Mathematical Methods wks 5,6: PDEs

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1 Mathematical Methos wks 5,6: PDEs John Magorrian, These are work-in-progress notes for the secon-year course on mathematical methos. The most up-to-ate version is available from 9 PDEs: introuction Most unergrauate physics problems boil own to solving ifferential equations, often partial ifferential equations. For example, aplace s equation, ψ =, 9.1 occurs in electrostatics, steay-state hyroynamics an heat flow. The time-epenent iffusion equation, ψ = D ψ, 9. t occurs in problems involving heat flow, nuclear reactions, approximations of ranom walks, among others. In many problems the iffusion coefficient Dψ, r is constant an 9. becomes the heat equation ψ t = D ψ. 9.3 Ug separation of variables ψr, t = φrt t in the familiar wave equation, leas to one sign choice in the Helmholtz equation, Finally, Schröinger s equation, is familiar from quantum mechanics. ψ c ψ =, 9.4 t φ ± k φ =. 9.5 m ψ + V rψ = i ψ t, 9.6 These equations are all linear, secon-orer homogenous PDEs: they can all be written in the form ˆDψ = g, 9.7 where ˆD is a linear secon-orer ifferential operator in x, t an gx, t =. inear homogenous equations have the convenient property that we can superpose solutions: if ψ 1 x, t an ψ x, t are two solutions, then so too is α 1 ψ 1 + α ψ. The next few sections will give some examples that illustrate how to solve such PDEs in practice. The general proceure is: Ientify the bounary conitions physics; Decie on a suitable coorinate system for the problem; 1 Obtain the general solution for the PDE to be solve; Fin the specific solution that satisfies the bounary conitions. In these lectures I ignore steps an an concentrate on steps 1 an. Mathematical Methos wks 5,6: PDEs V Equations of the form 9.7 in which g are inhomogenous. An example of an inhomogenous equation is Poisson s equation, ψ = ρ/ɛ. Such equations can be solve by the metho of Green s functions. Before going on to the etaile examples, note that there are many important PDEs that are not secon-orer linear equations. For example, the continuity equation, ρ + ρv =, 9.8 t is a linear first-orer equation if we happen to know ur, t, but nonlinear if neither ρ nor u are known. The Navier Stokes equation u ρ + u u = p ρ Φ + µ u, 9.9 t is an obviously non-linear first-orer inhomogenous PDE for ur, t. Both of these equations can be obtaine by taking moments of the Boltzmann equation, f t + v f f Φ x v = Γ cf, 9.1 which is actually an integro-ifferential equation for the phase-space ensity fx, v, t, because the collision term Γ c f involves a complicate integral over fx, v, t. See kinetic theory lectures for one way of approximating this Γ c. Here we merely note that if we assume that Γ c = then the Boltzmann equation becomes a first-orer homogenous PDE. 1 aplace s equation by separation of variables The erivations presente in the next few sections are slow an mostly careful, proceeing only in baby steps. They ll probably put you to sleep. Nevertheless, they might prove useful if there are etails of the proceure you on t unerstan. 1.1 Example: aplace s equation in Cartesian co-orinates The steay-state temperature istribution T x, y within a semi-infinite metal sheet of with satisfies aplace s equation, T =, with bounary conitions i the temperature at the eges T, y = T, y =, ii T x, y as y, an iii T x, = T. What is T x, y within the plate? General solution We try a solution of the form in which case aplace s equation becomes Diviing both sies by XY gives T x, y = XxY y, 1.1 Y X x 1 X X x + 1 Y + X Y =. 1. y Y y =, 1.3

2 V 3 Mathematical Methos wks 5,6: PDEs in which the first term epens only on x, an the secon only on y. Therefore we must have 1 X X x = 1 Y Y y = k, 1.4 where k is in general some complex constant. We coul write this separation constant as +k or even just k, but choog k simplifies the following. We are left with two ODEs both eigenvalue equations X x = k X, Y y = +k Y, for which the k-epenent general solutions are A + B X k x = x, if k =, A k cos kx + B k kx, otherwise, C + D Y k y = y, if k =, C k e ky + D k e ky, otherwise, where A k, B k, C k an D k are some possibly complex constants. aplace s equation is linear an homogeneous. Therefore a more general solution to T = is T x, y = k X k xy k y = A + B xc + D y + k A k cos kx + B k kxc k e ky + D k e ky. 1.7 Comments: i If we ha chosen +K instea of k as our separation constant then X K x woul be a sum of exponentials an Y K y woul involve es an coes. Both sets of solutions are equivalent because k = ±ik. ooking at the bcs, we guess that T will ecay exponentially as y an has to vanish at x = an x =, suggesting trigonometric series in x. Therefore to keep the subsequent algebra as simple as possible, we label our constant k. ii The separation constant is k. This means that if we inclue a term with, say k = in 1.7 we o not nee the term with k = : any A can be absorbe into A, any C into D etc. We therefore assume that for any X k Y k in the series 1.7 the corresponing X k Y k = : in other wors, each possible value of k appears at most once. iii The X k x an the Y k y are then I. That is, the only solution to k c kx k x = or k c ky k y = is if all c k =. Application of bounary conitions First, let us use the conition that the temperature at the eges T, y = T, y =. Setting x = in equation 1.7 gives = k X k Y k y = k which, ce the Y k y are I, is satisfie only if all A k =. For x = we have that = k X k Y k y = k A k Y k y, 1.8 B k ky k y, 1.9 which, by the same reasoning, requires that A k k =. We satisfy this by impog k = nπ/, where n is an integer. If we were to choose any k nπ/ we woul have to set the corresponing B k = in orer to satisfy the bc, so we have simply that X k x = when k nπ/. Substituting these results into 1.7, the solution subject to the bcs on x = an x = is T x, y = B n C n e nπy/ + D n e nπy/ = C n e nπy/ + D n e nπy/, Mathematical Methos wks 5,6: PDEs V where in the last line we have absorbe the constant B n into C n an D n. We will show later that this is in fact the most general solution to T = subject to the conition that T vanishes on the eges of the sheet, T, y = T, y =. Next we use the bc that T, y as y. From 1.1, = lim y = C n e nπy/ + D n e nπy/ lim y C n e nπy/ + D n e nπy/ = lim C n e nπy/, y giving all C n =, because the functions nπx/ are I. Finally, we apply the conition that T = T on the y = en. Ug 1.1 again, 1.11 T = T x, = D n. 1.1 In geometrical terms, this says that the coefficients D n } are the co-orinates of the function T = T expresse in terms of the orthogonal basis nπx/. To fin, say, D m we simply take the scalar prouct of 1.1 with mπx/: T mπ mπx T x = cos mπx D n = T mπ 1 1m = D m D m = 4T πm D n δ mn m o, m even. mπx Substituting this into 1.1, our final solution subject to all the bcs is in which n = m 1. T x, y = 4T m 1πx πm 1 exp x 1.13 m 1πy, 1.14 Alternative metho Just in case you re curious, here s one way of showing that 1.1 is inee the most general solution to our problem. You coul use the following metho as an alternative to separation of variables when solving problems, but I on t recommen it: see the comments below for why.

3 V 5 Mathematical Methos wks 5,6: PDEs The problem: we want to fin T x, y in the sheet < x <, y <. Ug the property of completeness of Fourier series, we can take a horizontal slice, y = const, across the sheet an write the temperature profile along the slice as T x, y = a + a n y cos + b n y, 1.15 in which the coefficients a n an b n epen on which horizontal slice we re looking at. The values of a n an b n at ifferent heights are relate through aplace s equation T =. Substituting this T x, y into aplace s equation we obtain = a n T = y n π a n cos + b n y n π b n = The contents of each of the square brackets must vanish because the es an coes are I. So we have a n y n π a n =, 1.17 which means a n y = F n expnπy/ + G n exp nπy/, where F n an G n are constants of integration. Similarly, b n y = H n expnπy/ + I n exp nπy/. Therefore our general solution to T = for < x < is T x, y = constant+ F n exp + nπy + G n exp H n exp nπy nπy + I n exp cos nπy Applying the bc T = at x = an x = remember that the constant an the exponentials form an I set gives eq 1.1. Comments 1. ecall that the usual Fourier-series expansion assumes perioicity i.e., that the function is efine along the circumference of a circle. Therefore we woul have been in trouble ha our bc require that T x, T x,. Take a look at 1. of HB to see one way of circumventing this.. To avoi this problem we coul try to be clever an expan the temperature along our constant-y slice as, e.g., a egenre series: x + T x, y = a l yp l, 1.19 l= where the argument x + / to the egenre polynomial P l lies in the range 1, 1. It is far from easy, however, to solve for a l y when this T x, y is substitute into aplace s equation. In constrast, the metho of separation of variables usually reuces the problem to familiar ODEs that we know how to solve. 1. aplace s equation in plane polar co-orinates In polar co-orinates, φ aplace s equation is 1 V + 1 V =. 1. φ Substituting a trial solution of the form V, φ = V V φ φ into 1. gives Multiply by /V V φ V φ V V where m is a separation constant. The equation for V φ φ, has solutions V V V + V 1 Mathematical Methos wks 5,6: PDEs V 6 V φ φ = + 1 V φ V φ φ = = 1 V φ V φ φ = m, 1.1 V φ φ = m V φ, 1. V m A + B φ φ = φ, if m =, A m cos mφ + B m mφ, if m. 1.3 We require that V φ φ = V φ φ + π because φ is an angular co-orinate. ooking at 1.3, this means that if m = then we must have B =, whereas if m then m must be an integer. As the separation constant is m, we nee only consier one sign of m. Therefore, without loss of generality, we have that for m =, 1,,.... By inspection, the solutions to the raial equation, are given by V m φ φ = A m cos mφ + B m mφ 1.4 V = m V, 1.5 V m = C + D log, if m =, C m m + D m m, if m >. Therefore, in plane polar co-orinates the general solution V, φ to aplace s equation, V =, is V, φ = V m m= m V φ φ = C + D log + Cm m + D m m A m cos mφ + B m mφ. Example: earthe ro in uniform electric fiel An infinite ro of raius a is earthe an place in a uniform electric fiel E x, E y, E z = E,,, with the axis of the ro coincient with the Oz axis. The bounary conitions on the unknown potential V, φ are i V, φ E cos φ as an ii V, φ = on = a. Taking the general solution 1.7 an applying the first bc, we have that C + D log + C m m A m cos mφ + B m mφ E cos φ, 1.8

4 V 7 Mathematical Methos wks 5,6: PDEs in which we ve roppe any term in V that ecreases as. There are two ways to fin the coefficients A m, B m etc. One is to rearrange 1.8 as a sum of I basis functions i.e., nφ an cos nφ an then to argue that the coefficient multiplying each basis function must be zero. The more powerful alternative is to project 1.8 along each of the basis functions to pick off the coefficients one by one. Taking the scalar prouct of both sies of 1.8 with nφ for n = 1,, 3,..., gives π C m m B m nφ mφ φ, C m m B m πδ nm, or 1.9 which can only be satisfie if all C m B m =. Next take the scalar prouct of 1.8 with cos nφ for n =, 1,,.... For n = we have that C + D log, 1.3 so that D =. For n 1, π C m m A m cos nφ cos mφ E π C m m A m πδ nm Eπδ m1, cos nφ cos φ φ, or 1.31 so that all C m A m =, except for C 1 A 1 = E. Substituting these results into 1.7, our potential has become V, φ = C E cos φ + D m A m m cos mφ + D m m B m mφ 1.3 Applying the bc V a, φ =, we have that = V a, φ = C Ea cos φ + D m A m a m cos mφ + D m a m B m mφ Taking the scalar prouct of 1.33 with cos nφ tells us that C = from the n = case an that all D m A m = except for D 1 A 1 = Ea. Similarly, the scalar prouct with nφ gives all D m B m =. Our solution to aplace s equation subject to both bcs is therefore V, φ = E a cos φ aplace s equation in spherical polar co-orinates Finally, what is the most general V r, θ, φ that solves aplace s equation in spherical polar co-orinates, V = 1 r r V 1 V + r r r θ φ + 1 r θ V =? 1.35 θ θ θ As usual, we first try a solution of the form V = V r rv θ θv φ φ. Substituting this trial solution into aplace s equation 1.35, multiplying by r θ/v r V θ V φ an rearranging, we fin that θ r V r + θ θ V θ = 1 V φ V r r r V θ θ θ V φ φ }} = m, 1.36 }} function of r, θ only fn φ only Mathematical Methos wks 5,6: PDEs V 8 where m is a separation constant. By the same reasoning we use earlier in the plane-polar case, we impose V φ φ + π = V φ φ, which means that the general solution to V φ φ is where m =, 1,, 3,... is a non-negative integer. V φ φ = A cos mφ + B mφ, 1.37 To fin V r r an V θ θ, ivie 1.36 by θ an rearrange slightly to obtain 1 r V r 1 = θ V θ m V r r r V θ θ θ θ = ll + 1, 1.38 θ }}}} only r only θ in which ll + 1 is an inspire choice of separation constant. The equation for V θ is therefore 1 θ θ x θ V θ θ 1 x x m θ V θ = ll + 1V θ, m 1 x V θ = ll + 1V θ, or 1.39 where in the last line we ve substitute x = cos θ, so that θ = θ x. This equation for V θ is the associate egenre equation of 8.. The eigenfunctions i.e., the V θ are associate egenre functions, Pl m cos θ, an exist only for l =, 1,..., an m l. Finally, the V r equation becomes which, by inspection, has a general solution Putting this together, we have that r V r = ll + 1V r, 1.4 r r V r r = Cr l + Dr l V r V θ V φ = Cr l + Dr l+1 Pl m cos θa cos mφ + B mφ 1.4 is one solution to aplace s equation for l =, 1,,... an m =, 1,,..., l. Because aplace s equation is linear an homogeneous, we can superpose solutions an write l V r, θ, φ = l= m= C lm r l + D lm r l+1 P m l cos θa lm cos mφ + B lm mφ, 1.43 in which the constants of integration A, B, C, D epen on the choice of l an m. Exercise: Show that 1.43 is the most general solution to aplace s equation. Hint: first use the property of completeness of S eigenfunctions to show that any V r = const, θ, φ can be expresse as lm P l m cos θf lm r cos mφ + G lm r mφ. Then substitute this expression into aplace s equation to obtain ODEs for F lm r an G lm r. In problems with axisymmetry, V = V r, θ, we have that all B lm = an the only A lm that are non zero are those with m =. Then the solution to aplace s equation becomes V, θ = C l r l + D l r l+1 P l cos θ 1.44 l=

5 V 9 Mathematical Methos wks 5,6: PDEs because P l x = P lx, the l th -orer egenre polynomial see 8.1. Example: Earthe sphere in a uniform electric fiel An earthe sphere is place in a uniform electric fiel E = Eẑ. There are no charges outsie the sphere, so the electric potential V satisfies aplace s equation, V =, with bcs i V Er cos θ as r an ii V r, θ = on the surface of the sphere r = a. The first bc can also be written V ErP 1 cos θ as r. From the general solution 1.44 we than have that, as r, C l r l P l cos θ ErP 1 cos θ l= To fin the coefficients C i, take the scalar prouct of both sies with P m cos θ: that is, multiply both sies by P m cos θ an integrate cos θ. The result is 1 1 C l r l P m cos θp l cos θ cos θ Er P m cos θp 1 cos θ cos θ, l= 1 1 C l r l m + 1 δ lm Er m + 1 δ m1, l= where in the last line we have use the orthogonality relation for the P l, namely, 1 1 Therefore all C l =, except for C 1 = E. Similarly, applying the secon bc, V r, θ = on r = a, gives 1.46 P l xp m x x = l + 1 δ lm = Ea + D 1 a P 1 cos θ + D l a l+1 P l cos θ }} l 1 l=1 As the P l are linearly inepenent, we must have that D 1 = a 3 E an all other D l =. Our final solution for the potential outsie the sphere is V r, θ = Er 1 a3 P 1 cos θ Spherical harmonics Spherical harmonics are efine for l =, 1,,... an m l via l+1 l m! Y lm θ, φ = 4π l+m! P l m cos θe imφ, m, 1 m Yl m θ, φ, m <. 1.5 Notice that they re linear combinations of the angular part, r 3 P m l cos θa cos mφ + B mφ, 1.51 of the solutions 1.4 to aplace s equation. Therefore another way of writing the general solution to aplace s equation V =, eq. 1.43, in spherical polar co-orinates is V r, θ, φ = l l= m= l C lm r l + D lm r l+1 Y lm θ, φ. 1.5 An alternative metho is of course to exploit the linear inepenence of the P l. Examples For reference, here are the first few spherical harmonics: Y = ; Y 1, 1 = 4π 8π θe iφ ; Y 1 = cos θ; 4π Y Some other properties The Y lm θ, φ are orthonormal, with Mathematical Methos wks 5,6: PDEs V = 8π θeiφ Y l m θ, φy lmθ, φ θθφ = δ l lδ m m Exercise: use equation 8.5 to show that i the Y lm are orthogonal an ii normalize. The Y lm θ, φ are also complete: any well-behave function fθ, φ efine on the surface of a sphere can be expresse as l fθ, φ = c lm Y lm θ, φ, 1.55 where the coefficients, are the projections of f onto each Y lm. l= m= l c lm = Ylmθ, φfθ, φ θθφ, Helmholtz/wave equation: vibrations of a circular rum Here is another example that involves two separation constants. The vertical isplacement u, φ, t of the surface of a circular rum satisfies the wave equation u 1 u c =, 11.1 t with the bounary conition that u = on the ege = a of the rum. What are the normal moes of the rum? et us separate variables in two steps. First we write u, φ, t = U, φt t. Substituting this into the wave equation an separating variables, we obtain T t = ω T, U + ω c U =, 11. ug ω as the separation constant. The time-epenent factor in our assume u, φ, t is clearly T t e ±iωt, while the spatial part U, φ is given by the solution of the Helmholtz equation subject to the bounary conition that U = for = a. Making the further substitution U, φ = U U φ φ to separate variables in the Helmholtz equation gives U φ U + U U φ φ + ω c U U φ =. 11.3

6 V 11 Mathematical Methos wks 5,6: PDEs Multiplying by /U U φ, gives U which separates into the two equations U + 1 U φ U φ φ + ω c =, 11.4 U φ φ + m U φ =, U ω + c m U =, 11.5 where m is another separation constant. The first of these has solution U φ e imφ. The perioicity conition U φ φ + π = U φ φ restricts m to be an integer: m =, ±1, ±,... Exercise: The separation constant in these equations is m, not m. Why o we nevertheless nee to inclue both signs of m in our general solution for U φ φ? The secon equation is Bessel s equation etting x ω/c, it becomes x x U + x m U = x x The solutions to this equation that are well-behave at the origin x = are the Bessel functions U x = J m x. General solution By the linearity an homogeneity of the wave equation, we can superpose solutions with ifferent separation constants m an ω to obtain the general solution u, φ, t = m= ω ω A ω,m J m c where the coefficients A ω,m are set by the bounary conitions of the problem. e imφ e ±iωt, 11.7 Bounary conitions The treatment of the bounary conition in most problems involving Bessel functions iffers slightly from the previous problems we have tackle. In the present problem we have that u = a, φ, t =, which means that = m= ω ωa A ω,m J m e imφ e ±iωt c The linear inepenence of the e imφ an e ±iωt factors means that this can hol only if J m ωa/c =. The general solution subject to the bounary conition that u = on the ege = a is therefore u, φ, t = m= ωmn J m e imφ A + c mne iωmnt + A mne iωmnt 11.9 where ω mn is the n th solution to J m ω mn a/c =. That is ω mn = cα mn /a, where the coefficients α mn enumerate the zeros J m α mn = of the m th Bessel function. Exercise: Explain why any choice of initial isplacement U, φ for which U a, φ = can be represente by the series 11.9 with a suitable choice of coefficients A + mn an A mn. Fin expressions for these coefficients in terms of U, φ. 1 Cool own: heat equation The temperature Θx, t of a semi-infinite wall x > satisfies the heat equation Mathematical Methos wks 5,6: PDEs V 1 Θ t = D Θ x, 1.1 where D is a constant, subject to the bounary conition that the temperature at the surface x = varies with time as Θx =, t = T + T 1 cos ωt. How oes Θx, t vary insie the wall? To answer this, let us try a solution of the form Θx, t = XxT t. Substituting this into the heat equation gives X T t = DT X x. 1. Diviing by DTX results in 1 T DT t = 1 X X x, 1.3 in which the HS epens only on t, not x, an the HS epens only on x, not t. The only way this can happen is if they are both equal to some constant, λ, where λ may be complex. Therefore the PDE 1.1 separates into the two ODEs, T t = λ DT, 1.4 X x = λ X. The general solutions to these are T λ t exp λ Dt an Xx = A λ e iλx. Comment: Notice that we have slightly simplifie the expression for these solutions by writing the separation constant as λ instea of, say, λ. For any choice of separation constant λ there are two possible values of λ: one of these gives a solution Xx = A +λ e +λx, the other to Xx = A λ e λx. Absorbing the constant of proportionality in this T λ t into A λ, the λ-epenent solution is Θ λ x, t = A λ e iλx λ Dt. 1.5 The heat equation is linear an homogeneous, so we can superpose solutions, giving the more general solution Θx, t = λ A λ expiλx λ Dt. 1.6 Now we turn to bounary conitions. The conition that Θx =, t = T + T 1 cos ωt means that λ A λ e λdt = T + T 1 e iωt + e iωt 1.7 The HS is clearly perioic. Impog this same perioicity on the HS implies that λ D = inω, where n Z. Therefore nω λ n = ±1 i D. 1.8 We rely on an implicit bounary conition to choose the correct sign in this expression: we expect Θx as x. This means that the real part of iλ must be less than, so that n ω 1 + i D, if n >, iλ n = 1.9 n ω 1 i D, if n <.

7 V 13 Mathematical Methos wks 5,6: PDEs Now equation 1.7 becomes n= A n e inωt = T + T 1 e iωt + e iωt. 1.1 Exploiting the linear inepenence of the e inωt for ifferent n, we have that A = T an A 1 = A 1 = T 1 /, with all other A n =. The full solution 1.6 is Θx, t = T + T 1 exp = T + T 1 e x/δ cos 1 + i δ ωt x δ x + iωt, + exp 1 i x iωt δ 1.11 where the skin epth δ D/ω.