Math The Laplacian. 1 Green s Identities, Fundamental Solution
|
|
- Augustus Holt
- 5 years ago
- Views:
Transcription
1 Math. 209 The Laplacian Green s Identities, Fundamental Solution Let be a bounded open set in R n, n 2, with smooth boundary. The fact that the boundary is smooth means that at each point x the external unit normal vector ν(x) is a smooth function of x. (If the boundary is C k then this function is C k. Locally, is an embedded hypersurface - a manifold of codimension ). Green s identities are (v u + u v)dx = (v ν u)ds () and (v u u v)dx = (v ν u u ν v)ds. (2) In the two identities above, u and v are real valued functions twice continiously differentiable in, (u, v C 2 ()), with first derivatives that have continuous extensions to. The Laplacian is u = n j= 2 u x 2 j = n j 2 u, j= the gradient is the normal derivative is u = ( u,..., n u), ν u = ν u where x y = n j= x jy j. The surface measure is ds = dh n. Recall that (2) follows from () and that () follows from the divergence theorem ( Φ) dx = (Φ ν)ds (3)
2 where Φ is a C () vector field (n functions) that has a continuous extension to. Recall also that the Laplacian commutes with rotations: (f(ox)) = ( f)(ox), where O is a O(n) matrix, i.e. OO = O O = I with O the transpose. Consequently, rotation invariant functions (i.e., radial functions, functions that depend only on r = x ) are mapped to rotation invariant functions. For radial functions, f = f rr + n f r r where r = x, f r = x r f. The equation f = 0 in the whole plane has affine solutions. If we seek solutions that are radial, that is f rr + n f r = 0, for r > 0, r then, multiplying by r n, we see that r n f r should be constant. There are only two possibilities: this constant is zero, and then f itself must be a constant, or this constant is not zero, and then f is a multiple of r 2 n (plus a constant) if n > 2 or a multiple of log r (plus a constant). The fundamental solution is N(x) = { 2π log x, if n = 2, (2 n)ω n x 2 n, if n 3. Here ω n is the area of the unit sphere in R n. Note that N is radial, and it is singular at x = 0. Proposition Let f C0(R 2 n ), and let u(x) = N(x y)f(y)dy. (5) R n Then u C 2 (R n ) and u = f Lemma Let N L loc (Rn ) and let φ C0(R n ). Then u = N φ is in C (R n ) and u(x) = N(x y) φ(y)dy (4) 2
3 Idea of proof of the lemma. First of all, the notation: L p loc is the space of functions that are locally in L p and that means that their restrictions to compacts are in L p. The space C0 is the space of functions with continuous derivatives of first order, and having compact support. Because φ has compact support u(x) = N(x y)φ(y)dy = N(y)φ(x y)dy is well defined. Fix x R n and take h R n with h. Note that ( ) h u(x + h) u(x) h N(x y) φ(y)dy = N(y) (φ(x + h y) φ(x y) h φ(x y)) dy h The functions y (φ(x + h y) φ(x y) h φ(x y)) for fixed x h and h are supported all in the same compact, are uniformly bounded, and converge to zero as h 0. Because of Lebesgue dominated, it follows that u is differentiable at x and that the derivative is given by the desired expression. Because u(x) = N(y) φ(x y)dy and φ is continuous, it follows that u is continuous. This finishes the proof of the lemma. Idea of proof of the Proposition. By the Lemma, u is C 2 and u(x) = N(x y) f(y)dy Fix x. The function y N(x y) f(y) is in L (R n ) and compactly supported in x y < R for a large enough R that we ll keep fixed. Therefore u(x) = lim N(x y) f(y)dy ɛ 0 {y;ɛ< x y <R} We will use Green s identities for the domains R ɛ = {y; ɛ < x y < R}. This is legitimate because the function y N(x y) is C 2 in a neighborhood of R ɛ. Note that, because of our choice of R, f(y) vanishes identically near 3
4 the outer boundary x y = R. Note also that y N(x y) = 0 for y R ɛ. From (2) we have N(x y) f(y)dy = {y;ɛ< x y <R} N(x y) x y =ɛ νf(y)ds f(y) x y =ɛ νn(x y)ds The external unit normal at the boundary is ν = (x y)/ x y. The first integral vanishes in the limit because f is bounded, N(x y) diverges like ɛ 2 n (or log ɛ) and the area of boundary vanishes like ɛ n : N(x y) ν f(y)ds Cɛ f x y =ɛ in n > 2, and the same thing replacing ɛ by ɛ log ɛ in n = 2. The second integral is more amusing, and it is here that it will become clear why the constants are chosen as they are in (4). We start by noting carefully that two minuses make a plus, and ν N(x y) = ω n x y n. Therefore, in view of the fact that x y = ɛ on the boundary we have f(y) ν N(x y)ds = f(y)ds ɛ n ω n x y =ɛ x y =ɛ Passing to polar coordinates centered at x we see that f(y) ν N(x y)ds = f(x + ɛz)ds ω n x y =ɛ z = and we do have lim f(x + ɛz)ds = f(x) ɛ 0 ω n z = because f is continuous. Remark The fundamental solution solves N = δ. This is rigorously true in the sense of distributions, but formally it can be appreciated without knowledge of distributions by using without justification the fact that δ is the identity for convolution δ f = f, the rule (N f) = N f and the Proposition above. 4
5 2 Harmonic functions Definition We say that a function u is harmonic in the open set R n if u C 2 () and if u = 0 holds in. We denote by B(x, r) the ball centered at x of radius r, by A f (x, r) the surface average A f (x, r) = f(y)ds ω n r n B(x,r) and by V f (x, r) the volume average V f (x, r) = n ω n r n B(x,r) f(y)dy Proposition 2 Let be an open set, let u be harmonic in and let B(x, r). Then holds. u(x) = A u (x, r) = V u (x, r) Idea of proof. Let ρ r and apply (2) with domain B(x, ρ) and functions v = and u. We obtain ν u ds = 0 On the other hand, because B(x,ρ) A u (x, ρ) = ω n z = u(x + ρz)ds it follows that d dρ A u(x, ρ) = ω n z = z u(x + ρz) ds = ρ n ω n B(x,ρ) ν u ds = 0 So A u (x, ρ) does not depend on ρ for 0 < ρ r. But, because the fiunction u is continuous, lim ρ 0 A u (x, ρ) = u(x), so that proves u(x) = A u (x, r). 5
6 The relation r V f (x, r) = n ρ n A f (x, ρ)dρ, 0 valid for any integrable function, implies the second equality, because A u (x.ρ) = u(x) does not depend on ρ. Remark 2 The converse of the mean value theorem holds. If u is C 2 and u(x) = A u (x, r) for each x and r sufficiently small, then u is harmonic. Indeed, in view of the above, the integral u(y)dy must vanish for B(x,r) each x and r small enough, and that implies that there cannot exist a point x where u(x) does not vanish. Theorem (Weak maximum principle.) Let be open, bounded. Let u C 2 () C 0 () satisfy u(x) 0, x. Then max u(x) = max u(x). x x Idea of proof. If u > 0 in then clearly u cannot have an interior maximum, so its maximum must be achieved on the boundary. If u 0, then a useful trick is to add ɛ x 2. The function u(x) + ɛ x 2 achieves its maximum on the boundary, for any positive ɛ. Therefore, max x u(x) + ɛ min x x 2 max u(x) + ɛ max x x x 2 and the result follws by taking the limit ɛ 0. Remark 3 If u is harmonic, then by applying the previous result to both u and u we deduce that max u(x) = max u(x) x x Definition 2 A continuous function u C 0 () is subharmonic in if x, r > 0 so that u(x) A u (x, ρ) holds ρ, 0 < ρ r. 6
7 Exercise Let u C 2 (), and assume that u(x) 0 holds for any x. Prove that u is subharmonic. Theorem 2 (Strong maximum principle) Let be open, bounded and connected. Let u C 0 () be subharmonic. Then, either u is constant, or holds. u(x) < sup u(x) x Idea of proof. Let M = sup x u(x). Consider the sets S = {x ; u(x) < M} and S 2 = {x ; u(x) = M}. The two sets are disjoint, and S is open. We show that S 2 is open as well. Indeed, take x S. Then 0 A u (x, ρ) M = (u(y) M)dS 0. ρ n ω n B(x,ρ) Because the integrand is non-positive we deduce that u(y) = M for all y so that y x = ρ, with 0 < ρ < r. This means that S 2 is open. Remark 4 The result implies that, if u C 2 () C 0 () satisfies u 0 in the bounded open connected domain, then either u is constant, or holds. u(x) < max x u(x) 3 Green s functions, Poisson kernel Exercise 2 Let be open, bounded, with smooth boundary. Let u C 2 () with first derivatives that are continuous up to the boundary u C(). Let x Then u(x) = (u ν N N ν u) ds + N(x y) u(y)dy (6) is true. The function N in the boundary integral is computed at x y, with y ; the normal derivative refers to the external normal at y. 7
8 Hint: Take a small ball B(x, r) and write (2) in \ B(x, r), with functions u and N(x y). Then use the calculation from the proof of the Proposition. Let us consider now the inhomogeneous Dirichlet problem { u = f, u = g where f is some given function in and g is a continuous function on. Suppose that we can find, for each x, a harmonic function n (x) (y) such that { y n (x) (y) = 0 n (x) (y) = N(x y), for y. Then, applying (2) we have 0 = n (x) (y) u(y)dy + Adding to (6) we deduce that G(x, y) = N(x y) n (x) (y) (N ν u u ν n (x) )ds. provides the solution to the Dirichlet problem, u(x) = G(x, y)f(y)dy + g(y) ν G(x, y)ds (7) Note that the Green s function solves { y G(x, y) = δ(x y) G(x, y) = 0 for y with δ(x y) the δ function concentrated at x. When f = 0 we obtain the representation of harmonic functions u(x) for x that satisfy u(y) = g(y) for y : u(x) = P (x, y)g(y)ds (8) with P (x, y) = ν G(x, y) (9) the Poisson kernel. The representations (7) and (8) are sublime, but the Green s function and the Poisson kernel for a general domain are hard to 8
9 obtain explicitly. Two important examples that can be computed are the half plane and the ball. The main idea, in both cases, is to reflect the singularity away. The reflection is rather straightforward for the half-plane, less so for the ball. Let = {x R n, x n > 0}. Set, G(x, y) = N(x y) N(x y) where x = (x,..., x n, x n ). This is a Green s function for. It is convenient to write a point in as x = (x, x n ) with x R n and x n > 0. Then x = (x, x n ), the Poisson kernel is a function of x y and x n, where y = (y, 0) represents a point on the boundary, P (x y, x n ) = 2x n ω n ( ) x y 2 + x 2 n 2 n We ought to cast a suspicious eye on the calculations leading to the Poisson kernel for the half-space, and prove rigorously that the result works: Proposition 3 Let g be a continuous bounded function of n variables, and let n >. The function u(x) = P (x y, x n )g(y )dy R n is harmonic in x n > 0 and the limit lim xn 0 u(x, x n ) = g(x ) holds. The calculations for a ball follow. Let = B(0, R). Let x and set x = R2 x (yes, trouble when x = 0, ignore for a moment). Note that if x 2 y then x y x y = R x does not depend on y. For n > 2 take the fundamental solutions N(x y) and N(x y) and write ( ) 2 n x G(x, y) = N(x y) N(x y) R Clearly, for y, G(x, y) vanishes. The second term is not singular in y, so it is harmonic in y. The Poisson kernel is This works even in n = 2. P (x, y) = Rω n R 2 x 2 x y n (0) 9
10 Proposition 4 Let g be a continuous function on B(0, R) in R n, n 2. The function { P (x, y)g(y)ds, for x < R B(0,R) u(x) = u(x) = g(x), for x = R given by the Poisson kernel (0) is harmonic in x < R, and continuous in x R. The proof uses the following properties of the Poisson kernel: P (x, y) C, P (x, y) > 0 for x < R, y = R, x P (x, y) = 0, for x < R, y = R, P (x, y)ds = x, x < R, B(0,R) lim x z, x <R P (x, y) = 0, uniformly for y = z = R, z y δ > 0. 4 Dirichlet principle, variational solutions Let be a bounded open set in R n. Let f C(), g C( ) and let A = {w C 2 (); w = g}. Let I[w] = ( ) 2 w 2 + wf dx. () Proposition 5 u C 2 () solves { u = f in, u = g on (2) if, and only if u = arg min w A I[w]. Note carefully that this is not an existence theorem, rather, it states the equivalence of two possible existence theorems. One theorem asserts that the Poisson problem with data f and g has a solution u with the desired smoothness. The other theorem asserts that one can minimize the integral I[w] and find a true minimum, in the class of admissible functin s A. The 0
11 variational method considers the minimization program. The program consists in two steps. The first step is to establish the existence of a minimum. Unfortunately, the natural function spaces for I[w] are not spaces of continuous functions, but rather Sobolev spaces based on L 2. This presents an opportunity to generalize: the right-hand side f will be allowed to be in L 2, because the method does not require more. This comes at a price: the minimum thus obtained is not smooth. The second step is to show that if the function f is smooth then the solution is smooth. Here, the natural smoothness requirement for f is more stringent than C(), it is the Hölder space C 0,α (). If is smooth, then the solution obtained is smoother than just C 2 (), it is C 2,α (). So, the innocent looking proposition above is both a begining and an end: it is the begining of a search for more general solutions, and it is the end of the classical problem in C H 0() Definition 3 Let R n be an open bounded set with smooth boundary. Th e space H () is the completion of C (R n ) with norm u 2 H () = ( u 2 + u 2 )dx. The space H 0() is the closure of C 0 () in H (). Similar definitions are given for the whole space and the torus. In those cases H 0 = H. However, in bounded domains H 0 is strictly smaller: it repres ents functions that vanish at the boundary, in a weak sense. Exercise 3 Show that H 0((0, )) H ((0, )). Lemma 2 (Poincaré Inequality) There exists a constant C depending on the bounded domain such that u 2 C u 2 dx holds for all u H 0(). Proof. Because is bounded in the direction x, we know that there exists an interval [a, b] such that t [a, b] holds for all t such that there exists
12 x 2,... x n so that x = (t, x 2,..., x n ). Because both sides of the inequality are continuous in H and because of the definition of H0, we may assume, WLOG that u C0 (). Then By Schwartz: u(x,... x n ) 2 = 2 x a u(t, x 2,..., x n )( u)(t, x 2,..., x n )dt { b } { u(x,... x n ) 2 2 u 2 2 b } (t, x 2,..., x n )dt u(t, x 2,..., x n ) 2 2 dt a a We keep x fixed, integrate dx 2... dx n and use Schwartz again: u 2 (x, x 2,... x n )dx 2... dx n 2 u L 2 u L 2. We integrate dx on [a, b], noting that the RHS is independent of x : u 2 L 2 2(b a) u L 2 u L 2. Dividing by u L 2 we obtain the inequality with C = 4(b a) 2. It is clear from this proof that we do not need to use the boundedness of, only the existence of some direction in which is bounded. Exercise 4 Show that the Poincaré inequality fails in H () if is bounded. Show that the Poincaré inequality fails in some unbounded domains: for instance in H 0(R). The Poincaré inequality implies that the scalar product (u, v) = u vdx is equivalent (gives the same topology) with the scalar product in H0(): < u, v >= (uv + u v)dx 2
13 Theorem 3 Let be a bounded open set with smooth boundary. Let f L 2 (). Then, there exists a unique u H 0() that solves where I[w] = I[u] = min I[w] w H0 () ( ) 2 w 2 + wf dx. The function u satisfies the variational formulation of the problem (2) with g = 0: (u, v) + vfdx = 0 v H0() (3) Proof. The function I[w] is bounded below. Indeed, using Schwartz we have I[w] w 2 dx f L 2 w L 2 2 and using the Poincaré inequality I[w] 2C w 2 L 2 f L 2 w L 2 and that is bounded below. Thus, the infimum exists: m = inf I[w]. w H0 () Let w k be a minimizing sequence, I[w k ] m. The sequence is bounded in H 0() because (w k, w k ) are bounded. Therefore there exists a subsequence (denoted again w k ) and an element u in H 0() such that w k converges weakly to u, lim k (w k, v) = (u, v) v H 0(). Then, because v vfdx is continouous in H 0, hence weakly continuous, it follows that ufdx = lim w k fdx k Exercise 5 In a Hilbert space, the square of the norm is weakly lower semicontinuous. 3
14 By the above exercise, 2 u 2 dx 2 lim inf k This shows that u achieves the minimum. I(u) = m. w k 2 dx The variational formulation (3) follows by looking at the function q(t) = I(u + tv) for fixed, but arbitrary v H0() and t real. q(t) is a quadratic polynomial in t, { } q(t) = I(u) + t (u, v) + vfdx + t2 (v, v) 2 that has a minimum at t = 0. The variational formulation is equivalent to the fact that q (0) = 0. The uniqueness of u follows immediately from variational formulation with v = u, the inequality thus u 2 C(u, u) 2C f L 2 u L 2, u L 2 2C f L 2, and because the variational formulation is linear. 4
Laplace s Equation. Chapter Mean Value Formulas
Chapter 1 Laplace s Equation Let be an open set in R n. A function u C 2 () is called harmonic in if it satisfies Laplace s equation n (1.1) u := D ii u = 0 in. i=1 A function u C 2 () is called subharmonic
More informationPartial Differential Equations
Part II Partial Differential Equations Year 2015 2014 2013 2012 2011 2010 2009 2008 2007 2006 2005 2015 Paper 4, Section II 29E Partial Differential Equations 72 (a) Show that the Cauchy problem for u(x,
More informationThe Dirichlet s P rinciple. In this lecture we discuss an alternative formulation of the Dirichlet problem for the Laplace equation:
Oct. 1 The Dirichlet s P rinciple In this lecture we discuss an alternative formulation of the Dirichlet problem for the Laplace equation: 1. Dirichlet s Principle. u = in, u = g on. ( 1 ) If we multiply
More informationSingular Integrals. 1 Calderon-Zygmund decomposition
Singular Integrals Analysis III Calderon-Zygmund decomposition Let f be an integrable function f dx 0, f = g + b with g Cα almost everywhere, with b
More informationGreen s Functions and Distributions
CHAPTER 9 Green s Functions and Distributions 9.1. Boundary Value Problems We would like to study, and solve if possible, boundary value problems such as the following: (1.1) u = f in U u = g on U, where
More informationMATH 425, FINAL EXAM SOLUTIONS
MATH 425, FINAL EXAM SOLUTIONS Each exercise is worth 50 points. Exercise. a The operator L is defined on smooth functions of (x, y by: Is the operator L linear? Prove your answer. L (u := arctan(xy u
More informationPerron method for the Dirichlet problem.
Introduzione alle equazioni alle derivate parziali, Laurea Magistrale in Matematica Perron method for the Dirichlet problem. We approach the question of existence of solution u C (Ω) C(Ω) of the Dirichlet
More informationSOLUTION OF POISSON S EQUATION. Contents
SOLUTION OF POISSON S EQUATION CRISTIAN E. GUTIÉRREZ OCTOBER 5, 2013 Contents 1. Differentiation under the integral sign 1 2. The Newtonian potential is C 1 2 3. The Newtonian potential from the 3rd Green
More informationRegularity for Poisson Equation
Regularity for Poisson Equation OcMountain Daylight Time. 4, 20 Intuitively, the solution u to the Poisson equation u= f () should have better regularity than the right hand side f. In particular one expects
More information14 Divergence, flux, Laplacian
Tel Aviv University, 204/5 Analysis-III,IV 240 4 Divergence, flux, Laplacian 4a What is the problem................ 240 4b Integral of derivative (again)........... 242 4c Divergence and flux.................
More information4 Divergence theorem and its consequences
Tel Aviv University, 205/6 Analysis-IV 65 4 Divergence theorem and its consequences 4a Divergence and flux................. 65 4b Piecewise smooth case............... 67 4c Divergence of gradient: Laplacian........
More informationis a weak solution with the a ij,b i,c2 C 1 ( )
Thus @u @x i PDE 69 is a weak solution with the RHS @f @x i L. Thus u W 3, loc (). Iterating further, and using a generalized Sobolev imbedding gives that u is smooth. Theorem 3.33 (Local smoothness).
More informationRegularizations of Singular Integral Operators (joint work with C. Liaw)
1 Outline Regularizations of Singular Integral Operators (joint work with C. Liaw) Sergei Treil Department of Mathematics Brown University April 4, 2014 2 Outline 1 Examples of Calderón Zygmund operators
More informationBoundedness, Harnack inequality and Hölder continuity for weak solutions
Boundedness, Harnack inequality and Hölder continuity for weak solutions Intoduction to PDE We describe results for weak solutions of elliptic equations with bounded coefficients in divergence-form. The
More informationSome lecture notes for Math 6050E: PDEs, Fall 2016
Some lecture notes for Math 65E: PDEs, Fall 216 Tianling Jin December 1, 216 1 Variational methods We discuss an example of the use of variational methods in obtaining existence of solutions. Theorem 1.1.
More informationMeasure and Integration: Solutions of CW2
Measure and Integration: s of CW2 Fall 206 [G. Holzegel] December 9, 206 Problem of Sheet 5 a) Left (f n ) and (g n ) be sequences of integrable functions with f n (x) f (x) and g n (x) g (x) for almost
More information4 Riesz Kernels. Since the functions k i (ξ) = ξ i. are bounded functions it is clear that R
4 Riesz Kernels. A natural generalization of the Hilbert transform to higher dimension is mutiplication of the Fourier Transform by homogeneous functions of degree 0, the simplest ones being R i f(ξ) =
More informationErrata Applied Analysis
Errata Applied Analysis p. 9: line 2 from the bottom: 2 instead of 2. p. 10: Last sentence should read: The lim sup of a sequence whose terms are bounded from above is finite or, and the lim inf of a sequence
More informationMAT 578 FUNCTIONAL ANALYSIS EXERCISES
MAT 578 FUNCTIONAL ANALYSIS EXERCISES JOHN QUIGG Exercise 1. Prove that if A is bounded in a topological vector space, then for every neighborhood V of 0 there exists c > 0 such that tv A for all t > c.
More informationTHE INVERSE FUNCTION THEOREM
THE INVERSE FUNCTION THEOREM W. PATRICK HOOPER The implicit function theorem is the following result: Theorem 1. Let f be a C 1 function from a neighborhood of a point a R n into R n. Suppose A = Df(a)
More informationHARMONIC ANALYSIS. Date:
HARMONIC ANALYSIS Contents. Introduction 2. Hardy-Littlewood maximal function 3. Approximation by convolution 4. Muckenhaupt weights 4.. Calderón-Zygmund decomposition 5. Fourier transform 6. BMO (bounded
More informationIn this chapter we study elliptical PDEs. That is, PDEs of the form. 2 u = lots,
Chapter 8 Elliptic PDEs In this chapter we study elliptical PDEs. That is, PDEs of the form 2 u = lots, where lots means lower-order terms (u x, u y,..., u, f). Here are some ways to think about the physical
More informationSobolev Spaces. Chapter 10
Chapter 1 Sobolev Spaces We now define spaces H 1,p (R n ), known as Sobolev spaces. For u to belong to H 1,p (R n ), we require that u L p (R n ) and that u have weak derivatives of first order in L p
More informationSolution Sheet 3. Solution Consider. with the metric. We also define a subset. and thus for any x, y X 0
Solution Sheet Throughout this sheet denotes a domain of R n with sufficiently smooth boundary. 1. Let 1 p
More informationS chauder Theory. x 2. = log( x 1 + x 2 ) + 1 ( x 1 + x 2 ) 2. ( 5) x 1 + x 2 x 1 + x 2. 2 = 2 x 1. x 1 x 2. 1 x 1.
Sep. 1 9 Intuitively, the solution u to the Poisson equation S chauder Theory u = f 1 should have better regularity than the right hand side f. In particular one expects u to be twice more differentiable
More informationElliptic PDEs of 2nd Order, Gilbarg and Trudinger
Elliptic PDEs of 2nd Order, Gilbarg and Trudinger Chapter 2 Laplace Equation Yung-Hsiang Huang 207.07.07. Mimic the proof for Theorem 3.. 2. Proof. I think we should assume u C 2 (Ω Γ). Let W be an open
More informationTHE DIRICHLET PROBLEM FOR THE LAPLACE OPERATOR
THE DIRICHLET PROBLEM FOR THE LAPLACE OPERATOR Stefano Meda Università di Milano-Bicocca c Stefano Meda 2013 ii A Francesco Contents I The Dirichlet problem via Perron s method 1 1 The classical Dirichlet
More information2. Function spaces and approximation
2.1 2. Function spaces and approximation 2.1. The space of test functions. Notation and prerequisites are collected in Appendix A. Let Ω be an open subset of R n. The space C0 (Ω), consisting of the C
More informationINTEGRATION ON MANIFOLDS and GAUSS-GREEN THEOREM
INTEGRATION ON MANIFOLS and GAUSS-GREEN THEOREM 1. Schwarz s paradox. Recall that for curves one defines length via polygonal approximation by line segments: a continuous curve γ : [a, b] R n is rectifiable
More informationSHARP BOUNDARY TRACE INEQUALITIES. 1. Introduction
SHARP BOUNDARY TRACE INEQUALITIES GILES AUCHMUTY Abstract. This paper describes sharp inequalities for the trace of Sobolev functions on the boundary of a bounded region R N. The inequalities bound (semi-)norms
More informationOverview of normed linear spaces
20 Chapter 2 Overview of normed linear spaces Starting from this chapter, we begin examining linear spaces with at least one extra structure (topology or geometry). We assume linearity; this is a natural
More informationFriedrich symmetric systems
viii CHAPTER 8 Friedrich symmetric systems In this chapter, we describe a theory due to Friedrich [13] for positive symmetric systems, which gives the existence and uniqueness of weak solutions of boundary
More informationv( x) u( y) dy for any r > 0, B r ( x) Ω, or equivalently u( w) ds for any r > 0, B r ( x) Ω, or ( not really) equivalently if v exists, v 0.
Sep. 26 The Perron Method In this lecture we show that one can show existence of solutions using maximum principle alone.. The Perron method. Recall in the last lecture we have shown the existence of solutions
More informationHeat kernels of some Schrödinger operators
Heat kernels of some Schrödinger operators Alexander Grigor yan Tsinghua University 28 September 2016 Consider an elliptic Schrödinger operator H = Δ + Φ, where Δ = n 2 i=1 is the Laplace operator in R
More informationPartial Differential Equations. William G. Faris
Partial Differential Equations William G. Faris May 17, 1999 Contents 1 The Laplace equation 5 1.1 Gradient and divergence....................... 5 1.2 Equilibrium conservation laws....................
More informationTHE L 2 -HODGE THEORY AND REPRESENTATION ON R n
THE L 2 -HODGE THEORY AND REPRESENTATION ON R n BAISHENG YAN Abstract. We present an elementary L 2 -Hodge theory on whole R n based on the minimization principle of the calculus of variations and some
More informationMATH COURSE NOTES - CLASS MEETING # Introduction to PDEs, Spring 2018 Professor: Jared Speck
MATH 8.52 COURSE NOTES - CLASS MEETING # 6 8.52 Introduction to PDEs, Spring 208 Professor: Jared Speck Class Meeting # 6: Laplace s and Poisson s Equations We will now study the Laplace and Poisson equations
More informationMATH COURSE NOTES - CLASS MEETING # Introduction to PDEs, Fall 2011 Professor: Jared Speck
MATH 8.52 COURSE NOTES - CLASS MEETING # 6 8.52 Introduction to PDEs, Fall 20 Professor: Jared Speck Class Meeting # 6: Laplace s and Poisson s Equations We will now study the Laplace and Poisson equations
More informationSuggested Solution to Assignment 7
MATH 422 (25-6) partial diferential equations Suggested Solution to Assignment 7 Exercise 7.. Suppose there exists one non-constant harmonic function u in, which attains its maximum M at x. Then by the
More informationMATH 205C: STATIONARY PHASE LEMMA
MATH 205C: STATIONARY PHASE LEMMA For ω, consider an integral of the form I(ω) = e iωf(x) u(x) dx, where u Cc (R n ) complex valued, with support in a compact set K, and f C (R n ) real valued. Thus, I(ω)
More informationELLIPTIC PARTIAL DIFFERENTIAL EQUATIONS EXERCISES I (HARMONIC FUNCTIONS)
ELLIPTIC PARTIAL DIFFERENTIAL EQUATIONS EXERCISES I (HARMONIC FUNCTIONS) MATANIA BEN-ARTZI. BOOKS [CH] R. Courant and D. Hilbert, Methods of Mathematical Physics, Vol. Interscience Publ. 962. II, [E] L.
More informationPARTIAL DIFFERENTIAL EQUATIONS. Lecturer: D.M.A. Stuart MT 2007
PARTIAL DIFFERENTIAL EQUATIONS Lecturer: D.M.A. Stuart MT 2007 In addition to the sets of lecture notes written by previous lecturers ([1, 2]) the books [4, 7] are very good for the PDE topics in the course.
More informationBoot camp - Problem set
Boot camp - Problem set Luis Silvestre September 29, 2017 In the summer of 2017, I led an intensive study group with four undergraduate students at the University of Chicago (Matthew Correia, David Lind,
More informationTHE HARDY LITTLEWOOD MAXIMAL FUNCTION OF A SOBOLEV FUNCTION. Juha Kinnunen. 1 f(y) dy, B(x, r) B(x,r)
Appeared in Israel J. Math. 00 (997), 7 24 THE HARDY LITTLEWOOD MAXIMAL FUNCTION OF A SOBOLEV FUNCTION Juha Kinnunen Abstract. We prove that the Hardy Littlewood maximal operator is bounded in the Sobolev
More informationSolutions: Problem Set 4 Math 201B, Winter 2007
Solutions: Problem Set 4 Math 2B, Winter 27 Problem. (a Define f : by { x /2 if < x
More informationLECTURE # 0 BASIC NOTATIONS AND CONCEPTS IN THE THEORY OF PARTIAL DIFFERENTIAL EQUATIONS (PDES)
LECTURE # 0 BASIC NOTATIONS AND CONCEPTS IN THE THEORY OF PARTIAL DIFFERENTIAL EQUATIONS (PDES) RAYTCHO LAZAROV 1 Notations and Basic Functional Spaces Scalar function in R d, d 1 will be denoted by u,
More informationProblem Set 5. 2 n k. Then a nk (x) = 1+( 1)k
Problem Set 5 1. (Folland 2.43) For x [, 1), let 1 a n (x)2 n (a n (x) = or 1) be the base-2 expansion of x. (If x is a dyadic rational, choose the expansion such that a n (x) = for large n.) Then the
More information= 2 x x 2 n It was introduced by P.-S.Laplace in a 1784 paper in connection with gravitational
Lecture 1, 9/8/2010 The Laplace operator in R n is = 2 x 2 + + 2 1 x 2 n It was introduced by P.-S.Laplace in a 1784 paper in connection with gravitational potentials. Recall that if we have masses M 1,...,
More informationIntegral Representation Formula, Boundary Integral Operators and Calderón projection
Integral Representation Formula, Boundary Integral Operators and Calderón projection Seminar BEM on Wave Scattering Franziska Weber ETH Zürich October 22, 2010 Outline Integral Representation Formula Newton
More informationTD M1 EDP 2018 no 2 Elliptic equations: regularity, maximum principle
TD M EDP 08 no Elliptic equations: regularity, maximum principle Estimates in the sup-norm I Let be an open bounded subset of R d of class C. Let A = (a ij ) be a symmetric matrix of functions of class
More informationMath 328 Course Notes
Math 328 Course Notes Ian Robertson March 3, 2006 3 Properties of C[0, 1]: Sup-norm and Completeness In this chapter we are going to examine the vector space of all continuous functions defined on the
More informationEXPOSITORY NOTES ON DISTRIBUTION THEORY, FALL 2018
EXPOSITORY NOTES ON DISTRIBUTION THEORY, FALL 2018 While these notes are under construction, I expect there will be many typos. The main reference for this is volume 1 of Hörmander, The analysis of liner
More informationSobolev spaces. May 18
Sobolev spaces May 18 2015 1 Weak derivatives The purpose of these notes is to give a very basic introduction to Sobolev spaces. More extensive treatments can e.g. be found in the classical references
More information2 A Model, Harmonic Map, Problem
ELLIPTIC SYSTEMS JOHN E. HUTCHINSON Department of Mathematics School of Mathematical Sciences, A.N.U. 1 Introduction Elliptic equations model the behaviour of scalar quantities u, such as temperature or
More informationRegularity of local minimizers of the interaction energy via obstacle problems
Regularity of local minimizers of the interaction energy via obstacle problems J. A. Carrillo, M. G. Delgadino, A. Mellet September 22, 2014 Abstract The repulsion strength at the origin for repulsive/attractive
More informationContinuous Functions on Metric Spaces
Continuous Functions on Metric Spaces Math 201A, Fall 2016 1 Continuous functions Definition 1. Let (X, d X ) and (Y, d Y ) be metric spaces. A function f : X Y is continuous at a X if for every ɛ > 0
More informationThe oblique derivative problem for general elliptic systems in Lipschitz domains
M. MITREA The oblique derivative problem for general elliptic systems in Lipschitz domains Let M be a smooth, oriented, connected, compact, boundaryless manifold of real dimension m, and let T M and T
More informationSolutions Final Exam May. 14, 2014
Solutions Final Exam May. 14, 2014 1. Determine whether the following statements are true or false. Justify your answer (i.e., prove the claim, derive a contradiction or give a counter-example). (a) (10
More informationTHE GREEN FUNCTION. Contents
THE GREEN FUNCTION CRISTIAN E. GUTIÉRREZ NOVEMBER 5, 203 Contents. Third Green s formula 2. The Green function 2.. Estimates of the Green function near the pole 2 2.2. Symmetry of the Green function 3
More informationMeasurable functions are approximately nice, even if look terrible.
Tel Aviv University, 2015 Functions of real variables 74 7 Approximation 7a A terrible integrable function........... 74 7b Approximation of sets................ 76 7c Approximation of functions............
More informationWeek 6 Notes, Math 865, Tanveer
Week 6 Notes, Math 865, Tanveer. Energy Methods for Euler and Navier-Stokes Equation We will consider this week basic energy estimates. These are estimates on the L 2 spatial norms of the solution u(x,
More informationMetric Spaces and Topology
Chapter 2 Metric Spaces and Topology From an engineering perspective, the most important way to construct a topology on a set is to define the topology in terms of a metric on the set. This approach underlies
More informationReal Analysis Problems
Real Analysis Problems Cristian E. Gutiérrez September 14, 29 1 1 CONTINUITY 1 Continuity Problem 1.1 Let r n be the sequence of rational numbers and Prove that f(x) = 1. f is continuous on the irrationals.
More informationLebesgue s Differentiation Theorem via Maximal Functions
Lebesgue s Differentiation Theorem via Maximal Functions Parth Soneji LMU München Hütteseminar, December 2013 Parth Soneji Lebesgue s Differentiation Theorem via Maximal Functions 1/12 Philosophy behind
More informationJUHA KINNUNEN. Harmonic Analysis
JUHA KINNUNEN Harmonic Analysis Department of Mathematics and Systems Analysis, Aalto University 27 Contents Calderón-Zygmund decomposition. Dyadic subcubes of a cube.........................2 Dyadic cubes
More informationRecall that if X is a compact metric space, C(X), the space of continuous (real-valued) functions on X, is a Banach space with the norm
Chapter 13 Radon Measures Recall that if X is a compact metric space, C(X), the space of continuous (real-valued) functions on X, is a Banach space with the norm (13.1) f = sup x X f(x). We want to identify
More informationFall f(x)g(x) dx. The starting place for the theory of Fourier series is that the family of functions {e inx } n= is orthonormal, that is
18.103 Fall 2013 1. Fourier Series, Part 1. We will consider several function spaces during our study of Fourier series. When we talk about L p ((, π)), it will be convenient to include the factor 1/ in
More informationLecture No 1 Introduction to Diffusion equations The heat equat
Lecture No 1 Introduction to Diffusion equations The heat equation Columbia University IAS summer program June, 2009 Outline of the lectures We will discuss some basic models of diffusion equations and
More informationCalculus of Variations. Final Examination
Université Paris-Saclay M AMS and Optimization January 18th, 018 Calculus of Variations Final Examination Duration : 3h ; all kind of paper documents (notes, books...) are authorized. The total score of
More informationLecture Notes in Advanced Calculus 1 (80315) Raz Kupferman Institute of Mathematics The Hebrew University
Lecture Notes in Advanced Calculus 1 (80315) Raz Kupferman Institute of Mathematics The Hebrew University February 7, 2007 2 Contents 1 Metric Spaces 1 1.1 Basic definitions...........................
More informationAnalysis in weighted spaces : preliminary version
Analysis in weighted spaces : preliminary version Frank Pacard To cite this version: Frank Pacard. Analysis in weighted spaces : preliminary version. 3rd cycle. Téhéran (Iran, 2006, pp.75.
More information3. (a) What is a simple function? What is an integrable function? How is f dµ defined? Define it first
Math 632/6321: Theory of Functions of a Real Variable Sample Preinary Exam Questions 1. Let (, M, µ) be a measure space. (a) Prove that if µ() < and if 1 p < q
More informationREVIEW OF ESSENTIAL MATH 346 TOPICS
REVIEW OF ESSENTIAL MATH 346 TOPICS 1. AXIOMATIC STRUCTURE OF R Doğan Çömez The real number system is a complete ordered field, i.e., it is a set R which is endowed with addition and multiplication operations
More informationReal Analysis Notes. Thomas Goller
Real Analysis Notes Thomas Goller September 4, 2011 Contents 1 Abstract Measure Spaces 2 1.1 Basic Definitions........................... 2 1.2 Measurable Functions........................ 2 1.3 Integration..............................
More informationP(E t, Ω)dt, (2) 4t has an advantage with respect. to the compactly supported mollifiers, i.e., the function W (t)f satisfies a semigroup law:
Introduction Functions of bounded variation, usually denoted by BV, have had and have an important role in several problems of calculus of variations. The main features that make BV functions suitable
More informationProbability and Measure
Part II Year 2018 2017 2016 2015 2014 2013 2012 2011 2010 2009 2008 2007 2006 2005 2018 84 Paper 4, Section II 26J Let (X, A) be a measurable space. Let T : X X be a measurable map, and µ a probability
More informationMathematics for Economists
Mathematics for Economists Victor Filipe Sao Paulo School of Economics FGV Metric Spaces: Basic Definitions Victor Filipe (EESP/FGV) Mathematics for Economists Jan.-Feb. 2017 1 / 34 Definitions and Examples
More informationON WEAKLY NONLINEAR BACKWARD PARABOLIC PROBLEM
ON WEAKLY NONLINEAR BACKWARD PARABOLIC PROBLEM OLEG ZUBELEVICH DEPARTMENT OF MATHEMATICS THE BUDGET AND TREASURY ACADEMY OF THE MINISTRY OF FINANCE OF THE RUSSIAN FEDERATION 7, ZLATOUSTINSKY MALIY PER.,
More informationANALYSIS QUALIFYING EXAM FALL 2017: SOLUTIONS. 1 cos(nx) lim. n 2 x 2. g n (x) = 1 cos(nx) n 2 x 2. x 2.
ANALYSIS QUALIFYING EXAM FALL 27: SOLUTIONS Problem. Determine, with justification, the it cos(nx) n 2 x 2 dx. Solution. For an integer n >, define g n : (, ) R by Also define g : (, ) R by g(x) = g n
More informationMATH 411 NOTES (UNDER CONSTRUCTION)
MATH 411 NOTES (NDE CONSTCTION 1. Notes on compact sets. This is similar to ideas you learned in Math 410, except open sets had not yet been defined. Definition 1.1. K n is compact if for every covering
More informationReal Analysis Math 131AH Rudin, Chapter #1. Dominique Abdi
Real Analysis Math 3AH Rudin, Chapter # Dominique Abdi.. If r is rational (r 0) and x is irrational, prove that r + x and rx are irrational. Solution. Assume the contrary, that r+x and rx are rational.
More informationu( x) = g( y) ds y ( 1 ) U solves u = 0 in U; u = 0 on U. ( 3)
M ath 5 2 7 Fall 2 0 0 9 L ecture 4 ( S ep. 6, 2 0 0 9 ) Properties and Estimates of Laplace s and Poisson s Equations In our last lecture we derived the formulas for the solutions of Poisson s equation
More informationReview of Multi-Calculus (Study Guide for Spivak s CHAPTER ONE TO THREE)
Review of Multi-Calculus (Study Guide for Spivak s CHPTER ONE TO THREE) This material is for June 9 to 16 (Monday to Monday) Chapter I: Functions on R n Dot product and norm for vectors in R n : Let X
More informationBased on the Appendix to B. Hasselblatt and A. Katok, A First Course in Dynamics, Cambridge University press,
NOTE ON ABSTRACT RIEMANN INTEGRAL Based on the Appendix to B. Hasselblatt and A. Katok, A First Course in Dynamics, Cambridge University press, 2003. a. Definitions. 1. Metric spaces DEFINITION 1.1. If
More informationA metric space X is a non-empty set endowed with a metric ρ (x, y):
Chapter 1 Preliminaries References: Troianiello, G.M., 1987, Elliptic differential equations and obstacle problems, Plenum Press, New York. Friedman, A., 1982, Variational principles and free-boundary
More informationCHAPTER 6. Differentiation
CHPTER 6 Differentiation The generalization from elementary calculus of differentiation in measure theory is less obvious than that of integration, and the methods of treating it are somewhat involved.
More informationFunctional Analysis. Franck Sueur Metric spaces Definitions Completeness Compactness Separability...
Functional Analysis Franck Sueur 2018-2019 Contents 1 Metric spaces 1 1.1 Definitions........................................ 1 1.2 Completeness...................................... 3 1.3 Compactness......................................
More informationWeak Convergence Methods for Energy Minimization
Weak Convergence Methods for Energy Minimization Bo Li Department of Mathematics University of California, San Diego E-mail: bli@math.ucsd.edu June 3, 2007 Introduction This compact set of notes present
More informationTopological properties
CHAPTER 4 Topological properties 1. Connectedness Definitions and examples Basic properties Connected components Connected versus path connected, again 2. Compactness Definition and first examples Topological
More informationApplied Math Qualifying Exam 11 October Instructions: Work 2 out of 3 problems in each of the 3 parts for a total of 6 problems.
Printed Name: Signature: Applied Math Qualifying Exam 11 October 2014 Instructions: Work 2 out of 3 problems in each of the 3 parts for a total of 6 problems. 2 Part 1 (1) Let Ω be an open subset of R
More informationExercise 1. Let f be a nonnegative measurable function. Show that. where ϕ is taken over all simple functions with ϕ f. k 1.
Real Variables, Fall 2014 Problem set 3 Solution suggestions xercise 1. Let f be a nonnegative measurable function. Show that f = sup ϕ, where ϕ is taken over all simple functions with ϕ f. For each n
More informationChapter One. The Calderón-Zygmund Theory I: Ellipticity
Chapter One The Calderón-Zygmund Theory I: Ellipticity Our story begins with a classical situation: convolution with homogeneous, Calderón- Zygmund ( kernels on R n. Let S n 1 R n denote the unit sphere
More informationNumerical Solutions to Partial Differential Equations
Numerical Solutions to Partial Differential Equations Zhiping Li LMAM and School of Mathematical Sciences Peking University Sobolev Embedding Theorems Embedding Operators and the Sobolev Embedding Theorem
More informationAn introduction to Birkhoff normal form
An introduction to Birkhoff normal form Dario Bambusi Dipartimento di Matematica, Universitá di Milano via Saldini 50, 0133 Milano (Italy) 19.11.14 1 Introduction The aim of this note is to present an
More informationMath 209B Homework 2
Math 29B Homework 2 Edward Burkard Note: All vector spaces are over the field F = R or C 4.6. Two Compactness Theorems. 4. Point Set Topology Exercise 6 The product of countably many sequentally compact
More information1 Distributions (due January 22, 2009)
Distributions (due January 22, 29). The distribution derivative of the locally integrable function ln( x ) is the principal value distribution /x. We know that, φ = lim φ(x) dx. x ɛ x Show that x, φ =
More informationThe wave equation. Intoduction to PDE. 1 The Wave Equation in one dimension. 2 u. v = f(ξ 1 ) + g(ξ 2 )
The wave equation Intoduction to PDE The Wave Equation in one dimension The equation is u t u c =. () x Setting ξ = x + ct, ξ = x ct and looking at the function v(ξ, ξ ) = u ( ξ +ξ, ξ ξ ) c, we see that
More informationMcGill University Department of Mathematics and Statistics. Ph.D. preliminary examination, PART A. PURE AND APPLIED MATHEMATICS Paper BETA
McGill University Department of Mathematics and Statistics Ph.D. preliminary examination, PART A PURE AND APPLIED MATHEMATICS Paper BETA 17 August, 2018 1:00 p.m. - 5:00 p.m. INSTRUCTIONS: (i) This paper
More informationMath 342 Partial Differential Equations «Viktor Grigoryan
Math 342 Partial ifferential Equations «Viktor Grigoryan 3 Green s first identity Having studied Laplace s equation in regions with simple geometry, we now start developing some tools, which will lead
More informationPrinciple of Mathematical Induction
Advanced Calculus I. Math 451, Fall 2016, Prof. Vershynin Principle of Mathematical Induction 1. Prove that 1 + 2 + + n = 1 n(n + 1) for all n N. 2 2. Prove that 1 2 + 2 2 + + n 2 = 1 n(n + 1)(2n + 1)
More information