[St]: Consider a Function T(X) As Repre- Senting the Value of A

I. Introduction & Motivation

I.1. Physical motivation. From [St]: consider a function T (x) as repre- senting the value of a physical variable at a particular point x in space. Is this a realistic thing to do? What can you measure? Suppose T (x) represents temperature at a point x in a room. You can measure the temperature with a thermometer, placing the bulb at the point x. Unlike the point, the bulb has nonzero size, so what you actually measure is more an average temperature over a small region of space. So really, the thermometer measures T (x)ϕ(x) dx, Z where ϕ(x) depends on the nature of the thermometer and where you place it. ϕ(x) will tend to be “concentrated” near the location of the thermometer bulb and nearly 0 once you are suﬃciently far away from the bulb. To say this is an average requires ϕ(x) 0 for all x ≥ ϕ(x) dx = 1 Z So it may be more meaningful to discuss things like T (x)ϕ(x) dx than things like the value of T at a particular point x. R

I.2. Mathematical motivation. How to “diﬀerentiate” nondiﬀerentiable functions? IBP: b b T 0(x)ϕ(x) dx = T (x)ϕ0(x) dx, − Za Za (*) provided ϕ is nice, and the boundary terms vanish.

Heaviside’s operational calculus (c. 1900) → Sobolev (1930s): realized generalized functions were continuous linear →functionals over some space of test functions Schwartz (1950s): developed & articulated this view in the language of →modern fnl analysis with Th´eorie des Distributions

1 Consider the DE u0 = H, where H(x) is the Heaviside function: 1, x > 0 H(x) = 0, x 0 ( ≤

H(x)

Figure 1. The Heaviside function H(x). We would like to say that the soln is x, x > 0 x+ = 0, x 0, ( ≤ but this function is not a classical soln: it is not diﬀerential at 0.

Figure 2. The piecewise continuous function x+.

Further, what is the derivative of H? Classical theory has no satisfac- tory treatment of derivatives of pointwise constant functions. (e.g. Cantor- Lebesgue). We will use IBP to diﬀerentiate the nondiﬀerentiable.

2 II. Basics of the theory

Let us not require T to have a speciﬁc value at x. T is no longer a function— call it a generalized function. Think of T as acting on a “weighted open set” instead of on a point x. “T (x) T (ϕ)” Formalize: T 7→acts on “test functions” ϕ. Denote: T, ϕ = T (x)ϕ(x) dx. h i Z ϕ(x)

a b [ ]

Figure 3. A test function ϕ. Reqs for “nice” test functions ϕ :

(1) ϕ C∞, and (2) boundary∈ terms must vanish (ϕ( ) = ϕ( ) = 0). −∞ ∞ (a) ϕ Cc, i.e., ϕ has compact support, or ∈ x (b) ϕ(x) | |→∞ 0 quickly(with derivs) −−−−−→ Choosing (2a) leads to the theory of distributions a` la Laurent Schwartz. Choosing (2b) leads to the theory of tempered distributions.

Deﬁnition 1. The space of test functions is D (Ω) := Cc∞(Ω)

Note: (2a) allows more distributions than (2b). (larger class of test functions fewer distributions. ⇒ Reqs for a distribution : (1) T, ϕ must exist for ϕ D. (2) Linearith i y: T, a ϕ + a ∈ϕ = a T, ϕ + a T, ϕ . h 1 1 2 2i 1h 1i 2h 2i Taking a cue from Riesz1:

1Actually, D0(Ω) is a larger class than the class of Borel measures on Ω.

3 Deﬁnition 2. The space of distributions is the (topological) dual space

D0(Ω) = continuous linear functionals on D(Ω) . { } Examples

1. Any f L1 (Ω). ∈ loc T , ϕ = f(x)ϕ(x) dx h f i Z “The distribution f” really means Tf .

2. Any regular Borel measure µ on Ω T , ϕ = ϕ(x) dµ h µ i Z T is regular iff T = T for f L1 (Ω). Otherwise, T is singular. f ∈ loc The most famous singular distribution, Dirac-δ: δ, ϕ = ϕ(0) = δ(x)ϕ(x) dx = ϕ(x) dµ h i so Z Z , x = 0 “δ(x) = ∞ ” 0, x = 0 ( 6 for 1, 0 E µ(E) = ∈ . 0, 0 / E ( ∈ Some distributions do not even come from a measure. Example. δ0, ϕ := ϕ0(0). h i − This defines a cont lin fnl on D (in fact, on C m, m 1), but not on C0.2 ≥ Note: require smooth test functions allow rough distributions; differen- tiability of a distribution relies on differen⇒ tiability of test functions. To define δ, ϕ , ϕ must be C0. h i 1 To define δ0, ϕ , ϕ must be C . h i

2Riesz gives a bijection btwn contin lin fnls on C0 and complex locally ﬁnite regular Borel measures.

4 III. Differentiation of Distributions. Key point: how do we understand T ? Only by T, ϕ . h i So what does T 0 mean? Must understand T 0, ϕ . h i Working in Ω = R: ∞ T 0, ϕ = T 0(x)ϕ(x) dx h i Z−∞ ∞ = [T (x)ϕ(x)]∞ T (x)ϕ0(x) dx −∞ − Z−∞ = T, ϕ0 −h i

Definition 3. For any T D0(Ω), ∈ D T, ϕ := T, D ϕ , ϕ D. h k i −h k i ∈ More generally (induct): α α α D T, ϕ := ( 1)| | T, D ϕ , ϕ D. h i − h i ∈ Note: this formula shows every distribution is (infinitely) differentiable. Note: Dα+βT = Dα(DβT ).

Recall that x+ is not diﬀerentiable in the classical sense. x, x > 0 x+ = 0, x 0, ( ≤ But x+ can be diﬀerentiated as a distribution:

x0 , ϕ = x , ϕ0 h + i −h + i ∞ = xϕ0(x) dx − Z0 ∞ = [ xϕ(x)]∞ + ϕ(x) dx − 0 Z0 ∞ = 0 + H(x)ϕ(x) dx Z−∞ = H, ϕ h i So x+0 = H, as hoped. Similarly,

5 x00 , ϕ = H0, ϕ h + i h i = H, ϕ0 −h i ∞ = ϕ0(x) dx − Z0 = ϕ(0)

So x+00 = H0 = δ, as hoped.

(Motivation for ϕ C∞.) Note: it doesn’t matter how rough T is — simply apply IBP until all deriv∈ atives are on ϕ.

6 IV. A Toolbox for Distribution Theory

Deﬁnition 4. For ϕ ∞ D, ϕ 0 iﬀ { k}k=1 ⊆ k → (i) K Ω compact s.t. spt ϕ K, k, and ∃ ⊆ k ⊆ ∀ (ii) Dαϕ 0 uniformly on K, α. k → ∀

Deﬁnition 5. For T ∞ D0, { k}k=1 ⊆ T 0 T , ϕ 0 C, ϕ D. k → ⇐⇒ h k i → ∈ ∀ ∈ This is weak or distributional (or “pointwise”) convergence.

Theorem 6. Diﬀerentiation is linear & continuous. Proof. a)

(aT + bT )0, ϕ = aT + bT , ϕ0 h 1 2 i −h 1 2 i = a T , ϕ0 b T , ϕ0 − h 1 i − h 2 i = a T 0, ϕ + b T 0, ϕ h 1 i h 2 i b) Let T D0 converge to T in D0. { n} ⊆ Tn0 , ϕ = Tn, ϕ0 h n i − h i →∞ T, ϕ0 −−−−→ − h i = T 0, ϕ h i

Theorem 7. D0 is complete.

Theorem 8. D is dense in D0. (Given T D0, ϕ D such that ϕ T as distributions.) ∈ ∃{ k} ⊆ k →

Proposition 9. If f has classical derivative f 0 and f 0 is integrable on Ω, then

0 Tf0 = Tf .

7 Proof. In the case Ω = I = (a, b), we have x f(x) = f 0(t) dt + f(c) x, c I. ∈ Zc For ϕ D, (fϕ)0 = f 0ϕ + fϕ0. ∈ Also, I (fϕ)0 dx = 0 because fϕ vanishes outside a closed subinterval of I. So R f 0ϕ + fϕ0 = 0. I I Hence Z Z

T 0 , ϕ = T , ϕ0 h f i −h f i = fϕ0 − ZI = f 0ϕ ZI = T 0 , ϕ h f i

D0 1 ae 1 Theorem 10. fk Lloc, fk f, and fk g Lloc. Then fk f. D0 { } ⊆ −−→ | | ≤ ∈ −−−→ (i.e., T T ) fk −−−→ f

D0 Deﬁnition 11. A sequence of functions fk such that fk δ is a delta- convergent sequence , or δ-sequence. { } −−−→

Theorem 12. Let f 0 be integrable on Rn with f = 1. Deﬁne ≥ n x n x1 xn fλ(x) = λ− f λ = λ− f λ , . . R. , λ D0 for λ > 0. Then f δ as λ 0. λ −−−→ →

Example. dx 1 = π = deﬁne f(x) = . R 1 + x2 ⇒ π(1 + x2) Z Then obtain the delta-sequence 1 1 λ f = 2 = 2 2 . λ λ · π(1+(x/λ) ) π(x +λ )

8 Example.

2 x2 1/2 e−x e− = π = deﬁne f(x) = √π . R ⇒ Z

2 −x /λ Then obtain the δ-sequence f = e . Further, λ √πλ

n x 2 x2 e−| | dx = e− k dxk Rn Rn Z Z k=1 n Y x2 = e− k dxk Rn Yk=1 Z = πn/2 gives the n-dimensional δ-sequence

2 e x /λ f (x) = −| | . (1) λ (πλ)n/2

IV.1. Extension from test functions to distributions. There are two main ways to extend operations on test functions to operations on distributions: via approximation or via adjoint identity.

Approximation: For T D0, and an operator S defined on D, find arbitrary ϕ T and define ST =∈lim Sϕ . n → n

Example. (Translation) Define S = τh on D by τh(ϕ) = τh, ϕ = ϕ(x h). h i − D0 To define τhT : find an arb sequence ϕn D, with ϕn T . Then T, ϕ = T (x)ϕ(x) dx = lim {ϕ (}x⊆)ϕ(x) dx, so −−−→ h i n R R

9 τ T, ϕ = lim τ ϕ (x)ϕ(x) dx h h i h n Z = lim ϕ (x h)ϕ(x) dx n − Z = lim ϕn(x)ϕ(x + h) dx Z = lim ϕn(x)τ hϕ(x) dx − Z = T, τ hϕ h − i

Example. (Differentiation) d D R Define S = dx on ( ). This is the infinitesimal version of translation. h 0 Let I denote the identity, use τ → I. h −−−−→ d 1 Then = limh 0 (τh I), so dx → h − d 1 T, ϕ = lim ( τhT, ϕ T, ϕ ) hdx i h 0 h h i − h i → 1 = lim ( T, τ hϕ T, ϕ ) h 0 h h − i − h i → 1 = lim T, (τ hϕ ϕ) h 0h h − − i → 1 = T, lim (τ h I) ϕ h h 0 h − − i → = T, d ϕ h −dx i

Adjoint Identities: Let T D0 and let R be an operator such that Rϕ D. Deﬁne S as the operator whic∈ h satisﬁes ∈ ST, ϕ = T, Rϕ , i.e. h i h i ST (x)ϕ(x) dx = T (x)Rϕ(x) dx. There is no general Zformula for S in termsZ of T .

Deﬁnition 13. S is the adjoint of R. If S = R, we say R is self-adjoint. Example: The Laplacian ∆. ∆T, ϕ = T, ∆ϕ . h i h i

10 Example. (Translation) Deﬁne S = τh by Sψ, ϕ := ψ, τ hϕ . h i h − i

τhψ(x)ϕ(x) dx = ψ(x)τ hϕ(x) dx − Z Z

Example. (Diﬀerentiation) Deﬁne S = d by Sψ, ϕ := ψ, d ϕ . dx h i −h dx i d ψ(x) ϕ(x) dx = ψ(x) d ϕ(x) dx dx − dx Z Z

Example. (Multiplication) Define S by Sψ, ϕ := ψ, f ϕ . h i h · i (f(x)ψ(x)) ϕ(x) dx = ψ(x) (f(x)ϕ(x)) dx Z Z Note: this only works when f ϕ D! So require f C ∞. Example of trouble: let f(x·) =∈sgn(x) so f is discon∈ tinuous at 0. Then f δ cannot be defined: · f δ, ϕ = δ, f ϕ = f(0)ϕ(0) h · i h · i but f(0) is undefined. Thus, the product of two arbitrary distributions is undefined.

IV.2. Multiplication. There is no way of deﬁning the product of two distributions as a natural extension of the product of two functions. a..

Deﬁnition 14. For T D0, f C∞, can deﬁne their product fT as the linear functional ∈ ∈ fT, ϕ = T, fϕ , ϕ D. h i h i ∀ ∈ The RHS makes sense because the product fϕ is in D, and ϕk 0 implies fϕ 0. Thus, → k → fT, ϕ = T, fϕ 0, h ki h ki → and fT is continuous.

11 When T is regular, fT , ϕ = T , fϕ h g i h g i = gfϕ Z = fg, ϕ h i 1 since fg is also in Lloc. Thus fTg = Tfg in this case.

Example. (Leibniz Rule) D (fT ), ϕ = fT, D ϕ h k i −h k i = T, fD ϕ −h k i = T, D (fϕ) (D f)ϕ −h k − k i = T, D (fϕ) + T, (D f)ϕ −h k i h k i = fD T, ϕ + (D f)T, ϕ h k i h k i Thus, Dk(fT ) = fDkT + (Dkf)T.

Proposition 15.

α α β α β D (fT ) = (D f)(D − T ) β β α X≤

12 V. Review: Items from the previous talk

A distribution is a continuous linear functional on Cc∞(Ω) and often written T, ϕ = T (ϕ) := T (x)ϕ(x) dx. h i Z Distributions are differentiated by α α α T 0, ϕ = T, ϕ0 , ∂ T, ϕ = ( 1)| | T, ∂ ϕ . h i −h i h i − h i The Dirac δ is defined by δ, ϕ = ϕ(0) h i and has derivative δ0, ϕ = ϕ0(0). h i − For x = max 0, x and the Heaviside function H(x), we have + { } x+00 = H0 = δ. The translation operator τ , ϕ = τ (ϕ) = ϕ(x h) h h i h − is defined on distributions by

τhT, ϕ = T, τ hϕ . h i h − i The multiplication-by-a-smooth-function operator (f C ∞) ∈ f T, ϕ = T, f ϕ . h · i h · i Adjoint Identities: Let T D0 and let R be an operator such that Rϕ D. Deﬁne S as the operator whic∈ h satisﬁes ∈ ST, ϕ = T, Rϕ . h i h i

13 VI. Convolutions

Convolution is a smoothing process: convolve anything with a C m function and the result is Cm, even if m = . ∞ Strategy: T may be nasty, but T ϕ is lovely (C ∞), so work with it instead. ∗ Even better, ﬁnd ϕk such that T ϕk T and use it to extend the usual rules of calculus & DEs.{ } ∗ → Need ϕ(x y) to discuss convolutions, so consider functions of 2 variables for a momen−t: ϕ(x, y) D(Ω Ω ). ∈ 1 × 2 Let T be a distribution on Ω (T D(Ω )). i i i ∈ i For ﬁxed y Ω2, the function ϕ( , y) is in D(Ω1), and T1 maps ϕ( , y) to the number ∈ · · T , ϕ( , y) = T (ϕ)(y). h 1 · i 1 Theorem 16. For ϕ, T as above, T (ϕ) D(Ω ) and ∂βT (ϕ) = T ∂βϕ . i 1 ∈ 2 y 1 1 y D So Ti : ϕ Ti(ϕ) preserves the smoothness of ϕ . 7→ ∈ Corollary 17. If ϕ C∞(Ω Ω ) has compact support as a function of x ∈ 1 × 2 and y separately, then T (ϕ) C∞(Ω ) for every T D0(Ω ). 1 ∈ 2 1 ∈ 1 y

(0,1)

spt ϕ(x+y)

(-1,0) (1,0) x

(0,-1) R2

Figure 4. For ϕ D(R) with support spt(ϕ) = [ 1, 1], the function ϕ(x + y) is deﬁned on R2 and does not ha∈ve compact support. −

14 1 Deﬁnition 18. Convolution of a Cc∞ function ϕ with an Lloc function f is (ϕ f)(x) := ϕ(x y)f(y) dy = ϕ(y)f(x y) dy. ∗ − − Z Z Now extend convolution to distributions: (T T )(ϕ) = T T , ϕ := T (T (ϕ(x + y)), ϕ D(Rn). 1 ∗ 2 h 1 ∗ 2 i 1 2 ∈ Suppose T are regular, and deﬁned by f L1 (Rn), where at least one of i i ∈ loc the fi has compact support. Then (T T )(ϕ) = T T , ϕ 1 ∗ 2 h 1 ∗ 2 i = T , T , ϕ(x + y) 1 h 2 i = f1(x) f2(y)ϕ( x + y) dy dx Z Z = f (x y)f (y)ϕ(x) dy dx x x y 1 − 2 7→ − Z Z = f f , ϕ h 1 ∗ 2 i with f1 f2 as above. Note: f∗ f = f f = T T = T T . 1 ∗ 2 2 ∗ 1 ⇒ 1 ∗ 2 2 ∗ 1 VI.1. Properties of the convolution.

n n Let T D0(R ) and ϕ D(R ). Then ∈ ∈ δ T, ϕ = T δ, ϕ h ∗ i h ∗ i = T, δ, ϕ(x + y) h i = T, ϕ(y) h i shows δ T = T δ = T . Additionally∗ , ∗ (Dαδ) T, ϕ = T, (Dαδ), ϕ(x + y) h ∗ i h i α α = T, ( 1)| |D ϕ(y) h − i = DαT, ϕ h i The Magic Property: (Dαδ) T = DαT = Dα(δ T ) = δ DαT. ∗ ∗ ∗ Further properties of :3 ∗ 3 Assuming one of the Ti has compact support.

15 1. spt(T T ) spt T + spt T . 1 ∗ 2 ⊆ 1 2 2. T (T T ) = (T T ) T = T T T . 1 ∗ 2 ∗ 3 1 ∗ 2 ∗ 3 1 ∗ 2 ∗ 3 3. Dα(T T ) = (Dαδ) T T 1 ∗ 2 ∗ 1 ∗ 2 = (DαT ) T = T (DαT ). 1 ∗ 2 1 ∗ 2 4. τ (T T ) = δ T T h 1 ∗ 2 h ∗ 1 ∗ 2 = (τ T ) T = T (τ T ). h 1 ∗ 2 1 ∗ h 2

To see the last, note that τhδ = δh by

τhδ, ϕ = δ, τ hϕ h i h − i = δ, ϕ(x + h) h i = ϕ(h) = δ , ϕ , h h i so δ T = τ T by h ∗ h δ T, ϕ = T, δ , ϕ(x + y) h h ∗ i h h i = DT, ϕ(x + h) E h i = T, τ hϕ h − i = τ T, ϕ . h h i Conclusion: (D0, +, ) is a commutative algebra with unit δ. ∗

Example. Let S H = δ. Then ∗ δ0 = (S H)0 = S H0 = S δ = S, ∗ ∗ ∗ 1 1 shows H− = δ0. Similarly, (δ0)− = H.

Deﬁnition 19. For f on Rn, its reﬂection in 0 is f˜(x) = f( x). For distributions, we extend the defn by − T˜, ϕ = T, ϕ˜ . h i h i n Theorem 20. The convolution (T ψ)(x) = T (τ ψ˜) is in C∞(R ). ∗ x

16 Proof. For any ϕ D(R), we have ∈ T ψ, ϕ = T, ψ(y), ϕ(x + y) . h ∗ i h Thus ψ(y), ϕ(x + y) = ψ(y)ϕ(x + y) dy h i Z = ψ(ξ x)ϕ(ξ) dξ − Z = ψ(ξ x), ϕ(ξ) h − i = ψ˜(x ξ), ϕ(ξ) h − i = τ ψ˜(ξ), ϕ(ξ) h ξ i shows T ψ, ϕ = T, τ ψ˜(x), ϕ(ξ) h ∗ i h ξ = DT (τ ψ˜), ϕ(ξ) E h ξ i = T (τ ψ˜), ϕ . h x i Finally, note that (T ψ)(x) = T (τ ψ˜) = T (ψ(x y)) ∗ x − is smooth by Cor. 17.

Corollary 21. T (ϕ) = (T ϕ˜)(0). ∗

VI.2. Applications of the convolution.

The C∞ function 1 exp 1 x 2 , x < 1 α(x) = − −| | | | (0, x 1 | | ≥ has support in the closed unit ball B = B(0, 1) of Rn, so the function 1 − β(x) = α(x) α(y) dy Z

17 is another C∞ function with support in the closed unit ball B which satisﬁes β = 1. Then take the δ-sequence 1 x R β (x) = β . λ λn λ

4 Theorem 22. The C∞ function T β converges strongly to T as λ 0. ∗ λ →

Deﬁnition 23. The convolution of f (or T ) with βλ is called a regularization of f (or T ). T β = T γ ∗ 1/k ∗ k is a regularizing sequence for the distribution T D0. ∈

Proposition 24. Suppose T 0 = 0. Then T = c R. ae ∈

Proof. Let γk be a regularizing sequence for δ. Then the C ∞ function T γk satisﬁes ∗

(T γ )0 = T 0 γ = 0 ∗ k ∗ k for every k, so T γk = ck. D0 ∗ c = T γ T , but still must show c converges in C. k ∗ k −−−→ { k} Pick ϕ D with ϕ = 1. Then ck = ck, ϕ converges in C; hence its limit c = lim∈c coincides with T . h i k R

D0 pw Note: in general, f f does not imply f f (it doesn’t even k −−−→ k −−−→ imply f is a function!). fk = constant is a special case.

Proposition 25. Suppose T 00 = 0. Then T is linear a.e.

Proof. For ϕ D, T ϕ C∞, and ∈ ∗ ∈ (T ϕ)00 = T 00 ϕ = 0. ∗ ∗

4 unif hT ∗ βλ, ϕi −−−−−→ hT, ϕi on every bounded subset of D.

18 Thus T ϕ is a linear function of the form (T ϕ)(x) = ax + b. Let h(x)∗= ax + b. Then ∗

(h β)(x) = h(x y)β(y) dy ∗ − Z = [a(x y) + b]β(y) dy − Z = ax + b since yβ(y) dy = 0. Thus h β = h, so ∗ (T β) γ = (T γ ) β = T γ . R ∗ ∗ k ∗ k ∗ ∗ k As k , this gives T β = T . → ∞ ∗ ae Note: this generalizes immediately to T (n) = 0 = T = P (x) = a + a x + + a xm, ⇒ ae 0 1 · · · m for m < n.

Deﬁnition 26. For T D0, a distribution S satisfying DkS(ϕ) = T (ϕ) for all ϕ D is called a primitive∈ (antiderivative) of T . ∈

Theorem 27. Any distribution in D0(R) has a primitive which is unique up to an additive constant.

Another notion of regularization.

Deﬁnition 28. If f has a pole at x0, the Cauchy principal value of the divergent integral f(x) dx is

R pv f(x) dx = lim f(x) dx. ε 0 x x0 ε Z → Z| − |≥

1 Deﬁnition 29. Obtaining the distributional derivative of f Lloc from a divergent integral by taking the principal value is called regularizing∈ the in- 1 α 1 α tegral. If f L but D f / L , then D T is a regularization of T α . ∈ loc ∈ loc f D f

19 Example. The function 1/x, x > 0, f(x) = 0, x 0 ( ≤ is not integrable on any neighborhood of 0, hence does not deﬁne a distribution on R via the standard formula

Tf (ϕ) = f, ϕ = f(x)ϕ(x) dx. h i R 1 Z However, f R+ L and so deﬁnes a regular distribution. | ∈ loc As classical derivatives, d log x = 1 , x = 0 dx | | x 6 so what is the relation of the distributional derivative of log x to 1/x? | |

(log x )0 = pv 1/x. | | is the distributional derivative of log x , and is not a function. | |

20 VII. Solving Differential Equations with Distributions

Consider a DE Lu = f, (2) where L is some linear5 diﬀerential operator of order m.

Definition 30. A classical solution to (2) is an m-times differentiable u defined on Ω which satisfies the equation in the sense of equality between functions. If u Cm(Ω) (so Lu is continuous) then u is a strong solution . ∈

Deﬁnition 31. A weak solution to (2) is u D0(Ω) which satisﬁes the equation in the sense of distributions: ∈ Lu, ϕ = f, ϕ , ϕ D(Ω). h i h i ∀ ∈ Note: every strong soln is a weak soln, but converse is false.

Example. For Ω = R, consider

xu0 = 0.

Strong soln: u = c1. Weak soln: consider u = c2H.

u0 = c2δ

xu0, ϕ = c xδ, ϕ h i 2h i = c δ, xϕ 2h i = 0 ϕ D. ∀ ∈ So u = c2H is also a weak soln, and

u = c1 + c2H. is the (weak) general solution (but not a strong solution). What else is odd about this example?

5Restr to linear is nec because we cannot deﬁne mult in D0 as a natural extn of mult of functions.

21 Example. Solve u00 = δ on R. We found the solution x+ = xH(x). Any other solution will satisfy the homogeneous equation D2 [u xH(x)] = 0, x − so will have the form ζ(x) = xH(x) + ax + b. Boundary conditions will determine the constants, e.g., ζ(0) = 0, ζ(1) = 1 = ζ(x) = xH(x). ⇒ , We use ζ to denote the general solution of u00 = δ. → Now use this to solve more general equations. Example. For Ω = (0, 1), solve

u00 = f. (3) Consider f = 0 outside (0, 1) so f L1(R). Then ∈ (f ζ)00 = f ζ00 ∗ ∗ = f δ ∗ = f. So one solution to (3) is u(x) = (f ζ)(x) 1∗ = (x ξ)H(x ξ)f(ξ) dξ 0 − − Z x = (x ξ)f(ξ) dξ 0 x 1, − ≤ ≤ Z0 and the general solution is thus x u(x) = (x ξ)f(ξ) dξ + ax + b. − Z0 If we have initial conditions

u0(0) = a, u(0) = b, this becomes x u(x) = (x ξ)f(ξ) dξ + u0(0)x + u(0). − Z0

22 Alternatively, boundary conditions at x = 0 and x = 1 can be expressed u(0) = b 1 u(1) = (1 ξ)f(ξ) dξ + a + b. − Z0

Definition 32. ζ is the fundamental solution of the operator D2. E D0 is the fundamental solution of the differential operator ∈ α L = cα(x)D α m |X|≤ iff LE = δ. Reason: L(f E) = f LE ∗ ∗ = f δ ∗ = f shows f E is a solution to Lu = f. ∗

log x L1 (R2), so compute the (dist) Laplacian ∆ log x . | | ∈ loc 0 | | ∆ log x , ϕ = log x , ∆ϕ h | | i h | | i = log x ∆ϕ(x) dx R2 | | Z = lim log x ∆ϕ(x) dx ε 0 x ε | | → Z| |≥

23 Now choose Ω to contain spt ϕ and B(0, ε), for ε > 0. Use Green’s formula6 on the open set . Ω = Ω B(0, ε) = x Ω . x > ε ε \ { ∈ | | } to get

∂Ω

Ω spt(ϕ) ε

D |x| = ε ν

Figure 5. The domains Ω and Ωε.

But ϕ and Dν ϕ vanish on (and outside) ∂Ω, so

log x ∆ϕ(x) dx = ϕ(x)∆ log x dx x ε | | x ε | | Z| |≥ Z| |≥

+ [log x Dνϕ(x) ϕ(x)Dν log x ] dσ x =ε | | − | |

Z| |

6 Ω(u∆v − v∆u) = ∂Ω(uDν v − vDν u).

24 By (4), ∆ log x = 0 and the ﬁrst integral on the right drops out. | | 1/2 With x = x2 + x2 = r, we have D = D on the circle x = ε; and so | | 1 2 ν − r | | log x ∆ϕ(x) dx x ε | | Z| |≥ ϕ(x) = log εDrϕ(x) + ε dσ x =ε − Z| | h i Since ϕ D(R2), D ϕ M on R2. Thus ∈ | r | ≤

log εDrϕ(x) dσ log ε M 2πε x =ε ≤| | · · Z| | ε 0 → 0. −−− −→ The other integral is

25 VIII. Review: Items from the previous talk

Convolution (ϕ f)(x) := ϕ(x y)f(y) dy = ϕ(y)f(x y) dy. ∗ − − Z Z (T T )(ϕ) = T T , ϕ := T (T (ϕ(x + y)), ϕ D(Rn). 1 ∗ 2 h 1 ∗ 2 i 1 2 ∈ Key properties: δ T = T δ = T. ∗ ∗ T T = T T . 1 ∗ 2 2 ∗ 1 T (T T ) = (T T ) T . 1 ∗ 2 ∗ 3 1 ∗ 2 ∗ 3 Dα(T T ) = (DαT ) T = T (DαT ). 1 ∗ 2 1 ∗ 2 1 ∗ 2 (T ψ)(x) = T (τxψ˜) is a C∞ function, where f˜(x) = f( x) and τxf(y) := f(x ∗y). − −

Theorem 33. Any distribution T D0(R) has a primitive (i.e., a distribution ∈ S D0 satisfying DkS(ϕ) = T (ϕ) for all ϕ D) which is unique up to an additiv∈ e constant. ∈

Deﬁnition 34. If f has a pole at x0, the Cauchy principal value of the divergent integral f(x) dx is

R pv f(x) dx = lim f(x) dx. ε 0 x x0 ε Z → Z| − |≥

Definition 35. E D0 is the fundamental solution of the differential operator ∈ α L = cα(x)D α m |X|≤ iff LE = δ. Reason: L(f E) = f LE = f δ = f ∗ ∗ ∗ shows f E is a solution to Lu = f. ∗ 1 log x is a fundamental solution of ∆ in R2, i.e., ∆ log x = 2πδ. 2π | | | |

26 3 1 Example. On R , is x − a weak solution to | | ∆u = 0?

For x = 0, 6 1 2 2 2 2 2 2 1/2 ∆ x − = (D + D + D )(x + x + x )− | | 1 2 3 1 2 3 3 x = D − j j (x2 + x2 + x2)3/2 j=1 1 2 3 X 3 3x2 1 = j (5) (x2 + x2 + x2)5/2 − (x2 + x2 + x2)3/2 j=1 1 2 3 1 2 3 ! X 3 3 x2 1 = 3 j (x2 + x2 + x2)5/2 − (x2 + x2 + x2)3/2 j=1 1 2 3 j=1 1 2 3 X X = 0, for x = 0 6 so one might think so . . .

1 1 3 Since x − L (R ), | | ∈ loc 1 1 ∆ x − , ϕ = x − , ∆ϕ | | | | 1 = lim x − ∆ϕ(x) dx ε 0 x ε | | → Z| |≥ 1 = lim ∆ x − ϕ(x) dx ε 0 x ε | | → Z| |≥ 1 1 + x − Dνϕ(x) Dν x − ϕ(x) dσ x =ε | | − | | Z| | by Green’s formula. Again, the ﬁrst integral vanishes by (5). With Dν = D , − r 1 x − ∆ϕ(x) dx x ε| | Z| |≥ 1 1 = Drϕ(x) dσ 2 ϕ(x) dσ. −ε x =ε − ε x =ε Z| | Z| |

27 Since D ϕ M on R3, | r | ≤ 1 M ε 0 Drϕ(x) dσ dσ = 4πεM → 0. ε x =ε ≤ ε x =ε −−−−→ Z| | Z| |

1 1 1 Thus ∆ x − , ϕ = 4πϕ(0) shows ∆ = 4πδ, and hence is a h | | i − x − − 4π x fundamental solution of ∆ on R3. | | | | By the preceding results, the Poisson equation ∆u = f has a solution given by 1 u = f ∗ −4π x when the conv is well-deﬁned. Recall, this is| |because for such a u, 1 ∆u = ∆ f ∗ −4π x | | 1 = f ∆ ∗ −4π x | | = f δ ∗ = f The solution may be interpreted physically as the potential generated by f, e.g., gravitational potential due to a mass density distribution f. 1 When f LK, ∈ 1 f(ξ) u(x) = dξ. −4π R3 x ξ Z | − |

28 Example. Temperature distribution on a slender, inﬁnite conducting bar is described by

ut = uxx (u(x, 0) = ϕ(x), where ϕ(x) is the initial heat distribution at t = 0. To get the general solution u = f E, we must ﬁnd the fundamental solution E satisfying ∗ (D D2)E(x, t) = 0 (6) t − x on the upper half plane R R , and × + E(x, 0) = δ(x,0) (7) on the boundary t = 0, x R. ∈ Such an E is given by Fourier theory:

1 x2/4t E(x, t) = e− . √4πt This satisﬁes (6) by x2/4t 2 e− x 1 DtE = √4πt 4t2 − 2t 2 2 e x /4t x e x /4t x2 1 D2E = − = − . x − √4πt 2t √4πt 4t2 − 2t

In order to satisfy (7), it suﬃces to show E(x, t) is a δ-sequence as t 0+, but we did this in (1). →

Example. The motion of an inﬁnite vibrating string solves D2u = D2u, where x R, t > 0. t x ∈ Suppose the string is released with initial shape u0 and initial velocity u1. Let E = 1 [H(x + t) H(x t)] 0 2 − − E = D E = 1 [δ(x + t) + δ(x t)] . 1 t 0 2 −

29 Then (D2 D2)E = 0 t − x 0 (D2 D2)E = 0 t − x 1 clearly hold in the upper half plane. When t = 0,

E0 = 0, E1 = δ, DtE1 = 0. Consequently, u = u E + u E 0 ∗ 1 1 ∗ 0 satisﬁes the initial conditions: u(x, 0) = u δ + u 0 = u 0 ∗ 1 ∗ 0 u (x, 0) = u D E + u D E = u 0 + u δ = u t 0 ∗ t 1 1 ∗ t 0 0 ∗ 1 ∗ 1 When u C2 and u C1 L1, 0 ∈ 1 ∈ ∩ u = 1 [u (x t) + u (x + t)] 2 0 − 1 + 1 u (ξ) [H(x + t ξ) H(x t ξ)] dξ 2 1 − − − − Z x+t 1 1 = 2 [u0(x t) + u1(x + t)] + 2 u1(ξ) dξ − x t Z −

H(x+t) H(x−t) H(x+t)−H(x−t)

−t t t

Figure 6. Construction of H(x + t) H(x t), which is convolved against u1. − −

The more general wave equation 2 2 2 Dt u = c Dxu is solved from this one via change of coordinates: x+ct 1 1 u = 2 [u0(x ct) + u1(x + ct)] + 2c u1(ξ) dξ. − x ct Z −

If the string is released from rest (u1 = 0), the solution is the average of two travelling waves u0(x ct), u1(x ct), both having the same shape u0 but travelling in opposite directions− with− velocity c.

30 IX. The Descent Method

The “Descent Method” was so coined in [La-vF] and is reminiscent of the method used by Schwartz to establish the convergence of the Fourier series associated to a periodic distribution. General idea of the Descent Method: (1) Begin with a formula you would like to manipulate, but cannot due to some issue of convergence. (2) Prove that the pointwise manipulations hold under some sufficiently restrictive conditions. (3) Integrate multiple times (say q times), until these conditions are met, so everything is sufficiently smooth and converges nicely. (4) Perform the desired manipulations. (5) Differentiate distributionally (q times) until you obtain the formula you need.

Resulting identity will hold distributionally, but may not make sense pointwise. We will be using periodic distributions: n n . n n D0(R ) := T D0(R ) . τ T = T, κ Z = D0(T ), per { ∈ κ ∈ } where Tn = Rn/Zn. Properties true for Rn will remain true on Tn when of a local nature; convolution product has all the familiar properties.

Theorem 36. If T D0 has compact support, then there is a continuous function f and a multi-index∈ α Nn such that ∈ T, ϕ = Dαf, ϕ , ϕ D. h i h i ∀ ∈

Theorem 37. (Dirichlet) If a periodic function f has a point x0 for which both

f(x0 ) = lim− f(x) and f(x0+) = lim f(x) − x x x x+ → 0 → 0 exist and are ﬁnite, then its Fourier series converges at x0 to the value 1[f(x ) + f(x +)]. 2 0− 0 In particular, if f is continuous on [a, b], then the Fourier series converges pointwise to the value of the function on [a, b].

31 Theorem 38. The Fourier series associated to a periodic distribution converges if and only if the Fourier coeﬃcients are of slow growth7

Combining these results, we know that we will be able to integrate any periodic distribution until it becomes a continuous fn, whence we can integrate it until it is Cm. This is the key that allows the descent method.

Example. Suppose you have the Fourier series of two periodic distributions:

Pα, Qβ. α Z β Z X∈ X∈ What is the product

P Q ? α  β α Z ! β Z X∈ X∈ The product will contain a coeﬃcientsR foreach point (α, β) Z2.

α,β ∈

¡£¢¥¤

¡£¢©¨ ¡¦¢¥§

Figure 7. Writing the Cauchy product of doubly inﬁnite series.

We’d like to say

P Q = R . α  β  α,β α Z ! β Z N N α + β =N X∈ X∈ X∈ | | X| | If the series do not converge absolutely , this rearrangement is not justiﬁed: need the Descent Method!

α N(α) 7“Slow growth” means that they do not grow faster than polynomially, i.e., D ϕ(x)| ≤ Cα(1 + |x|) , ∀α.

32 (1) Start with

P Q = R . α  β α,β α Z ! β Z (α,β) Z2 X∈ X∈ X∈ (2) We know that such rearrangemen ts are valid when the series is absolutely convergent. (3) Integrate the series term-by-term, q times. For large enough q, the series will converge absolutely, and even normally. (4) Rearrange the terms of the integrated series so that they are indexed/ordered in the concentric form mentioned above. (5) Diﬀerentiate the reordered series, term-by-term, q times.

We have just shown that as distributions,

P Q = R , α  β  α,β α Z ! β Z N N α + β =N X∈ X∈ X∈ | | X| | even though the sum on the right may not con verge pointwise.

Example. Suppose for x [0, y] you have an expression of the form ∈ ∞ a ( 1)f1(n,m)b xn c ( 1)f2(n,m)d xn m − n,m − m − n,m m=0 n N n N ! X X∈ X∈ and you would like to factor out the powers of x. Pointwise, such a manipulation would have no justiﬁcation unless we had some strong convergence conditions, positivity of the terms, etc. However, we interpret the series as a distribution and apply the descent method. Then we get the series

∞ a ( 1)f1(n,m)b c ( 1)f2(n,m)d xn m − n,m − m − n,m n N "m=0 # X∈ X which is equal, as a distribution, to the original expression, and shows the coeﬃcients of the xn much more clearly.

33 References

[AG] M. A. Al-Gwaiz, Theory of Distributions, Pure and Applied Mathe- matics 159, Marcel Dekker, Inc., New York, 1992.

[Ev] Lawrence C. Evans, Partial Diﬀerential Equations, Graduate Studies in Mathematics volume 19, American Mathematical Society, Provi- dence, 1998.

[La-vF] M. L. Lapidus and M. van Frankenhuysen, Fractal Geometry and Number Theory: Complex dimensions of fractal strings and zeros of zeta functions, Birkh¨auser, Boston, 2000 (Second rev. and enl. ed. to appear in 2005.)

[St] Robert Strichartz, A Guide to Distribution Theory and Fourier Transforms, Studies in Advanced Mathematics, CRC Press, 1994.

[Ru] Walter Rudin, Functional Analysis, 2nd ed., International Series in Pure and Applied Mathematics, McGraw-Hill, Inc., 1973.