Connection to Laplacian in spherical coordinates (Chapter 13)

Transcription

1 Connection to Laplacian in spherical coordinates (Chapter 13) We might often encounter the Laplace equation and spherical coordinates might be the most convenient 2 u(r, θ, φ) = 0 We already saw in Chapter 10 how to write the Laplacian operator in spherical coordinates, 2 φ = 1 ( r 2 r 2 u ) + 1 r r r 2 sin θ ( sin θ u ) + θ θ 1 2 u r 2 sin 2 θ φ 2 This is a partial differential equation we will solve by what will become a standard approach of separation of variables We take u(r, θ, φ) = R(r)Θ(θ)Φ(φ), which we substitute in and then multipy by r 2 RΘΦ

2 Separation of variables for the Laplace equation Because of this separation we now wind up with total derivatives ( 1 d r 2 dr ) + 1 ( 1 d sin θ dθ ) d 2 Φ R dr dr Θ sin θ dθ dθ Φ sin 2 θ dφ 2 = 0 Because the first two terms do not depend on φ, we must have from the last term 1 d 2 Φ Φ dφ 2 = m2 The functions Φ(φ) must be periodic with period 2π, and this suggest that m is an integer and Φ = sin mφ or Φ = cos mφ

3 Separation of variables for the Laplace equation, continued Now we can replace 1 d 2 Φ Φ with the constant m 2 dφ 2 1 R ( d r 2 dr ) dr dr Θ sin θ d dθ ( sin θ dθ dθ ) m2 sin 2 θ = 0 The first term is a function of only r, and the last two terms are now functions of only θ, so we can take ( 1 d r 2 dr ) = k R dr dr Here k is just a constant that we will later take k = l(l + 1) Then we have the final equation for the θ-dependent terms, ( 1 d sin θ dθ ) ) + (k m2 sin θ dθ dθ sin 2 Θ = 0 θ

4 Separation of variables for the Laplace equation, continued We solved for Φ, but we still need to solve for R and Θ For the moment, let s focus only on the Θ function that solves ( 1 d sin θ dθ ) ) + (k m2 sin θ dθ dθ sin 2 Θ = 0 θ Make a change of variables to x = cos θ and then Θ(θ) y(x) Then we have d dθ = dx d dθ dx = sin θ d dx, and 1 sin 2 θ = 1 1 x 2 We can then obtain the associated Legendre equation ] y 2xy + [l(l + 1) m2 1 x 2 y = 0 So we have found the associated Legendre equation from Laplace equation in spherical coordinates! Hence we know Θ(θ) = P m l (cos θ)

5 Laplace equation in spherical coordinates, continued We will see later that R(r) = r l or R(r) = r l 1 Recall, we found Φ(φ) = cos(mφ) or Φ(φ) = sin(mφ) Finally u(r, θ, φ) = R(r)Θ(θ)Φ(φ) So our solutions are u = r l Pl m (cos θ) sin mφ, u = r l Pl m (cos θ) cos mφ, u = r l 1 Pl m (cos θ) sin mφ, u = r l 1 Pl m (cos θ) cos mφ Superposition applies! So in general, a solution might be a linear combination of these solutions for different l and m Will depend on boundary conditions! For example, maybe we are interested in solutions near r = 0 where r l 1 diverges, then l u(r, θ, φ) = r l Pl m m= l l=0 (cos θ) [a lm cos mφ + b lm sin mφ]

6 Steady-state temperature in a sphere Heat travels by diffusion, so we have 2 T = 1 α 2 T t The constant 1 depends on the properties of the medium, in α 2 one can show that 1 = C α 2 κ, where C is the specific heat capacity and κ is thermal conductivity This can be derived from Fourier s Law J = κ T (For an isotropic medium, actually the conductivity is a rank 2 tensor!) and the continuity equation u t = J = κ 2 T and u t = C T t In steady state T t = 0, so T = 0

7 Steady-state temperature in a sphere Consider a sphere of radius r = 1, with the temperature T = 100 on the top half (z > 0 or 0 < θ < π/2) and T = 0 on the bottom half (z < 0 or π/2 < θ < π) We know that our solution is a solution to Laplace equation 2 T = 0 most conveniently in spherical coordinates l T (r, θ, φ) = r l Pl m m= l l=0 (cos θ) [a lm cos mφ + b lm sin mφ] Based on the problem, we see there is no φ dependence, so we only require m = 0 and cos mφ = 1, so we simplify T (r, θ) = l=0 c l r l P m=0 l (cos θ)

8 Steady-state temperature in a sphere Solution is a series in the Legendre polynomials! At r = 1 we know T (r = 1, θ) T (r = 1, θ) = c l P l (cos θ) Find the coefficents c l! With T (r = 1, x = cos θ) = 100f (x), we have f (x) = 0 for 1 < x < 0 and f (x) = 1 for 0 < x < 1, and then l=0 ( ) 2l c l = 100 f (x)p l (x)dx 2 1 ( ) 1 1 ( ) 1 c 0 = 100 dx = ( ) 3 1 ( ) 3 c 1 = 100 xdx =

9 Steady-state temperature in a sphere, continued ( ) 5 1 ( 3 c 2 = x ( ) 7 1 ( 5 c 3 = x x 0 ) dx = 0 ) dx = 100 ( 7 ) 16 The integral for c 4 = 0... we can even see this by symmetry... so we can go to c 5 c 5 = 100 ( 11 2 ) 1 0 ( ) 1 (63x 5 70x x ) ( ) 11 dx =

10 Steady-state temperature in a sphere, continued Now that we have our Legendre coefficients, c 0 = 100 ( 1 2), c 1 = 100 ( ) 3 4, c2 = 0, c 3 = 100 ( 7 ) 16,c4 = 0, c 5 = 100 ( 11 32), etc., we can write a series solution T (r, θ) = c l r l P l (cos θ) l=0 [ 1 T (r, θ) = r cos θ 7 ( 5 16 r 3 2 cos3 θ 3 ) ] 2 cos θ +... Notice that we write T (r, θ) since we determined from the beginning that the solution is independent of φ It is crucial to remember that we solved this only for r < 1! Our solution does not apply outside of this region

11 Laplace equation in Cartesian coordinates The Laplace equation is written 2 φ = 0 For example, let us work in two dimensions so we have to find φ(x, y) from, 2 φ x φ y 2 = 0 We use the method of separation of variables and write φ(x, y) = X (x)y (y)

12 Laplace equation in Cartesian coordinates The Laplace equation is written 2 φ = 0 For example, let us work in two dimensions so we have to find φ(x, y) from, 2 φ x φ y 2 = 0 We use the method of separation of variables and write φ(x, y) = X (x)y (y) X X + Y Y = 0

13 Laplace equation in Cartesian coordinates, continued Again we have two terms that only depend on one independent variable, so Y Y = k2 This is called a Helmholtz equation (we ve seen in before), and we can write it Y + k 2 Y = 0 The we have another equation to solve, X k 2 X = 0 Here k is real and k 0 Could we have done this a different way? Yes!

14 Laplace equation in Cartesian coordiates, continued We could have a different sign for the constant, and then Y k 2 Y = 0 The we have another equation to solve, X + k 2 X = 0 We will see that the choice will determine the nature of the solutions, which in turn will depend on the boundary conditions

15 Steady-state temperature in a semi-infinite plate Imagine a metal plate bounded at y = 0 but extending to infinity in the +y direction, and from x = 0 to x = 10 Hold the y = 0 surface at T = 100, and as y +, T = 0 Surface at x = 0 and x = 10 are both held at T = 0 Find the scalar temperature field T (x, y) inside the plate Again we solve 2 T = 0, or 2 T x T y 2 = 0 We apply separation of variables T (x, y) = X (x)y (y), and choose the sign of the constant carefully! X X + Y Y = 0

16 Steady-state temperature in semi-infinite plate, continued We choose the sign of the constant to give us reasonable behavior based on the boundary conditions X + k 2 X = 0 Y k 2 Y = 0 From the first ordinary differential equation, we get X (x) = sin kx and X (x) = cos kx Since T = 0 at x = 0, the cos kx solution does not work Boundary condition T = 0 at x = 10 means we have solutions X n (x) = sin nπx 10, with n = 1, 2, 3,...

17 Steady-state temperature in semi-infinite plate, continued Now the equation for Y (y), taking care that we now have k 2 n = ( nπ 10 )2 Y k 2 ny = 0 We get solutions Y n (y) = e kny and Y n (y) = e kny Since T = 0 as y, we have Y n (y) = e kny = e nπy/10 We can then write the general solutions that satisfy the boundary conditions as y, and at x = 0 and x = 10 T (x, y) = n=1 b n e nπy/10 sin nπx 10

18 Steady-state temperature in semi-infinite plate, continued To determine b n coefficients, use that T = 100 at y = 0 T (x, y = 0) = n=1 A Fourier series! We find the b n from b n = sin nπx dx = (20) 10 b n sin nπx 10 = 100 ( 10 nπ ) ( cos nπx ) We find b n = 400 nπ for odd n, and b n = 0 for even n T = 400 π ( e πy/10 sin πx e 3πy/10 sin 3πx e πy/2 sin πx )

19 Time-dependent diffusion or heat flow If 2 T 0, then the temperature field becomes time-dependent Imagine a strip of metal, extending to ± in the y direction, but bounded at x = 0 and x = l At t = 0, the system is in steady-state with u = 0 at x = 0, and u = 100 at x = l The temperature profile at t = 0 satisfies 2 u = 0, or d2 u = 0 dx 2 We find initially a temperature profile u = 100 x l Now abruptly set u = 0 at x = 0 and x = l The temperature profile is now time-dependent d 2 u dx 2 = 1 du α 2 dt

20 Time-dependent diffusion or heat flow, continued We apply separation of variables u(x, t) = T (t)x (x) X X = 1 α 2 T T = k2 We find T = e k2 α 2 t From the boundary conditions u = 0 at x = 0 and x = l, we find X = sin nπx l with n = 1, 2, 3,... u(x, t) = n=1 b n sin nπx e l nπx ( ) l 2 t We find the b n from the initial condition u(x, t = 0) = 100 x l

21 Time-dependent diffusion or heat flow, initial conditions We can solve for the initial conditions (t = 0) u(x, t = 0) = 100 x = l n=1 b n sin nπx l Notice compared to before, when we expanded periodic functions, the y 0 (x) and v 0 (x) are not periodic with period l However, we can show that the sin nπx l make a complete, orthogonal set over the interval 0 < x < l We find the b n then, using that l 0 sin mπx l We find that b n = 200 ( 1) n 1 π n sin nπx dx = l l 2 δ m,n

22 Time-dependent diffusion or heat flow, final answer Then we can finally write the series solution u = 200 π [ e (πα/l)2t sin πx l e (2πα/l)2t sin 2πx l + e (3πα/l)2t sin 3πx l. Notice that as t, we reach equilibrium T = 0 everywhere Another example, slightly different... final conditions involve a temperature gradient