ecture 22 igher-dimensional PDEs Relevant section of text: Chapter 7 We now examine some PDEs in higher dimensions, i.e., R 2 and R 3. In general, the heat and wave equations in higher dimensions are given by u t = k 2 u (heat equation, (1 2 u t 2 = c2 2 u (wave equation. (2 The higher-dimensional form of the heat equation comes from the application of the conservation of thermal energy to regions in R n, along with Fourier s law of heat conduction and the Divergence Theorem. A derivation of the wave equation in higher dimensions would require a detailed discussion of stress and strain in solids, which we forego. In what follows, we examine the two-dimensional wave equation, ce it leads to some interesting and quite visualizable solutions. Vibrating rectangular, homogeneous membrane The vibration of a thin membrane is, for small displacements, modelled well by the two-dimensional wave equation 2 ( u 2 t 2 = u c2 x 2 + 2 u y 2. (3 ere u(x, y, t represents the displacement of the membrane from its equilibrium position, which is assumed to be the surface u(x,y = 0. The simplest problem is that of a rectangular membrane, represented by the region 0 x, 0 y, (4 which we shall simply denote as D R 2. Sometimes, we shall also write the above region as [0,] [0,]. 151
We ll assume that the membrane is clamped, leading to the following boundary conditions on u(x,y,t: u(x, 0, t = 0, u(, y, t = 0, u(x,, t = 0, u(0, y, t = 0. (5 We ll also need initial conditions: u(x, y, 0 = α(x, y (initial displacement u (x,y,0 = β(x,y (initial velocity. (6 t The above boundary conditions are homogeneous if two solutions u 1 (x,y,t and u 2 (x,y,t satisfy the BCs, then a linear combination also satisfies the BCs. This suggest that we can try the separation of variables method to produce an infinite set of solutions. We ll first try a separation of spatial and time-dependent solutions: u(x,y,t = φ(x,yh(t. (7 Substitution into Eq. (3 yields We may separate solutions as follows: φh = c 2 φ xx h + c 2 φ yy h. (8 h c 2 h = φ xx + φ yy φ = λ, λ > 0, (9 where we have introduced the negative separation constant λ, based upon past results. The timedependent equation for h(t, h + λc 2 h = 0, (10 has oscillatory solutions, which is desirable. (We have skipped a longer analysis of the problem which would have involved an examination of all cases, i.e., (i λ = 0, (ii λ < 0 and (iii λ > 0. The spatial function φ satisfies the equation 2 φ x 2 + 2 φ = λφ, (11 y2 152
or, simply, 2 φ + λφ = 0, (12 which is a two-dimensional regular Sturm-iouville eigenvalue equation, with boundary conditions, φ(0,y = 0, φ(x,0 = 0, φ(,y = 0, φ(x, = 0. (13 These are homogeneous BCs, suggesting that we can attempt a separation-of-variables-type solution for φ(x,y, i.e., φ(x,y = f(xg(y. (14 (Note that this implies that our original function u(x,y,t has been separated into three functions, u(x, y, t = f(xg(yh(t. (15 We could have started with this form, but it probably would not have been clear why it should work. Substitution of (14 into (11 yields f g + fg = λfg. (16 In an attempt to separate variables, we now divide by the term fg to give f f + g g = λ. (17 We have separated variables: the first term involves x and the second term involves y, but this doesn t allow us to solve for either f or g. What we have to do is to perform another separation a separation of x and y terms as follows, f f = λ g g. (18 Now we have all x-dependent parts on the left and y-dependent parts on the right. We now introduce a second separation constant: f f = λ g g which yields the following equations for f and g: = µ, µ > 0, (19 d 2 f + µf = 0, dx2 f(0 = f( = 0, d 2 g + (λ µg = 0, dy2 g(0 = g( = 0. (20 This is a system of two Sturm-iouville eigenvalue problems: 153
1. An equation for f with eigenvalue µ, 2. An equation for g with eigenvalue λ µ. In fact, the system is coupled because of the appearance of µ in both eigenvalues. It might seem somewhat confug that we have to deal with another separation constant. owever, all will be fine if we proceed systematically. are We can solve the eigenvalue equation for f very easily we ve done it many times! The solutions f n (x = ( nπx, µ n = ( nπ 2. (21 For each value of µ n, n = 1,2,, the equation for g becomes the eigenvalue problem For simplicity, we ll define so that the above equation becomes For each ν n, n = 1,2,, the solutions are known: with eigenvalues d 2 g dy 2 + (λ µ ng = 0, g(0 = g( = 0. (22 ν n = λ µ n, (23 d 2 g dy 2 + ν ng = 0, g(0 = g( = 0. (24 ν nm = g nm (y = ( mπy, (25 ( mπ 2, m = 1,2,. (26 Note that we had to introduce another index, m, so that the eigenvalue ν is now doubly indexed. We now resubstitute for ν nm in terms of λ: λ µ n = so that the eigenvalues λ of the 2D Sturm-iouville equation (11 become λ nm = ( nπ ( mπ 2, (27 2 ( mπ 2 +. (28 154
The associated eigenfunctions will be given by φ nm (x,y = f n (xg nm (y ( nπx = We now go back to the time-dependent equation for h(t: ( mπy. (29 h + λc 2 h = 0. (30 The functions h nm (t associated with the spatial functions φ nm (x,y will satisfy the equation h nm + λ nm c 2 h nm (t = 0. (31 The general solution is given by h nm (t = a nm cos( λ nm ct + b nm ( λ nm ct. (32 The product solutions, the normal modes of vibration of the membrane, are given by u nm (x,y,t = φ nm (x,yh nm (t ( nπx = ( mπy [ a nm cos( λ nm ct + b nm ( ] λ nm ct, n,m = 1,2, (33. Each mode is composed of a spatial profile φ nm (x that is modulated in time with frequency ω nm = c [ (nπ 2 ( mπ ] 2 1/2 λ mn = c +. (34 We now examine the first few modes nice pictures of these modes are to be found in the textbook. 1. n = m = 1. This is the mode with the lowest frequency, The spatial profile, [ (π 2 ( π ] 2 1/2 ω 11 = c +. (35 φ 11 (x,y = ( πx ( πy, (36 has no zeros in the interior of the rectangle. As a result, this mode has no nodes the entire membrane executes a uniform up-and-down motion. 155
2. n = 2,m = 1. The spatial profile ( 2πx φ 21 (x,y = ( πy, (37 has a zero at x = /2, which implies that the vibrational mode has a node along the line x = /2, 0 y. As a result, the membrane is divided into two regions, the displacements of which will lie on opposite sides of the xy-plane. The frequency of this mode is 3. n = 1,m = 2. The spatial profile ω 21 = c [ (2π φ 12 (x,y = ] 2 ( π 1/2 2 +. (38 ( πx ( 2πy, (39 has a zero at y = /2, which implies that the vibrational mode has a node along the line y = /2, 0 x. As a result, the membrane is divided into two regions, the displacements of which will lie on opposite sides of the xy-plane. The frequency of this mode is ω 12 = c [ (π 2 + ( 2π 2 ] 1/2. (40 Note that if <, then ω 21 > ω 12 and vice versa. If =, then ω 21 = ω 12 and the two modes have the same frequency of vibration. In this case, the membrane is a square, and the two modes are identical, up to a rotation of π/2 around the origin. 4. n = 2,m = 2. As in the previous case, the spatial profile ( 2πx φ 22 (x,y = ( 2πy, (41 has a zero at y = /2, which implies that the vibrational mode has a node along the line y = /2, 0 x. But it also has a zero at x = /2, which implies that the mode has a node along this line. As a result, the membrane is divided into four regions, the displacements of which will lie on opposite sides of the xy-plane. The frequency of this mode is [ (2π 2 ω 12 = c + ( ] 2π 2 1/2. (42 Note that the frequency of this mode, ω 22, is greater than the frequencies ω 12 and ω 21. 156
ecture 23 Vibrating rectangular membrane (conclusion In the previous lecture, ug separation of variables, we obtained the normal modes of vibration for this problem, u nm (x,y,t = φ nm (x,yh nm (t ( nπx = ( mπy [ a nm cos( λ nm ct + b nm ( ] λ nm ct, n,m = 1,2, (43. Each mode is composed of a spatial profile φ nm (x that is modulated in time with frequency ω nm = c [ (nπ 2 ( mπ ] 2 1/2 λ mn = c +. (44 Because of the homogeneous boundary conditions, the general solution to the vibrating membrane problem may be written as a superposition of these normal mode solutions, u(x,y,t = = n=1 m=1 n=1 m=1 φ nm (x,yh nm (t ( nπx ( mπy [ a nm cos( λ nm ct + b nm ( ] λ nm ct. (45 Imposition of the initial condition u(x,y,0 = α(x,y implies that α(x, y = = n=1 m=1 n=1 m=1 a nm ( nπx ( mπy a mn φ nm (x,y. (46 We now make use of the fact that the product basis φ nm (x,y forms an orthogonal set over the region D = [0,] [0,], i.e., φ n1,m 1 (x,yφ n2,m 2 (x,y dxdy = D = = ( n1 πx 0 0 [ ( n1 πx 0 2 2, (n 1,m 1 = (n 2,m 2 0, (n 1,m 1 (n 2,m 2. ( m1 πy ( n2 πx ( m2 πy dxdy ( n2 πx ][ ( m1 πy ( m2 πy dx 0 ] dy (47 157
We also state, without proof, that this product basis sometimes called a tensor product basis is complete in the function space 2 [D], the set of functions f : D R (or C that are square-integrable on D, i.e., D f(x,y 2 da <. (48 As such, we may multiply both sides of Eq. (46 by φ kl (x,y for a k 1 and l 1, and integrate over D to give ikewise, the initial velocity condition, yields b kl = a kl = 4 α(x,yφ nm (x,y dxdy. (49 D u (x,y,0 = β(x,y, (50 t 1 c 4 β(x,yφ nm (x,y dxdy. (51 λ nm D This concludes our discussion of the (clamped rectangular vibrating membrane problem. The relevance of the vibrating membrane problem to quantum mechanics The 1D vibrating string and 2D rectangular vibrating membrane problems are quite relevant to quantum mechanics, as we now discuss briefly. In one-dimension, the time-dependent Schrödinger equation for a particle in a box is given by the eigenvalue/bvp, 2 d 2 2m dx2ψ = Eψ, ψ(0 = ψ( = 0. (52 The wavefunction ψ(x provides a description of a particle that is confined inside a potential We rewrite Eq. (52 as The solutions of this BVP are well known to us: d 2 2mE dx2ψ + ψ = 0, ψ(0 = ψ( = 0. (53 2 λ n = 2mE 2 = n2 π 2 2, ψ n(x = λ n x. (54 This implies the existence of a discrete set of energy eigenvalues, E n = 2 2m ( nπ 2, n = 1,2,. (55 158
In two dimensions, we consider a particle confined in the two-dimensional region [0,] [0,] which is surrounded by potential walls of infinite height. The time-dependent Schrödinger equation for the particle is 2 2m 2 ψ = Eψ, (56 with the same boundary conditions as the clamped rectangular membrane. It may be rewritten as We have derived the solutions earlier: λ nm = 2mE nm 2 = ( nπ The energy eigenvalues of these wavefunctions are 2 ψ + 2mE ψ = 0. (57 2 2 ( mπ 2 +, ψnm (x,y = E nm = 2 2m [ (nπ ( nπx ( mπ. (58 2 ( mπ ] 2 +. (59 159
Vibrating circular membrane Relevant section of text: 7.7 We now consider the problem of a clamped, vibrating circular membrane, i.e., a drum. The vertical displacement satisfies the 2D wave equation, 2 u t 2 = c2 2 u. (60 Since the drum is circular, it is convenient to use polar coordinates, i.e., u = u(r,θ. We assume that the drum has radius a > 0. The clamping of the drum along the outer boundary implies the following boundary condition u(a,θ,t = 0, π/2 θ π/2, t 0. (61 As was the case for aplace s equation on a circular region, we can impose only one boundary condition at this time. velocity: In order to determine a unique solution, we ll need initial conditions, i.e., initial displacement and u(r,θ,0 = α(r,θ, u (r,θ,0 = β(r,θ. (62 t Since the boundary condition is homogeneous, we ll try separation of variables, i.e., u(r,θ,t = φ(r,θh(t. (63 Substitution into (60 yields φh = c 2 ( 2 φh. (64 Separating variables: This yields the following ODEs: h c 2 h = 2 φ = λ. (65 φ h + λc 2 h = 0, (66 and 2 φ + λφ = 0. (67 Note that these ODEs have the same form as those for the rectangular vibrating membrane. If λ > 0, the solutions for h(t are oscillatory, i.e., h(t = c 1 cos(c λt + c 2 (c λt. (68 160
We now examine the eigenvalue equation (67 for φ(r, θ. In polar coordinates, the aplacian operator has the form 2 = 1 r so that the eigenvalue equation assumes the form 1 r r ( r φ r We now assume a separation-of-variables solution for φ, i.e., ( r + 1r 2 r r 2 θ2, (69 + 1 2 φ r 2 + λφ = 0. (70 θ2 φ(r,θ = f(rg(θ, φ(a,θ = 0. (71 Substitution into (70 yields Now divide by fg and multiply by r 2 : g r ( d r df dr dr We haven t quite separated variables, so we do so now: + f d 2 g r 2 + λfg = 0. (72 dθ2 ( r d r df + 1 d 2 g f dr dr g dθ 2 + λr2 = 0. (73 1 g d 2 g dθ 2 = r ( d r df + λr 2 = µ, (74 f dr dr where we have introduced another separation constant, µ. This yields the two equations, and r d dr d 2 g + µg = 0, (75 dθ2 ( r df + (λr 2 µf = 0, f(a = 0. (76 dr The equation for g(θ will have the following periodicity conditions, g( π = g(π dg dg ( π = (π. (77 dθ dθ Periodic solutions to (75 exist for µ 0: g m (θ = cos(mθ, m 0 and (mθ, m 1. (78 The eigenvalues µ are given by µ m = m 2. 161
ecture 24 Vibrating circular membrane (cont d Relevant section of text: 7.7 We now turn to the radial equation (76 for f(r. If we divide by r, the equation is cast into Sturm-iouville eigenvalue form: d dr ( r df dr m2 f + λrf = 0, 0 < r a. (79 r ere, p(r = r, q(r = m 2 /r and σ(r = r. Note that r = 0 is a gular point because q(r is undefined there. As such this Sturm-iouville eigenvalue problem is not regular. Nevertheless, we claim that solutions exist for this problem as is the case for regular S- problems, that is, 1. A set of eigenvalues λ nm, n = 1,2,3,, m = 0,1,2,, 2. A set of corresponding eigenfunctions f nm : For each fixed m, the functions f nm (r are orthogonal with respect to the weight function σ(r = r, i.e., a 0 f nm (rf n m(r rdr = 0, for n n. (80 To produce these solutions, we make the change of variable z = λr in Eq. (79. The result is (Exercise: z 2d2 F dz 2 + zdf dz + (z2 m 2 F = 0, m = 0,1,2,. (81 where F(z = f(z/ λ. (In the lecture, I wrote, erroneously, F(z = f( λr. Sorry about that! Note that this removes the parameter λ from the ODE of course, it is now hidden in the independent variable z. You may recognize Eq. (81 it is Bessel s equation, which is studied in AMAT 351. Very briefly, z = 0 is a regular gular point. The standard practice is to assume a Frobenius series solution of the form z p a n z n = n=0 a n+p z n+p, a 0 0, (82 n=0 where p is to be determined. One substitutes the Frobenius series into Eq. (81, collects like terms in z k, and demands that the series begins with a term that starts with a nonzero term involving a 0. The result is a quadratic equation in p. For Eq. (81, the indicial equation for p is p 2 m 2 = 0 p = ±m. (83 162
For m 0, the two distinct values of p, i.e., p 1 = m and p 2 = m, yield two linearly independent series solutions, J m (z = z m [a m,0 + a m,1 z + ], Y m (z = J m (z = z m [b m,0 + b m,1 z + ]. (84 The actual values of the coefficients a m,i and b m,i are not important for our discussion. For m = 0, we have p 1 = p 2 = 0, so only one solution is produced by the series solution method: J 0 (z = 1 + a 0,1 z +. (85 A second, linearly independent solution may be produced by the variation of parameters applied to J 0 (z. The result is the following function Y 0 (z = 2 ln z +. (86 π It is also important to mention that for all m 0, the functions J m (z are oscillatory for z > 0. (Another important result that is shown in AMAT 351 is that as z, the spacing of consecutive zeros of J m (z approaches π. For m = 0,1,2,, the general solution to the Bessel equation (81 will have the form, F m (z = c 1 J m (z + c 2 Y m (z. (87 We then transform back to the r variable to produce the general solution f(r to Eq. (76, f m (r = c 1 J m ( λr + c 2 Y m ( λr. (88 We now apply these solutions to the vibrating circular membrane problem. Recall that the membrane being clamped at r = a, i.e., f(a = 0, gave us one boundary condition. Before considering that boundary condition, we consider another condition which deals with the gular nature of the point r = 0: It is the condition f must be finite at r = 0, i.e., f(0 <. (89 Recall that we imposed such a condition on solutions to aplace s equation on the circular disk. The point r = 0 corresponds to the point z = 0. From a look at the behaviour solutions J m (z above, we note that, in all cases, Y m (z as z 0. (90 163
This implies that these functions must be excluded from the general solution f m (r in Eq. (88, i.e., c 2 = 0 so that f(r = c 1 J m ( λr. (91 We now return to the boundary condition f(a = 0. It becomes the condition, J m ( λa = 0. (92 Given that the functions J m (z are oscillatory for z > 0, it follows that the argument λa is a zero of the function J m (z. In other words, λa = zmn > 0, (93 where z mn denotes the nth zero of the Bessel function J m (z. The zeros of the Bessel functions are presented in standard mathematical tables. (Indeed, the results of this lecture give an idea of why these zeros are so important. For example, approximate values of the first three zeros of J 0 (z are z 0,1 = 2.40482, z 0,2 = 5.52007, z 0,3 = 8.65372. (94 The graph of J 0 (x for 0 x 20 is presented below. (Just for interest s sake, the numerical values used to plot the graph were computed ug the series expansion for J 0 (z. Bessel function J_0(x 1 0.75 0.5 0.25 0-0.25-0.5-0.75-1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 x For each m 0, the above conditions produce an infinite sequence of eigenvalues λ mn given by λ mn = ( zmn 2, n = 1,2,. (95 a 164
The resulting radial functions, have n positive zeros, with the nth zero located at r = a. f mn (r = J m ( λ mn r, (96 Note: Before going on to construct the full solutions to the clamped vibrating membrane problem, let us stop for a moment and look again at the clamping condition (93 that produces the eigenvalues λ mn in (95. This condition is analogous to the boundary condition ( λ = 0, (97 encountered in the 1D wave equation ( clamped string, zero-endpoint conditions, which implied the result, λ = nπ, n = 1,2,, (98 yielding the eigenvalue condition, λ n = ( nπ 2, n = 1,2,. (99 This was a relatively straightforward result, ce the zeros of the function are so well known. Comparing Eqs. (99 with (95, 1. the zeros z mn of J m (z correspond to the zeros z n = nπ of (z and 2. a corresponds to. The solutions to the clamped vibrating membrane will then have the form u mn (r,θ,t = f mn (rg m (θh mn (t, m = 0,1,2,, (100 where f mn (r = J m ( λ mn r, g m (θ = a m cos mθ + b m mθ, h mn (t = c mn cos( λ mn ct + d mn ( λ mn ct. (101 These solutions correpond to the normal modes of vibration of the circular membrane. We shall return to them in the next lecture. 165
The general solution to the vibration problem will be given by u(r,θ,t = u mn (r,θ,t. (102 m=0 n=1 If we assume that the membrane is initially at rest, i.e., u/ t = 0, then the ( λ mn ct terms vanish, so that the general solution can be expressed in the following form, u(r,θ,t = A mn J m ( λ mn rcos(mθcos( λ mn ct m=0 n=1 + B mn J m ( λ mn r(mθcos( λ mn ct. (103 m=0 n=1 166