Today in Physics 217: separation of variables
Introduction to the method, in Cartesian coordinates. y Example solution for the potential in an infinite V=0 slot, arbitrary V at the a bottom, in which we V(y) x introduce two common V=0 features of separation z solutions: • Completeness and orthogonality of sines • Fourier’s trick
11 October 2002 Physics 217, Fall 2002 1 Introduction 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 ∂∂Vn. 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 ∂∂∂222VVV ∇=2V + + =0 ∂∂∂xyz222 which can’t be integrated directly like the 1-D case.
11 October 2002 Physics 217, Fall 2002 2 Introduction to separation of variables (continued)
Consider solutions of the form Vxyz(),, = XxYyZz()()()
Then, the Laplace equation is dX222 dY dZ YZ++= XZ XY 0 dx222 dy dz or, dividing through by XYZ, 111dX222 dY dZ ++=0 XYZdx222 dy dz i.e. fx()++= gy() hz()0
11 October 2002 Physics 217, Fall 2002 3 Introduction 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: 111dX222 dY dZ ++=+−+=AB() AB 0 XYZdx222 dy dz
So this is a way to trade one partial differential equation for three ordinary differential equations, which can often be solved much more easily:
dX222 dY dZ −=AX00 −= BY ++=() A B Z 0 dx222 dy dz
11 October 2002 Physics 217, Fall 2002 4 Introduction 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 5 Example: the infinite slot
Griffiths, example 3.3: Two y infinite, grounded, metal plates lie parallel to the x-z V=0 plane, one at y = 0, the other a at y = a. The left end, at x = 0, V(y) x is closed off with an infinite V=0 strip insulated from the two z plates and maintained at a specified potential V0(y). Semi-infinite conducting Find the potential inside this plates: slot. zxya=−∞→∞,0 = →∞ ,0,. = Bottom of the slot: plate No z dependence, so insulated from the other VVx= (), y two, withVV= 0 ()y . 11 October 2002 Physics 217, Fall 2002 6 The infinite slot (continued) ∂∂∂222VVV ∂∂ 22 VV ∇=2V + + = + =0 ∂∂∂∂∂xyz222 xy 22 Suppose Vxy(),= XxYy()() ; then dX22 dY11 dX 2 dY 2 YX+=⇒+==−00 kk22 dx22 dyXY dx 2 dy 2 or: dX22 dY I.−=kX22 0 II. += kY 0 dx22 dy We chose k 2 rather than, say, A, to indicate that this constant is non-negative, and that the other one ( − k 2 ) is non-positive. Why? We’ll see in a minute. 11 October 2002 Physics 217, Fall 2002 7 The infinite slot (continued)
Boundary conditions: 1.Vx→→∞ 0 as (reference point at infinity) 2.Vy== 0 at 0 3.Vya== 0 at
4.VVy==0 () at x 0
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.
dX2 = kX2 dx2
11 October 2002 Physics 217, Fall 2002 8 The infinite slot (continued)
Let vdXdx = , and multiply through by vdXdx= : dv dX vkX= 2 dx dx dv dX Integrate over x: vdxkXdx= 2 ∫∫dx dx vX22 =+kS2 22 Here both X and dX/dx must be zero for all x at y = 0 and a, so the constant S = 0: dX =−kX or kX dx
11 October 2002 Physics 217, Fall 2002 9 The infinite slot (continued)
Separate and integrate: dX =−kdx or kdx ∫∫X ∫ lnXkxT=+ or −+ kxU Xe==kx+−+− T Ae kx or e kx U = Be kx , where A and B are constants. The general solution is a linear combination of these two particular solutions: Xx()=+ Aekx Be− kx
11 October 2002 Physics 217, Fall 2002 10 The infinite slot (continued)
Similarly, for equation II, dYd 222 y =− kY , a more general solution is Yy()=+ Ce′′iky De− iky or, using eeiiαα−+−− ee ii αα sinαα== , cos , 22i we have Y ()y =+Cksiny Dk cosy , and kx− kx Vxy(),sincos.=+() Ae Be() C kyD + ky We’re ready to determine the (four) constants A-D by plugging the (four) boundary conditions in and solving for them.
11 October 2002 Physics 217, Fall 2002 11 The infinite slot (continued)
Applying the boundary conditions: 1.Vx→→∞→∞→∞ 0 as , but ekx as x ⇒=A 0. − 2.Vy== 0 at 0, so 0 = BeCDkx ()() 0 + (1) ⇒=D 0. −−kx kx So far, Vx(),sinsin.y ==BCe ky Ge ky (See now why we chose the separation constants to be ± k 2 ?) 3.V == 0 at y aGeka , so 0 =−kx sin nπ ⇒=⇒==sinka 0 k , n 0,1,2,… a
11 October 2002 Physics 217, Fall 2002 12 The infinite slot (continued)
This actually gives us an infinite number of solutions (leaving out n = 0, since this would make V = 0 everywhere: a trivial solution): nyπ VxyGe(),==−nxaπ sin , n 1,2,3,… nn a The remaining boundary condition (4) needs to be used to determine the Gn. Note first that this solution won’t work unless V0 itself is sinusoidal. BUT, if the Vxyn (), are all solutions, then a linear combination of them is a solution too: ∞ Vxy(),,= ∑ Vn xy () n=1 ∞∞() () 22 2 because ∇=∇VVxyVxy∑∑nn, = ∇ , = 0 nn==11 11 October 2002 Physics 217, Fall 2002 13 The infinite slot (continued)
And this über-general solution can match arbitrary functions
V0(y) at x = 0 (boundary condition 4), because the expression of this, ∞∞ −naπ ()0 nyππ ny Vy0 ()==∑∑ Gennsin G sin nn==11aa is just the Fourier sine series representation of V0(y). In Math 281 you will learn that the sines form a complete set of functions, for which any arbitrary function of of y can be expressed as a series like this. Thus this solution works for any specified V0(y), and all we have left to do is to determine the Fourier coefficients Gn. We do this with Fourier’s trick:
11 October 2002 Physics 217, Fall 2002 14 The infinite slot (continued)
Take sin myam π , = 1,2,3, … and multiply it on both sides of the expression for boundary condition 4: myπππ∞ ny my sinVy0 ()= ∑ Gn sin sin aaan=1 Then integrate both sides over y, from zero to a: aa myπππ∞ ny my V() ysin dy= G sin sin dy ∫∫0 aaa∑ n 00n=1 (This is basically to Fourier-transform both sides.) Now focus on the integral within the sum.
11 October 2002 Physics 217, Fall 2002 15 Orthogonality of sines
Integrate by parts twice: aaa myππ nyam my ππ ny my ππ ny sin sindy =− sin cos + cos cos dy ∫∫aa naanaaπ 000 a a mamyππ ny m my ππ ny =+cos sin sin sin dy nnπ a a n∫ a a 0 0 a m2 myππ ny Thus 1sinsin0.−=dy 2 ∫ aa n 0
There are two possibilities here; either m = n or mn ≠ . If the latter is the case, then a myππ ny sin sindy =≠ 0()mn . ∫ aa 0 11 October 2002 Physics 217, Fall 2002 16 Orthogonality of sines (continued)
If on the other hand m = n, then aaa nyππππa ny ny ny sin22dy =− sin cos + cos dy ∫∫anaaπ a 000 aaaa nyπππ ny ny 2sin222dy =+== sindy cos dy dy a ∫∫∫∫aaa 0000 So, a a The functions myππ ny ()mn= sin sin dy = 2 sin()nyaπ , n= 1,2,3,… ∫ aa ()are orthogonal 0 0 mn≠ on the range a = δ ya=→0. 2 mn
11 October 2002 Physics 217, Fall 2002 17 The infinite slot (continued)
Now back to boundary condition 4: aa myπππ∞∞ ny my a Vy()sin dy== G sin sin dy Gδ ∫∫0 aaa∑∑nnmn2 00nn==11 a = G m 2 And we’re done: the complete solution is ∞ −nxaπ nyπ Vxy(),sin,where= ∑ Gen n=1 a a 2 nyπ GVydy= ()sin . n aa∫ 0 0
11 October 2002 Physics 217, Fall 2002 18