Connection to Laplacian in spherical coordinates (Chapter 13)

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

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 ) + 1 1 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

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 + 1 1 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

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)

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]

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

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)

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 + 1 1 c l = 100 f (x)p l (x)dx 2 1 ( ) 1 1 ( ) 1 c 0 = 100 dx = 100 2 0 2 ( ) 3 1 ( ) 3 c 1 = 100 xdx = 100 2 4 0

Steady-state temperature in a sphere, continued ( ) 5 1 ( 3 c 2 = 100 2 0 2 x 2 1 2 ( ) 7 1 ( 5 c 3 = 100 2 2 x 3 3 2 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 3 + 15x ) ( ) 11 dx = 100 8 32

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) = 100 2 + 3 4 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

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 2 + 2 φ y 2 = 0 We use the method of separation of variables and write φ(x, y) = X (x)y (y)

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 2 + 2 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

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!

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

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 2 + 2 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

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,...

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

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 = 2 10 10 0 100 sin nπx dx = (20) 10 b n sin nπx 10 = 100 ( 10 nπ ) ( cos nπx ) 10 0 10 We find b n = 400 nπ for odd n, and b n = 0 for even n T = 400 π ( e πy/10 sin πx 10 + 1 3 e 3πy/10 sin 3πx 10 + 1 5 e πy/2 sin πx ) 2 +...

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

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

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

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