^T2 +^∂_∂y²^T₂ =0 (Laplaces equation).

For the heat equation, the usual initial condition isu(x, 0) = f(x)∈ S(_R)_.

3.1 Existence and uniqueness of solutions of the heat equation

Proposition 3.1.1 (Existence forut =uxx)

Let f ∈ S(_R). Define u(x,t) = (f ∗Ht)(x)where Ht(x) = √¹

4πte^−x²^/4tfor t>0theHeat Kernel, i.e.

u(x,t) = √¹ 4πt

Z∞

−∞

f(y)e^−(x−y)²^/4tdy Then

(i) ut=uxxif t>0 (ii) limt→0⁺u(x,t) = f(x)

Proof: (i) By Leibniz’s rule (Prop 1.9) ut(x,t) = =

Z∞

−∞

f(y)^∂

∂t √1

4πte^−(x−y)²^/4t

dy

& similarly uxx(t) = Z∞

−∞

f(y) ^∂

2

∂x² 1

√4πte^−(x−y)²^/4t

dy

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3.1 Existence and uniqueness for the heat equation 3 THE HEAT EQUATION AND LAPLACES EQUATION

Now, sincex−yis independent oft

∂

∂t 1

√te^−(x−y)²^/4t

= ^1/2

t^1/2e^−(x−y)²^/4t+(x−y)²

4t^5/2 e^−(x−y)²^/4t

∂

∂x 1

√te^−(x−y)²^/4t

= (x−y)

2t^3/2 e^−(x−y)²^/4t =⇒ ^∂

2

∂x² 1

√te^−(x−y)²^/4t

= ^∂

∂t 1

√te^−(x−y)²^/4t

Hence result.

(ii) SinceHt(x) =K2t(x), (ii) is an application of Theorem 2.2.6.

Remark The expressionu(x,t) = √¹ 4πt

Z∞

−∞

f(y)e^−(x−y)²^/4tdyis an (integral) superposition of solutions of the form

√1

4πte^−(x−y)²^/4t.

Remark 3.1.2 For fixedt > 0,uis a convolution of 2 functions inS(R)and so is in S(R)(see problem 17). Also u(x, 0)∈ S(R).

Proposition 3.1.3 (Uniqueness for the heat equation) Let u(x,t), v(x,t)have the following properties:

(i) both are defined for x∈R, t≥0and are continuous

(ii) both solve the heat equation in the region x∈R, t>0with u(x, 0) =v(x, 0). (iii) both have the property in Remark 3.1.2 (inS(R))

Then u=v.

Proof: Letw=u−v. Thenwhas properties (i)–(iii) withw(x, 0) =0. DefineE[w](t) =R_∞

−∞|w(x,t)|²dx. Then E[w]⁰(t) =

Z _∞

−∞

∂

∂t|w(x,t)|²dx Z _∞

−∞(wtw¯+ww¯t)x= Z _∞

−∞(wxxw¯+ww¯xx)dx

= Z _∞

−∞((ww¯)_xx−2wxw¯x)dx=

(ww¯)_x ∞

−∞

| {z }

=0 asw∈ S(R)

−2 Z _∞

−∞|wx|²dx≤0

ThereforeE[w]is a decreasing function oft. Hence∀t≥0 0≤E[w](t)≤E[w](0) =

Z _∞

−∞|w(x, 0)|²dx=0 HenceE[w](t) =R_∞

−∞|w(x,t)|²dx=and sow(x,t) =0. Henceu(x,t) =v(x,t).

Remark 3.1.5 Though we assumed f ∈ S(R)the formulau(x,t) = √¹ 4πt

Z _∞

−∞ f(y)e^−(x−y)²^/4tdyis also correct for many f ∈ S/ (_R).

Example 3.1.6 Take f =e^3x∈ S/ (_R). Thenu(x,t) = √¹ 4πt

Z _∞

−∞e^3ye^−(x−y)²^/4tdywhere the exponent is 3y1

4t(x²−2xy+y²) =−¹

4t(y²−2y(x+6t) +x²) =−¹

4t((y−(x+6t))²−x²−12xt−36t²+x²)

=−¹

4t(y−x−6t)²+3x+9t u(x,t) = √¹

4πt Z _∞

−∞e⁻^4t¹^(y−x−6t)²^+3x+9tdy= √¹

4πte^3x+9t Z _∞

−∞e⁻^4t¹^(y−x−6t)²dy

= √¹

4πte^3x+9t Z _∞

−∞e^−y⁰²^/4tdy⁰ (y⁰ =y−x−6t)

=e^3x+9t by Prop 2.2.3, since the integrand isK2t(y⁰)

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3.2 Laplace’s equation 3 THE HEAT EQUATION AND LAPLACES EQUATION

3.2 Laplace’s equation u

xx

+ u

yy

= 0

A setΩ ⊆_R²is said to be open iff for every(x,y)∈ Ω, there is an r>0, s.t. the disc D((x,y),r), i.e. the disc of radius r, centered at(x,y), is contained inΩ.

LetΩ⊆_R²be open. A function u:Ω→_R(orC) is said to beharmoniciff it is twice differentiable and s.t. uxx+uyy=0.

Example 3.2.3 If f(z) is holomorphic with real and imaginary parts u,v, then u,v satisfy the Cauchy–Riemann equationux=vy,uy=−vxand souxx+uyy=vyx−vxy=0. Similarlyvxx+vyy=0, i.e.uandvare harmonic.

f(z) =

∑

∞ 0

anzⁿ=

∑

∞ 0

anrⁿe^inθ =

∑

∞ 0

anrⁿ(cosnθ+isinnθ) zⁿ =rⁿcosnθ+irⁿsinnθ

3.3 Laplace equation in the upper half–plane

Ω={(x,y):y,x∈_R,y>0}=_R²₊and we have∇²u=0,u= f(x).

Consider the following boundary value problem: given a functionf, find a functionu(x,y)s.t.uxx+uyy=0 inR²₊ andu(x, 0) = f(x).

ThePoisson kernelis defined to be Py(x) = _π¹_x₂_+y^y ₂, x∈R, y>0

Remark 3.3.2 For fixedy >0,Py(x)is infinitely differentiable, but it doesn’t decay rapidly enough as|x| →∞and is not inS(R).

The Poisson kernel has the properties (i) ∀y>0R_∞

−∞Py(x)dx=1 (ii) F(P_y)(ξ) =P^ˆy(ξ) =e^−y|^ξ^| (iii) for fixedδ>0,Ry

x>δPy(x)dx→0as y→0⁺. Proof: (i) 1

π Z∞

−∞

y

y²+x²dx= ¹ π

tan⁻¹ x

y _∞

−∞

= ¹ π

π

2 −−^π 2

=1

(ii) tutorial Ex 2.1 (iii) By symmetry,

Z

|x|>δ

Py(x)dx= ² π

Z∞

δ

y

x²+y²dx= ² π

tanx

y ∞

δ

= ² π

π

2 −tan⁻¹δ y

→ ² π

π 2 −^π

2

=0

asy→0⁺.

Theorem 3.3.6 Let f ∈ S(_R)_{. Define}u(x,y) = (f∗Py)(x)_,x ∈R,y>_{0. Then}

(i) u(x,y)is twice differentiable in the upper half planey>0 anduxx+uyy=0 (i.e.uis harmonic) (ii) lim

y→0⁺u(x,y) = f(x)for allx∈_R

(iii) if we extendu(x,y)tox∈_R,y≥0 by defining it to be f(x)wheny=0, the extendeduis continuous.

Proof: Similar to the heat equation (Prop. 3.1.1).

Again, this formula to applies to many f ∈ S/ (_R).

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3.4 The maximum principle 3 THE HEAT EQUATION AND LAPLACES EQUATION

Example 3.3.7 Letu(x, 0) =H(x) =

(1 ifx>0

0 ifx<0 (Heaviside step function). Then

u(x,y) = (H∗Py)(x) = Z∞

−∞

H(z)Py(x−y)dz= ^y π

Z∞

−∞

H(z) ^dz

y²+ (x−2)² ^{Def 3.3.1}

= ^y π

Z∞ 0

dz

y²+ (2−x)² = ¹ π

tan⁻¹2−x y

_∞

0

= ¹ π

π

2 −tan⁻¹

−^x y

= ¹ π

π

2 +tan⁻¹x y

z=x+iy

= ¹ 2+ ¹

π π

2 −argz

=1− ¹

πargz=Im

i− ¹ πlnz

Ex 3.2.3

3.4 The maximum principle

In this section,Ω⊆R²is open and bounded. We define theclosureofΩby ¯Ω= Ω∪∂Ωwhere∂Ωis the set of all boundary points ofΩ. The set ¯Ωis closed and bounded. (Closed means its complement is open.)

Such sets in the plane are calledcompact. It can be shown that a continuous function on a compact set always has a maximum and a minimum.

Lemma 3.4.1

Let u: ¯Ω→_Rbe continuous onΩ¯ and such that uxx+uyy≥0inΩ. Then the max. of u onΩ¯ is attained on∂Ω.

Remark such a function is calledsubharmonic.

Proof: Suppose that the max is attained at an interior point(x0,y0)∈Ω. Thenuxx(x0,y0)≤0 anduyy(x0,y0)≤ 0. Henceuxx(x0,y0) +uyy(x0,y0)≤0. This is a contradiction.

Theorem 3.4.2 (Max Principle) Letu: ¯Ω→Rbe continuous and harmonic inΩ. Then the max ofuon ¯Ωis attained on the boundary∂Ω.

Remark It may also be attained at an interior point, e.g.u=constant.

Proof: Letε > 0. Definev(x,y) = u(x,y) +ε(x²+y²). Thenvx = ux+2εx,vxx = uxx+2ε, similarlyvyy = uyy+2ε. Hencevxx+vyy=4ε>0 By Lemma 3.4.1 the max ofvis attended on∂Ω. Hence for any(x0,y0)∈_Ω, v(x0,y0)≤max_∂Ωv(x,y)i.e.

u(x0,y0) +ε(x²₀+y²₀)≤max

(x,y)∈∂Ω

n

u(x,y) +ε(x²+y²)^o≤max

∂Ω u(x,y) +εM

where M > 0 is a bound for x²+y²on∂Ω. Henceu(x0,y0) ≤ max_∂Ωu(x,y) +ε(M−(x²₀+y²₀)). Now let

ε→0⁺:u(x0,y0≤max_∂Ωu(x,y).

Corollary 3.4.3

Under the same assumptions the minimum of u is attained on∂Ω.

Proof: Apply Theorem 3.4.2 to−u.

3.5 Uniqueness for Laplace’s equation in the upper half plane

Given f(x)∈ S(_R), letu(x,y) = (f∗Py)(x). Then by Theorem 3.3.6u(x,y)satisfies (i) uxx+uyy=0 fory>0

(ii) u(x,y)→ f(x)asy→0⁺

Remark (iii) It is also true thatu(x,y)→0 asx²+y²→_∞,y>0 (Problem 33) Theorem 3.5.1 u(x,y)satisfying (i)-(iii) is unique.

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3.6 Mean Value Theorem 3 THE HEAT EQUATION AND LAPLACES EQUATION

Proof: Given f, suppose∃two suchu, i.e.u1(x,y)andu2(x,y). Letw=u1−u2. Thenwxx+wyy=0 inR²₊ = {(x,y)∈ _R²,y >0}. w(x, 0) = u1(x, 0)−u2(x, 0) = f(x)− f(x) =0. wis continuous in{(x,y)∈_R²,y ≥0} and lim_x2+y²→∞w(x,y) =0−0=0.

We want to showw=0. Suppose on the contrary that∃(x0,y0)∈R²₊s.t.w(x0,y0)6=0, sayw(x0,y0)>0. (For w(x0,y0)<0, the proof is similar). PickRlarge enough so that

(i) x²₀+y²₀<R²

(ii) ∀(x,y)on the semicirclex²+y²=R²,y≥0,w(x,y)≤ ¹₂w(x0,y0). Apply maximum principle towon the setΩR={(x,y):x²+y²<R,y>0}

max

x²+y²=R²,y>0

≤ ¹₂w(x₀,y₀) max

AB w(x, 0) =0 whereABis the straight part of the semicircle max∂ΩR

≤ ^w₍x0,y0) =⇒ w(x0,y0)≤ ¹₂w(x0,y0)<w(x0,y0)

This clearly is a contradiction.

3.6 Mean Value Theorem

Theorem 3.6.1 SupposeΩ ⊆R²is open and bounded andu:Ω →Ris harmonic. Let(x0,y0)∈Ωand letR>0 be such thatD((x0,y0),R)⊆Ω. Then for anyr∈[0,R]we have

u(x₀,y₀) = ¹ 2π

Z _2π

0 u(x₀+rcosθ,y₀+rsinθ)dθ Proof: Define

f(r) = ¹ 2π

Z _2π

0 u(x₀+rcosθ,y₀+rsinθ)

| {z }

U(r,θ)

dθ

r,θare polar coords with origin(x0,y0). Thus sinceuis harmonic, ^∂_∂r²^U2 +¹_r^∂U_∂r +_r¹₂^∂²^U

∂θ² =0 d f

dr = ¹ 2π

d dr

Z _2π

0 U(r,θ)dθ= ¹ 2π

Z _2π 0

∂U

∂r(r,θ)dθ (1) d²f

dr² = ¹ 2π

Z _2π 0

∂²U

∂r²(r,θ)dθ Hence rd f

dr +r²d²f dr² = ¹

2π Z _2π

0

r∂U

∂r +r²∂²U

∂r²

dθ

=− ¹ 2π

Z _2π 0

∂²U

∂θ² dθ=− ¹ 2π

∂U

∂θ 2π

0

=0 (2)

SinceU(r,θ) = U(r,θ+2π) =⇒ ^∂U_∂θ(r,θ) = ^∂U_∂θ(r,θ+2π). Hence ^{d f}_dr +r^d_dr²^f = 0 =⇒ _dr^d(r^{d f}_dr) = 0 =⇒ r^{d f}_dr =const. (1) =⇒ ^{d f}_dr is bounded inD((x0,y0),R). Taking lim asr→0 we find the constraint is 0 and so

d f

dr =0 =⇒ f =constc.

Evaluating atr=0, f(r) = _2π¹ R2π

0 u(x0,y0)dθ=u(x0,y0).

Example 3.6.2 Suppose u(x,y) is cts in x²+y² ≤ 1, harmonic in x²+y² > 1 andu(x,y) = 2x²+3y² on the boundaryx²+y²=1. Its value at the origin isu(0, 0) =. . .= ⁵₂.

Remark u(_cos_{θ, sin}θ) = _{2 cos}²θ+_{3 sin}²θ = ₂+_sin²_{θ. max}_x2+y²=1u(x,y) = _{3 and min}u(x,y) = _{2. Hence} 2≤u(x,y)≤3 inx²+y²≤1 by Maximum principle.

PAA by Douglas Heggie 2011 Notes of Paul Boeck