Partial Differential Equations

Chapter Partial Differential Equations. Introuction Have solve orinary ifferential equations, i.e. ones where there is one inepenent an one epenent variable. Only orinary ifferentiation is therefore involve. As the worl is three-imensional, most ifferential equations are functions of three spatial variables, eg x, y, z, an maybe time t also. Typical example is Laplace equation 2 Vr=0, where Vr is the electrostatic potential in region where there is no charge. The operator 2, calle the Laplacian, was introuce last year. In Cartesian coorinates 2 V= 2 V V V x 2 + 2 y 2+ 2 z.. 2 Another important example is the time-inepenent Schröinger equation for particle 2 2m 2 Ψr+VrΨr=EΨr,.2 for the quantum-mechanical motion of a particle of mass m in a potential Vr. Ψr is the particle s wave function an =h/2π, where h is Planck s constant. There are many more examples that you will come across later in your egree programme..2 Classification of Differential Equations Before consiering various ifferential equations DE in etail it is worth efining some of the terms use to classify these equations into ifferent types. The following terms are use: Orer. The orer of a DE is the orer of its highest erivative, so a n x n y x n+ a n x n y x n +...a 0xy=0.3

2 CHAPTER. PARTIAL DIFFERENTIAL EQUATIONS is a DE of orer n. This efinition hols even if there are several variables, so 3 y y = 0,.4 x 3+ 2 t2 is a thir-orer. Linearity. A linear DE can be written entirely as a linear function. i.e. no powers above the first power, of the unknown function an its erivative. So a n x n y x n+ a n x n y x n +...a 0xy=bx.5 is linear if the a i s an b are functions of x only. It is non-linear if any of the a i s epen on y. Example: a penulum, whose equation of motion is given by 2 θ t + g sinθ=0.6 2 l is non-linear inθ. However ifθis small then sinθ θ an the DE 2 θ t + g θ=0.7 2 l is linear. Linear DE s are important because they are easier to solve. Orinary/Partial. If an unknown function, eg y, is a function of only one variable, eg x, then one gets orinary DEs such as y = c..8 x If y is function of more than one variable, eg x an t then one gets a partial DE eg yx, t yx, t + = c.9 x t provie the variables, x an t, are inepenent. If the variables are epenent, eg x= f s, t, then it is necessary to specify which are hel constant yx, t x = cs, t.0 t Such constructions are familiar from thermoynamics where P pressure, V volume an T temperature are all inter-relate eg by the ieal gas equation PV=nRT an many functions, such as entropy S, have to written as partial DEs. This means that S T S P T. V Homogeneous. Means slightly ifferent things for linear an non-linear DEs. Will only consier the linear DE case. A matrix equation such as Ax=b is homogeneous if b=0. Similarly, a secon-orer DE Px 2 y x2+ Qxy+ Rxy=Gx.2 x

.3. SEPARATION OF VARIABLES 3 is homogeneous if Gx=0 an is inhomogenous if Gx 0. Solving the homogeneous DE is usually the first step in solving an inhomogenous DE. We will restrict ourselves to homogenous DEs. Solutions. By a solution of an orinary DE Fx, yx, y x, 2 y x 2,...=0 we mean some function y=ux in the range a<x<b for which the problem is efine. This solution can always be verfie by irect substitution. Does Fx, ux, u x, 2 u x 2,...=0? Uniqueness. A DE in general will have more than one solution because:. There are unknown constants which can only be etermine by the bounary conitions. Bounary conitions give information about the unknown function or its erivatives at some point. Eg y=0 at x= is a bounary conition. n bounary conitions are require to etermine constants for an n th -orer equation. So a secon-orer DE requires 2 bounary conitions. 2. For an n th -orer DE there are usually n inepenent functions, ux, satisfying the DE. So a secon-orer DE has 2 solutions. Which solution is correct is often etermine by the physics of the problem. Existence. There is no guarantee that a DE will have a solution of the form ux..2. Superposition Principle If V an V 2 are two solutions of any linear, homogeneous DE such as 2 Vr= 0, then V=c V + c 2 V 2, where c an c 2 are arbitrary constants, is another solution. Use extensively for orinary DEs, eg simple harmonic motion problem; is equally vali for partial DEs. This ability to a solutions is calle the Superposition Principle. Of funamental importance in Quantum Mechanics. Will exploit the superposition principle extensively when solving partial DEs..3 Separation of variables Most DEs that characterise physical problems epen on many variables an cannot be irectly solve.sometimes can solve these multi-imensional problems by separation of variables which turns a partial DE in n variables into n orinary DEs each in one variable. Take an n=2example ax, y 2 u x 2+ bx, u y 2 y2= 0..3 If this is separable we can write ux, y=xxyy which gives ax, yyy 2 X x 2+ bx, Y yxx2 y2= 0,.4

4 CHAPTER. PARTIAL DIFFERENTIAL EQUATIONS or, iviing through by XY an re-arranging: ax, y Xx 2 X x bx, y = 2 Yy 2 Y y2..5 This equation is separable provie that the left-han sie can be written totally in terms of x an the right-han sie totally in terms of y. This may require some re-arrangement between ax, y an bx, y to give Ax an By, respectively functions of x an y only. If eq. is separable, then have relationship of form f x= gy. Since relationship hols for all values of x an y, must mean that f x=c= gy, where c is some constant, often for convenience written as a square eg l 2. Can solve separately two equations Ax 2 X Xx x = c, By 2 Y = c..6 2 Yy y2 Note that separability epens on the coorinates chosen, it may be necessary to change coorinates..3. Laplace s equation in Cartesian coorinates Let us illustrate this with a physical example. Consier two infinitely large conucting plates. The one at z=0 is earthe while that at z=lis kept at a constant voltage V 0. V 0 L What is the potential between the two plates? You all know that the answer must be V = V 0 z/l but we are going to erive this by solving the partial ifferential equation. This will emonstrate the techniques to be use in more complex cases. Between the two plates, there is no charge an so the potential in this region satisfies Laplace s equation 2 Vr=0. The bounary conitions to be applie are that, inepenent of the values of x an y, on the plates V=0 at z=0, V= V 0 at z=l..7

.3. SEPARATION OF VARIABLES 5 Since the bounary conitions are expresse easily in terms of Cartesian coorinates, it makes obvious sense to attack the problem in this coorinate system. [Coul also use cylinrical polar coorinates.] In this system, Laplace s equation becomes 2 V= 2 V V V x 2 + 2 y 2+ 2 z = 0. 2 Let us try for a solution of the form Vx, y, z=function of x function of y function of z, Vx, y, z= Xx Yy Zz..8 At the moment we are just trying to get a single solution of the equation. If there is no solution of this kin then we will have to try something else but of course there will be! Substituting the prouct form of Eq..8 into Laplace s equation, we get Y Z 2 X x 2+ X Z 2 Y y 2+ X Y 2 Z = 0..9 z2 Note that we now have complete ifferentials straight s because X is a function of only one variable x, an similarly for Y an Z. Now ivie through the equation by the prouct V= X Y Z to get 2 X + 2 Y + 2 Z = 0..20 X x 2 Y y 2 Z z 2 Now the first term in Eq..20 is a function only of x, the secon only of y, an the thir only of z. BUT x, y, an z are inepenent variables. This means that we coul keep y an z fixe an vary just x. In so oing, the secon an thir terms remain fixe because they only epen upon y an z respectively. Hence the first term must also remain fixe even if x changes. That is, the first term is a constant, as are the secon an thir. Thus with X Y Z 2 X x 2 2 Y y 2 2 Z z 2 = l 2, = m 2, = +n 2..2 n 2 =l 2 + m 2..22 Note that n 2,l 2 an m 2 are as yet arbitrary constants an coul be negative. l, m, n are not necessarily integers. Have to solve 2 X x 2 = l2 X..23 For reall 0, this is the simple harmonic oscillator equation X=a l coslx+b l sinlx,.24

6 CHAPTER. PARTIAL DIFFERENTIAL EQUATIONS where a l an b l are arbitrary constants which must be fixe by the bounary conitions. For special casel=0, solution simplifies to X=a 0 + b 0 x..25 Ifl 2 is negative, putl=il; the cos lx an sinlx become coshlx an i sinhlx. Have seen such changes before when stuying the ampe oscillator in B27. Solutions for Y are similar to those for X, but with m replacing l. For Z have Has solutions 2 Z =+n 2 Z..26 z 2 Z= e n cosh nz+ f n sinh nz n 0, = e 0 + f 0 z n=0..27 As a consequence, solutions of the separable form o exist. For example, one solution woul be withl=3, m=4, an n=5. Vx, y, z= Constant sin 3x cos 4y sinh 5z is a solution of Laplace s equation, but many more with ifferent values of l, m, n exist. Most general solution is Vx, y, z= Constant { sinlx coslx } { sin my cos my } { sinh nz cosh nz with constraint n 2 =l 2 + m 2. By the superposition principle, any linear combination of such solutions is also a solution. The most general superposition is Vx, y, z = {a lm coslx+b lm sinlx} { c lm cos my+ lm sin my } l,m { e lm cosh nz+ f lm sinh nz }..28 For any choice oflan m, with n= l 2 + m 2, the above prouct is a solution. Hence the sum is also a solution. Notelan m o not have to be integers an so the above nee not be a iscrete sum. Also note that ifl 0, cosine is replace by an sine by x. } Imposing bounary conitions Solution Eq..28 is quite general, nee to relate it potential problem of two parallel plates: have to impose the bounary conitions. At z=0, Vz=0= e lm {a lm coslx+b lm sinlx} { c lm cos my+ lm sin my } = 0 lm

.4. LAPLACE EQUATION IN SPHERICAL POLAR COORDINATES 7 for all values of x an y. Hence e lm = 0 for alllan m. Most general solution simplifies to Vx, y, z= sinh nz {a lm coslx+b lm sinlx} { c lm cos my+ lm sin my }, lm.29 where coefficient f lm has been absorbe into reefine a lm an b lm. At z=l, Vz=L= sinh nl {a lm coslx+b lm sinlx} { c lm cos my+ lm sin my } = V 0, lm for all x an y. Clearly, only solution which gives something inepenent of x an y is the special case ofl=m=n=0. Write this explicitly as At z=l, Vx, y, z= z{a+bx}{c+y}..30 V 0 = L{a+bx}{c+y} for all x, y so that b==0 an ac=v 0 /L. The final solution is, from Eq..30, the expecte V= V 0z L.3.2 Comments. Metho of solution is Separation of Variables: look for a solution which is a prouct of a function of x times a function of y times a function of z. Reuces problem to that of solving three orinary ifferential equations in x, y an z. 2. Have foun an infinite number of solutions of the Laplace equation, but have not shown that we have foun them all. 3. In more complicate examples the orinary ifferential equations may be very much harer to solve than the simple oscillator equations here. 4. Unlike the present case, in general you cannot guess the final answer at the start!.4 Laplace equation in spherical polar coorinates The electrostatic an gravitational potentials satisfy, in the absence of sources, Laplace equation: 2 V= 0,.3 where V is the potential an 2 is the Laplacian operator see Chapter. Moreover, this equation is also central to quantum mechanics, because its solutions are strictly relate to the solutions of Schröinger equation. Often, in the stuy of physical systems, situations arise where, besies Eq..3, one also knows that the system is spherically symmetric a prominent

8 CHAPTER. PARTIAL DIFFERENTIAL EQUATIONS example being the stuy of atoms. In such cases, it is convenient to solve Laplace equation in spherical polar coorinates. The explicit expression of the previous equation in spherical polar coorinates is r 2 r in the form r 2 V r + r 2 sinθ θ sinθ V θ 2 V + = 0.32 r 2 sin 2 θ φ 2 Vr,θ,φ=Rr Θθ Φφ..33 Involves functions which epen purely upon one variable each, viz r, θ an φ. Inserting this into Laplace s equation ΘΦ r 2 R + RΦ sinθ Θ 2 Φ + RΘ = 0. r 2 r r r 2 sinθ θ θ r 2 sin 2 θ φ 2 After iviing by RΘΦ an multiplying by r 2 sin 2 θ, fin sin 2 θ r 2 R + R r r Θ sinθ sinθ Θ + 2 Φ = 0. θ θ Φ φ 2 First two terms here epen upon r anθbut thir is function purely of azimuthal angle φ. Since r, θ an φ are inepenent variables, means that thir term must be some constant, enote by m 2. Hence 2 Φ φ 2= m2 Φ,.34 which has solutions e ±imφ or, alternatively, cos mφ an sin mφ. As far as DE concerne, m coul have any value, even complex. However Physics imposes a fairly general bounary conition. When φ increases by 2π, the vector position returns to the same point; expect same physical solution. ThusΦφ+2π=Φφ. Can only be accomplishe if m is a real integer. Then Φφ is clearly a perioic function. The remainer of the equation can be manipulate into R r r 2 R r = m2 sin 2 θ Θ sinθ θ sinθ Θ θ Left han sie is function only of r, while right han sie epens only onθ. Means that both sies must be equal to some constant, enote byλ. Results in two orinary DEs: r 2 R = λ R,.35 r r sinθ Θ + λ sinθ m2 Θ = 0..36 θ θ sinθ Now look at the raial equation of Eq..35, rewritten as r 2 2 R R + 2r λ R=0..37 r 2 r.

.4. LAPLACE EQUATION IN SPHERICAL POLAR COORDINATES 9 This is a special kin of homogeneous equation which is unchange if the r- variable is scale as r α r, whereαis some constant. Try for a solution of the form Rr r β, since this also stays in same form uner the r αr scaling. Hence ββ r β + 2β r β λ r β = 0 Cancelling out the r β factor, which cannot vanish, givesβ 2 +β=λ, has solutions β= ± +4λ /2. To make things look a bit simpler, efine separation constant to beλ ll +, where l is not necessarily an integer. Then β = ± +4ll+ /2 = l or l. Most general form of the raial solution is Rr=A r l + B rl+..38 In orer not to interchange the two solutions, aopt the conventionl 2. Left only with theθequation which, with new separation constantll+, becomes sinθ Θ + ll+ sinθ m2 Θ=0,.39 θ θ sinθ which oes not look very attractive. A little more tractable with the variable µ=cosθ rather thanθ. Then µ/θ= sinθ an Hence = sinθ θ µ = µ 2 µ. [ µ 2 Θ ] [ ] + ll+ m2 Θ=0..40 µ µ µ 2 This is the famous Legenre ifferential equation important for quantum mechanics. Legenre iscovere his equation when trying to interpret planetary gravitational fiels, Recherches sur la figure es planètes 784. This footnote ties in with the series solution metho we will learn in the next Chapter. Get exactly the same result by trying for the more general series solution. Stanar manipulation leas to a n {n+ kn+k+ λ} r n+k = 0 n=0 The inicial equation leas to exactly the same result with β replace by k. For higher values of n have a n {n+kn+k+ λ}=a n n2k+=0. But 2k+=2β+=± + 4λ oesn t vanish. Hence a n = 0 for n an get back to the single-term solution erive above.