Partial Differential Equations

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16 1 The Fundamental Partial Differential Equations Theorem 1.2.16. Let g ∈ C(∂B(0; r)). Then, the function u: B(0; r) → R, defined by the Poisson formula (1.12), has the following properties: (i) u ∈ C ∞ (B(0; r)). (ii) ∆u = 0 on B(0; r). (iii) For every x 0 ∈ ∂B(0; r), we have lim u(x) = g(x 0 ). x→x 0 x∈B(0;r) Proof. Idea: Show first that the Poisson formula (1.12) can be differentiated under the integral to prove (i) and (ii). For (iii) one needs an explicit estimate for |u(x) − g(x 0 )|. A modification of Example 1.2.14 shows that { ( Φ(y − x) − Φ |x| r G r (x, y) = (y − r2˜x) ) for x ≠ 0, Φ(y) − Φ(r y |y| ) for x = 0 is a Green’s function for the ball B(0; r) such that ∂G r ∂ν (x, y) = K r(x, y) for x ∈ B(0; r) and y ∈ ∂B(0; r) (check this!). Note that the right hand side defines a smooth function on M := {(x, y) ∈ R n × R n | x ≠ y} which is harmonic in the variable y. For any fixed x ∈ B(0; r), the function y ↦→ G r (x, y) is harmonic on B(0; r) \ {x}. By Lemma 1.2.15, the function x ↦→ G r (x, y) is also ∂G harmonic on B(0; r) \ {y} for every y ∈ B(0; r). This in turn shows that the functions x ↦→ y r j ∂y j (x, y) and x ↦→ ∂Gr ∂Gr ∂ν (x, y) are harmonic on B(0; r) \ {y}. Since the function (x, y) ↦→ ∂ν (x, y) is smooth on M, we get that ∂Gr ∂G ∂ν (·, y) −→ r y→y 0 ∂ν (·, y0 ) locally uniformly on B(0; r) for y 0 ∈ ∂B(0; r). Thus, ∂Gr ∂ν (·, y0 ) also is harmonic, and finally, we obtain that the Poisson kernel K r (x, y) in x is harmonic on B(0; r). Note that g is bounded on ∂B(0; r). Moreover, for x ≠ yfor x ∈ B(0; r) the map x ↦→ DK r (x, y) is uniformly bounded in y, since K r is C 1 away from the diagonal. Therefore, y ↦→ DK r (x, y)g(y) is locally uniformly bounded in x, so we may differentiate ∫ u(x) = K r (x, y)g(y) dS ∂B(0;r) (y) ∂B(0;r) under the integral. Since K r is infinitely many times differentiable at x, and since all partial derivatives with respect to x also are continuous, we can repeat this argument, and we get u ∈ C ∞ (B(0; r)). In particular, we can calculate ∫ ∆u(x) = ∆ x K r (x, y)g(y) dS ∂B(0;r) (y) = 0. ∂B(0;r) Thus, (i) and (ii) are proven. To show (iii), we note first that applying the integral formula from Lemma 1.2.12 to the constant function v(x) = 1, we obtain the Poisson integral ∫ ∫ ∂G r 1 = − ∂ν (x, y) dS ∂B(0;r)(y) = K r (x, y) dS ∂B(0;r) (y). (1.13) ∂B(0;r) ∂B(0;r) Now fix x 0 ∈ ∂B(0; r). Since g is continuous, for every ε > 0, there is a δ > 0 with ∀y ∈ B(x 0 ; 2δ) ∩ ∂B(0; r) : Using ‖g‖ ∞ := sup y∈∂B(0;r) |g(y)| and K r ≥ 0, we now calculate ∣ g(y) − g(x 0 ) ∣ ∣ ≤ ε 2 .
1.2 The Laplace Equation 17 y 2δ δ x x 0 0 r ∫ |u(x) − g(x 0 )| = ∣ ∫ ≤ ∫ = ∂B(0;r) ∂B(0;r) Fig. 1.8. K r (x, y) ( g(y) − g(x 0 ) ) dS ∂B(0;r) (y) ∣ K r (x, y) ∣ ∣ g(y) − g(x 0 ) ∣ ∣ dS∂B(0;r) (y) ∂B(0;r)∩B(x 0 ;2δ) ∫ + ∂B(0;r)\B(x 0 ;2δ) K r (x, y) ∣ ∣ g(y) − g(x 0 ) ∣ ∣ dS∂B(0;r) (y) ≤ ε 2 + 2‖g‖ r 2 − |x| 2 ∞ rδ n vol(S n−1 )r n−1 K r (x, y) ∣ ∣ g(y) − g(x 0 ) ∣ ∣ dS∂B(0;r) (y) for x ∈ B(x 0 1 r ; δ) ∩ B(0; r), where, apart from K r (x, y) = 2 −|x| 2 nα(n)r |y−x| , we used that |x − y| ≥ δ for n y ∈ ∂B(0; r) \ B(x 0 ; 2δ) holds because of x ∈ B(x 0 ; δ). Now, if x tends to x 0 , then the second summand of the estimate tends to zero as well, and the claim follows. Corollary 1.2.17. Harmonic functions are real analytic. Proof. One represents the harmonic function by the Poisson formula (1.12) on small balls in the domain of definition. One observes that the integrand is real analytic, and one uses uniform convergence on small balls to interchange the power series expansion with the integral. Exercise 1.2.18. Let R n + := {x = (x 1 , . . . , x n ) ∈ R n | x n > 0}. (a) For x ∈ R n +, define ϕ x : R n + → R, y ↦→ Φ(y 1 − x 1 , . . . , y n−1 − x n−1 , y n + x n ), where Φ is the fundamental solution for the Laplace equation ∆u = 0 on R n . Show: (i) ϕ x is harmonic. (ii) ϕ x has a continuous extension to R n + = {x = (x 1 , . . . , x n ) ∈ R n | x n ≥ 0} such that (b) Define a function Show: 2 (x, y) = (i) ∂G ∂ν x n nα(n) |y−x| . n ϕ x (y) = Φ(y − x) ∀y ∈ ∂R n +. G: {(x, y) ∈ R n + × R n + | x ≠ y} → R, (x, y) ↦→ Φ(y − x) − ϕ x (y).
Page 1 and 2: Partial Differential Equations Summ
Page 3: Preface Let U ⊆ R n be an open se
Page 6 and 7: 4 1 The Fundamental Partial Differe
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Page 45 and 46: 2.1 Complete Integrals and Envelopi
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66 3 Various Solution Techniques fo
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84 3 Various Solution Techniques
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86 4 Linear Differential Operators
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A Function Spaces A.1 L p -spaces a
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A.1 L p -spaces 95 By this proposit
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A.1 L p -spaces 97 Proposition A.1.
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A.1 L p -spaces 99 (ii) For p < ∞
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A.1 L p -spaces 101 Lemma A.1.16 (Y
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A.2 Topological Vector Spaces 103 R
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1. Case: p αi (x) = 0 i = 1, . . .
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A.3 Spaces of Differentiable Functi
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A.4 The Fourier Transform 113 (iii)
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A.4 The Fourier Transform 115 ( Bec
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B Distributions B.1 Tempered Distri
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B.1 Tempered Distributions 119 Proo
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B.1 Tempered Distributions 121 ξ 0
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Proof. (i) The implication “⇐
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B.2 Distributions 125 Proof. Let ϕ
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B.2 Distributions 127 where ψ ∈
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B.2 Distributions 129 and the chara
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B.3 Convergence of Distributions 13
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134 C Sobolev spaces For s ∈ R, w
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136 C Sobolev spaces Theorem C.1.8.
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138 C Sobolev spaces Proof. By Prop
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140 C Sobolev spaces Theorem C.1.16
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142 C Sobolev spaces is bounded for
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144 C Sobolev spaces Fig. C.1. Theo
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146 C Sobolev spaces V m ~ U U V 0
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148 C Sobolev spaces Theorem C.3.4.
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150 C Sobolev spaces [∆ α h, ϕ]
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152 Index of the Schwartz space, 11
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154 Index singular support, 124 smo
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Partial Differential Equations

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