The Laplacian in Polar Coordinates

Transcription

1 The Laplacian in Polar Coordinates R. C. Trinity University Partial Differential Equations March 17, 15

2 To solve boundary value problems on circular regions, it is convenient to switch from rectangular (x,y) to polar (r,θ) spatial coordinates: x = r cosθ, r x θ y y = r sinθ, x +y = r. This requires us to express the rectangular Laplacian u = u xx +u yy in terms of derivatives with respect to r and θ.

3 The chain rule For any function f(r,θ), we have the familiar tree diagram and chain rule formulae: f x = f r r x + f θ θ x f f y = f r r y + f θ θ y r θ or x y x y f x = f r r x +f θ θ x f y = f r r y +f θ θ y

4 First take f = u to obtain u x = u r r x +u θ θ x u xx = u r r xx +(u r ) x r x +u θ θ xx +(u θ ) x θ x. Applying the chain rule with f = u r and then with f = u θ yields u xx = u r r xx +(u rr r x +u rθ θ x )r x +u θ θ xx +(u θr r x +u θθ θ x )θ x = u r r xx +u rr rx +u rθr x θ x +u θ θ xx +u θθ θx. An entirely similar computation using y instead of x also gives u yy = u r r yy +u rr ry +u rθr y θ y +u θ θ yy +u θθ θy. If we add these expressions and collect like terms we get ( u =u r (r xx +r yy )+u rr r x +ry) +urθ (r x θ x +r y θ y ) ( +u θ (θ xx +θ yy ) +u θθ θ x +θy).

5 Differentiate x +y = r with respect to x and then y: x = rr x r x = x r y = rr y r y = y r Now differentiate tanθ = y x r xx = r xr x r = r x r 3 r yy = r yr y r = r y r 3 with respect to x and then y: = y r 3, = x r 3. sec θθ x = y x θ x = y cos θ x = y r θ xx = y r 3 r x = xy r 4, sec θθ y = 1 x θ y = cos θ x = x r θ yy = x r 3 r y = xy r 4.

6 Together these yield r xx +r yy = y +x r 3 = 1 r, r x +ry = x +y r = 1. θ xx +θ yy = xy r 4 + xy r 4 =, θ x +θ y = y +x r 4 = 1 r, r x θ x +r y θ y = xy r 3 + yx r 3 =, and we finally obtain ( u =u r (r xx +r yy )+u rr r x +ry) +urθ (r x θ x +r y θ y ) ( +u θ (θ xx +θ yy ) +u θθ θ x +θy ) = 1 r u r +u rr + 1 r u θθ = u rr + 1 r u r + 1 r u θθ.

7 Example Use polar coordinates to show that the function u(x,y) = is harmonic. We need to show that u =. In polar coordinates we have so that and thus u(r,θ) = r sinθ r = sinθ r u r = sinθ r, u rr = sinθ r 3, u θθ = sinθ, r u = u rr + 1 r u r + 1 r u θθ = sinθ r 3 sinθ r 3 sinθ r 3 =. y x +y

8 The Dirichlet problem on a disk Goal: Solve the Dirichlet problem on a disk of radius a, centered at the origin. In polar coordinates this has the form u = u rr + 1 r u r + 1 r u θθ =, r < a, u = f y x u(a,θ) = f(θ), θ π. a Remarks: We will require that f is π-periodic. Likewise, we require that u(r,θ) is π-periodic in θ.

9 Separation of variables If we assume that u(r,θ) = R(r)Θ(θ) and plug into u =, we get R Θ+ 1 r R Θ+ 1 r RΘ = r R R +rr R + Θ Θ = r R R +rr R = Θ Θ = λ. This yields the pair of separated ODEs r R +rr λr = and Θ +λθ =. We also have the boundary conditions Θ is π-periodic and R(+) is finite.

10 Solving for Θ The solutions of Θ +λθ = are periodic only if λ = µ Θ = acos(µθ)+bsin(µθ). In order for the period to be π we also need 1 = cos(µ) = cos(πµ) πµ = πn µ = n N. Hence λ = n and Θ = Θ n = a n cos(nθ)+b n sin(nθ), n N. It follows that R satisfies r R +rr n R =, which is called an Euler equation.

11 Interlude Euler equations An Euler equation is a second order ODE of the form x y +αxy +βy =. Its solutions are determined by the roots of its indicial equation ρ +(α 1)ρ+β =. Case 1: If the roots are ρ 1 ρ, then the general solution is y = c 1 x ρ 1 +c x ρ. Case : If there is only one root ρ 1, then the general solution is y = c 1 x ρ 1 +c x ρ 1 lnx.

12 Solving for R The indicial equation of r R +rr n R = is ρ +(1 1)ρ n = ρ n = ρ = ±n. This means that R = c 1 r n +c r n (n ), R = c 1 +c lnr (n = ). These will be finite at r = only if c =. Setting c 1 = a n gives ( r ) n, R = R n = n N. a

13 Separated solutions and superposition We therefore obtain the separated solutions ( r ) n(an u n (r,θ) = R n (r)θ n (θ) = cos(nθ)+b n sin(nθ)), n N. a Noting that ( r ) (a u (r,θ) = cos+b sin) = a, a superposition gives the general solution u(r,θ) = a + ( r ) n (an cos(nθ)+b n sin(nθ)). a n=1

14 Boundary values and conclusion Imposing our Dirichlet boundary conditions gives f(θ) = u(a,θ) = a + (a n cos(nθ)+b n sin(nθ)), n=1 which is just the ordinary π-periodic Fourier series for f! Theorem The solution of the Dirichlet problem on the disk of radius a centered at the origin, with boundary condition u(a,θ) = f(θ) is u(r,θ) = a + n=1( r a) n(an cos(nθ)+b n sin(nθ)), where a n = 1 π π a = 1 π π f(θ)dθ, f(θ)cos(nθ)dθ, b n = 1 π π f(θ)sin(nθ)dθ.

15 Example Find the solution to the Dirichlet problem on a disk of radius 3 with boundary values given by 3 π (π +θ) if π θ <, f(θ) = 3 π (π θ) if θ < π, π if θ < 3π. We have a = 3. The graph of f is

16 Changing to polar coordinates The Dirichlet problem on a disk According to exercise.3.8 (with p = π, c = 3 and d = π/): 15 1 X 1 cos(nπ/) f (θ) = + cos(nθ). π n n=1 Hence, the solution to the Dirichlet problem is u(r, θ) = 15 1 X r n 1 cos(nπ/) + cos(nθ). π 3 n n= Examples

17 Example Solve the Dirichlet problem on a disk of radius with boundary values given by f(θ) = cos θ. Express your answer in cartesian coordinates. We have a = and f(θ) = cos θ = 1+cos(θ) = cos(θ), which is a finite π-periodic Fourier series (i.e. a = 1/, a = 1/, and all other coefficients are zero). Hence u(r,θ) = 1 ( r ) + 1 cos(θ) = 1 + r cos(θ). 8

18 Changing to polar coordinates The Dirichlet problem on a disk Since cos(θ) = cos θ sin θ, we find that r cos(θ) = r cos θ r sin θ = x y and hence u= 1 r cos(θ) 1 x y + = Examples

19 Example Solve the Dirichlet problem on a disk of radius 1 if the boundary value is 5 in the first quadrant, and zero elsewhere. We are given a = 1, f(θ) = 5 for θ π/ and f(θ) = otherwise. The Fourier coefficients of f are so that a = 1 π a n = 1 π b n = 1 π u(r,θ) = 5 +5 π π/ π/ π/ n=1 5dθ = 5, 5cos(nθ)dθ = 5sin(nπ/), nπ 5sin(nθ)dθ = 5(1 cos(nπ/)), nπ r n ( sin(nπ/) n cos(nθ)+ (1 cos(nπ/)) n ) sin(nθ).

20 Remarks: One can frequently use identities like (valid for r < 1) n=1 n=1 r n cos(nθ) n r n sin(nθ) n = 1 ln( 1 r cosθ +r ), ( ) r sinθ = arctan, 1 r cosθ to convert series solutions in polar coordinates to cartesian expressions. Using the second identity, one can show that the solution in the preceding example is u(x,y) = ( ( ) ( )) y x arctan +arctan. π 1 x 1 y