APPENDIX E., where the boundary values on the sector are given by = 0. n=1. a 00 n + 1 r a0 n, n2

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1 APPENDIX E Solutions to Problem Set 5. èproblem in textè èaè Use a series expansion to nd èèr;è satisfying èe.è è rr + r è r + r 2 è =, r 2 cosèè in the sector éré3, 0 éé 2, where the boundary values on the sector are given by èe.2è èè;è = 0 èè3;è = 2 3 cosèè èèr;, 0è = r, r è r; 2 = 0 We begin by representing èèr;è in terms of its Fourier series with respect to : èe.3è èèr;è= 2 a 0èrè+ a n èrè cosènè+ b n èrè sinènè In view of the last boundary condition, è must vanish along the line = 2. Let us then immediately set all the coecients b n èrè = 0 èif we run into a consistency problem latter on we mayhave togoback and relax this condition. But so long as this assumption works for us we will not loose any solution; because if a solution exists it unique.è Thus, we take èe.4è èèr;è= 2 a 0èrè+ a n èrè cosènè : Plugging this expression into èe.è we obtain a èe.5è 2 r a0 0 + a 00n + r a0n, n 2 a n cosènè =, cos r 2 : Multiplying both sides of èe.5è by cosèmè, integrating between 0 and 2, and employing the othogonality properties of the cosine functions we obtain èe.6è a r a0 0 =0 èe.7è a 00 + r a0, r 2 a =, r 2 èe.8è The general solution of èe.6è is èe.9è a 00 n + r a0 n, n2 r 2 a n =0 ; n =2; 3; 4; 5; a 0 èrè =A 0 + B 0 ln jrj : The equations in èe.8è are homogeneous Euler type equations with general solutions of the form èe.0è a n èrè =A n r,n + B n r n : 20

2 . èproblem IN TEXTè 2 Equation èe.7è is a non-homogeneous Euler-type equation. Its general solution can be determined by the Method of Undetermined Coecients. The two linearly independent solutions of the corresponding homogeneous problem are and a 00 + r a0, r 2 a y èrè = r, y 2 èrè = r W ëy ;y 2 ë=r, èè,,,r,2 r = 2 r : And so the general solution of èe.7è will be èe.è a èrè = r, A, R èrèè, r 2 è dr + r, 2 r, = r, A + 2 r + r B + = +A r, + B r : 2r B + R è rèè, Replacing the coecients a n èrè, n =0; ;:::, in èe.4è by the solutions èe.9è, èe.0è, and èe.2è we obtain èe.2è èèr;è= 2 A B 0 ln jrj + cosèè+ 2 r r 2 è dr, An r,n + n B n r cosènè : To x the constants A n ;B n, n =0; ; 2;:::,wenow impose the remaining boundary conditions. The rst boundary condition leads to 2 A 0 + cosèè+ èè;è=0 èa n + B n è cosèè =0 : Applying the orthogonality properties of the cosine functions we thus obtain èe.3è Utilizing these relations we thus obtain èe.4è The second boundary condition now leads to 2 èèr;è = A 0 = 0 +A + B = 0 A n + B n = 0 2 B 0 +, +A r,, è + A è r cosèè P + n=2 A n èr,n, r n è cosènè èè3;è= 2 3 cosèè 3 cosèè =, 2 B 0 ln j3j + +A 3,, è + A è3 cosèè, A n 3,n, 3 n cosènè which in view of the orthogonality properties of the cosine functions implies èe.5è + n=2 B 0 = A, 3A, 3 = 2 3 A n = 0

3 The second equation in èe.5è is equivalent to. èproblem IN TEXTè 22 thus, èe.4è becomes èe.6è A =, ; èèr;è=,, r, cosèè : This is the desired solution of èe.è and èe.2è. èbè Formulate the Dirichlet problem for the sector 0 érér,0éé èwhere R and are givenè, involving Poisson's equation, and obtain a general series solution. The general Dirichlet problem on the sector dened above would be the PDEèBVP èe.7è èe.8è è rr + r è r + r 2 è = fèr;è èèr; 0è = pèrè èèr; è = qèrè èèr; è = tèè AsaFourier expansion of the solution of this PDEèBVP we can use èe.9è èèr;è= 2 a 0èrè+ a n èrè cosènè+ b n èrè sinènè : since the functions cosènè and sinènè provide a complete set of functions on the interval è0; 2è, and so also on the subinterval è0; è. However, this set of basis functions is not the most natural for our given boundary conditions. A better choice would be an expansion based on eigenfunctions of the form n èe.20è n èxè = sin associated with to the Sturm-Louisville problem èe.2è èe.22è èe.23è y 00, 2 y = 0 yè0è = 0 yèè = 0 Note that the ODE in èe.2è corresponds the ODE with respect to that arises when we apply Separation of Variables to Laplace's equation. However, unlike the boundary conditions for our general Dirichlet problem, the boundary conditions èe.22è and èe.23è are homogeneous èas they must be for Sturm-Louisville problemsè. We will eventually apply the method of Lecture 5 to handle the inhomogeneous boundary conditions of the PDE; but rst let us concentrate on solving the following simpler problem: èe.24è èe.25è èe.26è èe.27è We begin by writing èe.28è rr + r r + r 2 = F èx; yè èr;è= èr; 0è = 0 èr; è = 0 èr; è = T èè a n èrè sin n

4 and èe.29è where coecients f n èrè are determined by èe.30è. èproblem IN TEXTè 23 fèr;è= f n èrè = 2 f n èrè sin Z fèr;èsin 0 n n d and the coecients a n èrè are to be determined by the requirement that èr;è satises èe.2è - èe.23è. Plugging èe.28è and èe.29è, respectively, into the left and right hands sides of èe.2è we obtain a 00n + r a0n, n2 2 n n r 2 2 a n sin = f n èrè sin or a 00n + r a0n, n2 2 r 2 2 a n, f n èrè sin n =0: From this we can conclude that the coecients function a n èrè obeys the following dierential equation: a 00 n + r a0 n, n2 èe.3è 2 r 2 2 a n = f n èrè: Now the homogeneous equation corresponding to èe.3è is the Euler-type equation r 2 R 00 + rr 0 n 2 + R =0: This equation has the following pair of linearly independent solutions R èrè = r, n R 2 èrè = r n Therefore the Method of Variation of Parameters yields Z a n èrè =r, n A n, r Z s n f n èsèds + r n B n + r èe.32è s, n f n èsèds : 2n R 2n R In order to ensure that the coecient functions are well-behaved as r! 0, we must take Z A n = 0 èe.33è s n f n èsèds 2n R, n We can also impose èe.23è which after expanding T èè in terms of the eigenfunctions sin leads to n n èe.34è a n èrè sin = t n sin where, of course, the coecients t n are given by èe.35è From èe.34è we can conclude that èe.36è t n = 2 Z 0 T èè sin a n èrè =t n : n d: Evaluating èe.32è at r = R and using èe.36è and èe.33è we obtain or èe.37è t n = R, n A n + R n B n B n = R, n t n, R, 2n A n

5 We now have formulas for all the ingredients of the expansion èe.28è. In summary, the solution to. èproblem IN TEXTè 24 èe.38è èe.39è èe.40è èe.4è is given by èe.42è where èe.43è èe.44è èe.45è èe.46è rr + r r + r 2 = F èx; yè èr;è= èr; 0è = 0 èr; è = 0 èr; è = T èè a n èrè sin Z a n èrè = r, n A n, r s n f n èsèds 2n R Z 0 A n = s n f n èsèds 2n R B n = R, n t n, R, 2n A n Z t n = 2 0 T èè sin n d n + r n B n + 2n Z r R s, n f n èsèds We will now ènallyè address the solution of Laplace's equation with the inhomogeneous boundary conditions èe.8è. So as we did in Lecture 5, we relate solutions of èe.47è to solutions of èe.48è If we set then we obviously have è rr + r è r + r è 2 = fèr;è èèr; 0è = pèrè èèr; è = qèrè èèr; è = tèè rr + r r + r 2 = F èr;è èr; 0è = 0 èr; è = 0 èr; è = T èè èr;è=èèr;è, cos pèrè, sin qèrè 2 2 èr; 0è = èèr; 0è, pèrè =0 èr; è = èèr; è, qèrè = 0 But we also have èr; è = èèr;è, cos 2 = tèèè, cos 2 pèrè, sin pèrè, sin 2 2 qèrè qèrè

6 2. èproblem IN TEXTè 25 and rr + r r + r 2 = è rr + r è r + r 2 è,, pèrè 00 + r pèrè0, r 2 2 qèrè 00 + r qèrè0, r pèrè cos 2 qèrè sin = fèr;è, pèrè 00 + r pèrè0, r 2, qèrè 00 + r qèrè0, r 2 2 Therefore, the solution èèr;è of èe.47è can be constructed by setting èèr;è=èr;è + cos pèrè + sin 2 where èr;è is the solution of èe.48è with F èr;è = fèr;è,, T èè = tèèè, cos pèrè 00 + r pèrè0, r 2 qèrè 00 + r qèrè0, r 2 2 pèrè, sin qèrè sin pèrè cos qèrè 2 pèrè cos 2 qèrè sin 2 qèrè The solution of èe.48è can be calculated via the formulas èe.42è - èe.46è èproblem in textè Discuss the properties of the mapping function èe.49è w = 2 z + z What happens to circles and straight-line rays in the z plane? Can this mapping function, together with the result of Problem 4.5.6ècè, be used to solve the exterior Dirichlet problem for a region outside an ellipse? If so, provide a recipe. A circle èwith center z = 0è in the z plane can be represented as a curve of the form zètè =e it : The image of such a circle under the mapping èe.49è is just wètè = 2 e it +, e it = 2 2 e it +,it e èe.50è = 2 = 2 + 2,, 2 + cosètè+i, 2, sinètè cosètè+i 2, 2 sinètè

7 2. èproblem IN TEXTè 26 This is the equation of an ellipse in the èu; vè-plane with major and minor axes given by èe.5è r max = 2 + r min = 2, Note that in the limit! the image of the circle of radius under the map! collapses to the line segment ë,; +ë on the real line. A straight line ray in the z plane can be represented by a curve of the form zètè =te io : The image of such aray under the mapping èe.49è is t 2 + t 2, èe.52è wètè = cos è o è+i sin è o è 2t 2t As t! +, this curve asymptotically approaches the straight line èe.53è w ètè = 2 ècos è oè+i sin è o èè t ; which is just a straight line with slope o with respect to the real axis. As t!, the curve èe.52è tends to the point èe.54è w = cos è o è As t! 0, the curve èe.52è approaches the curve èe.55è w 0 ètè = 2 ècos è oè, i sin è o èè t which is a straight line heading to innity in a direction perpendicular to the line èe.53è. Let us look again at the elliptical images of circles under the map w. From èe.5è we see that if é then r max èè and r min èè are monotonically increasing. This implies that the exterior of a circle of radius éis mapped : onto the exterior of an ellipse with semi-major axis 2 + and semi-minor axix,. 2 When 0 é ér max and r min are both monotonically decreasing. Thus, the interior of a circle or radius é is mapped onto the exterior of an ellipse with semi-major axis 2 + and semi-minor axis 2,. Suppose now we seek a solution of u xx + u yy = 0 in region exterior exterior to the ellipse 4y, 2 R, R 2 = ; 0 éré ; 4x, 2 R R satisfying u = fèè on the boundary of the ellipse. We can nd such a solution by rst solving the following Dirichlet problem and then applying the map!: nd a solution of Laplace's equation in the region interior to the circle of radius R satisfying èèr; è =fè,è : èthe minus sign is needed since the arg èwèzèè =,argèzè when jzj é. See Eq. èe.50è.è To obtain a appropriate solution of the Dirichlet problem on the circle we can simply employ the Poisson integral formula èe.56è èèr;è= 2 Z 2 0 R fè,è 2, r 2 R 2 + r 2, 2Rr cosè, è d :

8 3. èproblem IN TEXTè 27 We now pull this solution back to the original ellipse using the inverse mapping z = w, p w 2, : ènote: the map w is actually 2:, this is the inverse that maps : from the region outside the ellipse onto the region inside the corresponding circle.è Thus, the desired solution will be with èèzè determined by èe.56è. è w, p w 2, 3. èproblem in textè With w = u + iv, z = x + iy, let w = fèzè be analytic in a region of the z plane. Show that the èx; yè-plane curves u = const, v = const intersect orthogonally. Show also that the directional derivatives of u in any direction is equal to the directional derivative of v in the orthogonal direction, and use this fact to explain how a Neumann problem for a nite region R can be transformed into a Dirichlet problem for a conjugate function. First let us recall some simple facts from Analytic Geometry. Two parameterized curves : I R! R 2 and : J R! R 2 will intersect orthogonally at a point èx; yè 2 R 2 if there exist s o 2 I, t o 2 J such that ès o è=èt o è and èe.57è 0= d ds d so dt to = d x dt èt oè d x ds ès oè+ d y dt èt oè d y ds ès oè : If we represent the curves and locally as the graphs of two functions f and g of x èe.58è ètè = èt; fètèè ; èsè = ès; gètèè ; then the condition for intersection when t = s =0is or èe.59è 0=è;f 0 è0èè è;g 0 è0èè = + f 0 è0èg 0 è0è f 0 è0è =, g 0 è0è Now let wèzè = uèx; yè + ivèx; yè be an analytic function. The Cauchy-Riemann conditions then require = Let fèxè be a function whose graph corresponds, in a neighborhood of a point èx o ;y 0 è=èx o ;fèx o èè, to a curve uèx; yè =c.we then have the following identity èe.6è Dierentiating this equation with respect to x yields èe.62è u èx; f èxèè = c + f0 =0 : Similarly, let gèxè be a function whose graph corresponds to a curve vèx; yè =c 2 in a neighborhood of the point èx o ;y o è=èx o ;gèx o èè. We then g0 =0 : :

9 Using èe.60è we can rewrite èe.63è as èe.64è Using èe.64è to eliminate èe.65è in èe.62è we obtain 3. èproblem IN TEXTè 28 g0 : è + f 0 g 0 =0 which so long 6= 0 implies that the two curves uèx; yè =c and vèx; yè =c 2 intersect orthogonally. The directional derivative of a function fèx; yè with respect to the direction e =èe x ;e y èisgiven r e f = rf e = e + y Let e? =è,e y ;e x è, so that e e? = 0. Then if wèx; yè =uèx; yè+ivèx; yèwe have r e u = e + e = e x, = r e?v èin the second step we simply employed the Cauchy-Riemann conditionsè. Now suppose C = fz = ètè j t 2 I Rg is a closed curve and consider the following Neumann 2 2 u 2 = 0 unj C = 0 where un is the directional derivative of u along C in the direction normal to C. Assume the parameterization ètè ofc is such that d dt = is èwe can always nd such a parameterizationè. The tangent vector to the curve C at the point ètè will be and so d dt ètè = dx dt ètè; d y dt ètè nètè = dy dt ètè;, d x dt ètè will be a unit vector that always be orthogonal to d dt. The Neumann boundary condition thus takes the form èe.66è 0= unj C = r uj C = d dt :, d dt We know that the real and imaginary part of an analytic function will always satisfy Laplace's equation. The converse is also true, if u satises Laplace's equation in a region R then u can be regarded as the real èor imaginaryè part of an analytic function on R. Let v be a harmonic conjugate of u; i.e., v is a function such that wèx; yè = uèx; yè + ivèx; yè is analytic. Then the Cauchy-Riemann conditions @x :

10 Applying these relations to the far right hand side of èe.66è we obtain èe.67è But now èe.67è implies èe.68è 3. èproblem IN TEXTè 29 d y dt d x = d èv dt dt vèx; yèj C = const: Also since vèx; yè is the imaginary part of an analytic function 2 v 2 =0 2 : Thus, the harmonic conjugate of a solution of a Neumann problem on C is the solution of a Dirichlet problem on C.