Mathematical methods in the physical sciences 3nd edition Mary L. Boas Chapter 13 Partial differential equations Lecture 13 Laplace, diffusion, and wave equations 1 1. Introduction (partial differential equation) ex 1) Laplace equation 2 2 2 ux, y, z 0 2 u 2 u 2 u 0 x y z 2 : gravitational potential, electrostatic potential, steady-state temperature with no source 2u f x, y, z ex 2) Poisson’s equation: : with sources (=f(x,y,z)) ex 3) Diffusion or heat flow equation 2u x, y , z , t 1 u x, y, z, t ( : diffusivit y) 2 t 2 1 2u ex 4) Wave equation u 2 2 v t 2 ex 5) Helmholtz equation 2 F k 2 F 0 : space part of the solution of either the diffusion or the wave equation 3 2. Laplace’s equation: steady-state temperature in a rectangular plate (2D) In case of no heat source 2T 2T T 0 or 2 0 2 x y 2 T o solve this equation, we try a solution of the form, T x, y X x Y y , X ( x) : a function only of x, Y ( y ) : a function only of y. "Separation of variables" 4 2T 2T 2 XY 2 XY 2 0 0 2 2 2 x y x y d2X d 2Y 1 d 2 X 1 d 2Y XY Y X 2 0 0, 2 2 2 dx dy X dx Y dy 1 d2X 1 d 2Y 2 X dx Y dy2 t above equation with the variable- separated sides, T o satisfy he 1 d 2Y 1 d2X 2 (separationconstant k 0) k . const Y dy2 X dx2 X k 2 X and Y k 2Y . sin kx, X coskx, e ky , Y ky e , e ky sin kx e ky sin kx T XY ky e coskx ky e coskx 5 e ky sin kx e ky sin kx General solut ion: T XY ky e coskx ky e coskx 1) In the current problem, boundary conditions are i ) T 0 as y . discard e ky ii) T 0 when x 0. discard coskx iii) T 0 when x 10. sin 10k 0 k n / 10 iv) T 10 when y 0. nx nx T exp[ny / 10]sin T bn sin 100. 10 10 n 1 6 nx nx T exp[ny / 10] sin T bn sin 100. 10 10 n 1 Using theFourier series, 10 2 l nx 2 10 nx 10 nx bn f ( x) sin dx 100sin dx 20 cos l 0 l 10 0 10 n 10 0 400 , odd n, 200 n 1 1 n n 0, even n. T 400 x 1 3y / 10 3x sin . exp[ny / 10] sin e 10 3 10 ex. At x 5 and y 5, T 26.1 7 2) How about changing the boundary condition? Let us consider a finite plate of height 30 cm with the top edge at T=0. T=0 at 30 cm In this case, e^ky can not be discarded. e ky ae ky beky T 0 at y 30 12 e k 30 y 12 e k 30 y ( thatis, a 12 e30 k , b 12 e 30 k ) sinh k 30 y T Bn sinh n 1 n 30 y sin nx 10 10 nx nx 100 Bn sinh 3n sin bn sin , where bn Bn sinh 3n . 10 10 n 1 n 1 Ty 0 T odd n 1 n 400 30 y sin nx sinh 10 10 n sinh 3n 8 - To be considered I e kx sin ky kx e sin ky We could have another general solution,T XY , for k 2 . kx e cosky kx e cosky This is correct, but makes the problem more complicated. (Please check the boundary condition.) - To be considered II In case that the two adjacent sides are held at 100 (ex. C=D=100), the solution can be the combination of C=100 solutions (A, B, D: 0) and D=100 (A, B, C: 0) solutions. 9 -. Summary of separation of variables. 1) 2) 3) 4) 5) A solution is a product of functions of the independent variables. Separate partial equation into several independent ordinary equation. Solve the ordinary differential eq. Linear combination of these basic solutions Boundary condition (boundary value problem) 10 3. Diffusion or heat flow equation; heat flow in a bar or slab - Heat flow : 2u u F x, y, z T t T 2 F 1 2 F 1 u . 2 t "Separation of variables" dT 1 1 1 dT 2 F 2 k 2 . dt F T dt 1 2 1 1 dT 2 F k 2 , k F 2 T dt F k F , 2 2 dT 2 2 k 2 2 t k T T e dt cf. “Why do we need to choose –k^2, not +k^2?” 11 Let’s take a look at one example. At t=0, T=0 for x=0 and T=100 x=l. From t=0 on, T=0 for x=l. In case of this 1 - D problem, sin kx, 1 2 d 2F 2 2 F k 2 k F 0 F F dx coskx e k t sin kx u 2 2 k t coskx e 2 2 For T(x=0)=0 and T(x=l)=100 at t=0, the initial steady-state temperature distribution: d 2u0 u0 0 (no heat source) 0 2 dx 100 u0 ax b u0 x l 2 12 1) u 0 when x 0. discard cos kx - Using Boundary condition 2)u 0 when x l. sin kl 0 kl n ue n / l 2 t 2 nx nx sin u bn e n / l t sin l l n 1 At t 0, u u0 u bn sin n 1 nx 100 u0 x. l l 100 2l 1 1n 1 200 1 bn l n n n 1 u 200 / l 2 t x 1 2 / l 2 t 2x 1 3 / l 2 t 3x e sin e sin e sin . l 2 l 3 l 13 For some variation, when T0, we need to consider uf.as the final state, maybe a linear In this case, we can write down the solution simply like this. u bn e n / l t sin n 1 2 nx uf . l nx nx u0 bn sin u f u0 u f bn sin . l l n 1 n 1 14 4. Wave equation; vibrating string node Under the assumption that the string is not stretched, 2 y 1 2 y x 2 v 2 t 2 x=0 x=l y X x T t 1 d2X 1 1 d 2T 2 2 2 2 const . k X k X 0 and T k v T 0. 2 2 2 X dx v T dt frequency(sec-1 ), wavelength 2 angular frequency(radians) k 2 2 wave number. v v 15 sin kx, X coskx, sin kx sin t sin kvt, sin kx cost T y XT , where kv. coskvt, coskx sin t coskx cost Boundarycondition: y 0 at x 0, x l. nvt nx sin sin l l y sin nx cos nvt . l l 16 nvt nx sin sin l l y sin nx cos nvt l l nvt nx sin cos l l y sin nx sin nvt l l 1) case 1 For y (t 0) f ( x) and y ' 0, sin nvt / l should be discarded. y bb sin n 1 y0 bn sin n 1 nx nvt cos . l l nx f x . l After finding the Fourier series coefficients, y 8h x vt 1 3x 3vt sin cos sin cos 9 2 l l l l 17 nvt nx sin sin l l y sin nx cos nvt . l l 2) case 2 For y (t 0) 0 and y0 ' 0, cosnvt / l should be discarded. y Bn sin n 1 nx nvt sin . l l nv nx nx y B sin b sin V x (Use Fourier series expansion.) n n l l l t t 0 n 1 n 1 18 3) Eigenfunctions nx nvt sin : a characteristic function or eigenfunction l l n nv / l 2 n : characteristic frequency y sin A vibration with a pure freuquencyis called thenormalmode of vibration. first harmonic, fundamental second harmonic third fourth 19 Mathematical methods in the physical sciences 2nd edition Mary L. Boas Chapter 13 Partial differential equations Lecture 14 Using Bessel equation 20 5. Steady-state temperature in a cylinder For this problem, cylindrical coordinate (r, , z) is more us 1 u 1 2u 2u u 2 0 r 2 2 r r r r z 2 u Rr Z z P utting this into the equation, and dividing with u RZ , 1 1 d dR 1 1 d 2 1 d 2 Z 0 r 2 2 2 R r dr dr r d Z dz 21 1 1 d dR 1 1 d 2 1 d 2 Z 0 r 2 2 2 R r dr dr r d Z dz In order to say that a term is constant, 1) function of only one variable 2) variable does not elsewhere in the equation. - 1st step 1 1 d dR 1 1 d 2 1 d 2Z k 2 r 2 2 2 R r dr dr r d Z dz e kz 1 d 2Z 2 k ,Z . 2 kz Z dz e Because u ~ 0 at z , u ~ e kz . 22 - 2nd step 1 1 d dR 1 1 d 2 r d dR 1 d 2 r 2 2 k 0 r 2k 2 0 r r 2 2 2 R r dr dr r d R dr dr d Here, sin n , 1 d 2 2 n , 2 cosn . d - 3rd step r d dR 2 d dR 2 2 2 2 2 2 2 2 2 r n k r 0 r r k r n R 0 r R rR k r n R 0 R dr dr dr dr " Bessel equation": x xy x 2 p 2 y x 2 y xy x 2 p 2 y 0 Solut ion of the Bessel equations: J n kr (Bessel), N n kr (Neumannor Weber). Here, we take only R ~ J n (kr), because N n kr at r 0. - Boundarycondition: Rr 1 J n k 0. J n kr sin ne kr u RZ , where k is a zero of J n . J kr cosne kr n 23 In azimuthalsymmetry,n 0 J 0 km 0 ( J n : oscillating function) u cm J 0 km r e k m z . n 1 Boundarycondition: u z 0 cm J 0 km r 100. n 1 Here, using theorthogonal ity of the Bessel function, 1 0 rJ 0 k r cm J 0 km r dr 100 rJ 0 k r dr 1 0 n 1 Only if m, we have the non - zero termin the left side. cm r J 0 km r dr 100rJ 0 km r dr. 1 2 0 1 0 left term: rJ 0 km r dr 12 J12 km . 1 2 0 J p (a ) 0 J p (b) 0, if a b 0 xJ p axJ p bxdx 1 J p2 1 a 1 J p2 1 a 1 J p21 a , if a b 2 2 2 1 24 cm rJ 0 km r dr 100rJ 0 km r dr. 1 2 0 1 0 For the right term: rJ 0 km r d xJ1 x xJ0 x (recusion relation) dx 1 d km rJ1 km r km rJ 0 km r , For x km r , km dr 1 d rJ1 km r km dr 1 1 1 rJ k r dr rJ k r J1 km . 1 m 0 0 m km k m 0 1 Combiningtwo results, cm r J 0 km r dr 100rJ 0 km r dr cm 12 J12 km 1 0 cm 2 1 0 100J1 km km 100J1 km 2 200 2 km J1 km km J1 km 25 If thegiven temperatureof thebase of thecylinderis f r , , 1, but an cosn bn sin n . u J n kmn r amn cosn bmn sin n e k mn z . m 1 n 0 u z 0 J n kmn r amn cosn bmn sin n f r , . m 1 n 0 Because of theorthogonality, 1 2 0 0 f r , J k r cosrdrd a 1 2 a 12 J21 k . Similarly, b 2 J 0 0 2 1 1 2 k 0 0 J k r cos2 rdrd 2 f r , J k r sinrdrd . 26 Bessel’s equation - Bessel’s equation 1) Equation and solution - named equation which have been studied extensively. - “Bessel function”: solution of a special differential equation. xxy x 2 p 2 y x 2 y xy x 2 p 2 y 0 - being something like damped sines and cosines. - many applications. ex) problems involving cylindrical symmetry (cf. cylinder function); motion of pendulum whose length increases steadily; small oscillations of a flexible chain; railway transition curves; stability of a vertical wire or beam; Fresnel integral in optics; current distribution in a conductor; Fourier series for the arc of a circle. 27 Bessel’s equation xxy x 2 p 2 y x 2 y xy x 2 p 2 y 0 - Solution : 2n p n 1 x p 1 x J p x , where p x e dx p 1!, 0 n 0 n 1n p 1 2 cosp J p x J p x N p x (Neumann or Weber) p 0. sin p generalsolution y AJ p x BN p x - Graph 28 Bessel’s equation 2) Recursion relations 4) d p x J p x x p J p 1 x dx d p x J p x x p J p 1 x dx 2p J p 1 x J p 1 x J p x x J p 1 x J p 1 x 2 J p x 5) J p x 1) 2) 3) p p J p x J p 1 x J p x J p 1 x x x 29 Bessel’s equation 3) Orthogonality cf. Comparison J p x , N p x consider just J p x for one value of p x a, b,, J p x 0 x 1, J p ax 0, J p bx 0, 2 For y sin nx, y n y 0 For J p ax, xxy a 2 x 2 p 2 y 0 sin x, cos x consider just sin x x nx, sin x 0 x 1, sin nx 0 1 sin nx sin mxdx 0, 0 for n m, xJ axJ bxdx 0, 1 0 p p for a b 30 Bessel’s equation xxy x 2 p 2 y x 2 y xy x 2 p 2 y 0 xxy a 2 x 2 p 2 y 0 J p (a) 0 J p (b) 0, if a b 0 xJ p axJ p bxdx 12 J p21 a 12 J p21 a 12 J p21 a , if a b 1 31 6. Vibration of a circular membrane (just like drum) 1 2 z z 2 2 v t 2 z F x, y T t 2 F k 2 F 0 and T k 2v 2T 0 1 F 1 2 F In polarcoordinate, k 2 F 0. r 2 2 r r r r F Rr . P utting this form into the above equation, and then dividing with F , 1 1 d dR 1 1 d 2 k2 0 r 2 2 r R dr dr r d 1 d 2 d 2 2 2 n n 0 (simple harmonic equation) 2 2 d d r d dR 2 2 2 r n k r R 0 (Bessel's equation) dr dr sin n sin kvt sin n sin kvt sin n coskvt J n kr z Rr T t J n kr cosn coskvt cosn sin kvt cosn coskvt 32 sin n sin kvt sin n coskvt z Rr T t J n kr cosn sin kvt cosn coskvt Boundarycondit ion: z (r 1) 0 J n k 0. kmn : zerosof J n (double series) Characteristic freuqency of thenormal modes: mn kmn v / 2 cf. They are not integral multiples of the fundamental as is true for the string (characteristics of the bessel function). This is why a drum is less musical than a violin. 33 ex) z J n kmn r cosn coskmn vt 34 (m,n)=(1,0) (2,0) (1,1) American Journal of Physics, 35, 1029 (1967) 35 Mathematical methods in the physical sciences 2nd edition Mary L. Boas Chapter 13 Partial differential equations Lecture 15 Using Legendre equation 36 7. Steady-state temperature in a sphere - Sphere of radius 1 where the surface of upper half is 100, the other is 0 degree 1 2 u 1 u 1 2u u 2 r 0 2 sin 2 2 2 r r r r sin r sin 2 1 d 2 dR 1 1 d d 1 1 d 2 For u Rr , 0 r sin R dr dr sin d d sin 2 d 2 sin m 1 d 2 2 m , d 2 cosm 1 d 2 dR 1 1 d d m2 0. r sin R dr dr sin d d sin 2 r l 1 d 2 dR R r k R dr dr 1 / r l 1 1 d d m2 sin 2 k 0 (Associat ed Legendre equat ion k l l 1) sin d d sin Pl m cos (Legendre polynomial ). 37 1) For the interior of the sphere, no divergence. T herefore,discard 1/ r l 1. sin m u r Pl cos cosm l m 2) Due to the azimuthal symmetry,m 0 u cl r l Pl cos l 0 100, 0 / 2 3) ur 1 cl Pl cos /2 l 0 0, 0, 1 x 0, or ur 1 cl Pl cos 100 f x , where f x l 0 1, 0 x 1. In thiscase, we can expandthisin a series of Legendrepolynomina ls f x 12 P0 x 34 P1 x 167 P3 x 11 P x (Here, x cos ) 32 5 c P cos 100 P cos l 0 1 2 l l 0 3 4 P1 cos 167 P3 cos 11 P cos 32 5 1 3 7 c0 100 , c1 100 , c2 100 ,. 2 4 16 5 u x 100 12 P0 cos 34 rP1 cos 167 r 3 P3 cos 11 r P5 cos 32 38 Legendre’s equation - Legendre’s equation 1) Equation and solution i. .m 0 (Legendre equation) l l 1 d 2 1 x 2 y 2 xy l l 1 y 0 Solution : Pl x x 1 . 2!l! dx ii. m 0 (Associated Legendre equation) m2 1 x y 2 xy l l 1 y0 2 1 x 2 Solution : Pl x 1 x m For m l , Pl m 0 2 m/2 m d Pl x l m l . dx 1 d d m2 cf . l l 1 0 (Associated Legendre equation). sin sin d d sin 2 39 Legendre’s equation 1 d d m2 cf . l l 1 0 (Associated Legendre equation) sin sin d d sin 2 1 d d cos d d cos m2 l l 1 0 sin sin d cos x d d cos d sin 2 d 2 d m2 sin 2 l l 1 0 d cos x d cos sin d d m2 2 l l 1 0 1 cos d cos x d cos 1 cos2 d 2 d m2 1 cos 2 cos l l 1 0 d cos x 2 d cos 1 cos2 2 d 2 d m2 1 x 2x l l 1 0 2 2 dx dx 1 x 2 40 Legendre’s equation - Legendre polynomials - Associated Legendre polynomials P0 x 1 Pl m x 1 m P1 x x P2 x P10 x x P11 x 1 x 2 1 cf . x 3P1 x 2 P3 x 5 1/ 2 P2 2 x 3 1 P11 x P11 x 2 1 2 3x 1 2 1 P3 x 5 x 3 3x 2 1 P4 x 35x 4 30x 2 3 8 1 P5 x 63x 5 70x 3 15x 8 1 P6 x 231x 6 315x 4 105x 2 5 16 l m! P m x l m! l 1 2 P2 x 24 1 P21 x P21 x 6 1 P20 x 3 x 2 1 2 P21 x 3 x 1 x 2 P22 x 3 1 x 2 1/ 2 41 Legendre’s equation 2) Orthogonality 2 2 P x dx P cos sin d 1 l 0 l 1 P 1 1 l m 2 . 2l 1 2 2 l m! Pl m cos sin d . 2l 1 l m! x 2 dx 0 cf .x cos dx sin d 42 8. Poisson’s equation F 0 (F : conservative) F V 1) For gravit ational force, F 4G 2V 4G : Poisson equation cf. 0 2V 0 : Laplace equation 2) For electrostatic force, E 4 2V 4 (in Gaussian unit ). cf. 0 2V 0 In general, 2u x, y, z f x, y, z u 1 4 f x, y, z x x y y z z 2 2 2 dxdydz. 2 w 0 (Laplaceequation) 2 u w 2u 2 w f u w : solutionof Poissonequation 43 Example 1 2V 4 V x, y , z 1 4 4 x, y, z x x y y z z 2 2 2 dxdydz q x y z a 2 2 2 . In sphericalcoordinat e, Vq q r 2ar cos a 2 2 . r l m sin m Basic solution of Laplaceequation Pl cos r l 1Pl cos r l 1 cosm 1) zero at infinite discard r l 2) symmet ricabout z - axis solutionis indep.of . m 0, cosm 1 V Vq cl r l 1Pl cos l grounded sphere 44 Boundarycondition: V 0 for r R. Vr R Using cl R l 1Pl cos 0. q R 2 2aR cos a 2 l R l Pl cos q , l 1 2 2 a R 2aR cos a l q R l Pl cos q cl R l 1Pl cos l 1 a l l From this relation, cl R V qRl qR2l 1 l 1 or cl l 1 . a a R 2l 1r l 1Pl cos q 2 2 a l 1 r 2ar cos a l q q r 2ar cos a 2 2 q r 2ar cos a 2 l 1 2 q ( R / a ) R / a P cos 2 r l l l l 1 R / a q r R / a 2r R / a cos 2 2 2 2 ‘Method of the images’ . In thesecond term,charge: R / a q, position: 0,0,R 2 / a 45 cf. Electric multipoles monopole dipole quadrupole octopole 2) Expansion for the potential of an arbitrary localized charge distribution V r 1 4 0 1 rdr r r 2 r r 2 2 r r r r 2rr cos r 1 2 cos r r 2 2 r r r r r 1 , where 2 cos r r 1 1 1 1 3 5 1 / 2 1 1 2 3 r r r r 2 8 16 46 2 2 3 3 1 1 1 r r 5 r r 3 r r 1 2 cos 2 cos 2 cos r r r 2 r r 8 r r 16 r r 2 3 1 r r r 2 3 1 cos 3 cos 1 / 2 5 cos 3 cos / 2 r r r r 1 r Pn cos r n 0 r n V r - 1 4 0 n n n n = = = = 0 1 2 3 Legendrepolynomina l r r P cos rd multipoleexpansion n 0 : : : : 1 n 1 n n monopole contribution dipole quadrupole octopole 47