12.11 Laplace s Equation in Cylindrical and

Transcription

1 SEC. 2. Laplace s Equation in Cylinrical an Spherical Coorinates. Potential Laplace s Equation in Cylinrical an Spherical Coorinates. Potential One of the most important PDEs in physics an engineering applications is Laplace s equation, given by () 2 u u xx u yy u zz. Here, x, y, z are Cartesian coorinates in space (Fig. 67 in Sec. 9.), u xx 2 u>x 2, etc. The expression 2 u is calle the Laplacian of u. The theory of the solutions of () is calle potential theory. Solutions of () that have continuous secon partial erivatives are known as harmonic functions. Laplace s equation occurs mainly in gravitation, electrostatics (see Theorem 3, Sec. 9.7), steay-state heat flow (Sec. 2.5), an flui flow (to be iscusse in Sec. 8.4). Recall from Sec. 9.7 that the gravitational potential u(x, y, z) at a point (x, y, z) resulting from a single mass locate at a point (X, Y, Z) is (2) u (x, y, z) c r c (r ) 2(x X) 2 (y Y) 2 (z Z) 2 an u satisfies (). Similarly, if mass is istribute in a region T in space with ensity r (X, Y, Z), its potential at a point (x, y, z) not occupie by mass is (3) u (x, y, z) k It satisfies () because 2 (>r) (Sec. 9.7) an r is not a function of x, y, z. Practical problems involving Laplace s equation are bounary value problems in a region T in space with bounary surface S. Such problems can be groupe into three types (see also Sec. 2.6 for the two-imensional case): (I) First bounary value problem or Dirichlet problem if u is prescribe on S. (II) Secon bounary value problem or Neumann problem if the normal erivative u n u>n is prescribe on S. (III) Thir or mixe bounary value problem or Robin problem if u is prescribe on a portion of S an u n on the remaining portion of S. In general, when we want to solve a bounary value problem, we have to first select the appropriate coorinates in which the bounary surface S has a simple representation. Here are some examples followe by some applications. Laplacian in Cylinrical Coorinates T r (X, Y, Z) r X Y Z. The first step in solving a bounary value problem is generally the introuction of coorinates in which the bounary surface S has a simple representation. Cylinrical symmetry (a cyliner as a region T ) calls for cylinrical coorinates r, u, z relate to x, y, z by (4) x r cos u, y r sin u, z z (Fig. 3).

2 594 CHAP. 2 Partial Differential Equations (PDEs) z z (r, θ, z) φ (r, θ, φ ) z θ y r x Fig. 3. Cylinrical coorinates (r, u 2p) r θ y x Fig. 32. Spherical coorinates (r, u 2p, p) For these we get 2 u immeiately by aing to (5) in Sec. 2.; thus, u zz (5) 2 u 2 u r 2 r u r 2 u r 2 u 2 2 u z 2. Laplacian in Spherical Coorinates Spherical symmetry (a ball as region T boune by a sphere S) requires spherical coorinates r, u, relate to x, y, z by (6) x r cos u sin, y r sin u sin, z r cos (Fig. 32). Using the chain rule (as in Sec. 2.), we obtain 2 u in spherical coorinates (7) 2 u 2 u r 2 2 r u r 2 u cot r 2 2 u r 2 r 2 sin 2 2 u u 2. We leave the etails as an exercise. It is sometimes practical to write (7) in the form (7r) 2 u r 2 c r ar 2 u r b sin u asin b sin 2 2 u u 2. Remark on Notation. Equation (6) is use in calculus an extens the familiar notation for polar coorinates. Unfortunately, some books use u an interchange, an extension of the notation x r cos, y r sin for polar coorinates (use in some European countries). Bounary Value Problem in Spherical Coorinates We shall solve the following Dirichlet problem in spherical coorinates: (8) (9) 2 u r 2 c r ar 2 u r b sin u (R, ) f () u asin b. () lim r: u (r, ).

3 SEC. 2. Laplace s Equation in Cylinrical an Spherical Coorinates. Potential 595 The PDE (8) follows from (7) or (7r) by assuming that the solution u will not epen on u because the Dirichlet conition (9) is inepenent of u. This may be an electrostatic potential (or a temperature) f () at which the sphere S: r R is kept. Conition () means that the potential at infinity will be zero. Separating Variables by substituting u (r, ) G (r)h() into (8). Multiplying (8) by r 2, making the substitution an then iviing by GH, we obtain G r ar 2 G r b H sin H asin b. By the usual argument both sies must be equal to a constant k. Thus we get the two ODEs () an G r ar 2 G r b k or r 2 2 G r 2 G 2r r kg (2) sin H asin b kh. The solutions of () will take a simple form if we set Gr G>r, etc., we obtain k n (n ). Then, writing (3) r 2 Gs 2rGr n (n ) G. This is an Euler Cauchy equation. From Sec. 2.5 we know that it has solutions G r a. Substituting this an ropping the common factor r a gives a (a ) 2a n (n ). The roots are a n an n. Hence solutions are (4) G n (r) r n an G* n (r) r n. We now solve (2). Setting cos w, we have sin 2 w 2 an w sin w w. Consequently, (2) with k n (n ) takes the form (5) w c ( w 2 ) H w n (n )H. This is Legenre s equation (see Sec. 5.3), written out

4 596 CHAP. 2 Partial Differential Equations (PDEs) (5) ( w 2 ) 2 H H 2 2w w w n (n )H. For integer n,, Á the Legenre polynomials H P n (w) P n (cos ) n,, Á, are solutions of Legenre s equation (5). We thus obtain the following two sequences of solution u GH of Laplace s equation (8), with constant A n an B n, where n,, Á, (6) (a) u n (r, ) A n r n P n (cos ), (b) u* n (r, ) B n r P n n (cos ) Use of Fourier Legenre Series Interior Problem: Potential Within the Sphere S. (6a), We consier a series of terms from (7) u(r, ) a n A n r n P n (cos ) (r R). Since S is given by r R, for (7) to satisfy the Dirichlet conition (9) on the sphere S, we must have (8) u (R, ) a A n R n P n (cos ) f (); n that is, (8) must be the Fourier Legenre series of f (). From (7) in Sec. 5.8 we get the coefficients (9*) A n R n 2n 2 f (w) P n (w) w where f (w) enotes f () as a function of w cos. Since w sin, an the limits of integration an correspon to p an, respectively, we also obtain (9) 2n A n 2R p f ()P n n (cos ) sin, n,, Á. If f () an f r () are piecewise continuous on the interval p, then the series (7) with coefficients (9) solves our problem for points insie the sphere because it can be shown that uner these continuity assumptions the series (7) with coefficients (9) gives the erivatives occurring in (8) by termwise ifferentiation, thus justifying our erivation.

5 SEC. 2. Laplace s Equation in Cylinrical an Spherical Coorinates. Potential 597 Exterior Problem: Potential Outsie the Sphere S. Outsie the sphere we cannot use the functions u n in (6a) because they o not satisfy (). But we can use the u n * in (6b), which o satisfy () (but coul not be use insie S; why?). Proceeing as before leas to the solution of the exterior problem (2) u(r, ) a n B n r n P n(cos ) (r R) satisfying (8), (9), (), with coefficients (2) B n 2n 2 R n p f ()P n (cos ) sin. The next example illustrates all this for a sphere of raius consisting of two hemispheres that are separate by a small strip of insulating material along the equator, so that these hemispheres can be kept at ifferent potentials ( V an V). EXAMPLE Spherical Capacitor Fin the potential insie an outsie a spherical capacitor consisting of two metallic hemispheres of raius ft separate by a small slit for reasons of insulation, if the upper hemisphere is kept at V an the lower is groune (Fig. 33). Solution. The given bounary conition is (recall Fig. 32) Since R, we thus obtain from (9) where w cos. Hence P n (cos ) sin P n (w) w, we integrate from to, an we finally get ri of the minus by integrating from to. You can evaluate this integral by your CAS or continue by using () in Sec. 5.2, obtaining A n 55 (2n ) a M f () e if p>2 if p>2 p. p>2 2n A n 2 P n (cos ) sin 2n 2 P n (w) w m (2n 2m)! () m w n2m w 2 n m!(n m)!(n 2m)! where M n>2 for even n an M (n )>2 for o n. The integral equals >(n 2m ). Thus z volts x y Fig. 33. Spherical capacitor in Example

6 598 CHAP. 2 Partial Differential Equations (PDEs) (22) A n 55 (2n ) 2 n a M m (2n 2m)! () m m!(n m)!(n 2m )!. Taking n, we get A (since! ). For n, 2, 3, Á 55 we get A ! 65!!2! 2, A a 4! 2! b,!2!3!!!! A a 6! 4!!3!4!!2!2! b 385 8, etc. Hence the potential (7) insie the sphere is (since P ) (23) u (r, ) (Fig. 34) 2 r P 385 (cos ) 8 r 3 P 3 (cos ) Á with P, P 3, Á given by ( r), Sec Since R, we see from (9) an (2) in this section that B n A n, an (2) thus gives the potential outsie the sphere (24) u (r, ) 55 r 65 2r 2 P (cos ) 385 8r 4 P 3(cos ) Á. Partial sums of these series can now be use for computing approximate values of the inner an outer potential. Also, it is interesting to see that far away from the sphere the potential is approximately that of a point charge, namely, 55>r. (Compare with Theorem 3 in Sec. 9.7.) y π π t 2 Fig. 34. Partial sums of the first 4, 6, an nonzero terms of (23) for r R EXAMPLE 2 Simpler Cases. Help with Problems The technicalities encountere in cases that are similar to the one shown in Example can often be avoie. For instance, fin the potential insie the sphere S: r R when S is kept at the potential f () cos 2. (Can you see the potential on S? What is it at the North Pole? The equator? The South Pole?) Solution. w cos, cos 2 2 cos 2 2w 2 4 3P 2 (w) (3 2w 2 2 ) 3. Hence the potential in the interior of the sphere is u 4 3 r 2 P 2 (w) 3 4 3r 2 P 2 (cos ) 3 2 3r 2 (3 cos 2 ) 3. PROBLEM SET 2.. Spherical coorinates. Derive (7) from 2 u in 3. Sketch P n (cos u), u 2p, for n,, 2. (Use spherical coorinates. (r) in Sec. 5.2.) 2. Cylinrical coorinates. Verify (5) by transforming 4. Zero surfaces. Fin the surfaces on which u, u 2, u 3 2 u back into Cartesian coorinates. in (6) are zero.

7 SEC. 2. Laplace s Equation in Cylinrical an Spherical Coorinates. Potential CAS PROBLEM. Partial Sums. In Example in the text verify the values of an compute A 4, Á A, A, A 2, A 3, A. Try to fin out graphically how well the corresponing partial sums of (23) approximate the given bounary function. 6. CAS EXPERIMENT. Gibbs Phenomenon. Stuy the Gibbs phenomenon in Example (Fig. 34) graphically. 7. Verify that u n an u* n in (6) are solutions of (8). 8 5 POTENTIALS DEPENDING ONLY ON r 8. Dimension 3. Verify that the potential u c>r, r 2x 2 y 2 z 2 satisfies Laplace s equation in spherical coorinates. 9. Spherical symmetry. Show that the only solution of Laplace s equation epening only on r 2x 2 y 2 z 2 is u c>r k with constant c an k.. Cylinrical symmetry. Show that the only solution of Laplace s equation epening only on r 2x 2 y 2 is u c ln r k.. Verification. Substituting u (r) with r as in Prob. 9 into u xx u yy u zz, verify that us 2ur>r, in agreement with (7). 2. Dirichlet problem. Fin the electrostatic potential between coaxial cyliners of raii r 2 cm an r 2 4 cm kept at the potentials U 22 V an U 2 4 V, respectively. 3. Dirichlet problem. Fin the electrostatic potential between two concentric spheres of raii r 2 cm an r 2 4 cm kept at the potentials U 22 V an U 2 4 V, respectively. Sketch an compare the equipotential lines in Probs. 2 an 3. Comment. 4. Heat problem. If the surface of the ball r 2 x 2 y 2 z 2 R 2 is kept at temperature zero an the initial temperature in the ball is f (r), show that the temperature u (r, t) in the ball is a solution of u t c 2 (u rr 2u r >r) satisfying the conitions u (R, t), u (r, ) f (r). Show that setting v ru gives v t c 2 v rr, v (R, t), v (r, ) rf (r). Inclue the conition v (, t) (which hols because u must be boune at r ), an solve the resulting problem by separating variables. 5. What are the analogs of Probs. 2 an 3 in heat conuction? 6 2 BOUNDARY VALUE PROBLEMS IN SPHERICAL COORDINATES r, U, Fin the potential in the interior of the sphere r R if the interior is free of charges an the potential on the sphere is 6. f () cos 7. f () 8. f () cos 2 9. f () cos 2 2. f () cos 3 3 cos 2 5 cos 2. Point charge. Show that in Prob. 7 the potential exterior to the sphere is the same as that of a point charge at the origin. 22. Exterior potential. Fin the potentials exterior to the sphere in Probs. 6 an Plane intersections. Sketch the intersections of the equipotential surfaces in Prob. 6 with xz-plane. 24. TEAM PROJECT. Transmission Line an Relate PDEs. Consier a long cable or telephone wire (Fig. 35) that is imperfectly insulate, so that leaks occur along the entire length of the cable. The source S of the current i (x, t) in the cable is at x, the receiving en T at x l. The current flows from S to T an through the loa, an returns to the groun. Let the constants R, L, C, an G enote the resistance, inuctance, capacitance to groun, an conuctance to groun, respectively, of the cable per unit length. S x = Fig. 35. Transmission line T x = l Loa (a) Show that ( first transmission line equation ) u i Ri L x t where u (x, t) is the potential in the cable. Hint: Apply Kirchhoff s voltage law to a small portion of the cable between x an x x (ifference of the potentials at x an x x resistive rop inuctive rop). (b) Show that for the cable in (a) ( secon transmission line equation ), i x Gu C u t. Hint: Use Kirchhoff s current law (ifference of the currents at x an x x loss ue to leakage to groun capacitive loss). (c) Secon-orer PDEs. Show that elimination of i or u from the transmission line equations leas to u xx LCu tt (RC GL)u t RGu, i xx LCi tt (RC GL)i t RGi. () Telegraph equations. For a submarine cable, G is negligible an the frequencies are low. Show that this leas to the so-calle submarine cable equations or telegraph equations u xx RCu t, i xx RCi t.

8 6 CHAP. 2 Partial Differential Equations (PDEs) Fin the potential in a submarine cable with ens (x, x l) groune an initial voltage istribution U const. (e) High-frequency line equations. Show that in the case of alternating currents of high frequencies the equations in (c) can be approximate by the so-calle high-frequency line equations u xx LCu tt, i xx LCi tt. Solve the first of them, assuming that the initial potential is U sin (px>l), an u t (x, ) an u at the ens x an x l for all t. 25. Reflection in a sphere. Let r, u, be spherical coorinates. If u (r, u, ) satisfies 2 u, show that v (r, u, ) u (>r, u, )>r satisfies 2 v. 2.2 Solution of PDEs by Laplace Transforms Reaers familiar with Chap. 6 may woner whether Laplace transforms can also be use for solving partial ifferential equations. The answer is yes, particularly if one of the inepenent variables ranges over the positive axis. The steps to obtain a solution are similar to those in Chap. 6. For a PDE in two variables they are as follows.. Take the Laplace transform with respect to one of the two variables, usually t. This gives an ODE for the transform of the unknown function. This is so since the erivatives of this function with respect to the other variable slip into the transforme equation. The latter also incorporates the given bounary an initial conitions. 2. Solving that ODE, obtain the transform of the unknown function. 3. Taking the inverse transform, obtain the solution of the given problem. If the coefficients of the given equation o not epen on t, the use of Laplace transforms will simplify the problem. We explain the metho in terms of a typical example. EXAMPLE Semi-Infinite String Fin the isplacement w (x, t) of an elastic string subject to the following conitions. (We write w since we nee u to enote the unit step function.) (i) The string is initially at rest on the x-axis from x to ( semi-infinite string ). (ii) For t the left en of the string (x ) is move in a given fashion, namely, accoring to a single sine wave w (, t) f (t) e sin t if t 2p otherwise (Fig. 36). (iii) Furthermore, lim w (x, t) for t. x: f(t) Fig. 36. π 2π t Motion of the left en of the string in Example as a function of time t