MATH 124B: INTRODUCTION TO PDES AND FOURIER SERIES

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1 MATH 24B: INTROUCTION TO PES AN FOURIER SERIES SHO SETO Contents. Fourier Series.. Even, Odd, Period, and Complex functions 5.2. Orthogonality and General Fourier Series 6.3. Convergence and completeness 7.4. Proof of convergence 2. Harmonic functions aplace s equation Rectangles Poisson s Formula Circles, Wedges, and Annuli 9 3. Green s identities and Green s functions Green s second identity Green s function Half-space and sphere General Eigenvalue Problem Minimax principle 29 Appendix A. Some inear Algebra 3. Fourier Series Recall the homogeneous wave equation with irichlet boundary conditions u tt = c 2 u xx for < x < ( u(, t = u(, t = with initial conditions { u(x, = φ(x u t (x, = ψ(x. We look for special solutions to the above by using separation of variables, that is, assume the solution takes the form u(x, t = F (xg(t. Plugging into (, we obtain G (t c 2 G(t = F (x F (x ate: September 5, 26.

2 2 SHO SETO Since the right hand side is independent of the variable x and vice versa, we conclude that both sides are equal to a constant, say λ := β 2, β >. λ is positive since F (x = λf (x and we multiply both side by F (x and integrate from to so that λ (F (x 2 dx = F (xf (xdx = F (xf (x = (F (x 2 dx. (F (x 2 dx Where we use the irichlet boundary condition in the second line. Hence we obtain a pair of OEs to solve the separated solution. That is { F (x + β 2 F (x = G (t + c 2 β 2 G(t =. These can be solved by finding the roots to the characteristic polynomials so that F (x = A cos(βx + B sin(βx G(t = C cos(βct + sin(βct, where A, B, C, are constants. Now we plug in the irichlet conditions: and Assuming B, we have that β = nπ F ( = A = F ( = B sin(β =. for n Z. Thus ( nπ λ n = 2 are the possible eigenvalues and ( nπ F n (x = sin x are solutions. Note that each sine function may be multiplied by an arbitrary constant and remains a solution to the OE. For each n, we obtain a solution ( ( ( nπct nπct ( nπ u n (x, t = A n cos + B n sin sin x, where A n and B n are arbitrary constants Since the PE is linear, any finite sum u(x, t = n u n (x, t

3 MATH 24B: INTROUCTION TO PES AN FOURIER SERIES 3 is also a solution. If we can choose A n and B n such that (2 φ(x = ( nπ A n sin x n and ψ(x = n nπc ( nπ B n sin x, then we get a solution with the given initial conditions. The question now is, when is it possible to write φ and ψ in the given form? Infinite series of the form (2 are called Fourier sine series on (,. To obtain the coefficients, we use the following property for sine functions: emma.. 2 { ( nπ ( mπ sin x sin x if m = n dx = if m. Proof. Using the trig identity Then when 2 sin A sin B = (cos(a B cos(a + B. 2 ( nπ ( mπ sin x sin x dx = ( ( (n mπ (m + nπ cos x cos x dx Evaluating gives the result. Remark.. A similar proof shows orthogonality with the cosine functions as well. Therefore, integrating against φ, all but one term vanishes. This gives us a formula for the Fourier coefficients of the sine series A m = 2 We also define the Fourier cosine series as with the coefficient formula given by φ(x = 2 A + A m = 2 ( mπ φ(x sin x dx. ( nπ A n cos x, ( mπ φ(x cos x dx. Finally, summing the sine and cosine series, we obtain the full Fourier series of φ(x defined on < x <. φ(x = ( ( nπ ( nπ 2 A + A n cos x + B n sin x.

4 4 SHO SETO with the coefficients given by A n = B n = ( nπ φ(x cos x dx, (n =,, 2,... ( nπ φ(x sin x dx, (n =, 2,... Example.. et φ(x = in the interval [, ]. Computing the coefficients, we have Therefore, in (,. = 4 π A m = 2 ( mπ sin x dx = 2 mπ ( ( m. ( ( πx sin + ( 3πx 3 sin + ( 5πx 5 sin +. Example.2. et φ(x = x in the interval (,. Its Fourier sine coefficients are so that x = 2 π A m = 2 = ( ( mπ x sin x dx m+ 2 mπ ( ( πx sin ( 2πx 2 sin + ( 3πx 3 sin. Example.3. Next compute the cosine series for φ(x = x. The coefficients are given by Thus in (,, we have x = 2 4 π 2 A = 2 A m = 2 xdx = ( mπx x cos = 2 m 2 π 2 (( m. ( cos πx + ( 3πx 9 cos dx + ( 5πx 25 cos +. Example.4. Solve the problem u tt = c 2 u xx u(, t = = u(, t u(x, = x u t (x, =.

5 MATH 24B: INTROUCTION TO PES AN FOURIER SERIES 5 The expansion of a separated solution has the form u(x, t = ( ( ( nπct nπct ( nπ A n cos + B n sin sin x ifferentiating this with respect to t yields so that hence B n = and u t (x, t = nπc ( ( nπct A n sin u t (x, = x = u(x, = ( nπct + B n cos nπc ( nπx B n sin ( nπx A n sin. Comparing with the Fourier sine series of x, we have that n+ 2 A n = ( nπ ( nπ sin x and so the solution is given by u(x, t = 2 π ( n+ ( nπx sin cos n ( nπct... Even, Odd, Period, and Complex functions. efinition.. A function φ(x that is defined for x R is called periodic if there is a number p > such that φ(x + p = φ(x, for all x. The number p is called the period of φ(x. Proposition.. If φ(x has period p, then does not depend on a. Proof. By fundamental theorem of calculus, d da a+p a a+p a φ(xdx φ(xdx = φ(a + p φ(a = φ(a φ(a =.

6 6 SHO SETO If a function is defined only on an interval of length p, it can be extended to a function of period p by taking its periodic extension: et φ be defined on the interval < x <. Then define φ per (x by φ per (x = φ(x 2m for + 2m < x < + 2m for all integers m. Another symmetry property a function may possess are the following efinition.2. A function is even if and odd if φ(x = φ( x φ( x = φ(x. To ensure the definition makes sense, we require the function be defined on an interval (,. We define the even extension of a function defined on (, to be { φ(x for < x < φ even (x = φ( x for < x <. and its odd extension is φ(x for < x < φ odd (x = φ( x for < x < for x =. Since sin(nπx/ is odd, the Fourier sine series can be regarded as an expansion of an arbitrary function that is odd and has period 2 defined on the whole line R. ikewise, since cos(nπx/ is even so the Fourier cosine series can be regarded an an expansion of an arbitrary function which is even and has period 2 defined on the whole line R. Note that the irichlet boundary condition corresponds to an odd extension, the Neumann to an even extension, and periodic boundary conditions correspond to the periodic extension. Since we can express the sine and cosine functions as complex exponential functions, we have a way to express the Fourier series in complex form, namely φ(x = c n e inπx/ with coefficients given by n= c n = φ(xe inπx/ Orthogonality and General Fourier Series. Now we consider a more general setting. efinition.3. For two real valued functions f and g defined on an interval (a, b, define their ( 2 -inner product to be We also have the 2 norm (f, g = b a f(xg(xdx. b f 2 = (f, f = f 2 dx. 2 f and g are orthogonal (with respect to the inner product if (f, g =. The key property we use for Fourier series is that every other eigenfunction. a every eigenfunction is orthogonal to

7 .3. Convergence and completeness. MATH 24B: INTROUCTION TO PES AN FOURIER SERIES 7 efinition.4. We say that an infinite series f n (x converges to f(x pointwise in (a, b if it converges to f(x for each a < x < b, that is N f(x f n (x, as N. Note that the rate of convergence depends on x. efinition.5. We say that the series converges uniformly to f in [a, b] if N max [a,b] f(x f n (x as N. Note that the convergence is independent of the point x. efinition.6. We say the series converges in 2 to f if b N 2 f(x f n (x dx as N. In norm notation: a f N f n 2 as N. Example.5. et f n (x = ( xx n on < x <. Then N f n (x = N (x n x n = x N. Since < x <, x N as N. The convergence depends on x however so it is only pointwise. However, it does converge in the 2 sense since x N 2 dx = 2N +. Example.6. et n f n (x = + n 2 x n 2 + (n 2 x 2 in the interval < x <. Then its partial sum is given by N f n (x = N + N 2 x 2 as N when x >, hence converges pointwise. However in the 2 sense, ( N 2 N f n (x dx = N dy. ( + y 2 2 Furthermore, it does not converge uniformly since max (, N + N 2 x 2 = N

8 8 SHO SETO Now we present 3 theorems on convergence of Fourier series. Theorem. (Uniform Convergence. The Fourier series A n X n converges to f(x uniformly on [a, b] provided that ( f(x, f (x and f (x exist and are continuous for a x b. (2 f(x satisfies the appropriate boundary conditions. In fact, for the classical Fourier series, f is not required to exist. Theorem.2 ( 2 Convergence. The Fourier series converges to f(x in the 2 sense in (a, b if f(x 2 (a, b. Theorem.3 (Pointwise convergence for classical Fourier series. ( The classical Fourier series converges to f(x pointwise on (a, b provided that f(x is continuous on a x b and f (x is piecewise continuous on a x b, for example f(x = x on [, ]. (2 More generally, if f(x is only piecewise continuous on a b b and f (x is also piecewise continuous on a x b, then the classical Fourier series converges (as an infinite sum at every point x. The sum is given by n A n X n (x = 2 ( lim f(t + lim f(t t x + t x for all a < x < b. Outside the interval, it is given by the corresponding extensions of f. Example.7. The Fourier sine series of f(x = on (, π is f(x = n= 4 sin((2n + x. (2n + π It does not converge uniformly since the end points are zero. Notice that this does not contradict the convergence theorem since the sine series requires the irichlet boundary conditions. Note that this example shows that the terms cannot in general be differentiated term by term, since the derivative is f = however 4 cos((2n + x π is not a convergent series. n= Theorem.4 (east square approximation. et {X n } be any ( 2 orthogonal set of functions. et f <. et N be a fixed positive integer. Among all possible choices of c,... c N, the choice that minimizes N f c N X n is given by the Fourier coefficients c i = A i. Proof. For simplicity, assume we are working with real valued functions f and X n. We want to minimize E N = f n N c n X n 2 = (f, f 2 n N c n (f, X n + n,m c n c m (X n, X m.

9 MATH 24B: INTROUCTION TO PES AN FOURIER SERIES 9 By orthogonality, (X n, X m = except when n = m. Hence, E N = f 2 2 c n (f, X n + c 2 n X n 2. n N n N We compute the square in c n so that E N = n N ( X n 2 c n (f, X 2 n + f 2 (f, X n 2 X n 2 X n. 2 n N Then the term containing c n is minimized when it is zero since it is nonnegative, i.e. c n = (f, X n X n = A n. We can say a little more. We know that E N, hence E N = f 2 n N A 2 n X n 2. Since this holds for all N, writing out explicitly the 2 norm, we obtain Bessel s inequality A 2 n b a X n (x 2 dx Furthermore, if E N, we obtain the following: b a f(x 2 dx. Theorem.5 (Parseval s equality. The Fourier series of f(x converges in 2 if and only if b A n 2 X n (x 2 dx = a b a f(x 2 dx efinition.7. The infinite orthogonal set {X i (x} is called complete if Parseval s equality holds for all f <. Example.8. By applying Parseval s identity to the sine series for f(x =, we have Integrating, we have nodd ( 2 4 π π sin 2 (nxdx = dx nπ nodd ( 2 4 π nπ 2 = π which is another proof for obtaining the Basel sum ( n. n 2

10 SHO SETO.4. Proof of convergence. For convenience, let = π. For C function f which is 2π periodic. Then the Fourier series is f(x = 2 A + (A n cos(nx + B n sin(nx with coefficients given by and A n = π B n = π π π π π f(y cos(nydy f(y sin(nydy. Therefore the N-th partial sum, S N (x is given by S N (x = ( π N + 2 (cos(ny cos(nx + sin(ny sin(nx f(ydy. 2π π Using the cosine summation formula cos(a + B = cos(a cos(b sin(a sin(b, we rewrite the above as S N (x = π K N (x yf(ydy 2π where π K N (θ := + 2 N cos(nθ. The function K N (θ is called the irichlet kernel. Note that and in fact, can be expressed as 2π π π K N (θdθ = K N (θ = sin((n + 2 θ sin( 2 θ To see this, we rewrite as a finite geometric series K N (θ = + = N n= N N (e inθ + e inθ e inθ = e inθ e i(n+θ e iθ = e i(n+ 2 θ e i(n+ 2 θ e iθ 2 + e iθ 2 = sin((n + 2 θ sin( 2 θ.

11 MATH 24B: INTROUCTION TO PES AN FOURIER SERIES Proof of pointwise convergence of classical Fourier series. By changing variables to θ = y x and using the evenness of K N, we have S N (x = π K N (θf(x + θdθ. 2π π Using the fact that the integral of the irichlet kernel is, we have where S N (x f(x = π 2π π = π 2π g(θ = π K N (θ(f(x + θ f(xdθ g(θ sin((n + 2 θdθ, f(x + θ f(x sin( 2 θ. It remains to show that the integral tends to zero as N. Since the functions φ N (θ = sin((n + θ, N =, 2, 3,... 2 form an orthogonal set on the interval (, π, and therefore on ( π, π, Bessel s inequality holds: N= (g, φ N 2 φ N 2 g 2. Since φ N 2 = π, if g <, then the infinite sum converges, i.e. (g, φ N as N. To check that g is finite, we have g 2 = π π (f(x + θ f(x 2 sin 2 ( θ dθ. 2 Since as long as f exists. lim g(θ = 2f (x θ Proof of uniform convergence. Assume that f and f (x are continuous with period 2π. et A n and B n be the Fourier coefficients for f(x and A n and B n be the Fourier coefficients of f (x. They are related by { A n = n B n B n = n A n. It can be shown by Bessel s inequality that ( A n 2 + B n 2 f 2 < π

12 2 SHO SETO Therefore ( A n cos(nx + B n sin(nx ( A n + B n = n ( B n + A n ( /2 ( /2 2( A n n 2 + B n 2 <. 2 Therefore, from pointwise convergence, we have max f(x S N (x max A n cos(nx + B n sin(nx as N. N+ 2. Harmonic functions 2.. aplace s equation. The stationary heat or wave equation ( A n + B n N+ or in higher dimensions u xx =, u := div( u = is called the aplace equation, and its solutions are called harmonic functions. The inhomogeneous version u = f for a given function f is called Poisson s equation. Just like the heat and wave equations, we can give aplace equations The first thing we will prove is the maximum principle. Theorem 2. (Maximum principle. et be a connected bounded open set in R n. et u be harmonic in and continuous on. Then the maximum and minimum values of u are attained on, the boundary of and nowhere inside unless u is a constant. Proof. et ε > and define Then n i= 2 u x 2 i = v ε (x = u(x ε x 2, x R n. v ε = 2nε <. Suppose x is a local minimum. Then v ε, a contradiction, hence there is no local minimum in the interior of. et x be the minimum for v ε. Since holds for all x and for all ε, we have u(x v(x v ε (x = u(x ε x 2 u(x min u. Repeating the argument for v ε = u + ε x 2 gives us the maximum.

13 MATH 24B: INTROUCTION TO PES AN FOURIER SERIES Uniqueness. The maximum principle can be used to prove the uniqueness of solutions for the Poisson equation with irichlet condition { u = f in u = h on. Suppose that v is another solution to the Poisson equation with same boundary conditions. Then u v is harmonic and on the boundary, hence by the maximum principle u = v on Invariance. The aplace equation is invariant under rigid motion. We show this explicitly in 2 and 3 dimensions. For two dimensions, translation is obvious. Consider a rotation in the plane by angle θ. It can be given by the transformation (x, y = (x cos θ + y sin θ, x sin θ + y cos θ. Changing variables, we have u x = u x x x + u y y x = u x cos θ u y sin θ. and 2 u x = (u 2 x cos θ u ỹ sin θ x cos θ (u x cos θ u ỹ sin θ y sin θ. by a similar computation, we obtain u xx + u yy = (u x x + u y y (cos2 θ + sin 2 θ = u x x + u y y therefore, the aplacian is rotationally invariant. In polar coordinates, x = r cos θ, y = r sin θ, the aplacian is given by = 2 r + 2 r r + 2 r 2 θ 2 A rotationally symmetric solution, i.e. u(r, θ = u(r is given by u(r = c log r + c 2. In higher dimensions, we can show rotational invariance through some linear algebra. We define a rotation in R n, to be a transformation given by an orthogonal matrix, i.e. x = B x where B O(n = {B is an n n matrix such that B T B = BB T the trace of the Hessian matrix of u hence = I}. Then the aplacian is Tr(Hess u(x = Tr(B T Hess u(x B = Tr(Hess u(x B T B = Tr(Hess u(x. In spherical coordinates, x = s cos φ y = s sin φ z = r cos θ s = r sin θ,

14 4 SHO SETO we can view this as a change of coordinates from (x, y, z to cylindrical coordinates (s, φ, z, then to spherical (r, θ, φ. Since z is fixed in the first transformation, we can apply the 2 dimensional change of coordinates so that u xx + u yy = u ss + s u s + s 2 u φφ. In the second change of coordinates, φ is fixed so that Combining, we have By chain rule, so that u zz + u ss = u rr + r u r + r 2 u θθ. u xx + u yy + u zz = u rr + r u r + r 2 u θθ + s u s + s 2 u φφ. u s = u r r s + u θ θ s + u φ φ s s = u r r + u cos θ θ r u = u rr + 2 r u r + ( u r 2 θθ + (cot θu θ + sin 2 θ u φφ. The rotationally symmetric solutions in 3 dimensions is given by u = c r + c 2. Example 2.. If the domain that we are dealing with is rotationally symmetric, one strategy is to convert the aplace or Poisson equation into polar coordinates. For example, consider the irichlet problem { u xx + u yy = in r < a u = on r = a. Since the domain is rotationally symmetric with rotationally symmetric boundary condition, the solution should depend only on r. In polar coordinates, u rr + r u r =. We can view this as a first order equation of u r, hence multiplying by the integrating factor, we have (ru r r = r. Integrating twice, we have Inserting the boundary condition, we have u(r = 4 r2 + c ln(r + c 2. = a2 4 + c ln(a + c 2. hence we get a relation between the coefficients c and c 2.

15 MATH 24B: INTROUCTION TO PES AN FOURIER SERIES Rectangles. We will give explicit solutions over rectangles and cubes. et be a rectangle, that is = {(x, y R 2 < x < a, < y < b} Example 2.2. We will find harmonic functions on with the given boundary conditions For simplicity, assume h = k = j =. We will find a separated solution, so assume u(x, y = X(xY (y. Then the aplace equation can be rewritten as X (x X(x + Y (y Y (y =. Hence, there is a constant such that X + λx = and Y λy =. Since X( = X (a =, we can show that λ = β 2 >. In fact, the explicit eigenfunctions and eigenvalues are { ( (n+ Xn (x = sin Next we solve for β 2 n = λ n = ( n πx a 2 π 2 a 2. Y λy = with Y ( + Y ( =. Since this is a constant coefficient second order OE with λ >, we have Inserting the boundary conditions, we have Y (y = A cosh(β n y + B sinh(β n y. = Y ( + Y ( = Bβ n + A. For convenience, let B =, so that A = β n. Then Y (y = β n cosh(β n y sinh(β n y. Therefore, we can formally add the solutions X n Y n to obtain u(x, y = A n sin(β n x(β n cosh(β n y sinh(β n y. n= This satisfies the 3 boundary conditions. It remains to insert the last boundary condition: g(x = A n (β n cosh(β n b sinh(β n b sin(β n x. n= This is a Fourier sine series hence we can use the Fourier coefficient formula to get A n = 2 a a (β n cosh(β n b sinh(β n b g(x sin(β n xdx.

16 6 SHO SETO 2.3. Poisson s Formula. Next we discuss the irichlet problem for a circle, { u xx + u yy = for x 2 + y 2 < a 2 u = h(θ for x 2 + y 2 = a 2 We consider separated solutions u(r, θ = R(rΘ(θ so that = R Θ + r R Θ + r 2 RΘ. ividing by RΘ and multiplying by r 2, we obtain { Θ + λθ = r 2 R + rr λr =. For the angular component, we give periodic boundary conditions: Hence we get explicit solutions Θ(θ + 2π = Θ(θ Θ(θ = A cos(nθ + B sin(nθ with λ = n 2, (n =,, 2,.... In the radial direction, the OE is of Euler type hence we assume that the solution is of the form R(r = r α so that α(α r α + αr α n 2 r α =, hence α = ±n, so R(r = c r n + c 2 r n. Combining the angular and radial solutions, we have for n =, 2,.... For n =, we have u(r, θ = (c r n + c 2 r n (A cos(nθ + B sin(nθ u = c + c 2 log(r Since u is harmonic, and hence continuous on, we do not consider the solutions r n and log r which are discontinuous at r =. Hence, summing the solutions, we have u(r, θ = 2 A + r n (A n cos(nθ + B n sin(nθ. Inserting the irichlet boundary condition, we have h(θ = 2 A + a n (A n cos(nθ + B n sin(nθ. Since this is the Fourier series for h(θ, we have the formulas for the coefficients A n = πa n B n = πa n 2π 2π h(φ cos(nφdφ h(φ sin(nφdφ

17 MATH 24B: INTROUCTION TO PES AN FOURIER SERIES 7 Much like the irichlet kernel, we can find a closed form. formulas into the solution, we have Plugging in the Fourier coefficient u(r, θ = 2π h(φdφ + 2π = ( 2π h(φ + 2 2π r n 2π h(φ(cos(nφ cos(nθ + sin(nφ sin(nθdφ πa n ( r n cos(n(θ φ dφ a By expressing cosine in terms of the complex exponential, we have + ( r n e in(θ φ + a Inserting this, we obtain Poisson s formula ( r n e in(θ φ a = + rei(θ φ a re + re i(θ φ i(θ φ a re i(θ φ a 2 r 2 = a 2 2ar cos(θ φ + r. 2 u(r, θ = (a2 r 2 2π 2π h(φ a 2 2ar cos(θ φ + r 2 dφ. This formula can be used to express any harmonic function inside a circle in terms of its boundary values. In vector notation, viewing points as vectors originating from, u( x = a2 x 2 2πa y =a u( y x y 2 ds Theorem 2.2. et h(φ = u( y be any continuous function on the circle C =. Then the Poisson formula provides the only harmonic function in for which for all x C. Proof. We define the Poisson kernel P (r, θ = lim x x u(x = h(x a 2 r 2 a 2 2ar cos θ + r 2 so that u(r, θ = 2π P (r, θ φh(φdφ 2π for r < a. The Poisson kernel has the following properties ( P (r, θ > for r < a. (2 2π P (r, θdθ = 2π

18 8 SHO SETO (3 P is harmonic inside the circle. Now u(r, θ is harmonic since P is harmonic and we can differentiate under the integral sign in this situation (P (r, θ needs to be continuous in the compact interval. Now we need to show the limit, so fix an angle θ, and consider r near a, then u(r, θ h(θ = 2π 2π P (r, θ φ(h(φ h(θ dφ. First we will show that the kernel is concentrated at θ =, that is, for any ε >, and for each δ θ π δ, we can find an r sufficiently close to a so that 2 P (r, θ = Using this r, we have that a 2 r 2 a 2 2ar cos θ + r 2 = a 2 r 2 (a r 2 + 4ar sin 2 (θ/2 a 2 r 2 (a r 2 + 4ar sin 2 (δ < ε. u(r, θ h(θ ε 2π θ +δ θ δ ( + 2Hε, P (r, θ φdφ + ε h(φ h(θ dφ 2π φ θ >δ where δ > was chosen so that h(φ h(θ < ε whenever φ θ < δ and h H for some constant H, since h is continuous. We can apply the Poisson formula to obtain the mean value property and the maximum principle: Proposition 2. (Mean Value Property. et u be harmonic in a disk, continuous up to the boundary. Then the value of u at the center of equals the average of u on its circumference. Proof. et x = in the Poisson formula, then u( = 2πa y =a u(yds Proposition 2.2 ((Strong Maximum Principle. et u be harmonic in. If the maximum is attained in the interior, then u is a constant. et x M be the maximum point. Then u(x u(x M = M, for all x. Proof. By applying mean value property at this point, we have M = u(x M = average on circle M so that u = M on the circle. Repeating the argument, we can fill the domain with circles so that u is a constant on, if max is attained in the interior. Proposition 2.3 (ifferentiability. et u be a harmonic function in any open set of the plane. Then u( x = u(x, y possesses all partial derivatives of all orders in. Roughly, by the Poisson integral formula, we see that the integrand is differentiable and does not require the differentiability of the original function.

19 MATH 24B: INTROUCTION TO PES AN FOURIER SERIES Circles, Wedges, and Annuli. We define the following domains: Wedge: {(r, θ < θ < θ, < r < a} Annulus: {(r, θ < a < r < b} Exterior of a circle: {(r, θ a < r < } Wedge. We consider aplace equation with homogeneous irichlet condition on the straight sides and an inhomogeneous Neumann condition on the circular side: u = u(r, = u(r, β = u r (a, θ = h(θ. By separation of variables, we get the two OEs { r 2 R + rr λr = Θ + λθ =. Inserting the homogeneous conditions gives us Θ( = Θ(β =, hence we have a solution ( nπθ Θ n (θ = sin β ( 2 λ n =. In the radial direction, we have nπ β r 2 R + rr λr = This is of Euler type and has the solution R(r = r α, where α 2 = λ. Throwing out the singular solutions, we have the solution ( nπθ u(r, θ = A n r nπ/β sin. β With the inhomogenous boundary condition, we view it as a Fourier sine series with coefficients A n = a nπ/β 2 β ( nπθ h(θ sin dθ. nπ β Annulus. Consider the irichlet problem for the annulus in 2 dimensions, u xx + u yy = in < a 2 < x 2 + y 2 < b 2 u = g(θ for r = a u = h(θ for r = b. Since r, we use the solutions of the circle except that we can now include the singular (in the circle case terms so u(r, θ = 2 (C + log r + (C n r n + n r n cos(nθ + (A n r n + B n r n sin(nθ.

20 2 SHO SETO Plugging in the boundary conditions, and noting that {, cos(nθ, sin(nθ} are orthogonal on the interval (, 2π, we have the relations { A n a n + B n a n = 2π A n b n + B n b n = 2π 2π g(θ sin(nθdθ 2π h(θ sin(nθdθ and similarly for the constant and cosine terms. In matrix form, this is equivalent to solving for ( ( ( a n a n An αn = b n b n where a, b are given and α n, β n are the corresponding Fourier coefficients. The determinant of the matrix is nonzero as long as a b and has a unique solution. B n Exterior of a circle. We consider the irichlet problem for the exterior of a circle in 2 dimensions u xx + u yy = for x 2 + y 2 > a 2 u = h(θ u is bounded for r = a We can solve this in the same way as a circle again however, the condition u being bounded implies that there are no terms r n since r n as r. Hence we only keep the terms r n. Therefore, u(r, θ = 2 A + β n r n (A n cos(nθ + B n sin(nθ. Incorporating the boundary condition, we have and A n = an π B n = an π π π π π h(θ cos(nθdθ h(θ sin(nθdθ. Furthermore, there is a corresponding Poisson formula in this case given by u(r, θ = (r2 a 2 2π 2π h(φ a 2 2ar cos(θ φ + r 2 dφ. 3. Green s identities and Green s functions In this section, we present a way to solve aplace s equation on more general domains. Recall the divergence theorem, let be a bounded region and F a vector field defined on. Then div F dv = F nds where n is the unit outer normal vector of. We first prove Green s first identity, for u, v real valued functions, v u n ds = v udx + v udx.

21 MATH 24B: INTROUCTION TO PES AN FOURIER SERIES 2 This is a simple consequence of the divergence theorem and the product rule div(v u = v u + v u. Setting v =, we have u n ds = udx. We can apply this to obtain a necessary solution for the solution to exist for the Neumann problem in any domain : If { u = f in Then integrating over we have u = h n hds = on, Hence, for a solution to exist, the above must be satisfied between the boundary data and the inhomogeneous term Mean value property. We also show a mean value property for harmonic functions in 3 dimensions. et B be a ball of radius a. Suppose u = on B. On a sphere, we have the normal direction being the same as the radial direction, hence u n = u r. Therefore, plugging into Green s first identity, we obtain u ds =. r In spherical coordinates, the surface integral is given by 2π π B fdv. u r (a, θ, φa 2 sin(φdφdθ =. ividing out the constant by 4πa 2 and moving the r derivative outside the integral, we get ( 2π π u(r, θ, φ sin(φdφdθ =. r 4π Note that this holds for all r = a. Hence we see that the quantity 2π π u(r, θ, φ sin(φdφdθ = 4π A( B is independent of r. Taking the limit r, we get 4π 2π π r=a u( sin(φdφdθ = u(, hence uds = u(. A( B B This proves the mean value property in three dimensions. From the mean value property, we automatically have the maximum principle Proposition 3.. If is any solid region, a nonconstant harmonic function in cannot take its maximum value inside, but only on. B uds

22 22 SHO SETO Uniqueness of irichlet problem. Given two harmonic functions u and u 2 with the same boundary data, their difference u := u u 2 is harmonic and is zero at the boundary. Using Green s first identity with v = u, we obtain = u u n ds = u 2 dv. Hence u = and u is a constant in. Since the boundary is, we have that u =. This establishes the uniqueness. We define the term on the right as the following efinition 3.. efine the energy of u as We have the following E[u] = 2 u 2 dv. Proposition 3.2 (irichlet principle. et u be the unique harmonic function in with boundary data u = h(x on. et w be any other function with the same boundary data. Then the harmonic function u is the minimizer of the energy, i.e. E[w] E[u]. Proof. et v := u w. Then E[w] = (u v 2 dv 2 = E[u] + E[v] where the mixed term vanish by applying Green s first identity. 3.. Green s second identity. By subtracting Green s first identity, we have Green s second identity ( u v v udv = u v n v u ds. n 3... Representation formula. Much like the Poisson formula, we want to obtain a integral representation for any harmonic function. We have the following Theorem 3.. In dimension 3, if u = in, then u(x = ( u(x ( 4π n x x u + x x n ds. Proof. Choose polar coordinates centered at x. Apply Green s second identity with v = 4πr. et ε = B(x, ε.

23 MATH 24B: INTROUCTION TO PES AN FOURIER SERIES 23 Then u = and v = in ε. Apply Green s second identity to this pair to obtain ( u ( u ds =. n r n r ε Now the boundary ε consists of 2 parts, with the normal vector for the inner sphere pointing inward so that =, thus n r ( u ( u ( ds = u ( u ds. n r n r r=ε r r r r This holds for any small ε >. We now take the limit as ε. On {r = ε}, we have ( = r r r = 2 ε, 2 Hence ( u r=ε r ( r u r ds = ε r 2 u r ds uds + r=ε ε r=ε u = 4πū + 4πε r, where the bar denotes the average of the function on the sphere r = ε. et ε so that the last term equals 4πu(. This gives the integral formula Green s function. We investigate further the nature of the integral formulas. efinition 3.2. The Green s function G(x, x for the operator and the domain at the point x is a function defined for x such that ( G(x, x possesses continuous second derivatives and G = in except at the point x. (2 G = on. (3 The function G(x, x + 4π x x is finite at x and has continuous second derivatives everywhere and is harmonic at x. Theorem 3.2. If G(x, x is the Green s function, then the solution for the irichlet problem is given by the formula u(x = u(x G(x, x ds. n Proof. By the representation formula, we have ( u(x = u v n u n v ds, where v(x =. efine H(x := G(x, x 4π x x v(x. Then H is harmonic in by definition of the Green s function. Applying Green s second identity to u and H, we have = ( u H n u n H ds. Since G = H + v, adding the representation formula and the identity, we obtain ( u(x = u G n u n G ds. Since G vanishes on, the second term vanishes we we obtain the integral formula.

24 24 SHO SETO Symmetry of Green s function. Proposition 3.3. For any a, b, where is some bounded region, G(a, b = G(b, a. Proof. et u(x = G(x, a and v(x = G(x, b. et ε = B(a, ε B(b, ε. By Green s second identity, we have (u v v udx = ε + x a =ε x a =ε ( u v n v u ds n ( u v n v u n ds + x b =ε ( u v n v u ds. n Since the Green s function vanishes on and is harmonic in the deleted neighborhood, we have ( u v n v u ( ds + u v n n v u ds =. n x b =ε This holds for any ε. Now we focus on the a term. et r = x a, and on the sphere, we have so that n == r [( v r=ε 4πr + H n v ( n ( = v r=ε 4πr n + v ( n 4πr = v sin φdφdθ. 4π r=ε By mean value property, the last is ] 4πr + H r 2 sin φdφdθ + r 2 sin φdφdθ r=ε 2π π v lim ε 4πε 2 ε2 sin φdφdθ = v(a. ( H v n v H n r 2 sin φdφdθ and similarly, we have hence x b =ε ( u v n v u ds u(b n = v(a u(b = G(a, b G(b, a Half-space and sphere.

25 MATH 24B: INTROUCTION TO PES AN FOURIER SERIES Solutions on half-space. We will first determine the Green s function for a half-space in three dimensions. et := {x = (x, y, z R 3 z > }. efine a reflected point as x = (x, y, z. Then the Green s function for is given by G(x, x = 4π x x + 4π x x. We need to check the three properties of the Green s function. ( G is finite and differentiable except at x. Using polar coordinates centered at x, we see that G =. (2 By symmetry, we see that G = on. (3 Since G + 4π x x = 4π x x has no singularities, hence is harmonic at x. Now we use the Green s function to solve { u = for z > u(x, y, = h(x, y. Since the unit normal is (,,, we have G n = G z z=. Now G n = G z = ( z + z z= 4π x x z z 3 x x z= 3 = z 2π x x. 3 Therefore, we have the solution to the aplace equation on the half space as u(x, y, z = z h(x, ydxdy 2π R 2 ((x x 2 + (y y 2 + z Solutions on Sphere. We compute the Green s function on the ball = { x < a}. efine the reflected point of a ball by x = a2 x x 2. et x x = r and x x = r. We claim that the Green s function for the ball is for x, and G(x, x = 4πr + a 4π x r G(x, = 4π x + 4πa. For x, we check the three properties. The first and third properties are straightforward since x lies outside the circle, and we can use polar coordinates with different centers. For the second property, we notice that for x = a, we have r a x a x r = x x,

26 26 SHO SETO where r = x. Factoring out r a, we get r a r = r for any x = a, hence G = on the sphere x = a. Now we use the Green s function to give a formula for the solution of the irichlet problem in a ball in dimension 3, { u = in x < a u = h on x = a. First consider x. We need to calculate G on the boundary x = a. n Now r 2 = x x 2, hence 2r r = 2(x x so that r = x x and similarly for r. Also r computing, G = x x 4πr 3 a r x x 4π(r 3. Using the fact that x = a2 r 2 x and for x = a, r = a r r, we have G = ( x x 4πr 3 r2 a x + x 2 on the surface, hence G n = G x a = a2 r 2 4πar. 3 Inserting this in to the integral formula, we obtain u(x = a2 x 2 4πa In 2 dimensions, the Green function is given by x =a G(x, x = 2π log x x 2π log where the third property is replaced by 2π log x x. h(x x x 3 ds. ( x a x x, 4. General Eigenvalue Problem Throughout this section, we will use the 2 inner product and norm, (f, g = f(xg(xdx f 2 = (f, f. We now consider the irichlet eigenvalue problem { u + λu = u = in on.

27 MATH 24B: INTROUCTION TO PES AN FOURIER SERIES 27 where is an arbitrary domain (open in 3 dimensions, with smooth boundary. We denote the eigenvalues by < λ λ 2 λ 3 repeating up to multiplicity. Consider the minimum problem: { } w 2 m = min w = on, w. w 2 The quotient we minimize is called the Rayleigh quotient. Theorem 4.. Assume that u(x is a solution to the minimum problem. Then the value of the minimum equals the smallest eigenvalue λ of the irichlet problem and u(x is its eigenfunction, i.e. { } w 2 λ = min w = on, w, and u = λ w 2 u, in. Proof. Consider a test function w, which is any C 2 function w such that w = on and w. By assumption, we have m = u 2 w 2 u 2 w 2 for any test function w, where m is the minimum. et v be any test function and consider w := u + εv, ε is any constant. Then the function f(ε = (u + εv 2 = u 2 + 2ε( u, v + v 2 u + εv 2 u 2 + 2ε(u, v + v 2 has a minimum at ε =, hence f ( =. Computing, we have Therefore, we have = f ( = 2 u 2 ( u, v 2 u 2 (u, v u 4. ( u, v = m(u, v. Integrating by parts, or by Green s first identity, we have ( u + mu, v = which holds for all test functions v, therefore m is an eigenvalue for with eigenfunction u. To show that m is the smallest eigenvalue, let v j = λ j v j where λ j is any other eigenvalue. Then m v j 2 v 2 = λ j. Theorem 4.2. Suppose that λ,... λ n are already known, with the eigenfunctions v,..., v n, respectively. Then { } w 2 λ n = min w, w = on, w C 2, = (w, v w 2 = (w, v 2 = = (w, v n, assuming that the minimum exists. Furthermore, the minimizing function is the nth eigenfunction v n.

28 28 SHO SETO Proof. et u denote the minimizing function, which we assume exists. Note that u = on and u is orthogonal to the previous eigenfunctions. et m be the minimum value. et w = u + εv where v satisfies the constraints in the minimizing problem. By the same reasoning as above, ( u + muvdv = and ( u + muv j dv = u( v j + mv j dv = (m λ j uv j dv =, since u is orthogonal to v j. We also used the fact that u and v j are vanishing on the boundary. et h be an arbitrary trial function, which is C 2, h = on, and h. By Gram-Schmidt process, we let v(x = h(x n k= (h, v k (v k, v k v k(x. This v is not orthogonal to all the v k. Since each function satisfies the constraint conditions, v also satisfies as well. Hence ( u + muhdv =. since h = v + c k v k, where c k are the coefficients in the Gram-Schmidt decomposition. Since this holds for any test function, we get that u + mu =. It remains to show that m = λ n. Suppose m < λ n, then it is attained by some u. However, by induction, λ n = v n 2 which was the minimizer of the Rayleigh quotient with less v n 2 constraints, hence u would be smaller, which is a contradiction. So we know λ n m. By the same reason as the n = case, we can show that m λ n. Since u must be linearly independent to v n, we have that m = λ n, note that its value can be equal to λ n. We can then prove that the first eigenvalue is simple (multiplicity one so that in fact, our eigenvalues can be ordered as < λ < λ 2 λ 3 First we prove a maximum principle in this setting. emma 4.. The first eigenfunction can be chosen to be nonnegative, u. Proof. et u + and u be the nonnegative and negative parts of u. By the minimal characterization, we know that λ = min{ u + 2 u +, u 2 2 u } u 2 = λ 2 u 2 This would give a contradiction unless one of u + or u is zero. We choose u =. Next we use a maximum principle to show that emma 4.2. The first eigenfunction is positive, i.e. u >, in the interior of.

29 MATH 24B: INTROUCTION TO PES AN FOURIER SERIES Minimax principle. We start with an approximation scheme where we let w,..., w n be n arbitrary test functions (C 2 functions that vanish on. efine the matrix A = (a ij and B = (b ij, with entries given by a ij = ( w i, w j = w i w j dv b ij = (w i, w j = w i w j dv. Consider the characteristic equation det(a λb =. Note that it is a polynomial equation in λ. enote the roots by First we show that λ λ n. λ n = max c Ac c Bc c. emma 4.3. For A a real n n symmetric matrix, the largest root λ n of det(a λi = is given by Ac c λ n = max c c c. Proof. From linear algebra, a n n real symmetric matrix has an orthonormal eigenbasis consisting of eigenvectors {v i }. Then, for any v R n, we can write as v = c v + + c n v n. We have Av v v v = A(c v + + c n v n v v v = c2 λ v c 2 nλ n v n 2 v 2 λ n v 2 v 2 = λ n. The maximum is attained when v = v n. Now for B positive definite and symmetric, we can decompose as B = b 2, with b positive definite and symmetric and so B = b b. Note that det(a λb = det(b Ab λ det(b = hence we apply the lemma to b Ab λi, hence λ n = max c b Ab c c. c c et c = bv, then λ n = max v Av v Bv v.

30 3 SHO SETO Returning to the minimax principle, we see that the matrix B defined above is positive definite since n x T Bx = (x i w i, x j w j = (x i w i (x j w j = ( x i w i 2 >. i j i so that i,j= This leads to the minimax principle λ n = max c ( c j w j 2 c i w i 2 Theorem 4.3. If w,..., w n is an arbitrary set of test functions and λ n is defined as above, then the nth eigenvalue is given by λ n = min λ n where the minimum is taken over all possible choices of n test functions w,..., w n. Proof. Fix any choice of n test functions w,..., w n. We can then choose constants c,... c n not all zero so that for eigenfunctions v,... v n and w := c j w j, (w, v k = n c j (w j, v k =. j= so that w is orthogonal to the first n eigenfunctions. Then λ n w 2 w 2 λ n. Since this holds for any choice of n test functions, we can take the minimum on both sides. For the opposite inequality, we simply let w i = v i to obtain the minimum. Using the minimax principle, we can give an inequality relation about eigenvalues in different domains: Theorem 4.4. et 2. Then λ n ( λ n ( 2, where λ n ( is the nth eigenvalue of the irichlet problem on. Proof. et w,..., w n be test functions and c i be coefficients such that for w = c j w j, λ n( = w 2 w 2, that is, the Rayleigh quotient is maximized among the fixed test functions. Since 2, we can extend these test functions to be test functions on 2 so that λ n( = λ n( 2 Taking the minimum over test functions of, we have λ n ( = min λ test functions of n( = min λ test functions of n( 2 min λ test functions of 2 n( 2 = λ n ( 2.

31 MATH 24B: INTROUCTION TO PES AN FOURIER SERIES 3 Now we prove that the eigenvalues increase: Corollary 4.. For R n, the eigenvalues of the irichlet problem, { u = λu on u = on. form an infinite sequence λ n such that λ n as n. Appendix A. Some inear Algebra Theorem A.. If A is an n n real symmetric matrix, then it is diagonalizable under an orthonormal basis of eigenvectors. Corollary A.. If A is positive definite and symmetric, then it can be decomposed into its square root matrices A = a 2 where a is positive definite and symmetric. Proof. Since A is positive definite and symmetric, we can write as A = P P T = (P /2 P T (P /2 P T = a 2 where P is a matrix of (normalized eigenvectors, hence P T P = I, is a diagonal matrix of eigenvalues and /2 is the square root of the entries, which is possible since A is positive definite hence each eigenvalue is positive.