15 Solving the Laplace equation by Fourier method

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1 5 Solving the Laplace equation by Fouie method I aleady intoduced two o thee dimensional heat equation, when I deived it, ecall that it taes the fom u t = α 2 u + F, (5.) whee u: [0, ) D R, D R is the domain in which we conside the equation, α 2 is the diffusivity coefficient, F : [0, ) D R is the function that descibes the souces (F > 0) o sins (F < 0) of themal enegy, and is the Laplace opeato, which in Catesian coodinates taes the fom u = u xx + u yy, D R 2, o u = u xx + u yy + x zz, D R 3, if the pocesses ae studied in thee dimensional space. Of couse we need also the bounday conditions on D and the initial conditions in D. In a simila vein it can be poved that the wave equation in two o thee dimensions can be witten as u tt = c 2 u + F, (5.2) whee now c is the wave velocity, and F is an extenal foce. We also will need bounday and initial conditions. Vey often the pocesses descibed by the heat o wave equation appoach some equilibium if t. This means that the solution does not change with time and in paticula u t o u tt tend to zeo as t. Theefoe equations (5.) and (5.2) tun into u = f, (5.3) whee f = F/α 2 fo the heat equation and f = F/c 2 fo the wave equation. Equation (5.3) is called Poisson equation, and, in case if f = 0, u = 0, (5.4) Laplace equation, one of the most impotant equations in mathematics. Since I am taling about the equilibium (stationay) poblems (5.3) and (5.4) only bounday conditions ae elevant, in the equilibium state the system fogets about the initial conditions (it can be igoously poved that initial value poblem fo eithe Poisson o Laplace equations is ill posed). In the following I will use the sepaation of vaiables to solve the Laplace equation (5.4), we will loo into popeties of (5.3) in the fothcoming lectues. Only fo some special plane geometies of the domain D it is possible to use the sepaation of vaiables. Fist of all, in Catesian coodinates, these ae vaious ectangles, I will leave this case fo homewo poblems (see the textboo). It is also possible to use sepaation of vaiables in cicula - based domains, such as inteio of the dis, exteio of the dis, secto, annulus, and pat of an annulus. To do this I fist need to ewite the Laplace opeato in pola coodinates. Recall that Catesian coodinates (x, y) and pola coodinates (, θ) ae connected as x = cos θ, y = sin θ, Math 483/683: Patial Diffeential Equations by Atem Novozhilov atem.novozhilov@ndsu.edu. Sping 206

2 o 2 = x 2 + y 2, tan θ = y x. I have ( u(x, y) = u( cos θ, sin θ) = u(, θ) = v x 2 + y 2, actan x) y. Fo the following I will need x x = = cos θ, x 2 + y2 y y = = sin θ, x 2 + y2 y θ x = + (y/x) 2 x 2 = y x 2 + y 2 = sin θ, θ y = + (y/x) 2 x = x x 2 + y 2 = cos θ. Now I stat calculating the patial deivatives using the usual chain ule ( u x = v x + v θ θ x = v cos θ + v θ sin θ ), cos θ u y = v y + v θ θ y = v sin θ + v θ, u xx = (u x ) x + (u x ) θ θ y ( sin θ = v cos θ v θ + sin θ ) ( 2 v θ cos θ + v θ cos θ v sin θ sin θ v θθ cos θ ) ( v θ sin θ ), u yy = (u y ) y + (u y ) θ θ y ( cos θ = v sin θ + v θ sin θ ) ( 2 v θ sin θ + v θ sin θ + v cos θ + cos θ v θθ sin θ ) cos θ v θ. Now I add two last lines to find the Laplace opeato in pola coodinates (eplacing v with u) u = u + u + 2 u θθ. I note that in cylindical coodinated x = cos θ, y = sin θ, x the Laplace opeato is wheeas in the spheical coodinates u = u + u + 2 u θθ + u zz = 0, x = sin φ cos θ, y = sin φ sin θ, z = cos φ, the Laplace opeato is u = u + 2 u + 2 u φφ + cot θ 2 u φ + 2 sin 2 φ u θθ. 2

3 Example 5.. To show how the sepaation of vaiables wo fo the Laplace equation in pola coodinates, conside the following bounday value poblem u = 0, u(, θ) = g (θ), u( 2, θ) = g 2 (θ), that is conside the poblem inside the annulus < < 2, and on both boundaies Type I nonhomogeneous bounday conditions ae given. I stat with a usual assumption that u(, θ) = R()Θ(θ). Since my equation in pola coodinates taes the fom u + u + 2 u θθ = 0, I get, denoting with the pime the coesponding deivatives, o 2 R Θ + R Θ + RΘ = 0, 2 R + R = Θ R Θ. Since the left hand side depends only on and the ight hand side depends only on θ hence both sides must be equal to a constant, which I will denote λ. Using this constant I end up with two ODE and 2 R + R λr = 0, (5.5) Θ + λθ = 0. (5.6) At this point I must add some bounday conditions to one of these poblems so that in the end I get a Stum Liouville poblem, whose eigenfunctions I can use as building blocs fo my genealized Fouie seies. The oiginal bounday conditions fo u ae of no help hee since they ae non-homogeneous. Thee should be something else to the poblem. And indeed, afte some though, it is possible to guess that my solution must be peiodic function (solution must be continuously diffeentiable), which implies that u(, θ ) = u(, θ + ), u θ (, θ ) = u θ (, θ + ). This implies that my second equation (5.6) must be supplemented with Θ() = Θ(), Θ () = Θ(). Now we now that these peiodic bounday conditions plus (5.6) is an eigenvalue Stum Liouville poblem with the eigenvalues λ = 2, = 0,,... and eigenfunctions Θ (θ) = A cos θ + B sin θ. Now I can etun to (5.5), which can be witten as 2 R + R 2 R = 0. 3

4 I stat with the case = 0. Then the equation can be solved by substitution R () = S(): which finally gives me 2 R + R = 0 2 S + S = 0 = S() = B, R 0 () = a 0 + b 0 log. (I do not use the absolute value since 0.) If =, 2,... I have the so-called Cauchy Eule diffeential equation, which can be solved by the ansatz R() = µ. I get µ(µ ) + µ 2 = 0 = µ,2 = ±, and hence the geneal solution is given by R () = c + d, =, 2,... What I did is I poved that any function of the fom u (, θ) = R ()Θ (θ) solve the Laplace equation u = 0 (such functions ae called hamonic) and satisfies the peiodic bounday conditions. Since the Laplace equation is linea, I will use the pinciple of supeposition to ague that the function [ ] u(, θ) = A + B log + (C + D ) cos θ + (E + G ) sin θ solves the Laplace equation and satisfies the peiodic conditions. It seems that I have a lot of abitay constants to detemine fom the emaining two bounday conditions, but caeful analysis shows that I have enough. To wit, let my bounday conditions have the following Fouie seies (notice that I do not divide by 2 the fist coefficient) g (θ) = a () 0 + g 2 (θ) = a (2) 0 + a () a (2) cos θ + b() sin θ, cos θ + b(2) sin θ. Now, compaing these seies with the solution in the fom of the seies and invoing the bounday conditions, I get { A + B log = a () 0 = 2 g (θ) dθ, A + B log 2 = a (2) 0 = 2 g 2(θ) dθ, { C + D = a () = g (θ) cos θ dθ, C 2 + D 2 = a (2) = g =, 2,... 2(θ) cos θ dθ, { E + G = b () = E 2 + G 2 = b (2) = g (θ) sin θ dθ, g 2(θ) sin θ dθ, =, 2,... 4

5 and each system fo each is a system of two equations with two unnowns, which can be always (except fo some degeneate cases) solved. Fo example, assuming that g (θ) = a, g 2 (θ) = b implies B = b a log 2 /, A = a (b a) log log 2 /, and all othe constants ae zeo. Hence the solution is u(, θ) = A + B log. If I assume that g (θ) = a cos θ, g 2 (θ) = b cos θ then I end up with the system C + D = a, C 2 + D 2 = b, which is easy to solve. The solution to the bounday value poblem fo the Laplace equation hence u(, θ) = (C + D ) cos θ. Example 5.2 (Inteio Diichlet poblem fo the Laplace equation and Poisson fomula). Conside now the poblem u = 0, 0 < u(, θ) = g(θ), 0 θ < 2. To solve it I will do exactly the same steps as in the pevious example (assume that the solution can be pesented as a poduct, get two ODE, use the peiodic bounday conditions on θ, end up with the same eigenvalues and eigenfunctions, solve the ODE fo R) fist. Then I note that a significant pat of the solutions to the ODE fo R has no meaning since I am dealing also with the point = 0 and hence neithe log no mae sence at this point. Since these solutions have no physical meaning I dop them to end up with the function u(, θ) = a 0 a cos θ + b sin θ. By using the given type I o Diichlet bounday condition I immediately find that (note that I conveniently assumed that dis has adius, mae sue that you can solve the case fo an abitay adius) a = ˆ g(θ) cos θ dθ, b = ˆ g(θ) sin θ dθ. I solved my poblem, but it tuns out that I can ewite this solution in a closed neat fom. Intechanging the integals and sums in my solution I get u(, θ) = ˆ ( ) g(ϕ) (cos ϕ cos θ + sin ϕ sin θ) dϕ = ˆ ( ) g(ϕ) cos (θ ϕ) dϕ. 5

6 Now ( ) cos θ = Re z ( = Re 2 + z ) ( ) + z = Re z 2( z) ( ) ( ( + z)( z) z 2 ) + z z = Re 2 z 2 = Re 2 z 2 Theefoe, finally, I can conclude that = u(, θ) = 2 2 2( cos θ). ˆ which is called the Poisson integal fomula, and the expession 2 g(ϕ) + 2 dϕ, (5.7) 2 cos(θ ϕ) 2 2( cos(θ ϕ)) is called the Poisson enel. To emphasize, the Poisson integal fomula gives a closed fom solution fo the Diichlet bounday poblem fo the Laplace equation in a dis. Thee ae seveal immediate consequences of the Poisson fomula. The value of the solution at the cente of the dis is given by the aveage of its bounday value. u(0, θ) = 2 ˆ g(ϕ) dϕ. This is a paticula case of the following geneal fact: Let u(x, y) be hamonic (i.e., solves the Laplace equation) inside the dis of adius a with the cente (x 0, y 0 ), then u(x 0, y 0 ) = u ds, 2a whee γ is the bounday of the dis. This immediately follows fom the Poisson fomula afte shift and escaling. If u is a nonconstant hamonic function defined in D then it cannot achieve its local maximum o local minimum at any inteio point in D. This is tue since the aveage of a continuous eal function lies stictly between its minimal and maximal values, and hence, due to the pevious point, I cannot have a local minimum o local maximum at an inteio point. Immediately fom the pevious I have that if u is hamonic in D and if m and M ae minimal and maximal values of u on the bounday of D, then γ m u(x, y) M anywhee in D. This statement is called the maximum pinciple fo the Laplace equation. 6

7 If u and u 2 solve the same Poisson equation u = f on D with the same bounday conditions then u = u 2 within D, that is, the solution to the Diichlet bounday value poblem fo the Poisson equation is unique. This follows fom lineaity of the equation and the maximum pinciple. Indeed, by lineaity function v = u u 2 solves the Laplace equation u = 0 with homogeneous bounday conditions v = 0 on D. By the maximum pinciple this implies that v(x, y) = 0 fo all (x, y) D and hence u = u 2. 7