c2 2 x2. (1) t = c2 2 u, (2) 2 = 2 x x 2, (3)

Transcription

1 ecture 13 The wave equation - final comments Sections of text by Haberman u(x,t), In the previous lecture, we studied the so-called wave equation in one-dimension, i.e., for a function It was derived from the modelling of a vibrating string. u t = u c2 2 x2. (1) We mention here that are multidimensional forms of the wave equation: Here we are concerned with functions u = u(x 1,x 2,,x n,t) of n spatial variables and one time variable. The wave equation will take the form where 2 denotes the aplacian operator in R n, in Cartesian coordinates. u t = c2 2 u, (2) 2 = 2 x x 2, (3) n For the case n = 2, the wave equation could model the vibration of a thin two-dimensional membrane, e.g., a drum. We shall consider this problem a little later in the course. And for the case n = 3, the wave equation can model the vibration of a solid three-dimensional body. But the wave equation is can also model the propagation of sound waves in a medium, as well as the propagation of electromagnetic waves. In the case of higher dimensions, however, it is possible that the speed of propagation of waves may not be the same in different directions, so that the term c 2 2 u may have to be replaced with a more complicated one. 84

2 aplace s equation and its solution using separation of variables Section 2.5 of text by Haberman Recall that aplace s equation for a function u : R n R is given by 2 u =. (4) The solution of this equation, with appropriate boundary conditions, is important in the determination of steady-state or equilibrium temperature distributions for the heat equation. In R n, the heat equation with sources will assume the general form cρ u t = K 2 u + Q, (5) We also assume the existence of boundary conditions appropriate to the problem of concern. If there exists a time-independent, steady-state temperature distribution u(x,t) = u eq (x), then the right hand side of the above equation is zero, implying that u eq must satisfy the PDE (or ODE in R), 2 u = Q K. (6) This is known as Poisson s equation. In the case that there are no sources, i.e., Q(x) =, Poisson s equation becomes aplace s equation. The above discussion also applies to electrostatics. Recall that the electrostatic potential function V (r) associated with a charge distribution ρ(r) satisfies Poisson s equation: 2 V = ρ. (7) ǫ In the case that there is no charge, then V satisfies aplace s equation. For this reason, all the heat problems that we examine in this section can also be viewed as electrostatic potential problems. We have already considered aplace s equation in one-dimension, i.e., d 2 u =, (8) dx2 along with various boundary conditions (e.g., fixed endpoint temperatures, zero flux, mixed conditions), for the determination of steady-state temperature distributions on a rod. The determination of these distributions was relatively straightforward. It is more complicated in higher dimensions. In what follows, we consider some rather simple, yet illustrative cases in R 2. 85

3 aplace s equation over a rectangular region Here, u = u(x,y) and we consider the problem 2 u x u =, (9) y2 over the rectangular region x, y. As for boundary conditions, we consider the case of fixed boundary temperature distributions, i.e., u(x,) = f 1 (x), u(,y) = g 2 (y), u(x,h) = f 2 (x), u(,y) = g 1 (y). (1) This situation is sketched below: y H f 2(x) g 1(y) g 2(y) O f 1(x) x These four conditions are nonhomogeneous as a result, the technique of superposition of solutions/separation of variables will not work. However, there is a trick that will allow us to use S of V: We ll divide the solution u(x,y) into four components, i.e., u(x,y) = u 1 (x,y) + u 2 (x,y) + u 3 (x,y) + u 4 (x,y). (11) Each of the components u i will satisfy one non-zero BC and three zero BCs. For example, we ll let u 1 (x,y) be the solution to Eq. (9) that satisfies the following BCs: BC1: u 1 (x,) = f 1 (x), BC2: u 1 (,y) =, BC3: u 1 (x,h) =, BC4: u 1 (,y) =. (12) 86

4 We now apply separation of variables to each u i function. For u 1 (x,y), let u 1 (x,y) = h(x)φ(y). (13) We re adopting the notation used in the book, in an effort to reduce confusion. (By the way, the textbook provides the solution for u 4 (x,y).) The three homogeneous BCs from above will yield the following conditions on h and φ: BC2: h()φ(y) = h() =, BC3: h(x)φ(h) = φ(h) =, BC4: h()φ(y) = h() =. (14) We see that there are two BCs for h(x) and only one for φ(y). The situation is sketched below: y H O f 1(x) x Substitution of (13) into (9) yields which can be separated to h (x)φ(y) + h(x)φ (y) =, (15) h (x) h(x) = (y) φ = µ. (16) φ(y) We do not yet know whether µ should be positive or negative, so we just leave it for now. The separation yields the following problems for h and φ: and h µh =, h() =, h() =, (17) φ + µφ =, φ(h) =. (18) We ve seen the BVP for h(x) before: nontrivial solutions exist only for µ <, so we let µ = λ, λ > to give h + λh =, h() =, h() =, (19) 87

5 with eigenvalues and associated eigenfunctions λ = λ n = ( nπ ) 2, n = 1,2,, (2) h n (x) = sin ( nπ ). (21) We now consider the φ-equation, recalling that µ = λ, so that µ n = λ n. The associated φ n functions will satisfy the DE φ n(y) ( nπ ) 2 φn (y) =, φ(h) =. (22) This is not a boundary value problem but an initial value problem, with only one condition. The general solution could be written as φ n (y) = C 1 e nπy/ + C 2 e nπy/ (23) but it will be more convenient to use hyperbolic functions, i.e., φ n (y) = D 1 cosh(nπy/) + D 2 sinh( nπy/). (24) In order to impose the boundary condition at y = H, it is even more convenient to use shifted hyperbolic functions, which also satisfy the DE: φ n (y) = A 1 cosh nπ (y H) + A 2 sinh nπ (y H). (25) The condition φ n (H) = implies that A 1 = (Exercise) so that the φ n (x) functions associated with the h n (x) functions are φ n (y) = A 2 sinh nπ (y H). (26) As a result, the product solutions yielded by separation of variables are (up to a constant) u 1,n (x,y) = h n (x)φ n (y) ( nπx ) = sin sinh nπ (y H). (27) Note that these functions are oscillatory in the x-direction but nonoscillatory in the y-direction. 88

6 ecture 14 aplace s equation over a rectangular region (cont d) In the previous lecture, we used separation of variables to construct the following solutions for the function u 1 (x,y) which satisfied zero boundary conditions on three sides of the rectangle: It now remains to accomodate the nonzero BC u 1,n (x,y) = h n (x)φ n (y) ( nπx ) = sin sinh nπ (y H). (28) u 1 (x,) = f 1 (x), x. (29) As we have done before, we look for an appropriate superposition of these solutions, i.e., It follows that u 1 (x,y) = u 1 (x,) = = n=1 a n u 1,n (x,y) n=1 a n sin n=1 ( nπx ) sinh nπ (y H). (3) ( a n sinh nπh ) ( nπx ) sin = f(x). (31) This is just a Fourier expansion of f(x) in the complete basis φ n (x), where the coefficients are a n sinh( nπh/). If we multiply the middle and right sides of the equation by sin(kπx/) and integrate over [,], we obtain, by virtue of the orthogonality of the φ n (x), ( ) ( a k sinh kπh ) ( ) kπx = f(x)sin dx, k = 1,2,. (32) 2 We then rearrange to solve for the a k : a k = 2 sinh( kπh/) ( ) kπx f(x)sin dx, k = 1,2,. (33) The sinh function is odd, so we could write the above equations as follows, 2 ( ) kπx a k = f(x)sin dx, k = 1,2,. (34) sinh(kπh/) Now we repeat the procedure for u 2 (x,y), u 3 (x,y) and u 4 (x,y)! We can then construct the solution u(x,y) from Eq. (11). 89

7 Example: Here we determine the steady-state temperature distribution on [, ] [, H] which satisfies the boundary conditions, u(x,) = f 1 (x) = 1, u(,y) =, u(x,h) =, u(,y) =. (35) (This boundary value problem corresponds to the function u 1 (x,y) discussed in class.) The Fourier expansion coefficients for the constant function f 1 (x) = 1 were computed earlier in the course. But now they have to be modified by the sinh term in Eq. (33): 4 kπ sinh( kπ/h) a k =, k even,, k odd. (36) In the figure below, = H = 1 and M = 1 terms were used in the expansion of u(x,y), i.e., 1 ( nπx ) ( nπ ) u(x,y) a n sin sinh (y H). (37) n=1 Steady-state heat distribution u(x, y): = H = 1. The shading at a point (x, y) is proportional to its temperature u(x, y) 1. The darker the shade, the higher the temperature. As, the effects of the vertical boundaries at x = and x = will become negligible, at least for points sufficiently far away from these boundaries, so that the distribution in the interior should approach the one-dimensional case u(x,y) u eq (y) = 1 (H y). (38) H 9

8 Steady-state heat distribution u(x, y): = 1, H = 1. The distribution for = 1, H = 1 is shown above. aplace s equation over a circular region For physical problems with circular symmetry, e.g. a circular disk, it is convenient to work in planar polar coordinates (r,θ) so that u = u(r,θ). In polar coordinates, aplace s equation becomes 2 u = 1 ( r u ) u r r r r 2 =. (39) θ2 There are several conventions for the domain of definition of the angular coordinate θ. Here, we shall use θ π. Circular disk, radius a We shall solve aplace s equation 2 u = over the region r a, θ π. There is only one boundary for this region, the outer perimeter r = a. Over this boundary, we shall impose the boundary condition BC1: u(a,θ) = f(θ). (4) Of particular interest will be the case f(θ) = T, constant. The polar coordinates of any point (x,y) (,) are unique. This is not the case at (,), for which r = but θ is not unique. This is simply an illustration of the fact that polar coordinates are singular at r =. This is also reflected in the aplacian in Eq. (39) the point r = is a singular point of the differential equation in r that will result from separation of variables. Because of this singularity at r =, we ll also need a condition on solutions there: With an eye to physical applications, we impose the condition of boundedness, C2: u(, θ) <. (41) 91

9 We shall also need periodicity conditions on u that imply continuity and continuous differentiability across the ray θ = π = : C4: C3: u(r, ) = u(r, π) (42) u (r, ) = u θ (r,π) (43) θ (44) As a result, there are four conditions on the solution u(r,θ). Only the outer boundary condition in Eq. (4) is nonhomogeneous. (Exercise.) As a result, we shall try to use the separation of variables technique to construct solutions to (39) with the above boundary conditions. Following the notation in the book, we write u(r,θ) = G(r)φ(θ). (45) The conditions C2 C4 translate to the following: C2 : G()φ(θ) < G() <, (46) C3 : G(r)φ() = G(r)φ(π) φ() = φ(π), (47) C4 : G(r)φ () = G(r)φ (π) φ () = φ (π). (48) Substitution of (45) into (39) yields 1 r d dr ( r dg(r) ) dr φ(θ) + 1 φ r 2G(r)d2 =. (49) dθ2 We can separate variables by moving the second term on the HS to the RHS, multiplying by r 2 and dividing by G(r): 1 G r d ( r dg(r) ) = 1 d 2 φ = µ, (5) dr dr φ dθ2 where, once again, µ is the separation constant. We ll determine the restrictions on µ shortly. First, we examine the resulting φ-equation, φ + µφ =, φ(π) = φ(), φ (π) = φ (). (51) The general solution to this DE is φ(θ) = C 1 cos( µθ) + C 2 sin( µθ). (52) 92

10 The first condition (C3 ) implies that C 1 cos( µπ) + C 2 sin( µπ) = C 1 cos( µ()) + C 2 sin( µ()). (53) Since cos is an even function, we have that C 2 sin( µπ) =. (54) The second condition (C4 ) implies that C 1 sin( µπ) + C 2 cos( µπ) = C 1 sin( µ()) + C 2 cos( µ()). (55) Once again, since cos is an even function, we have that C 1 sin( µπ) =. (56) Therefore, conditions (54) and (56) must be satisfied simultaneously. Since C 1 and C 2 cannot be both zero (otherwise the solution is the trivial zero solution), it follows that µπ = nπ, n =, ±1, ±2,. (57) This, in turn, implies that the separation constant may assume the discrete values µ = µ n = n 2, n =,1,2,. (58) (The negative values of n yield the same values of µ.) We may separate the sin and cos solutions into the sets: sin(nθ), n = 1,2,, (59) and cos(nθ), n =,1,2,. (6) (Note that n = is excluded from the sin case since it yields the trivial zero solution). Another way to express this set is as follows: 1, cos(nθ), sin(nθ), n = 1,2,. (61) Note that these functions form an orthogonal set on [,π]. 93

11 ecture 15 aplace s equation over a circular region (cont d) Circular disk, radius a (cont d) We now return to the r-equation obtained by separation of variables: r d G dr ( r dg ) = µ = n 2, n =,1,2, (62) dr where we have substituted the values of the separation constant µ as determined from our analysis of the φ-equation. The above DE in G(r) may be rewritten as follows: r 2d2 G dr 2 + rdg dr n2 G =. (63) This DE happens to be a special one that you may have encountered in another DEs course it is known as Euler s equation. It is special because of the r 2, r 1 and r terms multiplying the second-, first- and zeroth order derivatives of G(r). In fact, one might guess that a suitable power of r could be a solution to this DE. For example, if you let G(r) = r p, then taking two derivatives and multiplying by r 2 yields, up to a coefficient, r p again. The same is true for the middle term. So let s substitute G(r) = r p into the above DE: p(p 1)r p + pr p n 2 r p = [p 2 n 2 ]r p =. (64) Since this relation must hold for all r, we have that p = ±n. There is a problem when n =, however only one solution, i.e., G(r) = r = C is obtained. We ll have to treat the case n = separately. For n 1, we may write the general solution to Eq. (63) as G n (r) = c 1 r n + c 2 r n, n = 1,2,. (65) et s now return to Eq. (63) to treat the case n =. We ll let H(r) = G (r) so that Eq. (63) becomes Integration yields r 2 H (r) + rh(r) = or H (r) H(r) = 1, r. (66) r ln H(r) = ln r + C = ln 1 r + C H(r) = A1 r. (67) Thus G(r) = Aln r. For n =, therefore, the general solution is G (r) = a 1 + a 2 ln r. (68) 94

12 After all of that work, we re going to have to do some cutting, because of the boundedness constraint discussed in the previous lecture: The point r = is a singular point of Eq. (63), which means that some solutions may not be continuous there. For solutions to be useful physically, demand that they be at least bounded at r =. As a result, the terms c 2 r n in G n (r) and a 2 ln r in G (r) must be omitted, by setting c 2 = a 2 =. The net result is that solutions to the r-equation (63) assume the form G n (r) = r n, n =,1,2,. (69) We now collect our results to construct the product solutions u(r, θ) = G(r)φ(θ) produced by the separation of variables method. These solutions will have the form u n (r,θ) = r n cos(nθ), n =,1,2,, v n (r,θ) = r n sin(nθ), n = 1,2,. (7) Each of these functions is a solution to aplace s equation on the circular disk r a. However, no single one of these functions will generally satisfy a given boundary condition u(r, θ) = f(θ). We shall need to construct an infinite superposition of these functions to accomodate arbitrary boundary conditions: u(r,θ) = which we shall write in the form, A n u n (r,θ) + B n v n (r,θ), (71) n= n=1 u(r,θ) = A + A n r n cos(nθ) + B n r n sin(nθ). (72) n= n=1 The boundary condition u(a,θ) = f(θ) (73) then becomes f(θ) = A + A n a n cos(nθ) + B n a n sin(nθ). (74) n= n=1 This is essentially the Fourier expansion of the boundary value function f(θ), θ π, in terms of the orthogonal set of functions {cos(nθ),sin(nθ)} n= over this interval. The minor difference between this and the classical Fourier expansion is the appearance of the multiplicative terms a n. We may extract the values of the expansion coefficients A n and B n by exploiting the orthogonality of the basis functions. 95

13 In order to obtain A ; we simply integrate both sides of the above equation with respect to θ over [, π] this is equivalent to multiplying by the function g(x) = 1 and integrating. All integrals in cos(nθ) and sin(nθ) will vanish, leaving A = 1 f(θ) dθ. (75) 2π In order to determine A k, we multiply both sides of Eq. (74) by cos(kθ) and integrate over θ: πa k a k = f(θ)cos(kθ) dθ A k = 1 a k f(θ)cos(kθ) dθ, k = 1,2,. (76) π Similarly, to determine the B k, multiply both sides of (74) by sin(kθ) and integrate over θ: πb k a k = f(θ)sin(kθ) dθ B k = 1 a k f(θ)sin(kθ) dθ, k = 1,2,. (77) π Mean-value property Before continuing with some examples, there is an important note regarding Eq. (75) for the A coefficient of the expansion. Since the temperature at the center of the disk is u(,θ) = A, we have u(,θ) = 1 f(θ) dθ 2π 1 π = f(θ) a dθ 2πa 1 = f(θ) ds (78) 2πa C a = f, (79) where C a denotes the circular boundary of radius a, center O, and f denotes the average value of f(θ) over this closed curve. In other words, the temperature at the center of the disk is the average value of the temperatures over the boundary curve. This is an illustration of a more general mean-value property that is satisfied by solutions of aplace s equation. We refer the reader to the textbook by Haberman, p. 83, for a more detailed discussion. Example 1: We now return to the circular disk problem, i.e., 2 u = for r a and θ π, with the boundary condition u(a,θ) = f(θ) = T (constant). (8) 96

14 In other words, the outer circular boundary is kept at a constant temperature T. In this case, the coefficients A n and B n are easy to compute: A = 1 2π A n = T a n π B n = T a n π T dθ = T, cos(nθ) dθ =, sin(nθ) dθ =. (81) Therefore, the solution to aplace s equation is u(r,θ) = T. In other words, the steady-state temperature distribution for this problem is the constant temperature function u eq (r,θ) = T. Example 2: We now consider the boundary condition, u(a,θ) = f(θ) = T cos θ, π θ π. (82) The temperature at a fixed point on the boundary is held constant in time, but the temperatures vary as we move over the boundary. Some sample values are sketched below. T/2 T/2 T T T/2 T/2 The reader might notice that the function f(θ) is already expressed in terms of its Fourier series, i.e., it consists of only one term, namely, cos(θ). We expect that only the term A 1 in Eq. (74) will be nonzero. That being said, let s formally compute the coefficients: The net result is A = 1 2π A 1 = T aπ A n = T a n π B n = T a n π T cos θ dθ = T 2π cos θ cos θ dθ = T a, cos θ dθ =, cos θ cos(nθ) dθ =, n > 1, cos θ sin(nθ) dθ =, n 1. (83) u(r,θ) = A 1 r cos θ = T r cos θ. (84) a 97

15 Just to check: u(a,θ) = T a acos θ = T cos θ, (85) so the boundary condition is satisfied. The level sets of u(r,θ), i.e., a contour diagram of the function, may give an idea of the qualitative behaviour of this steady-state temperature distribution. We might first look for (r,θ) such that u(r,θ) = T a r cos θ = C, constant. (86) But in this case, note that r cos θ = x, in other words T a x = C, (87) implying that lines of constant x-value parallel to the y-axis are contours of this steady-state temperature distribution. The situation is sketched below. T/2 T/2 T T 3T/4 3T/4 T/4 T/4 Example 3: We now consider an annular region instead of a circular one, i.e., the washer-shaped region < a r b, θ π. This region has two boundaries, an inner one at r = a and an outer one at r = b. et us impose the following boundary conditions, u(a,θ) = f(θ) = T a, u(b,θ) = g(θ) = T b. (88) The problem is to find the steady-state temperature distribution in the washer under these boundary conditions. We must solve aplace s equation 2 u = in this region subject to the above two conditions. You actually solved this problem recently in Problem Set No. 2 we wish to show how this problem is a special case of the general treatment formulated earlier. Because the singular point r = has now been removed from our region of concern, the previously troublesome terms, r n and ln r may now be included in the radial functions G n (r): G n (r) = c 1 r n + c 2 r n, n = 1,2, G (r) = a 1 + a 2 ln r. (89) 98

16 Since the boundary temperature distributions f(θ) and g(θ) are constant, all terms in cos(nθ) and sin(nθ) will be zero. The only contribution to u(r, θ) will be from the product solution corresponding to n =, i.e., u(r,θ) = u (r,θ) = G (r) = a 1 + a 2 ln r. (9) We see that the steady-state temperature distribution does not depend upon θ, because the boundary conditions are θ-independent. As a result, we may write u(r,θ) = u(r) = a 1 + a 2 ln r. (91) We now impose the two boundary conditions on u(r): u(a) = T a a 1 + a 2 ln a = T a u(b) = T b a 1 + a 2 ln b = T b (92) Subtracting the second equation from the first yields Solving for a 1 yields The net result is that a 2 ln(a/b) = T a T b a 2 = T b T a ln(b/a). (93) a 1 = T a T b T a ln a. (94) ln(b/a) u(r) = T a + T b T a ln(r/a). (95) ln(b/a) In other words, the interpolation between T a (inner) and T b (outer) is logarithmic and not linear, as in the one-dimensional case. 99