Laplace's Equation on a Disc

LECTURE 15

Laplace’s Equation on a Disc

Last time we solved the Dirichlet problem for Laplace’s equation on a rectangular region. Today we’ll look at the corresponding Dirichlet problem for a disc.

Thus, we consider a disc of radius a 2 2 2 2 (1) D = [x, y] ∈ R | x + y = a upon which the following Dirichlet problem is posed:

(2a) uxx + uyy = 0 , ∀ [x, y] ∈ D (2b) u (a cos θ, a sin θ) = h (θ) , 0 ≤ θ ≤ 2π

We shall solve this problem by ﬁrst rewriting Laplace’s equation in terms of a polar coordinates (which are most natural to the region D) and then separating variables and preceding as in Lecture 14.

Now under the change of variables x = r cos θ r = px2 + y2 ⇐⇒ y = r sin θ θ = tan−1 (y/x) we have ∂ ∂r ∂ ∂θ ∂ x ∂ y ∂ ∂ sin θ ∂ = + = − = cos θ − ∂x ∂x ∂r ∂x ∂θ px2 + y2 ∂r x2 + y2 ∂θ ∂r r ∂θ ∂ ∂r ∂ ∂θ ∂ y ∂ x ∂ ∂ cos θ ∂ = + = + = sin θ + ∂y ∂x ∂r ∂x ∂θ px2 + y2 ∂r x2 + y2 ∂θ ∂r r ∂θ After a short but tedious calculation one ﬁnds ∂2 ∂2 ∂2 1 ∂ 1 ∂2 (3) + = + + ∂x2 ∂y2 ∂r2 r ∂r r2 ∂θ2 and so in terms of polar coordinates Laplace’s equation becomes 1 1 (3) u + u + u = 0 . rr r r r2 θθ We’ll now apply Separation of Variables to this PDE. Setting u (r, θ) = R (r) T (θ) and plugging in we get R0 R R00T + T + T 00 = 0 r r2 Multiplying both sides by r2/RT we get r2 r T 00 R00 + R0 = − R R T Observing that each side depends only a variable that does not appear on the opposite side we conclude that both sides must be equal to a constant. Let’s denote this constant by λ2. We then have r2 r T 00 R00 + R0 = λ2 = − R R T 65 15. LAPLACE’S EQUATION ON A DISC 66

or the following pair of ordinary diﬀerential equations (4a) T 00 = −λ2T (4b) r2R00 + rR0 = λ2R

The ﬁrst equation (4a) should be quite familiar by now. It has as its general solution (5) T (θ) = A cos (λθ) + B sin (λθ)

The second equation (4b) is an Euler type equation. Such equations can (almost always) be solved using the ansatz R (r) = rm, regarding m as an adjustable parameter. Inserting rm in place of R in (4b) we get 0 = r2 m (m − 1) rm−2 + r mrm−1 − λ2 (rm) = m (m − 1) + m − n2 rm = m2 − λ2 rm Evidently, we must take m2 = λ2 which in turn requires m = ±λ. The general solution of (4b) is thus (6) R (r) = arλ + br−λ Err, except that something screwy happens when λ = 0; in that case we only get one linearly independent solution R (r) = a some constant. To get a second linearly independent solution for the λ = 0 case, we can employ Reduction of Order, or simply solve (7) r2R00 + rR0 = 0 afresh. The latter is easy enough. Dividing out by r and setting S = R0, (7) becomes 1 dS dS dr Z dS Z dr rS0 + S = 0 =⇒ = −r =⇒ = − =⇒ = − + C S dr S r S r =⇒ ln |S| = − ln |r| + C Setting C = ln |c| we then have c c ln |S| = − ln |r| + ln |c| = ln =⇒ S = r r Finally, we integrate S to recover R Z Z Z c R (r) = R0dr + d = Sdr + d = dr + d = c ln |r| + d (λ = 0 case) r In conclusion, the general solution of (4b) is crλ + dr−λ if λ 6= 0 (8) R (r) = c ln |r| + d if λ = 0 Putting (5) and (8) together, we obtain λ −λ (Aλ cos (λθ) + Bλ sin (λθ)) r + (Cλ cos (λθ) + Dλ sin (λθ)) r λ 6= 0 uλ (r, θ) = A0 + C0 ln |r| λ = 0 as Separation-of-Variables-type solutions to Laplace’s equation in polar coordinates (n.b. when λ = 0, cos (λθ) = 1 and sin (λθ) = 0).

Now we can whittle down this set of possible solutions even further by imposing some hidden boundary conditions (besides (2b)).

One thing we expect of any viable solution is that u (r, θ) = u (r, θ + 2π), for after all a point with coordinates [r cos (θ) , r sin (θ)] is the same as the point with coordinates [r cos (θ + 2π) , r sin (θ + 2π)]. But cos (λθ) = cos (λ (θ + 2π)) ⇐⇒ λ = n ∈ := {0, ±1, ±2, ±3,...} sin (λθ) = sin (λ (θ + 2π)) Z That is, λ must be an integer, we may as well take to be nonnegative since the solutions where λ = n and λ = −n are actually coincide (up to changing the signs of the arbitary constants B and D). 15. LAPLACE’S EQUATION ON A DISC 67

Secondly, we expect any viable solution to be continuous at r = 0. This will require us to throw out the solutions where C and D are non-zero; for both r−n and ln |r| become unbounded as r → 0. We thus have n n An cos (nθ) r + Bn sin (nθ) r n = 1, 2, 3,... un (r, θ) = A0 n = 0 or even more succinctly n n (9) un (r, θ) = An cos (nθ) r + Bn sin (nθ) r , n = 0, 1, 2, 3,...

Now let’s take a general linear combination of the solutions (9) and impose the boundary condition (2b). Thus, we set ∞ X n X n u (r, θ) = An cos (nθ) r + Bn sin (nθ) r n=0 n=1 and impose ∞ X n X n (10) h (θ) = u (a, θ) = An cos (nθ) a + Bn sin (nθ) a n=0 n=1

Now from Fourier theory we now that any continuous function on the circle has a unique Fourier expansion whose coeﬃcients can be explicitly determined in terms of certain Fourier integrals. Applying this theory to the case at hand we have ∞ ∞ a0 X X (11a) h (θ) = + a cos (nθ) + b sin (nθ) 2 n n n=1 n=1 1 Z 2π (11b) an = h (θ) cos (nθ) dθ π 0 1 Z 2π (11c) bn = h (θ) sin (nθ) dθ π 0 Comparing (10) with (11a) we conclude that our solution must be ∞ X n X n u (r, θ) = A0 + An cos (nθ) r + Bn sin (nθ) r n=1 n=1 with Z 2π a0 1 A0 = = h (θ) dθ 2 2π 0 1 Z 2π An = n h (θ) cos (nθ) dθ , n = 1, 2, 3,... a π 0 1 Z 2π Bn = n h (θ) sin (nθ) dθ , n = 1, 2, 3,... a π 0 1. POISSON SUM FORMULA 68

1. Poisson Sum Formula

We now have

∞ X n X n u (r, θ) = A0 + An cos (nθ) r + Bn sin (nθ) r n=1 n=1 1 Z 2π = h (θ) dθ 2π 0 ∞ X 1 Z 2π + h (θ0) cos (nθ0) cos (nθ) dθ0rn arπ n=1 0 X 1 Z 2π + h (θ0) sin (nθ0) sin (nθ) dθrn arπ n=1 0 or, interchanging the order of summation and integration (which we can do so long as the series converges in the ﬁrst place) ∞ ! 1 Z 2π X r n (13) u (r, θ) = h (θ0) 1 + 2 (cos (nθ) cos (nθ0) + sin (nθ) sin (nθ0)) dθ0 2π a 0 n=1 ∞ ! 1 Z 2π X r n = h (θ0) 1 + 2 cos (n (θ − θ0)) dθ 2π a 0 n=1 where we have employed the cosine angle sum formula (14) cos (α − β) = cos (α) cos (β) + sin (α) sin (β)

Next, inserting eiφ + e−iφ (15) cos (φ) = 2 into the series ∞ X (16) 1 + 2 cos (nφ) tn n=1 we get ∞ ∞ ∞ X X n X n 1 + einφ + e−inφ tn = 1 + eiφt + e−iφt n=1 n=1 n=1 Using the well-known geometric series ∞ X 1 (17) xn = 1 − x n=0 we obtain ∞ X 1 1 1 + 2 cos (nφ) tn = 1 + − 1 + − 1 1 − eiφt 1 − e−iφt n=1 1 − t2 = 1 − eiφt − e−iφt + t2 1 − t2 = 1 − 2 cos(φ)t + t2 or ∞ X 1 − t2 (18) 1 + 2 cos (nφ) tn = 1 − 2 cos(φ)t + t2 n=1 1. POISSON SUM FORMULA 69

Applying this last formula to ∞ ! X r n 1 + 2 cos (n (θ − θ0)) a n=1 we ﬁnd ∞ ! r 2 X r n 1 − a2 − r2 1 + 2 cos (n (θ − θ0)) = a = a 0 r r 2 a2 − cos (θ − θ0) ar + r2 n=1 1 − 2 cos (θ − θ ) a + a Thus, Z 2π 2 2 1 0 a − r 0 (20) u (r, θ) = h (θ ) 2 0 2 dθ 2π 0 a − 2 cos (θ − θ ) ar + r

Finally, we’ll convert this result into something that’s understandable in a coordinate free way. Set x = [r cos (θ) , r sin (θ)] y = [a cos (θ0) , a sin (θ0)] then kxk = r2 kyk = a2 and kx − yk2 = (r cos θ − a cos θ0)2 + (r sin θ − a sin θ0)2 = r2 cos2 θ − 2ar cos θ cos θ0 + a2 cos2 θ0 + r2 sin2 θ − 2ar sin θ sin θ0 + a2 sin2 θ0 = r2 cos2 θ + sin2 θ − 2ar (cos θ cos θ0 + sin θ sin θ0) + a2 cos2 θ0 + sin2 θ0 = r2 − 2ar cos (θ − θ0) + a2 and so we can write 1 Z a2 − kxk2 (21) u (x) = h (y) 2 ds 2πa y∈∂D kx − yk where ∂D is the circle bounding D. The additional factor of 1/a arises because adθ = ds is the correct expression for the inﬁnitesimal arc-length when we interprete (20) as a path integral about the boundary of D.