Physics 217, Fall 2002 11 October 2002(c) University of Rochester 111 October 2002 Physics 217, Fall 2002 1Today in Physics 217: separation of variables Introduction to the method, in Cartesian coordinates. Example solution for the potential in an infinite slot, arbitrary V at the bottom, in which we introduce two common features of separation solutions:• Completeness and orthogonality of sines•Fourier’s trickyzxV=0V=0V(y)a11 October 2002 Physics 217, Fall 2002 2Introduction to separation of variables If the method of images can’t be used on a problem you must solve that involves conductors, the next thing to try is direct solution of the Laplace equation, subject to boundary conditions on V and/or Separation of variables is the easiest direct solution technique. It works best with conductors for which the surfaces are well behaved (planes, spheres, cylinders, etc.). Here’s how it works, in Cartesian coordinates, in which the Laplace equation is which can’t be integrated directly like the 1-D case..Vn∂∂22222220VVVVxyz∂∂∂∇= + + =∂∂∂11 October 2002 Physics 217, Fall 2002 3Introduction to separation of variables (continued)Consider solutions of the formThen, the Laplace equation isor, dividing through by XYZ,i.e.()()()(),,Vxyz XxYyZz=2222220dX dY dZYZ XZ XYdx dy dz++=2222221110dX dY dZXYZdx dy dz++=()()()0f x gy hz++=Physics 217, Fall 2002 11 October 2002(c) University of Rochester 211 October 2002 Physics 217, Fall 2002 4Introduction to separation of variables (continued)The only way for this to be true for all x,y,z is for each term to be a constant, and for the three constants to add up to zero:So this is a way to trade one partial differential equation for three ordinary differential equations, which can often be solved much more easily:()2222221110dX dY dZAB ABXYZdx dy dz++=+−+=()22222200 0dX dY dZAX BY A B Zdx dy dz−= −= ++ =11 October 2002 Physics 217, Fall 2002 5Introduction to separation of variables (continued)Nothing guarantees that V will always factor into functions of x, y, and z alone. In fact, there are certainly many solutions to the Laplace equation that are not of this form. However, there are plenty of electrostatic problems for which the boundary conditions are specified on well-behaved surfaces, and do turn out to have solutions of this form, and the solutions to electrostatics problems are unique, so if separation of variables yields a solution at all, it’s guaranteed to be the correct one.Separation of variables is also a very useful PDE solution technique in quantum mechanics, where one finds many problems in which the boundary conditions are specified on regular, well-behaved surfaces.11 October 2002 Physics 217, Fall 2002 6Example: the infinite slotGriffiths, example 3.3:Two infinite, grounded, metal plates lie parallel to the x-zplane, one at y = 0, the other at y = a. The left end, at x = 0, is closed off with an infinite strip insulated from the two plates and maintained at a specified potential V0(y). Find the potential inside this slot.No z dependence, so(),VVxy=yzxV=0V=0V(y)aSemi-infinite conducting plates:Bottom of the slot: plate insulated from the other two, with,0 ,0,.zxya=−∞→∞ = →∞ =()0.VVy=Physics 217, Fall 2002 11 October 2002(c) University of Rochester 311 October 2002 Physics 217, Fall 2002 7The infinite slot (continued)We chose rather than, say, A, to indicate that this constant is non-negative, and that the other one ( ) is non-positive. Why? We’ll see in a minute.()()()222 222222 2222 2 22222 2 22222220Suppose , ; then1100or:I. 0 II. 0VVV VVVxyz xyVxy XxYydX dY dX dYYX kkXYdx dy dx dydX dYkX kYdx dy∂∂∂∂∂∇= + + = + =∂∂∂∂∂=+=⇒ +==−−= +=2k2k−11 October 2002 Physics 217, Fall 2002 8The infinite slot (continued)Boundary conditions:Solutions: It turns out that you know equations I and II very well from MTH 165. But the means by which they’re solved is too useful to forget, so I’ll remind you, first, with I.()01. 0 as 2. 0 at 03. 0 at 4. at 0VxVyVyaVVy x→→∞======(reference point at infinity)222dXkXdx=11 October 2002 Physics 217, Fall 2002 9The infinite slot (continued)Let and multiply through by Integrate over x:Here both X and dX/dx must be zero for all x at y = 0 and a, so the constant S = 0:2dv dXvkXdx dx=:vdXdx=,vdXdx=222222dv dXvdxkX dxdx dxvXkS==+∫∫ or dXkX kXdx=−Physics 217, Fall 2002 11 October 2002(c) University of Rochester 411 October 2002 Physics 217, Fall 2002 10The infinite slot (continued)Separate and integrate:where A and B are constants. The general solution is a linear combination of these two particular solutions: or ln or or ,kx T kx kx U kxdXkdx kdxXXkxT kxUXe Ae e Be+−+−=−=+ −+== =∫∫ ∫()kx kxXx Ae Be−=+11 October 2002 Physics 217, Fall 2002 11The infinite slot (continued)Similarly, for equation II, a more general solution isor, usingwe have andWe’re ready to determine the (four) constants A-D by plugging the (four) boundary conditions in and solving for them.222,dYdy kY=−()iky ikyYy Ce De−′′=+sin , cos ,22ii iiee eeiαα αααα−−−+==()sin cos ,Yy C ky D ky=+()()(),sincos.kx kxVxy Ae Be C ky D ky−=+ +11 October 2002 Physics 217, Fall 2002 12The infinite slot (continued)Applying the boundary conditions:So far,(See now why we chose the separation constants to be ?)()()1. 0 as , but as 0.2. 0 at 0, so 0 0 (1)0.kxkxVx e xAVy BeCDD−→→∞ →∞→∞⇒=== = +⇒=(),sinsin.kx kxVxyBCe kyGe ky−−==2k±3. 0 at , so 0 sinsin 0 , 0, 1, 2,kxVya Gekanka k naπ−== =⇒=⇒= =…Physics 217, Fall 2002 11 October 2002(c) University of Rochester 511 October 2002 Physics 217, Fall 2002 13The infinite slot (continued)This actually gives us an infinite number of solutions (leaving out n = 0, since this would make V = 0 everywhere: a trivialsolution):The remaining boundary condition (4) needs to be used to determine the Gn. Note first that this solution won’t work unless V0itself is sinusoidal.BUT, if theare all solutions, then a linear combination of them is a solution too: (), sin , 1,2,3,nxannnyVxy Ge naππ−==…(),nVxy()()()()122 211,,because , , 0nnnnnnVxy V xyVVxy Vxy∞=∞∞===∇=∇ = ∇ =∑∑∑11 October 2002 Physics 217, Fall 2002 14The infinite slot (continued)And this über-general solution can match arbitrary functions V0(y) at x = 0 (boundary condition 4), because the