Department of Mathematics

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1 INDIAN INSTITUTE OF TECHNOLOGY, BOMBAY Department of Mathematics MA 04 - Complex Analysis & PDE s Solutions to Tutorial No.13 Q. 1 (T) Assuming that term-wise differentiation is permissible, show that a solution of Laplace equation in the disc z < 1 satisfying the boundary condition u(1, θ) = f(θ) is given by u(r, θ) = a 0 + n=1 where a n, b n are the Fourier coefficients of f. r n (a n cos nθ + b n sin nθ) Sol: You may use Laplace equation in polar coordinates and verify directly that for each n r n cos nθ, r n sin nθ satisfy it. Alternately, you may observe that r n cos nθ, r n sin nθ are real and imaginary parts of the complex analytic function z n and hence they are harmonic. Hence, using term-wise differentiation, it follows that the infinite sum also satisfies the Laplace equation. Now the boundary condition gives f(θ) = u(1, θ) = a 0 + (a n cos nθ + b n sin nθ). n=1 Treating the function f (which may be defined only on 0 θ ) as a periodic function on the whole of the real line, we see that a n, b n are its Fourier coefficients. Therefore, a 0 = 1 f(θ)dθ, a n = 1 f(θ) cos nθ dθ, b n = 1 f(θ) sin nθ dθ. Q. (T) Find the electrostatic potential in the disc z < 1 assuming that the potential is kept at 50 volts on the upper-half of the circle and at 50 volts on the lower-half of the circle. Do the problem both using PDE s and using conformal mappings. Compare the two methods. Sol: Take f(θ) = 50, θ > 0 and = 50 for θ < 0 and apply the previous exercise. We have a n = 0 for all n and Therefore, b n = 0 50 sin nθdθ = u(r, θ) = 00 n=0 00 n, n odd 0 n even. sin(n + 1)θ. n + 1

2 To use conformal method, we transform the unit circle onto the upper half plane via the Cayley map: w = i 1 + z 1 z and observe that the upper-semi circle (resp. lower semi-circle) is mapped onto the negative (resp. positive) real axis. The transformed problem in the w-plane is then easily solved by taking a suitable linear function of Arg w viz., 100 Arg w 50 Therefore the required solution is Arg w 50 = Arg ( i 1+z ) 1 z 50 = 100 Arg ( ) 1+z 1 z = 100 = 100 [Arg(1 z) Arg(1 + z)] Im[Ln(1 + z) Ln(1 z)]. (It may be easily checked that this coincides with the result we have obtained in the earlier method.) The advantage of the second method is that it gives the answer in a closed formula. On the other hand, its success is restricted to cases wherein we can find a suitable and not too complicated conformal transformation. The first method has wider applicability. [Yet another method is to use directly, the Poisson integral formula: In this method, we simply use the following fact: (the proof which we have not covered.) Given a piecewise continuous function f on the boundary of a disc z R the Poisson integral formula, u(r, θ) = 1 0 R r f(r, α) dα (1) R Rr(cos(θ α) + r defines a harmonic function inside the disc, which coincides with f at all points on the boundary where f is continuous. (The first assertion is easy, we you remember that the integrand in (1) was the real part of a holomorphic function. The second part follows from a similar statement about Fourier transforms.] Q. 3 (S) Find steady state temperature u in a semi-circular thin plate 0 < r < a, 0 < θ < if the semi-circle r = a is kept at constant temperature u = u 0 and the bounding segment a < x < a is kept at temperature u = 0. Do the problem both using PDE s and using conformal mappings. Compare the two methods.

3 Sol: Similar to the above problem. Here again, a n = 0 for all n, because u(r, 0) = 0. Therefore Now u(a, θ) = u 0 implies that u(r, θ) = u 0 = n=1 n=1 Therefore, b n = 0 for n even and 4an b n r n cos nθ. b n a n sin nθ. for n odd. In the second method, we can use the same transformation as before except for a scaling factor, viz., consider w = i 1 + z/a 1 z/a which maps the upper-half of the disc z < a onto the second quadrant. The segment a < x < a is mapped to the positive y-axis and the upper semi-circular part onto the negative x-axis. Therefore, a suitable linear function of Arg w should be chosen so that it is 0 on the positive y-axis and equals u 0 on the negative real axis. Therefore we have u = u 0 = u 0 = u 0 [ Arg ( Arg 1+z/a 1 z/a i 1+z/a 1 z/a ) ] Im[Ln(1 + z/a) Ln(1 z/a)]. [This problem can also be done by using Schwartz reflection principle to convert it into the problem in Q., viz., define u(x, y) = u(x, y), for (x, y) in the lower-half disc. The principle says that the extended function u is then harmonic in the whole disc.] Q.4 (L) Using polar coordinates, determine the vibrations u(r, θ, t) of a circular membrane with the boundary condition and with symmetrical initial conditions u(r, θ, t) = 0 () u(r, θ, 0) = f(r); u t (r, θ, 0) = g(r). (3) Sol: Naturally, it should help us to solve the problem if we use ploar cooridnate form of the euqation: x = r cos θ, y = r sin θ. 3

4 u tt = c (u rr + u r r + u θθ r ) (4) The symmetrical initial conditions suggest that we hould try solutions which are independent of the variable θ. Thus the equation (4) reduces to u tt = c (u rr + u r ). (5) r Step I Putting u(r, t) = W (r)g(t) we obtain two ordinary differential equations: G tt c G = W rr + W r /r = k. (6) W (The first equation demands that the constant be negative.) The second can be reduced to Bessel s equation, if we put s = kr. W r = W s k; W rr = W ss k and therefore, (The paramenter ν = 0 here.) W ss + W s s + W = 0. (7) The solutions of this are Bessel functions J 0 and Y 0 of first and second kind respectively: J 0 (r) = ( 1) m r m (8) m=0 m (m!) Y 0 (r) = (J 0 (r) ln r ) + γ + ( 1) m 1 h m r m. (9) m (m!) Step II Since Y 0 (r) as r 0 we cannot use this function for the vibrations of the membrane. Therefore, we have W (r) = J 0 (kr) and its multiples as the only possibilities for W. Since W (R) = 0 it follows that the choices for k are such that kr are positive zeros of the J 0 which are infinitely many, say, kr = α m, k m = α m /R. Correspondingly, we get functions W m (r) = J 0 (α m r/r). Step III Corresponding to each m 1, we have the eigenvalues λ m = ck m and the time functions m=1 and the eigenfunctions G m (t) = A m cos λ m t + B m sin λ m t (10) u m (r, t) = W m (r)g m (t) = J 0 (k m r)(a m cos λ m t + B m sin λ m t) (11) which are called (m t h) normal modes. 4

5 The zeros of u m give nodal lines which are all concentric circles placed at irregular distances from the center. The irregularity of the placement of zeros of Bessel functions explains why the sound of a drum is so different from that of any string instrument. Step III The final initial value problem cn now be solved by taking an inifinite sum u((r, t) = W m (r)g m (t). (1) Putting t = 0 we get Therefore A m = m=1 f(r) = m=1 A m J 0 (λ m r). (13) R R J1 rf(r)j 0 (α m r/r) dr. (14) (α m ) 0 Differentiating (1) wrt t and putting t = 0 we get and hence B m = Q.5 (T) Show that the substitution g(r) = m=1 RJ 1 (α m )cα m B m λ m J 0 (λ m r). (15) R u(r, θ, t) = F (r, θ)g(t) 0 r(r)j 0 (α m r/r) dr. (16) into the wave equation u tt = c (u rr + 1 r u r + 1 r u θθ ) (17) yields G tt + c k G = 0; & F rr + 1 r F r + 1 r F θθ + k F = 0. (18) Further show that the substitution F = W (r)q(θ) in (18) yields, Q + n Q = 0; & r W + rw + (k r n )W = 0. (19) Sol: In tut sheet this was marked as a (S) problem. convert this into a (T) problem. Please Ex. 5-9 are on vibrations of a circular membrane which is fixed along its boundary and hence u 0 on the boundary of the membrane. This was not explicitly stated. 5

6 The suggested substitution yields after separation of variables, G tt c G = F rr + 1 r F r + 1 r F θθ = K F a constant. If this constant is zero, it means F is a harmonic function which vanishes on the boundary and hence leads to u 0. For K > 0 this means F is the solution of F = KF on the circular membrane. Just as in the case of rectangular membrane as seen in Tut 1 Q.4, F 0 and hence u 0. Therefore, K is negative. [Here are full details: 0 r = r = r ( 0 ( 0 Therefore, the function 0 kf + 1 r F θ + Fr ( 1 F F rr = ( r 0 F F r dθ ) r ) dθ r F F r + 1 r F F θθ + Fθ + Fr (F F r ) r + 1 ) r F F r) dθ α(r) = 0 rf F r dθ is monotonically increasing. But for r = 0 and r = R it vanishes. Therefore, its derivative vanishes everywhere... Alternatively, you can enclose the disc inside a rectangle, extend the function F to the whole of the rectangle by zero, and then work with Cartesian coordinates as discussed under Problem 4 of tut1.] Upon putting K = k, we obtain (17, 18). The substitution F = W (r)q(θ) in (18), upon separating the variables yields, ) dθ r W + rw + k r W = Q Q = K 1. The equations (19) follow upon putting K 1 = n. Q. 6 (T) Show that solutions Q of (19) are periodic with period. Deduce that n is an integer in (19)and hence Q n = cos nθ, Q n = sin nθ, W n = J n (kr) are solutions of (19). Sol: Observe that a solution of Q + K 1 Q = 0 will be defined on the entire of real line. But Q should make sense on the circle and hence we have Q( + θ) = Q(θ). Thus Q must be periodic. This requires that that K 1 = n and then Q is a linear combination of cos nθ, and sin nθ. 6

7 Therefore, n must be an integer and then Q is actually of period. Now it follows that W satisfies the modified Bessel s equation and hence W n = J n (kr) are solutions of the second part of (19). Q. 7 (T) Obtain solutions of (17) of the form and u m,n = (A m,n cos ck m,n t + B m,n sin ck m,n t)j n (k m,n r) cos nθ (0) u m,n = (A m,n cos ck m,n t + B m,n sin ck m,n t)j n (k m,n r) sin nθ (1) (here, J n are Bessel functions) which satisfy the boundary condition u(r, θ, t) = 0 () Sol: Bessel functions have infinitely many zeros. Let us denote the positive zeros of J n by α 1,n, α,n,..., The boundary condition u(r, θ, t) = 0 forces us to have J n (kr) = 0 for r = R. It follows that k = α m,n /R (for some m) which we shall denote by k m,n. Corresponding to this value of k, we obtain two independent solutions cos ck m,n t, sin ck m,n t of the equation G + c k = 0. The rest of the conclusion is easy. Q. 8 (S) Show that the initial condition in (1, ) leads to B m,n = 0 = B m,n. u t (r, θ, 0) = 0 (3) Sol: Just differentiate the two equations wrt t and put t = 0. Q. 9 S Determine the number of nodal lines of u m,n in (0). Sol: The nodal lines are those lines along which the membrane never moves. They are given by u(, t) = 0. Thus for u m,n they are simply the zeros J n (k m,n r) cos nθ. The first factor has precisely m 1 each of them defining a circle as a nodal line zeros and the second fact has n zeros, each of these defining a diameter as a nodal curve. So, in all there are m + n 1 nodal curves. Q. 11 (T) If u(r, θ) is harmonic, show that v(r, θ) = u(r 1, θ) is also harmonic. Give some examples. Sol: First observe that v(r, θ) is harmonic iff v(r, θ) is. Now write z = re iθ. Then z z 1 is analytic. Therefore, u(r 1, θ) = u(z 1 ) is harmonic. Therefore u(r 1, θ) is harmonic. 7

8 Typical examples are lnr, lnr, or r n cos nθ, r n cos nθ etc. Alternately, to avoid confusing notation, you can set up w = u(s, θ) with s = r 1 and show that w rr + w r r + w θθ r = 0 by simply working out through chain rule. Q. 1 (S) If u(r, θ, φ) satisfies u = 0 in spherical coordinates, show that v(r, θ, φ) = r 1 u(r 1, θ, φ) satisfies v = 0. Sol: In this case, we do not have any short-cut as in the previous exercise. So, we simply follow the chain rule. Put w == u(s, θ, φ), s = 1/r, v = w/r. Therefore, v r = w/r + w r /r = (wr + w s )/r 3. v rr = w/r 3 + 4w s /r 4 + w ss /r 5 = (wr + 4w s r + w ss )/r 5. Therefore v rr + v r r + v φφ r + cot φ r v φ + v θθ r sin φ = wr +4w s r+w ss r (wr+w s) 5 r 4 = s 5 ( w s s + w ss + w φφ s + w φφ r 3 cot φ s w φ + w θθ s sin φ + cot φ ) r w 3 φ + w θθ = 0. r 3 sin φ Q.13 (L) Discuss the potential in spherical region with the boundary condition independent of θ. Sol: We begin with the Laplace equation in spherical coordinates: u = 1 r (r u r ) r + (sin φu φ) φ sin φ + u θθ sin = 0. (4) φ In general the boundary values are give by a function of θ and φ. If the boundary values of u are independent of θ, we can expect that the solution u will also be independent of θ. In particular, u θθ = 0 and so we rewrite (4) as (r u r ) r + (sin φu φ) φ sin φ As usual, put u(r, φ) = G(r)H(φ) to obtain = 0. (5) (r G r ) r G = (sin φh φ) φ H sin φ = K. (6) 8

9 Consider the first equation which can be rewritten in the form r G + rg KG = 0. (7) We see that this is an Euler-Cauchy equation with its auxiliary equation (obtained by putting G = r m ) m(m 1) + m K = 0. (8) In order to get solutions in a nice form we put K = n(n + 1) and obtain Now consider the second equation G n (r) = r n ; G n(r) = 1 rn+1. (9) (sin φh φ ) φ H sin φ = n(n + 1). (30) Putting cos φ = w, an hence sin φ = 1 w, d dφ = sin φ d dw, we get the Legendre equation ((1 w )H w ) w + n(n + 1)H = 0. (31) So far, the value of n was totally arbitrary. In order to get continuous solutions u it can shown that we need to restrict to integer values of n only. For integers n (31) has Legendre polynomials H n = P n (w) = P n (cos φ) as solutions. P n (w) = [n/] m=0 ( 1) n (n m)! m (n m)!(n m)! wn m. (3) Therefore, we obtain two sequences of solutions of (5): u n (r, φ) = A n r n P n (cos φ), Here A n, B n are constants. u n(r, φ) = B n r n+1p n(cos φ), n = 0, 1,, 3,..., (33) Solution inside the sphere (interior problem) The solutions in the first sequence are valid inside the sphere, whereas those in the second sequence are not. Therefore we seek a particular solution of (5) of the form u(r, φ) = n=0 A n r n P n (cos φ) (34) 9

10 satisfying the boundary condition f(w) = f(φ) = u(r, φ) = n=0 A n R n P n (cos φ) (35) Therefore A n R n are Fourier-Legendre coefficients of f(φ) Therefore A n R n = n + 1 A n = n + 1 R n 1 1 f(w)p n (w)dw. (36) 0 f(φ)p n(cos φ) sin φdφ. (37) Solution Outside the Sphere (Exterior problem) The role of the two sequences interchanged if we move out of the sphere. All solutions in the second series make sense. The solutions in the second one explode as r and hence rejected for physical reasons. Therefore, we have a general solution of (5) outside the sphere given by u n(r, φ) = n=0 With the boundary condition as before, we have B n r n+1p n(cos φ). (38) B n = (n + 1)Rn+1 0 f(φ)p n(cos φ) sin φ dφ. (39) Q.14 (T) Find the potential in the interior of the sphere of radius R = 1 assuming that there is no charge in the interior of the sphere and the potential on the boundary surface is the constant 1. Also find the potential exterior to the sphere and compare with the potential of a point charge at the origin. Sol: Put R = 1, f(φ) 1 in (36). We get A n = n P n(cos φ) sin φdφ = 0 n > 1 1 n = 0. (40) Therefore, u 1 inside the sphere. This could have been derived directly by Dirichlet s problem that a harmonic function which is a constant on the boundary must be equal to the same constant inside as well. For the external problem we similarly have B 0 = 1 and B n = 0, n > 1. Therefore u (r, φ) = 1/r which coincides with the potential of a point charge kept at the origin. 10