Laplace s Equation on a Disc

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Brittney Riley
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1 LECTURE 15 Lplce s Eqution on Disc Lst time we solved the Diichlet poblem fo Lplce s eqution on ectngul egion. Tody we ll look t the coesponding Diichlet poblem fo disc. Thus, we conside disc of dius 1 D [x, y] R 2 x 2 y 2 2} upon which the following Diichlet poblem is posed: 2 2b u xx u yy, [x, y] D u cos θ, sin θ h θ, θ We shll solve this poblem by fist ewiting Lplce s eqution in tems of pol coodintes which e most ntul to the egion D nd then septing vibles nd peceding s in Lectue 14. Now unde the chnge of vibles we hve x cos θ y sin θ x x θ x θ y x θ x θ Afte shot but tedious clcultion one finds 3 } x x2 y 2 y x2 y 2 x2 y 2 θ tn 1 y/x y x 2 y 2 x x 2 y 2 2 x 2 2 y θ 2 nd so in tems of pol coodintes Lplce s eqution becomes 3 u 1 u 1 2 u θθ. θ cos θ sin θ θ sin θ cos θ We ll now pply Seption of Vibles to this PDE. Setting u, θ R T θ nd plugging in we get Multiplying both sides by 2 /RT we get R T R T R 2 T 2 R R R R T T Obseving tht ech side depends only vible tht does not ppe on the opposite side we conclude tht both sides must be equl to constnt. Let s denote this constnt by λ 2. We then hve 2 R R R R λ 2 T T 65 θ θ

Lplce s eqution in tems of pol coodintes which e most ntul to the egion D nd then septing vibles nd peceding s in Lectue 14.

2 15. LAPLACE S EQUATION ON A DISC 66 o the following pi of odiny diffeentil equtions 4 T λ 2 T 4b 2 R R λ 2 R The fist eqution 4 should be quite fmili by now. It hs s its genel solution 5 T θ A cos λθ B sin λθ The second eqution 4b is n Eule type eqution. Such equtions cn lmost lwys be solved using the nstz R m, egding m s n djustble pmete. Inseting m in plce of R in 4b we get 2 m m 1 m 2 m m 1 λ 2 m m m 1 m n 2 m m 2 λ 2 m Evidently, we must tke m 2 λ 2 which in tun equies m ±λ. The genel solution of 4b is thus 6 R λ b λ E, except tht something scewy hppens when λ ; in tht cse we only get one linely independent solution R some constnt. To get second linely independent solution fo the λ cse, we cn employ Reduction of Ode, o simply solve 7 2 R R fesh. The ltte is esy enough. Dividing out by nd setting S R, 7 becomes ds S S S 1 ds S d ds S d ln S ln C Setting C ln c we then hve ln S ln ln c ln c S c Finlly, we integte S to ecove R R R d d Sd d In conclusion, the genel solution of 4b is 8 R c d d c ln d c λ d λ if λ c ln d if λ d C λ cse Putting 5 nd 8 togethe, we obtin Aλ cos λθ B u λ, θ λ sin λθ λ C λ cos λθ D λ sin λθ λ λ A C ln λ s Seption-of-Vibles-type solutions to Lplce s eqution in pol coodintes n.b. when λ, cos λθ 1 nd sin λθ. Now we cn whittle down this set of possible solutions even futhe by imposing some hidden boundy conditions besides 2b. One thing we expect of ny vible solution is tht u, θ u, θ, fo fte ll point with coodintes [ cos θ, sin θ] is the sme s the point with coodintes [ cos θ, sin θ ]. But } cos λθ cos λ θ λ n Z :, ±1, ±2, ±3,...} sin λθ sin λ θ Tht is, λ must be n intege, we my s well tke to be nonnegtive since the solutions whee λ n nd λ n e ctully coincide up to chnging the signs of the bity constnts B nd D.

Inseting m in plce of R in 4b we get 2 m m 1 m 2 m m 1 λ 2 m m m 1 m n 2 m m 2 λ 2 m Evidently, we must tke m 2 λ 2 which in tun equies m ±λ.

3 15. LAPLACE S EQUATION ON A DISC 67 Secondly, we expect ny vible solution to be continuous t. This will equie us to thow out the solutions whee C nd D e non-zeo; fo both n nd ln become unbounded s. We thus hve An cos nθ u n, θ n B n sin nθ n n 1, 2, 3,... A n o even moe succinctly 9 u n, θ A n cos nθ n B n sin nθ n, n, 1, 2, 3,... Now let s tke genel line combintion of the solutions 9 nd impose the boundy condition 2b. Thus, we set u, θ B n sin nθ n nd impose 1 h θ u, θ n A n cos nθ n n A n cos nθ n B n sin nθ n Now fom Fouie theoy we now tht ny continuous function on the cicle hs unique Fouie expnsion whose coefficients cn be explicitly detemined in tems of cetin Fouie integls. Applying this theoy to the cse t hnd we hve h θ 2 11 n cos nθ b n sin nθ 11b 11c n 1 π b n 1 π h θ cos nθ dθ h θ sin nθ dθ Comping 1 with 11 we conclude tht ou solution must be u, θ A B n sin nθ n with A 2 1 A n 1 n π B n 1 n π A n cos nθ n h θ dθ h θ cos nθ dθ, n 1, 2, 3,... h θ sin nθ dθ, n 1, 2, 3,...

.. Now let s tke genel line combintion of the solutions 9 nd impose the boundy condition 2b.

4 1. POISSON SUM FORMULA Poisson Sum Fomul We now hve u, θ A B n sin nθ n A n cos nθ n 1 h θ dθ 1 h θ cos nθ cos nθ dθ n π 1 h θ sin nθ sin nθ dθ n π o, intechnging the ode of summtion nd integtion which we cn do so long s the seies conveges in the fist plce u, θ 1 n h θ cos nθ cos nθ sin nθ sin nθ dθ 13 1 h θ whee we hve employed the cosine ngle sum fomul n cos n θ θ dθ 14 cos α β cos α cos β sin α sin β Next, inseting 15 cos φ eiφ e iφ into the seies 16 we get Using the well-known geometic seies 17 we obtin o 2 cos nφ t n 1 e inφ e inφ t n 1 e iφ t n e iφ t n 18 n x n 1 1 x cos nφ t n e iφ t 1 1 t 2 1 e iφ t e iφ t t 2 1 t cos φ t 2 cos nφ t n 1 t cos φ t e iφ t 1

which we cn do so long s the seies conveges in the fist plce u, θ 1 n h θ cos nθ cos nθ sin nθ sin nθ dθ 13 1 h θ whee we hve employed the cosine ngle sum fomul n cos n θ θ dθ

5 1. POISSON SUM FORMULA 69 Applying this lst fomul to we find Thus, n cos n θ θ 2 u, θ 1 n cos n θ θ cos θ θ 2 h θ cos θ θ 2 dθ cos θ θ 2 Finlly, we ll convet this esult into something tht s undestndble in coodinte fee wy. Set then nd nd so we cn wite x [ cos θ, sin θ] y [ cos θ, sin θ ] x 2 y 2 x y 2 cos θ cos θ 2 sin θ sin θ 2 2 cos 2 θ 2 cos θ cos θ 2 cos 2 θ 2 sin 2 θ 2 sin θ sin θ 2 sin 2 θ 2 cos 2 θ sin 2 θ 2 cos θ cos θ sin θ sin θ 2 cos 2 θ sin 2 θ 2 2 cos θ θ 2 21 u x 1 y D h y 2 x 2 x y 2 ds whee D is the cicle bounding D. The dditionl fcto of 1/ ises becuse dθ ds is the coect expession fo the infinitesiml c-length when we intepete 2 s pth integl bout the boundy of D.

Set then nd nd so we cn wite x [ cos θ, sin θ] y [ cos θ, sin θ ] x 2 y 2 x y 2 cos θ cos θ 2 sin θ sin θ 2 2 cos 2 θ 2 cos θ cos θ 2 cos 2 θ 2 sin 2 θ 2 sin θ sin θ 2 sin 2 θ

2.016 Hydrodynamics Prof. A.H. Techet

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