Math 372 Syllabus and Lecture Schedule. Gautam Iyer, Spring 2012

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1 Math 37 Syllabus and ecture Schedule. Gautam Iyer, Spring ecture, Mon /6: Introduction and motivation. (Sec..) Introduction. A PE is a differential equation involving derivatives with respect to more than one variable. Ubiquitous in nature Heat equation: t u u = governs evolution of temperature in a conductor. Wave equation: t u u = governs propagation of waves. aplace equation: u =, satisfied by the stream function of an incompressible fluid. (Such functions are called Harmonic functions.) Steady states of the unforced heat/wave equations. Poisson equation: u = is the electrostatic potential distribution in a conductor with uniform charge density. Steady states of the forced heat/wave equation. No simple solution formula. inear OE s have explicit solutions. inear PE s usually do not. Elementary existence theorem for general OE s. Analogue for PE s involves the assumption of analyticity. Counter examples exist if this assumption is relaxed. (Sec..) Method of characteristics. Consider first a x u + b x u =, where a, b are functions of x, y. ( a b ) u = means the vector ( ab ) is tangential to level sets of u. ecture, Wed /8: Method of characteristics. Solve the OE dx a = dy b to find the level sets of u (called characteristic curves). By solving the above OE, find a function F so that F (x, y) = c describes all characteristic curves. u(x, y) = f(f (x, y)) for an arbitrary f is the general solution. E.g. general solution of y x u + x y u = is u = f(x + y ). Characteristics are circles with center the origin. For 3, same trick works. Consider the PE au x + bu y + cu z = Characteristics are given by the OE dx a = dy b = dz c. Solve these OE s, and write characteristics as in the form F (x, y, z) = c, and G(x, y, z) = c (where F, G are functions you should explicitly find), and c, c are constants. The general solution is of the form u(x, y, z) = f(f (x, y, z), G(x, y, z)), where f : R R is a general (differentiable) function. E.g. general solution of u x + u y + 3u z = is of the form u(x, y, z) = f(x y, 3x z), where f : R R is a general (differentiable) function. ecture 3, Fri /: erivation of PE s from physical principles. Solving first order, linear inhomogeneous equations by the method of characteristics. (See the handout for details.) (Sec..3) erivation of PE s from Physics. One dimensional heat equation. Heat is proportional to temperature. Rate of heat flow is proportional to the temperature gradient. Show d b dt a θ(x, t)cρ dx = α b a xθ(x, t) dx. ecture 4, Mon /3. Conclude t θ = κ xθ. Higher dimensional heat equation State the ivergence theorem. Above reasoning shows t θ = κ θ. One dimensional transport equation. u concentration of a pollutant in a () flowing pipe. c velocity of water flow (constant). Show d b dt a u(x, t) dx = cu(a, t) cu(b, t) = b a xu(x, t) dx. ecture 5, Wed /5. Conclude t u + c x u =. MOC: u(x, t) = f(x ct). One dimensional wave equation. u displacement of a particle from it s mean position. T Tension in the string Conclude b a ρ t u(x, t) dx = T (u x (b, t) u x (a, t)), and hence t u = c xu, for c = ρ T. Higher dimensional wave equation: t u u =. ecture 6, Fri /7: Boundary conditions. (Sec..4) Auxiliary condition: Specify u(x, y) along some curve. E.g. Solve x u + y y u = u, subject to the condition u(, y) = f(y). Characteristics: y = c e x. General solution u(x, y) = /(x + g(ye x )). Auxiliary condition gives g(y) = /f(y), and substitute back. Boundary conditions. irichlet: Specify u on the boundary of the domain.

2 Heat equation: u = on corresponds to holding the temperature of the boundary of the conductor constant (e.g. by immersing it in a bath of constant temperature). Wave equation: u = on the boundary corresponds to holding the endpoints of a guitar string fixed. Poisson equation: Specifying u on the boundary corresponds to specifying the voltage at the boundary. Neumann: Specify u ˆn (the normal derivative) on the boundary of the domain. Specifying the full derivative on is too much. eads to inconsistencies (proof later). Heat equation: u n = on corresponds to perfectly insulated boundaries (no heat exchanged with the surroundings). u Wave equation: n = on corresponds to no stress on the boundary. E.g. allowing one end of the string to move freely in a track perpendicular to the string. Poisson equation: Specifying u n on corresponds to specifying the current at the boundary. Initial conditions: Specify conditions at time. Heat equation: Enough to specify u(x, ) (initial temperature). Wave equation: Must specify both u(x, ) (initial position), and t u(x, ) (initial velocity). ecture 7, Mon /3: One dimensional wave equation. (Sec..) Wave equation on the line Say t u c xu =, for x R, t >. et v = ( t + c x )u. Then ( t c x )v =. M.O.C shows v(x, t) = h(x + ct). By M.O.C, deduce u(x, t) = f(x ct) + g(x + ct). Alembert s principle: If u(x, ) = ϕ(x) and t u(x, ) = ψ(x), then u(x, t) = (ϕ(x + ct) + ϕ(x ct)) + x+ct c x ct ψ. ecture 8, Wed /: Conservation of Energy and Causality. (Sec..) Causality for the wave equation. omain of dependence is the inverted cone, bounded by particles travelling with speed ±c. omain of influence is the upright cone, bounded by particles travelling with speed ±c. No information propagates faster than the speed c. (Sec..) Conservation of energy. The energy E = ( tu) + c ( x u) is constant in time. Uniqueness for initial value problem. Say u, u are solutions to the forced wave equation t u c xu = f, with the same initial position and initial velocity, and decay sufficiently at. Set v = u u. Then v is a solution to t v c xv =. Conservation of energy implies ( tv) + ( x v) is constant in time, and hence must be (the energy at time ). Thus x v = t v =, and so v is constant in both space and time. Since v = at time, v = for all time. Consequently u = u as desired. ecture 9, Fri /3: Maximum principle. (Sec..3) Heat equation. Physical meaning: Heat does not collect at hot points. emma: et R = (, ) (, T ). If t v κ xv <, then v does not attain a maximum in the interior of R, or on the top of R. At an interior maximum, t v = and xv, which gives a contradiction. (Weak) Maximum principle: If t u κ xu, then v attains a maximum on the sides or bottom of R. (The maximum, however, can also be attained at interior points of R or on the top of R.) Proof: et v(x, t) = u(x, t) + εx. Verify t v κ xv <, apply the lemma, and send ε. Note: The strong Maximum principle says that if v attains it s maximum at an interior point (x, t ), then v must be constant up to time t. The proof is much harder, however, is still accessible to you! ecture, Mon /6. Maximum principle in higher dimensions. (Complete statement and proof on your HW.) Uniqueness for the irichlet problem, initial value problem: If u, u are solutions to the forced heat equation t u κ u = in, with initial conditions u(x, ) = ϕ(x), and irichlet boundary conditions u(x, t) = g(x, t) for x. Then u = u. Maximum principle proof: Set v = u u. By the maximum principle, both the maximum and minimum of v are attained either at t =, or when x. By the given initial and boundary conditions, v = both when t = and x. Consequently, by the maximum principle, v = identically. ecture, Wed /8. Energy decay for the heat equation. et u satisfy the heat equation with Neumann (or irichlet) B.C. et E(t) = u(x, t) dx. Then d dte. de case: dt = u tu dx = κ u xu = κ ( xu).

3 3 Higher dimensional case: Observe first u u dv = u u dv = [ (u u) u ] dv = u dv + u u n ds. The last boundary integral is by the given boundary conditions. Consequently, de dt = κ u dv. (Sec..4) Heat equation on the line. et δ be a point source of heat, of strength, located at the origin. Suppose u solves the heat equation with initial data δ. Then v(x, t) = u(αx, α t) also solves the heat equation, with initial data a point source of heat. R v(x, t) dx = α u(x, t) dx. So the initial data for v should be a point source R with strength /α. ecture, Fri /. Consequently, w(x, t) = αu(αx, α t) is a solution to the heat equation with initial data δ. Consequently w = u. Consequently, u(x, t) = t u( x t, ). et r = x/ t, and f(r) = u(r, ). Then u(x, t) = / tf(x/ t). Heat equation reduces to the OE f + rf + f =. Solve this: (f + rf) = f + rf + f =. So f + rf = c. e r / is an integrating factor: f = e r / r c / ds + c es e r /. Since f at ±, must have c =. u(x, ) dx = f(r) dr = = c = π. Thus u(x, t) = G(x, t) = πt e x /t is the solution of the heat equation on the line, with κ = and initial data δ (a point source of strength, located at the origin). ecture 3, Mon /3. Translating we see u(x, t) = G(x y, t) is the solution to the heat equation with initial data δ y (a point source of strength located at y). Approximate f by δ yi f(y i )(y i+ y i ), and see that u(x, t) = f(y)g(x y, t) is the desired solution with initial data f. Check that this works: t xu = f(y)( t x)g(x y, t) dy =. Checking u(x, ) = f(x) is harder (will return to that next week). Rescaling time gives u(x, t) = f(y)g(x y, κt) dy to be the solution to t u κ xu = with initial data f. Formula above shows that for any t >, u is infinitely differentiable in both x and t, regardless of how differentiable f was. Infinite speed of propagation. omain of dependence of the point (x, t) is (, ). ecture 4, Wed /5. efine the error function erf(x) = π x e y dy. Write solutions in terms of the erf. E.g. f(x) = for x, and f(x) = for x >. u(x, t) = ( erf(x/ 4κt) solves the heat equation t u κ xu = with u(x, ) = f(x) (for all x ). Strong maximum principle: If f is continuous and not identically constant, u(x, t) < max f for any t >. Proof: u(x, t) = f(y)g(x y, κt) dy < (max f)g(x y, κt) dy = max f. (Crucially uses the fact that G(x, t) dx = and G(x, t) >.) ecture 5, Fri /7: Midterm. In class, closed book. Covers everything up to ecture (Sec..3). ecture 6, Mon /. (Sec..4) Comparison between the heat and wave equation. Initial data: Heat equation requires u(x, ), wave requires u(x, ) and t u(x, ). Smoothness of solutions: For t >, solutions to the heat equation are infinitely differentiable, no matter how many times differentiable (or not differentiable) the initial data is. Solutions to the wave equation are only as differentiable as the initial data. Time reversibility: If u is a solution of the wave equation, then v(x, t) = u(x, T t) is also a solution (i.e. reversing time, gets back a solution to the wave equation). For the heat equation, reversing time gets the backward heat equation which forces heat to collect at hot points. (Sec. 3.3 & 3.4) uhamel s principle OE version: ẏ Ay = g(t), with y() = x. et S(x, t) = e At x be the solution operator of the homogeneous equation ẏ = Ay, with y() = x. The solution to ẏ Ay = g(t) is S(x, t) + t S(g(s), t s) ds. ecture 7, Wed /. Heat equation: t u κ xu = g, u(x, ) = f(x). The solution operator S(, t) takes functions as the first argument, and outputs functions. et S(f, t)(x) = f(y)g(x y, κt) dy. Then u(x, t) = S(f, t)(x) + t S(g(, s), t s) ds is the desired solution. Explicitly u(x, t) = f(y)g(x y, κt) dy + t g(y, s)g(x y, t s) dy ds. nd Order OE: ÿ Ay = h(t), with y() = a and ẏ() = b. et S(x, t) solve t S AS =, with S(x, ) = and t S(x, ) = x. Then y(t) = t S(a, t) + S(b, t) + t S(g(s), t s) ds is the desired solution.

4 4 ecture 8, Fri /4. Wave equation: t u c xu = h, u(x, ) = f(x), t u(x, ) = g(x). et S(ψ, t)(x) = x+ct c ψ(y) dy. x ct Then u(x, t) = t S(ϕ, t)(x) + S(ψ, t)(x) + t Explicitly, u(x, t) = [ϕ(x + ct) + ϕ(x ct)] + c is the domain of dependence of (x, t). (Sec. 3.5) Continuity of the heat equation at t =. S(g(, s), t s) ds. x+ct x ct ψ + c G is an approximate identity. Namely, G has the following properties: G(x, t) G(x, t) dx = For any δ >, lim G(x, t) dx. t + x δ ecture 9, Mon /7. h, where If f is bounded, and continuous at x, then lim u(x, t) = f(x). t Pick any ε >. Then δ > such that whenever z < δ, we have f(z) f(x) < ε. u(x, t) f(x) = G(y, t) [f(x y) f(x)] dy = y <δ ( ) + y δ ( ). First term: y <δ ( ) ε G(y, t) dy ε y <δ Second term: y <δ ( ) (max f) G(y, t) dy as t. y <δ So for t small, can make u(x, t) f(x) < ε. QE. (Sec. 4.) Separation of variables. Wave Equation, irichlet B.C. u(x, t) = X(x)T (t). Then X X = T c T = λ. Hence X(x) = α sin( λx) + β cos( λx). X() = X() = = λ = n π. Solve for T : T (t) = A sin( ct) + B cos( ct). ecture, Wed /9. X n = sin( x), T n = A n cos( ct) + B n sin( ct), u n(x, t) = X n (x)t n (t). Frequency of note heard is nc. All frequencies heard are multiples of c/! u(x, t) = X nt n is a solution to the wave equation, with u(x, ) = A n sin( x), and c tu(x, ) = B n sin( x). Heat Equation, irichet B.C. If u(x, t) = X(x)T (t) is a separated solution, then X X = T c T = λ. As before, λ = n π, X(x) = sin( x). Solve for T : T (t) = A n e n π κt u(x, t) = X n(x)t n (t) = A ne n κt sin( x), solves the heat equation with irichlet B.C. and I.. u(x, ) = A n sin( x). π Positivity of λ: If X = λx, with X() = X() =, then λ = ecture, Fri 3/. (X ) X >. Heat equation, Neumann B.C.: If u(x, t) = X(x)T (t) is a separated solution, then X X = T c T = λ. As before, X(x) = α cos( λx) + β sin( λx). B.C. = X () = X () = = λ = with X(x) = α, or λ = n π with X(x) = α cos( x). Solve for T : T (t) = A n e n π κt π κt cos( x), solves the heat x). u(x, t) = X n(x)t n (t) = A + A ne n equation with Neumann B.C. and I.. u(x, ) = A + A n cos( Wave Equation, Neumann B.C.: Similar. (Sec. 5.) Fourier Series Goal: Find the coefficients in the Fourier Sine and Cosine series. emma (Symmetry): f() = g() = and f() = g() = implies f, g = f, g. Here h, h = h (x)h (x) dx. Proof: Integrate by parts twice. (Sec. 5.3) ecture, Mon 3/5. emma (Orthogonality): If X n () = X n () =, X n = λ n X n, then X m, X n = whenever λ m λ n. Proof: λ m X m, X n = X m, X n = X m, X n = λ n X m, X n. Consequently (λ m λ n ) X m, X n =. (Sec. 5.3) Fourier Sine series. et X n = sin( x), λ n = n π. If f = B nx n, then B n = f, X n X n, X n = f(x) sin( x) dx. Proof: f, X m = n= B n X n, X m = B m X n, X n, by the lemma. Fourier Cosine Series. et X n = cos( x), and X =. Verify the Symmetry and Orthogonality lemmas for functions with Neumann Boundary conditions (i.e. X () = X () = ). If f(x) = A X + A nx n, then A Consequently, A n = = f,x X,X, and A n = f,xn f(x) cos( x) dx, for n =,,,.... X n,x n. ecture 3, Wed 3/7: igression The Black Scholes formula.

5 5 Application of the Heat Equation to Option Pricing (by Kasper arsen). ecture 4, Mon 3/9. (Sec. 5.) Full Fourier series. Periodic boundary conditions: f(x + ) = f(x) for all x. Write f(x) = A + A n cos( x) + B n sin( x). The symmetry and orthogonality lemmas give mπ sin( x) sin( x) dx = cos( mπ x) cos( Explicitly compute sin( x) cos( x) dx =. Get A n = f(x) cos( x) dx and B n = (Sec. 5.) Complex Fourier series. Suppose f is periodic and complex valued. mπ sin( x) dx =. et e n (x) = exp(i x), and want f(x) = c ne n (x). f(x) sin( x) dx. x) cos( x) dx = efine g, h = g(x)h(x) dx. irectly check e n, e m = if n m. (This also follows from the symmetry, orthogonality lemmas). ecture 5, Wed 3/. Conclude c n = f, e n / e n, e n = f(x) exp( i x) dx. (Sec. 5.3/5.4) Convergence of Fourier series Pointwise, uniform and convergence. Uniform convergence implies pointwise convergence, but not conversely. Uniform convergence on a finite interval implies convergence, but not conversely. Pointwise need not imply convergence; convergence need not imply pointwise convergence. f n (x) = if x (n, n + ) and otherwise. Then (f n ) pointwise, but not uniformly or in. ecture 6, Fri 3/3: Midterm. In class, closed book. Covers everything from ecture to ecture. ecture 7, Mon 3/6. et f n (x) = / k if k n < k+, x [ n k, n+ k ), and f n (x) = otherwise. Then (f n ) in [, ], but not pointwise or uniformly. Proof that uniform convergence on a finite interval implies convergence. Pointwise, uniform and convergence of series of functions. These are defined as the respective convergence of the partial sums. Bessel s inequality: A n X n f. (Here f = f(x) dx, and X n are an orthogonal system, and A n = f,xn X n,x n.) Pythagoras theorem: If f, g =, then f + g = f + g. Consequently, S N f = N A n X n. (Here S N f = N A nx n ) ecture 8, Wed 3/8. f S N f, S N f = (since f S N f, X n = for all n N). Since f = (f S N f)+s N f, the above three bullets imply Bessel s inequality. def Amongst all functions of the form P N = N b nx N, S N f is the one that best approximates f in the norm Proof: et E N f = f S N f. Then P N, E N f =. Hence f P N = E N f + (S N f P N ). By above E N f, S N f P N =. By the Pythagoras theorem, f P N f S N. QE. Proposition: If P N is any sequence of functions of the form P N = N b n,n X n such that P N f in, then S N f f in. Proof: E N f f P N, and the RHS converges to. QE. Proposition: The series A nx n converges to f in if and only if f = A n X n (Parseval s identity.) Proof (reverse): Say f = A n X n. Then E N f = f S N f = f N A n X n. QE. ecture 9, Fri 3/3. Convergence of Cesàro sums. et S N f(x) = N N c ne i x, where c n = f(x)e i Show S N f(x) = f(y) N (x y) dy, where N (z) = Compute N (z) = sin( (N + ) π x) π sin ( x) N n= S Nf. efine σ N f = N Show σ N f(x) = f(y)k N (x y) dy, where K N (z) = N Can compute K N (z) = sin( N π N sin ( Can check x) π x) K N (z) dz =, K N, and x. N N [will be on Homework] lim N z >ε z [,] ei z. N N (z). K N (z) dz =, for any ε >. Hence if f is continuous at x, then lim N σ N f(x) = f(x) (on HW). Further, if f is continuous on [, ] then (σ N f) f uniformly on [, ]. Consequently, (σ N f) f on [, ], and from last time this implies (S N f) f in [, ].

6 6 ecture 3, Mon 4/. General convergence theorems. If f(x) dx <, then the Fourier Sine/Cosine series converges to f in. At any point x (, ) where f is continuous, then then Cesàro sums converge but the partial sums need not. Consequently, if f is continuous on [, ] and satisfies the boundary conditions, then then then Cesàro sums converge pointwise, converge but the partial sums need not. If f is differentiable at x (, ) then the Fourier Sine/Cosine series converges to f at x. Consequently, if f satisfies the boundary conditions, and is continuous and piecewise differentiable then the Fourier Sine/Cosine series converge pointwise to f. If further f (x) dx <, then the Fourier Sine/Cosine series converge uniformly. Analogous results hold for the full / complex Fourier series. Fourier coefficients of f. Suppose f is periodic, differentiable and f (x) dx <. et c n be the (complex) Fourier coefficients of f, and d n those of f. Then d n = i c n. Proof: ifferentiating term by term is NOT JUSTIFIABE! However, using the formula for d n and integrating by parts gives the desired relation. Sobolev embedding: If n s c n < for s >, then f is continuous! (If s > 3/, then f is differentiable, and f is continuous.) Proof: Try it yourself, if you konw the Cauchy-Schwartz inequality and the Weirstrauss M test. ecture 3, Wed 4/4. (Sec. 6.) aplace and poisson equation. aplace equation in upper half plane: t u + x =, for t > and x R. iffers from the wave equation by only a sign. aplace equation only reqires the initial position, OR the initial velocity. Wave equation requires both. aplace equation satisfies a maximum principle. Wave does not. Solutions to the aplace equation are smooth for any t >. Solutions to the wave equation are only as differentiable as the initial data. Wave equation has finite speed of propagation. The laplace equation does not. Motivation: Steady states of the heat equation. The equilibrium temperature in a conductor satisfies u = f, where f is the sources / sinks of heat. irichlet boundary conditions correspond to holding the temperature at the boundary constant. Neumann boundary conditions correspond to insulating the conductor. Electrodynamics: Electric electric potential. Maxwell s equations reduce to u = ρ, where u is the electric potential and ρ is proportional to the charge density. irichlet boundary conditions correspond to specifying the voltage at the boundary. Neumann boundary conditions correspond to specifying the current. Harmonic functions are functions that satisfy u =. The charge density in a perfect conductor is, so the electric potential is harmonic. Equilibrium temperature in a conductor (in the absence of sources / sinks) is also a harmonic function. Uniqueness: Suppose u, u are solutions of u = f in Ω with u = g on Ω. Then u = u. Proof: et v = u u. Then v = in Ω, and v = on Ω. By the divergence theorem Ω v v = Ω v v ˆn + Ω v. Consequently Ω v =, forcing v to be a constant. Since v = on Ω, we must have v = identically. ecture 3, Fri 4/6. aplacian in polar coordinates. Put x = r cos θ, y = r sin θ. Get r = x + y and θ = tan (y/x). Compute x r = cos θ, y r = sin θ, x θ = r sin θ, and yθ = r cos θ. Compute x u = cos θ r u r sin θ θu, and y u = sin θ r u + r cos θ θu Compute u = r u + r ru + r θ u. (Sec. 6.3) aplace equation in a disk. et be a disc with center and radius a. et u = in, with u = f on. Switch to polar coordinates and separate variables. et u(r, θ) = R(r)T (θ). Get r R +rr R = T T = λ Periodic boundary conditions on T gives T (θ) = cos(nθ), T (θ) = sin(θ), or T (θ) =, and λ = n for n {,,... }. ecture 33, Mon 4/9. Try R(r) = r α as a solution. Get α(α ) + α = n, and hence α = ±n. Reject α = n as X(r) = r n blows up at r =.

7 7 Thus separated solutions are of the form r n cos(nθ), r n sin(nθ) and constants. A n cos(nθ) + B n sin(nθ), If the full Fourier series of f is f(θ) = A + then u(r, θ) = A + A n rn r a cos(nθ) + B n n n Harmonic functions in a disk. Use complex notation. (x, y) = z = re iθ. Write f(θ) = c ne inθ. a n sin(nθ). Then u(x, y) = c n rn a e inθ. n Since c n = π π π f(φ)e inφ dφ, we have u(r, θ) = π where P (r, θ) = π Compute P (r, θ) = π [ + rn a n (e inθ + e inθ ) ]. a r a +r ar cos(θ) As r a, P behaves like an approximate identity: π P (r, θ) dθ = (Proof: P (r, θ) = π π ( + ( r a )n cos(nθ)).) P (r, θ) > for r < a. (Proof: cos θ, so P (r, θ) a r For any ε >, lim P (r, θ) =. (Proof: on HW). r a θ >ε ecture 34, Wed 4/. lim r a P (r, θ) = if θ, and otherwise. Proof: P (r, θ) = π Poissons formula: a r (a r) +ar( cos θ). π P (r, θ φ) f(θ) dθ, π(a+r) >.) If u = in some domain, and a disk with center x, and radius a is completely contained inside, then u(x) = a x x u(y) πa y x =a dy x y whenever x x < a. Mean value property: u =, then u(x) = πr u(y) dy. [Note, the RHS y x =r is a line integral over the circle with center x and radius a.] Proof: use the Poisson formula. Strong maximum principle: If u = in, then u attains it s maximum (and minimum) only on, unless u is constant. Proof: Use the mean value property. ecture 35, Fri 4/3. Weak Maximum principle including convection terms. et u = u + b u, with b bounded. Suppose u in a (bounded) domain, and u is continuous up to the boundary of. Then u attains it s maximum on (i.e., max u max u). emma: Suppose first v <. Then v has no interior maximum in. Proof: At an interior maximum x, u = and u, contradicting u <. Now let v = u + εe λx, for λ to be chosen later. Compute v = u + ε ( ) λ e λx + λb e λx. Know u. Choosing λ < max b will guarantee the second term is stricitly negative. So v <, and by the emma max v max v. As before, max u max v max v max u + εeλ, and send ε. ecture 36, Mon 4/6. (Sec. 7.) Greens Identities Greens first identity: R 3 bounded. u v + u v = u v n. Proof: ivergence theorem applied to (u v) = u v + u v. irichlet s principle: The function minimising the energy E(u) def = u subject to u = f on, is the harmonic function with boundary values f. Proof: et v be any function which is on. Must have d dεe(u + εv) = when ε =. Greens identity gives v u =. Since this is true for any v, must have u =. The above shows that if u minimises E, then we must have u =. We need to also check the converse: Namely if u =, then u minimises E. Proof: et u, v = f on, and suppose u = in. Set w = u v; then v = u w and w = on. Green s identity implies ( u) ( w) = w u ˆn w u =. Thus E(v) = u + w ( u) ( w) = E(u) + E(w) E(u). ecture 37, Wed 4/8. The Rayleigh quotient: et E(u) = ( u )/ u. functions u which are on. et ε R, and v be any function which is on. Suppose λ = min E, and is attained for the function ϕ. Minimise E, over all Green s identity shows d dε (ϕ + εv) ε= = iff v( ϕ λϕ) =. Eigenfunctions of the aplacian are candidates for the minimiser of E. et ϕ solve ϕ = λϕ, with ϕ = on, and ϕ > in. This is called the Principal eigenfunction. It s existence (positivity) is equivalent to the maximum principle! Claim: ϕ minimises E. Proof: et ε >, and u be any function with u = on. Compute λ u ϕ ϕ+ε = u ϕ+ε ϕ u (Greens identity, and completing the square). Send ε. QE.

8 8 ecture 38, Mon 4/3. (Sec. 7.) Greens second identity: u v ˆn = u v v u. ivergence theorem applied to (u v v u) = u v v u. Mean value property: If u = in B R, a 3 ball of radius R and center x. Then u(x ) = 4πR B R u(x) dx (the RHS is a surface integral). Proof: Without loss assume x =. et v(x) = x. Know (from HW) v = for x. For ε >, let ε = B R B ε = {x R 3 ε < x < R}. Greens identity implies ε u v ˆn =. ε ( ) = B R ( ) + B ε ( ). (The normal derivative points radially inward on B ε, and radially outward on B R.) Compute B R u v ˆn = R B R u(x) dx. Similarly B ε u v u ˆn v ˆn = ε B ε u(x) dx ε 4π u(), since u is continuous at. Since B R ( ) = B ε ( ), we get u() = 4πR B R u(x) dx. QE. ecture 39, Wed 4/5. Representation formula: If R 3, and u = in. Then u(x ) = [u(x) N ˆn (x x ) N(x x ) u (x)] dx, ˆn where N(x) = 4π x is the Newton potential. (Recall N = when x ). Remark: For a domain, the same formula is true with N(x) = π ln x. Proof: Without loss, x =. et ε = {x x > ε}, and B ε = {x x < ε}. Greens identity implies ε u N ˆn N u ˆn =. ε ( ) = ( ) + B ε ( ), where the normal derivative points radially inward on B ε, and outward to on. ( ) is the RHS we want. B ε ( ) ε u(x ), exactly as before. QE Corollary: Another proof of the Mean value property. Proof: Put = {x x x = R, in the representation formula. ecture 4, Fri 4/7. Physical intuition behind the Newton potential. N is the steady temperature obtained from a point sink of heat located at. Symmetry forces N(x) = f( x ). Computing heat flux through B R gives B R N ˆn =. Consequently 4πf (R) = for all R, forcing N(x) = 4π x, as we had. (Sec. 7.3) Greens functions. G(x, x ) is the greens function of a domain R 3 if it has the following properties: For x x, all second order partials of G (w.r.t. x) are continuous and G(x, x ) =. G(x, x ) = on H(x) = G(x, x ) N(x x ), for x x extends to a continuous harmonic function in. If u = in and u = f on, then u(x ) = G f(x) Proof: Know u(x ) = N u(x) ˆn (x x ) N(x x ) u ˆn. ˆn (x, x ). By greens identity, u H ˆn H u ˆn =. Since G(x, x ) = H(x) + N(x x ), adding the above two identites (and using G(x, x ) = for x ) finishes the proof. Symmetry of Greens functions: G(a, b) = G(b, a). Proof: Put u(x) = G(x, a) and v(x) = G(x, b). et ε = {x x a > ε & x b > ε}. Greens identity: ε u v ˆn =. ε = + B(a,ε) +, with the usual convention about normals. B(b,ε) Since u = v = on, ( ) =. Claim: B(a,ε) u v ˆn = v(a), and B(b,ε) u v ˆn = u(b). Claim finishes the proof, since v(a) = u(b) G(a, b) = G(b, a). ecture 4, Mon 4/3. Proof of claim. In B(a, ε), v is harmonic and u(x) = H(x) + N(x a) for some harmonic function H. Greens identity implies B(a,ε) H v ˆn v H ˆn =. Representation formula implies v N(x a) B(a,ε) ˆn v N ˆn (x a) = v(a). (The sign is reversed, since the normal vector is inward pointing). Adding gives B(a,ε) u v ˆn = v(a). QE. Physical interpretation: Steady temperature at a caused by a point sink at b is the same as a steady temperature at b caused by a point sink at a. ecture 4, Wed 5/. Greens function in Half-Space. = {(x, y, z) z > }. G(x, x ) = N(x x ) N(x x ), where x is the image of the point x reflected about the x-y plane.

9 9 Consequently the solution to u = in with u = f on is given by u(x) = f(y)g(x, y) dy = x 3 f(y, y, ) dy dy π [(x y ) + (x y ) + x. 3 ]3/ Greens function in a shpere = {x R 3 x < a}. et x, and x = a x x. Then G(x, x ) = N(x x ) + N( x a (x x )). Basic congruent triangles argument shows G(x, x ) = when x. Compute G ˆn = a x 4πa x x 3. 3 Poisson formula. If u = in and u = f on, then u(x ) = a x f(x) dx. (Surface integral over the sphere of radius a) x 4πa x x 3