Circular Membranes. Farlow, Lesson 30. November 21, Department of Mathematical Sciences Florida Atlantic University Boca Raton, FL 33431

Transcription

1 The Problem Polar coordinates Solving the Problem by Separation of Variables Circular Membranes Farlow, Lesson 30 Department of Mathematical Sciences Florida Atlantic University Boca Raton, FL November 21, 2008 Farlow, Lesson 30 CM

2 Outline 1 The Problem 2 Polar coordinates 3 Solving the Problem by Separation of Variables

3 Description We consider a circular membrane or drumhead of radius a > 0, xed around the edges, and able to vibrate. We set up coordinates so that the center of the membrane coincides with the origin. We denote by u(x; y ; t) the displacement from the horizontal at time t of the point of the membrane at (x; y). We want to get a formula or expression for u.

4 The model We model u as a solution of the acoustic equation u tt = c 2 r 2 u = c 2 (u xx + u yy ): We are led to consider the following initial values, boundary conditions problem: The equation: u tt = c 2 r 2 u; x 2 + y 2 a 2 ; t 0, The boundary conditions: u(x; y ; t) = 0; x 2 + y 2 = a 2 ; t 0. The initial conditions: u(x; y ; 0) = (x; y) u t (x; y ; 0) = (x; y):

5 The Laplacian in polar coordinates Because of the circular symmetry of the problem, we describe it and solve it in polar coordinates The inverse relations are x = r cos ; y = r sin r = p x 2 + y 2 = arctan y x : (The expression for is actually a bit more complicated, but what I wrote is good enough to get x ; y, which is all we really need).

6 Laplacian in polar coordinates, continued Now: u x = u r r x + u x ; r x = 1 2 (x 2 + y 2 ) 1=2 (2x) = x r ; 1 x = 1 + y x x y u x = u r u r r 2 ; x x u xx = (u r ) + x r u r r 2 y x 2 = = (u rr r x + u r x ) x r + u r x y r 2 ; (u ) x y r xr x r 2 (u r r x + u x ) y r 2 + u 2yr x r 3 r 2 u y r 2 x

7 Laplacian in polar coordinates, continued Using again the expressions for r x ; x obtained above, collecting terms, one ends with x 2 u xx = u rr r 2 2u xy r r 3 + u y 2 r 4 + u y 2 r r 3 + 2u xy r 4 : Similarly, using r y = y =r, y = x=r 2, one gets y 2 u yy = u rr r 2 + 2u xy r r 3 + u x 2 r 4 + u x 2 r r 3 2u xy r 4 : Adding, remembering that x 2 + y 2 = r 2 we get the Laplacian in polar coordinates r 2 u = u rr + 1 r u r + 1 r 2 u.

8 The problem in polar coordinates The equation: u tt = c 2 (u rr + 1 r u r + 1 r 2 u ); 0 r a; 0 2; t BC: u(a; ; t) = 0; 0 2; t 0; IC: u(r ; ; 0) = (r ; ); u t (r ; ; 0) = (r ; ) 0 r a; 0 2:

9 Steps We will separate variables in two steps. The rst step applies to any BCIC problem of the form u tt = c 2 r 2 u (x; y) 2 D; t 0 u(x; y ; t) = 0 (x; y) t 0; u(x; y ; 0) = (x; y) (x; y) 2 D; u t (x; y ; 0) = (x; y) (x; y) 2 D; where D is a domain (region) in the being the boundary of D. We look for solutions of the equation the form u(x; y ; t) = w(x; y)t (t) satisfying the boundary condition. Once we get all possible such solutions we obtain the solution satisfying the initial conditions by superposition.

10 The rst step Plugging u = wt into the equation, we get wt 00 = c 2 (r 2 w)t ; dividing by c 2 wt and using the fact that a function of the space variables can only equal one of time if they are both constants: T 00 c 2 T = r2 w w = : (We call the constant rather than because then it will be positive). The boundary conditions require that w be zero on the boundary of the region D (a circle of radius a in the case of the membrane). We get the following equation/bc Problem: T 00 + c 2 T = 0 (1) r 2 w + w = 0 (x; y) 2 D (2) w(x; y) = 0 (x; y)

11 The very long second step We now specialize to the case of the membrane, solving the BC problem (2) in this case. That means nding the eigenvalues for which that problem has a non-zero solution, then returning with these values to solve (1). We assume now that everything is described in polar coordinates, so w = w(r ; ) and (2) is w rr + 1 r w r + 1 r 2 w + w = 0; 0 r a; 0 2; t 0 w(a; ; t) = 0 0 2; t 0 We look for solutions w(r ; ) = R(r) (). The condition w(a; ) = 0 then translates to R(a) = 0. Plugging w = R into the equation, multiplying by r 2 =(R ), separating the variables, etc., gives r 2 R 00 + rr 0 + R R r 2 00 = = : Calling the constant makes sense because this way it will be positive. We must solve:

12 Second step, continued 00 + = 0 (3) r 2 R 00 + rr 0 + (r 2 )R = 0 (4) R(a) = 0 It seems we have no condition on the solution of (3). But there is a hidden one. For obvious physical reasons, must be periodic of period 2. If < 0, the solutions of (3) are exponentials, highly non periodic. If = 0, there is precisely one: The constant solution. If m > 0, the solutions are combinations of cos p, sin p, periodic of fundamental period 2= p. The only way that 2 can be a multiple of this period (so that we have period 2 is if p = n, n = 1; 2; : : :. The conclusion is that = n 2, n = 0; 1; 2; : : :, and the solutions of (3) have the form 0 () = a 0 ; n () = a n cos n + b n sin n; n = 1; 2; : : :

13 Slogging along with step 2 Now that we know that = n 2, n = 0; 1; 2; : : :, we can use it and rewrite (4). I will now state as a fact that one must have > 0. The best way one justies that is using either Green's Theorem or Gauss' divergence theorem. Details provided upon request. So we will fall back to the textbook's notation, replacing by 2 so that now (1) and (4) become T 00 + c 2 2 T = 0; (5) r 2 R 00 + rr 0 + ( 2 r 2 n 2 )R = 0; R(a) = 0: (6) Equation (6) is almost in the form of a very studied equation. To put it completely in that form we introduce a new function of r (and ) that we will call J; the denition is J(s) = R(s=) or R(r) = J(r). With s = r, the equation/bc problem for J is s 2 J 00 + sj 0 + (s 2 n 2 )J = 0; J(a) = 0: (7)

14 and slogging... The equation s 2 J 00 + sj 0 + (s 2 n 2 )J = 0 is known as Bessel's equation of order n; it is a linear ODE of order 2 and as such has two linearly independent solutions. Of these solutions, one is singular (goes to 1) at 0, so we discard it. Finite solutions are multiples of what is called the Bessel function of the rst kind of order n, denoted by J n. The end conclusion is that R(r) = J n (r).

15 Still in the second step = Here is a fact, valid for Bessel functions of the rst kind. The equation J n (s) = 0 has a sequence of positive solutions 0 < k n1 < k n2 < ; lim m!1 k nm = 1: Some values are given in the textbook. The condition 0 = R(a) = J n (a) translates to = k nm =a; m = 1; 2; : : :. From now on, for simplicity, I'll assume a = 1, so = k nm. The equation for T can be written in the form T 00 + c 2 k 2 nm T = 0; so that the solution is T (t) = c nm cos ck nm t + d nm sin ck nm t. This gives us now this lovely nal expression of our solution, after solving the equation for T u(r ; ; t) 1X 1X n=0 m=1 J n (k nm r)(c nm cos ck nm t + d nm sin ck nm t)(a n cos n + b n sin n):

16 The end For more details, one can now turn to the textbook or to my next set of notes (once they appear): Circular Membranes Part Deux.