Ordinary differential equations Notes for FYS3140

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1 Ordinary differential equations Notes for FYS3140 Susanne Viefers, Dept of Physics, University of Oslo April 4, 2018 Abstract Ordinary differential equations show up in many places in physics, and these notes aim to give a comprehensive overview of some of the important types of equations, and solution methods. There is obviously some overlap with the relevant chapters in the book. The lectures will follow these notes rather closely, but with additional examples. Contents 1 Introduction some basic definitions 2 2 Linear first order DEs 2 3 Ordinary second order DEs Homogeneous equation Variation of constants Homogeneous DEs with constant coefficients Euler-Cauchy equation Inhomogeneous DEs Constant coefficients method of undetermined coefficients Factorization Variation of parameters Greens functions Series expansion methods Power series Fröbenius method

2 1 Introduction some basic definitions A differential equation (DE) is an equation involving derivatives of the function we aim to solve for. An ordinary DE involves derivatives with respect to only one variable (as opposed to partial DEs which will be treated later in the course). A simple example would be y + 5y = 0 (1) where y = y(x) is the function we are to solve for, and the primes denote derivatives with respect to the variable x. The order of a DE is defined as the order of the highest derivative occurring in the equation. So Eq.(1) is second order, while the general form of a first order DE is F (x, y, y ) = 0. (2) Finally, a DE is linear if it contains only linear terms in y, y, y etc. In other words, no terms of the type yy, (y ) 2 and similar are allowed. The general solution of some DE usually contains unknown integration constants, which are fixed by specifying the boundary conditions (abbreviated as BCs) of the problem. BCs are generally given as the value of the function y(x 0 ) and/or its derivatives at some specified point(s). The number of BCs required, depends on the order of the DE. For example, we will see later that for second order DE s there will be two unknown constants, which are fixed by two BCs. Typically, these dictate the value of the function and its first derivative at some point, {y(x 0 ), y (x 0 )}, or the value of the solution at two different points, {y(x a ), y(x b )}. 2 Linear first order DEs In this section we will consider linear first order DEs of the general form y + P (x)y = Q(x). (3) For later reference, note that since y = dy dx, this can be equivalently written as dy + [P y Q]dx = 0. (4) Such equations may, for example, show up as part of the solution procedure of second order DEs, which are important in physics. The method we will use is called the method of integrating factors, which I will try to motivate in the following. For this we need the concept of an exact DE. Recall the definition of a total (or exact) differential of a function u(x, y), Now, a first order DE of the form du = u u dx + dy. (5) x y M(x, y)dx + N(x, y)dy = 0 (6) is called exact if its left-hand side is a total differential, i.e. M = u u x and N = y for some function u(x, y). Note that this implies the condition M y = N x. Then Eq.(6) simply says du = 0, i.e. u(x, y) = constant. In other words, if a first order DE was of this exact form, it would be trivial to solve. In most cases of course, a DE of the general form (4) will not be exact. However and this is the main idea of our solution method it can be made exact by multiplying it with an appropriately chosen so-called integrating factor, µ(x), µ(x)m(x, y)dx + µ(x)n(x, y)dy = 0 (7) 2

3 so that the condition for exactness is (µm) y = (µn) x. (8) For our case (4) this implies which has the solution In practice, the condition (9) is implemented by demanding µ(x)p (x) = dµ dx, (9) µ(x) = e P (x)dx. (10) d dx (µ(x)y(x)) = µ(x)y (x) + µ(x)p (x)y(x) (11) as can easily be checked directly. Combining this with the DE itself, i.e. with µ(x)y + µ(x)p (x)y = µ(x)q(x), (12) we get the second condition, d (µ(x)y(x)) = µ(x)q(x) (13) dx which gives the solution of our original DE, y(x) = 1 [ ] µ(x)q(x)dx + C (14) µ(x) where C is an integration constant, and µ(x) is found from (10). In the special case Q(x) = 0, i.e. homogeneous DEs, the solution is simply y(x) = 3 Ordinary second order DEs C µ(x) = P (x)dx Ce. (15) In the remainder of these notes, we will consider DEs of the general form y + P (x)y + Q(x)y = R(x). (16) We will start with some general theory before addressing the simplest case, homogeneous DEs (i.e. R(x) = 0), studying both the case with constant coefficients and with variable coefficients (Euler- Cauchy equations). After this we will turn to non-homogeneous equations, R(x) 0. As we will see, in this case the solution can be written as a sum of the solution of the corresponding homogeneous equation, and some particular solution valid for that particular R(x). Therefore we will need methods to determine particular solutions, and we will go through several such methods (method of undetermined coefficients, factorisation, variation of parameters). We then discuss the beautiful Greens function method, where one finds a full solution to the DE for given boundary conditions and left-hand side, but applicable to any right-hand side R(x). Finally, we will return once again to homogeneous DEs, and show that in difficult cases that cannot be solved by any of the methods of section 3.1, there exist rather powerful solution methods using a series expansion of the solution, and then determining the coefficients of the expansion. As is probably clear from this introduction, there exists no single, unique solution method for DEs of the type (16). Which method works, or is most efficient, depends on the DE at hand, and choosing the most convenient method is ultimately a matter of experience (or trial and error...). Assume now that we are given a DE of the type (16), along with fixed values of the solution and its derivative at some point x 0, y(x 0 ) and y (x 0 ). Further assume that the functions P, Q and R are 3

4 analytic at x 0 so that they have derivatives of all orders there. Then the value of y (x 0 ) is fixed by the DE itself (with x 0 inserted for x), and similarly all higher derivatives at x 0, y (n) (x 0 ) are in principle obtained consecutively by taking derivatives of the DE. This means that the Taylor series of the solution y(x) around x 0 is fixed, y(x) = 0 1 n! y(n) (x 0 )(x x 0 ) n. (17) In other words, we conclude that specifying the two boundary conditions y(x 0 ) and y (x 0 ), uniquely fixes the solution of our second order DE. This general result will be quite useful in what is to follow. 3.1 Homogeneous equation Consider Eq.(16) with zero right-hand side, R(x) = 0. We start by noting that if y 1 and y 2 are two solutions of the DE, then so is any linear combination c 1 y 1 + c 2 y 2, as is easily seen by inserting into the DE. Now, two solutions are linearly independent if neither of them is a multiple of the other, so c 1 y 1 (x) + c 2 y 2 (x) = 0 can never be satisfied for all x, unless c 1 = c 2 = 0. Arguing via the Taylor expansion as above, we see that if y 2 (x 0 ) = Ky 1 (x 0 ) and y 2(x 0 ) = Ky 1(x 0 ) (some x 0, same K), then this implies that y 2 (x) = Ky 1 (x), i.e. the two solutions are linearly dependent. Thus the condition of linear independence can be formulated as or equivalently y 1(x 0 ) y 2 (x 0) y 1(x 0 ) y 2 (x 0 ) (18) y 1(x 0 ) y 2 (x 0 ) y 1(x 0 ) y 2(x 0 ) W (x 0) 0. (19) This determinant is called the Wronskian. A general solution can then be written as y(x) = c 1 y 1 (x) + c 2 y 2 (x) (20) where y 1 and y 2 are linearly independent. The two coefficients c 1 and c 2 are fixed by the boundary conditions, i.e. given values for y(x 0 ) and y (x 0 ) Variation of constants This is a very useful general principle: If y 1 (x) is a solution of our DE, then a second, linearly independent solution can be found as y 2 (x) = C(x)y 1 (x). (21) So if, by some method, one solution to the DE has been found, one can insert the ansatz (21) back into the DE, which results in a DE for the unknown function C(x) itself. The latter is typically straightforward to solve, as it is of lower order than the original DE. In this case one gets a first order DE for C (x), which is straightforwardly solved and then integrated to find C(x) itself. Example: Consider the DE y + y x y = 0. (22) x2 It is given that y 1 = x is a solution, as you can quickly check. So try to write the second one as y 2 = C(x)y 1 = xc(x) and insert this into the DE. Since y 2 = C + xc and y 2 = xc + 2C,we get (check!) C + 3C = 0. (23) x Note that the terms C cancelled, so we are left with a first order DE for C. In this sense we have reduced the order of the DE. This is a common feature of making this type 4

5 of ansatz, and we will see several versions of this. To solve (23), we think of it as a first order DE in u(x) C (x), i.e. u = 3u x which has the solution (do this calculation for yourself!) (24) u(x) = C (x) = k 1 x 3 (25) where k 1 is an integration constant. From this we immediately find that C(x) = d 1 x 2 + d 2 (26) where d 1, d 2 are integration constants. To get to y 2 (x), recall that y 2 = xc(x). The constants d 1, d 2 can be chosen arbitrarily since we will anyway write the full solution as a linear combination of y 1 and y 2. In particular, note that the constant piece, d 2 will give a term x = y 1 which we already have. So we can safely set d 1 = 1 and d 2 = 0 to get y 2 (x) = xc(x) = 1 x. (27) The full, general solution is then a linear superposition with two unknown constants, y(x) = c 1 y 1 (x) + c 2 y 2 (x) = c 1 x + c 2 x. (28) We will use and refer to this method frequently, so make sure you understand this example Homogeneous DEs with constant coefficients The simplest second order homogeneous DEs are those with constant coefficients, y + ay + by = 0. (29) These always have a solution of the type e λx, and λ can be determined by insertion, which leads to the characteristic equation λ 2 + aλ + b = 0. (30) Depending on the solutions of this equation, we distinguish between three cases: Case 1: Two non-identical, real roots λ + and λ. These then give us two linearly independent solutions, and thus the general solution y(x) = c e λ x + c + e λ+x (31) Case 2: Two identical real roots, λ + = λ = λ. Then obviously, one solution is y 1 (x) = e λx. The second solution is found by variation of constants, i.e. from the ansatz y 2 (x) = C(x)y 1 (x). It is left as an exercise to the reader to show that the second solution is xe λx, i.e. the general solution takes the form y(x) = (c 1 x + c 2 )e λx. (32) Case 3: In general, the solution may be a pair of complex numbers, λ ± = µ ± iω. The two linearly independent solutions are then, again, e λ±x, and so the general solution may be written in three equivalent ways (according to taste or what is most convenient), y(x) = e µx ( c 1 e iωx + c 2 e iωx) (33) = e µx ( c 1 cos(ωx) + c 2 sin(ωx)) (34) = ce µx sin(ωx + δ). (35) It is left as an exercise to check how the various constants (c 1, c 2, c 1, c 2, c and δ) are related to one another. 5

6 3.1.3 Euler-Cauchy equation The Euler-Cauchy equation is an important example of a homogeneous second order equation whose coefficients are not constant. It is of the form x 2 y + a 1 xy + a 0 y = 0 (36) where a 1 and a 0 are constants. There are two (equivalent) approaches to solving such equations. One is to note that there will be a solution of the form x m, and solve for m. Here we present the other, which is used in the book (see page 434, setting f(x) = 0). It involves simplifying the equation by the following substitution, x e z (x > 0) (37) x = x = e z (x < 0). (38) As shown in the book and the lectures, this change of variables transforms our DE to d 2 y dz 2 + (a 1 1) dy dz + a 0y(z) = 0. (39) This is now an equation with constant coefficients, and can be solved as in section However, some care is required concerning the sign of x when substituting back to our original variable. As can be seen from writing (36) in standard form, y + a 1 x y + a 0 y = 0, (40) x2 the equation is singular at x = 0 and has no solution there. We must distinguish the regimes x < 0 and x > 0, and the solutions will in general be different in these two regimes. Let us illustrate this with a simple example: Example: The DE is transformed, with the substitution (38), into so For x > 0, we substitute back z = ln(x) and thus x 2 y + xy 9y = 0 (41) d 2 y 9y(z) = 0 (42) dz2 y(z) = Ae 3z + Be 3z. (43) y(x) = Ax 3 + B x 3 (x > 0). (44) At negative x the substitution (38) says z = ln( x) = ln x. So the solution does take the same form, but with, in general, different coefficients, As anticipated, these solutions do not exist at x = 0. y(x) = Ãx3 + B x 3 (x > 0). (45) Example: The previous example was of the simplest possible type as it involved two distinct, real root of the characteristic equation (cfr section 3.1.2). Let us see what happens in a more general case, x 2 y + 7xy + 13y = 0. (46) 6

7 In terms of the variable z, Eq.(39), we get Thus the characteristic equation is with complex solutions Thus, d 2 y dz 2 + 6dy + 13y(z) = 0. (47) dz λ 2 + 6λ + 13 = 0 (48) λ = 3 ± 2i. (49) y(z) = e 3z (A cos(2z) + B sin(2z)). (50) By keeping absolute value signs explicitly when transforming back to x, we get a form of the solution valid for both x > 0 and x < 0, y(x) = x 3 (A cos(2 ln x ) + B sin(2 ln x )). (51) However it should be kept in mind that the constants (A and B) can in general be different in the two regimes. Note that this solution is in fact rather complicated, given the relatively easy solution procedure. *** Finally, there is obviously the possibility of finding identical roots to the characteristic equation, giving only one solution y 1. A second solution can then be found using variation of parameters as discussed previously. 3.2 Inhomogeneous DEs Let us now discuss the inhomogeneous equation, y + P (x)y + Q(x)y = R(x). (52) Our experience in solving homogeneous equations will be very useful here, for the following reason: Note that if y h (x) = c 1 y 1 (x) + c 2 y 2 (x) (where y 1 and y 2 are linearly independent) is a solution of the homogeneous version of (52), and y p (x) is some, any, solution of the full DE (with no arbitrary constants), then y(x) = y h (x) + y p (x) = c 1 y 1 (x) + c 2 y 2 (x) + y p (x) (53) is also a solution of the full equation. We call y p a particular solution. In fact, since this solution contains two arbitrary constants (c 1, c 2, to be determined by boundary conditions), it is the general solution to the DE! We can double check that this solution can indeed be made to fit the boundary conditions, i.e. some fixed values for y(x 0 ) and y (x 0 ). Single out the terms from the homogeneous solution, i.e. c 1 y 1 (x 0 ) + c 2 y 2 (x 0 ) = y(x 0 ) y p (x 0 ) (54) c 1 y 1(x 0 ) + c 2 y 2(x 0 ) = y (x 0 ) y p(x 0 ) (55) This set of equations does have a solution for c 1 and c 2 since, by assumption, y 1 and y 2 are linearly independent, and thus W = y 1(x 0 ) y 2 (x 0 ) y 1(x 0 ) y 2(x 0 ) 0. (56) So the way to address DEs of the type (52) is to first find the general solution y h to the homogeneous equation. Then, use any method whatsoever to find a particular solution y p. Adding these two, y h + y p gives us the full solution to our DE. There are various ways of finding a particular solution, depending on the problem at hand. In the following we will go through some of these methods. 7

8 3.2.1 Constant coefficients method of undetermined coefficients We start with a class of DEs simple enough that the best solution method is qualified guessing. These are DEs with constant coefficients on the left-hand side, y + ay + by = R(x), (57) and where R(x) is a function whose derivatives resemble the function itself exponential functions, simple trigonometric functions, polynomials, or combinations of these. The strategy is to guess a particular solution of the same form as R(x), possibly times x or x 2, with an unknown coefficient. This unknown coefficient is then determined by inserting into the DE. If R(x) is a sum of several functions like the ones mentioned, then a particular solution is typically a linear combination (superposition) of those functions. To show how this works, let us list the various scenarios, and then illustrate by examples. Assume that we have solved the homogeneous equation, and that the roots of its characteristic equation are α and β, which may or may not be identical, real or complex. Case 1: R(x) = Ae kx If k α, β: Try y p = Be kx with B to be determined If k = α or k = β, and α β: Try y p = Cxe kx with C to be determined If k = α = β, try y p = Dx 2 e kx with D to be determined Case 2: R(x) = A sin(kx)ora cos(kx) y p will be of the form B sin(kx) + C cos(kx), possibly times x or x 2. This is really just Case 1 in disguise: It may be most efficient to replace the RHS by Ae ikx, solve as in Case 1, and take the real or imaginary part at the end (see Boas). Case 3: R(x) = e kx P n (x), where P n is a polynomial of order n. Note that this includes Case 1 (n = 0) and Case 2 (k complex), as well as pure polynomials (k = 0). If k α, β: Try y p = e kx Q n (x) If k = α or k = β, and α β: Try y p = xe kx Q n (x) If k = α = β, try y p = x 2 e kx Q n (x) In all of these, Q n is a polynomial of degree n, Q n (x) = q 0 + q 1 x + q 2 x q n x n. Example y + 2y + y = 2e x. (58) We start by solving the homogeneous equation, and find the roots of the characteristic equation to be identical, λ = 1. Thus, y h (x) = Ae x + Bxe x. (59) For the particular solution we then need to go up one more power, y p (x) = Cx 2 e x. (60) in order to determine C, calculate y p and y p, and insert into the full DE (58). One finds that the DE is fulfilled for C = 1 (try for yourself). Thus, the general solution is y(x) = ( A + Bx + x 2) e x. (61) Note that we are left with two unknown constants A and B, which would be fixed by a given set of boundary conditions. 8

9 3.2.2 Factorization This method is similar to Reduction of order discussed on page 434 in Boas, but for non-homogeneous DEs (16). The idea is that if one knows (at least) one of the solutions of the homogeneous equation, u(x), then this can be used to find a particular solution in the following way: Write an ansatz y p (x) = u(x)v(x) where v(x) is unknown. Inserting this into the DE, will give a first order DE for v(x). To see this, note that (suppressing the argument, x), and which, upon inserting into the DE (16) gives y p = u v + uv (62) y p = u v + 2u v + uv (63) v(u + P u + Qu) + v (2u + up ) + v u = R. (64) Notice what happens here: The terms multiplying v are just the LHS of the original DE. Since we assumed that u is a solution of the homogeneous DE, these by construction add up to zero. Thus we are left only with terms going as v and v, i.e. a first order DE for v. Thus the name reduction of order. Renaming v (x) as w(x), we thus get [ ] 2u w + u + P w = R (65) u which is solved as in section 2. The original function v is then finally found by another integration, v = wdx (66) and thus y p = uv has been found. Note that we do not need to include any integration constants since we are looking for a particular solution. Example, exam 2006: See lecture Variation of parameters Here is another, quite general way of finding a particular solution, once the solution to the homogeneous equation has been found. Again we consider the DE y + P (x)y + Q(x)y = R(x) (67) and assume two linearly independent solutions y 1 and y 2 of the homogeneous equation have been found. We start by stating the main result, which will be derived below: A particular solution is found in terms of y 1 and y 2 as follows: where W is the Wronskian, y p (x) = y 1 y2 R W dx + y 2 W = y 1(x) y 1(x) y1 R dx (68) W y 2 (x) y 2(x). (69) Note: In order for this to work, make sure you write the DE in standard form, i.e. only a factor 1 in front of y, before applying this formula. Derivation of (68): Since we have assumed that we already know the general solution of the homogeneous equation, y h (x) = c 1 y 1 (x) + c 2 y 2 (x) (70) 9

10 we use variation of parameters to make the following very general ansatz for the particular solution, y p (x) = f 1 (x)y 1 (x) + f 2 (x)y 2 (x). (71) Ultimately we want to insert this into the full DE, so we will need (suppressing arguments for simplicity) y p = f 1y 1 + f 1 y 1 + f 2y 2 + f 2 y 2. (72) Now, it turns out that we may simplify things by imposing the following condition, f 1y 1 + f 2y 2 = 0. (73) That this perhaps somewhat mysterious trick (due to Lagrange) is allowed, has to do with the fact that we introduced two unknown functions, f 1 (x) and f 2 (x). This in a sense gives too much freedom, since we are looking for a particular solution with no unknown parameters and the requirement that y p must satisfy the DE, only gives one condition. So there is freedom to impose another. Ultimately, the justification that this is a consistent thing to do, is that no inconsistencies show up in the following derivation. The reason that this particular constraint is very convenient, is that we avoid terms with second derivatives of the unknown functions f 1 and f 2 (which are the ones we seek to determine) when we insert our ansatz into the DE. First of all, Eq.(72) now becomes and thus y p = f 1 y 1 + f 2 y 2 (74) y p = f 1y 1 + f 2y 2 + f 1 y 1 + f 2 y 2. (75) Insert these into the full DE and use the fact that y 1 and y 2 are solutions of the homogeneous equation, i.e. y 1 + P y 1 + Qy 1 = 0 (76) and similarly for y 2. This will cancel all terms multiplying f 1 and f 2 and leave us with f 1y 1 + f 2y 2 = R. (77) This expression together with (73) constitutes a set of equations that can now be solved for the two unknown functions f 1 and f 2. After some straightforward algebra one finds f 1W = y 2 R (78) and Thus and and the result (68) follows. f 2W = y 1 R. (79) f 1 = f 2 = y2 R dx (80) W y1 R dx (81) W Example: Consider the DE y + y = 1 cos x. (82) It is written in standard form (no prefactor in front of y ), and so we read off R(x) = 1/ cos x. It is easily seen that the solution to the homogeneous equation, y = y, can be written as y h = c 1 cos x + c 2 sin x. (83) 10

11 The Wronskian is easy, so that and thus our particular solution is 3.3 Greens functions W = y 1 y 2 y 2 y 1 = cos 2 x + sin 2 x = 1 (84) sin x f 1 (x) = dx = ln cos x (85) cos x cos x f 2 (x) = dx = x (86) cos x y p = cos x ln cos x + x sin x. (87) In this method one finds the full solution to second order, inhomogeneous DEs by determining an object called the Green s function. The Green s function is, as we will see, uniquely determined by the LHS (homogeneous part) of the DE, along with the boundary conditions imposed on the solution but it is independent of the RHS R(x). This means that once the Green s function for a given LHS and set of boundary conditions has been found, it will give the full solution of the DE for any RHS R(x). We start very generally by recasting our DE as where D is the differential operator Dy(x) = R(x) (88) D = d2 dx 2 + P (x) d + Q(x). (89) dx Formally then, if the inverse of this differential operator exists (i.e. an operator D 1 such that D 1 D is just the identity), then we could act on the DE from the left with D 1 and thus write the solution of our DE simply as y(x) = D 1 R(x). (90) Indeed, there is such an inverse, and perhaps it does not come as a big surprise that (being the inverse of a differential operator) it is given by an integral equation: y(x) = b a G(x, z)r(z)dz; a x, z b. (91) The limits of this integral, a and b, basically define the interval on which boundary conditions are imposed see later. The function G(x, z) is what we call the Green s function, and we see that if it were known, this would give us the solution y(x). So let us see what we know about this function. First of all, if we apply the operator D on Eq.(91), we get Dy(x) = b a [DG(x, z)] R(z)dz = R(x). (92) From this we see that the part inside the square brackets must be the Dirac delta function, DG(x, z) = δ(x z) (93) with D given in (89). We will discuss the delta function, δ(x x ), later in the course, and many of you are probably familiar with it. If not, what we need to know at this stage is that it is not a proper function, but a distribution (singular at the point x = x ) with the important property that b a δ(x x )f(x )dx = f(x); (a x, x b) (94) 11

12 for any arbitrary function f(x) i.e. when integrated over, the delta function picks out the value of the function at x = x. Eq.(93) shows that the Green s function must obey our original DE, but with R(x) replaced by the delta function. This may not look like a major simplification compared to solving the original DE in the first place, but we will see now that we have some additional information for G(x, z), which will help us determine it for a given set of boundary conditions: Restrictions on G(x, z): First of all, we have not yet used that the solution y(x) must obey a given set of boundary conditions. In the following discussion we will restrict ourselves to so-called homogeneous boundary conditions, i.e. y(x) = 0 and/or y (x) = 0 at specified points. (The more general case can be handled but is more difficult.) These boundary conditions can be ensured by imposing the same boundary conditions on the Green s function (in the variable x). For example, the set of BC s y(a) = y(b) = 0 is automatically ensured if we demand G(a, z) = G(b, z) = 0 as is easily seen from (91). The other thing we can get further information from is that the DE (93) has a singularity at x = z. We now show that this implies a discontinuity in the first derivative dg/dx at this point. Consider the DE for the Green s function, and integrate it over the infinitesimally small interval x = z ɛ to x = z + ɛ (suppressing variables of G(x, z)): z+ɛ d 2 G dx 2 dx + z+ɛ P (x) dg dx dx + z+ɛ Q(x)Gdx = z+ɛ δ(x z)dx = 1 (95) where the last equality follows from the properties of the delta function. If we assume G(x, z) is continuous at x = z, it follows that the last two terms on the LHS go to zero as ɛ 0, and lim ɛ 0 z+ɛ P (x) dg dx z+ɛ lim ɛ 0 Q(x)G(x, z)dx = 0 (96) dx = lim [P (x)g(x, z)]z+ɛ ɛ 0 lim ɛ 0 z+ɛ G(x, z) dp dx = 0. (97) dx In the last step we used integration by parts. Eq.(95) then implies a finite discontinuity in dg/dx, as can be seen by integrating once, z+ɛ lim ɛ 0 d 2 G dx = lim dx2 ɛ 0 Let us summarise what we have obtained so far: The solution to our DE is formally y(x) = b a [ ] z+ɛ dg = 1. (98) dx G(x, z)r(z)dz (99) obeying some boundary conditions in a x b. The present method is useful mainly if these BCs are homogeneous. Our goal is thus to determine the Green s function G(x, z). G(x, z) obeys the original DE but with R(x) replaced by the Dirac delta function δ(x z). G(x, z) considered as a function of x obeys the same (homogeneous) BCs as the solution y(x) itself. G is continuous at x = z, while dg/dx has a discontinuity of 1 there. How to implement this in practice is best illustrated by working through an example. Thus consider the DE y + y = 1 (100) sin x 12

13 with (homogeneous) BCs y(0) = y(π/2) = 0. Note that this DE (its RHS) is singular at x = 0. Since the Green s function is independent of choice for R(x), we can nevertheless go ahead and find it but for this particular R(x) we thus expect to find a solution that is not valid at x = 0. We know that the Green s function must obey the DE d 2 G(x, z) dx 2 + G(x, z) = δ(x z). (101) In particular note that wherever x z, the delta function is zero, and G(x, z) thus obeys the homogeneous DE. It is easy to see that the solution to the homogenous DE is c 1 sin x + c 2 cos x (102) where the constants c 1 and c 2 are, in general, functions of z. Because of the singularity at x = z, one must expect to get different solutions for x < z and x > z, respectively. Thus we consider the two intervals (x < z and x > z) separately and have the solution G(x, z) = { A(z) sin x + B(z) cos x if x < z; C(z) sin x + D(z) cos x if x > z. (103) What is left, then, is to determine A(z), B(z), C(z) and D(z). This is done by implementing the boundary conditions, and the constraint that G must be continuous at x = z, while dg/dx has a finite discontinuity. First of all, the boundary conditions say that G(0, z) = G(π/2, z) = 0. Since both x and z are defined in the interval between 0 and π/2 (and we are excluding the singularity x = z), x = 0 must correspond to the case x < z, and x = π/2 means x > z. From Eq.(103) we thus see that G(0, z) = 0 implies B(z) = 0, while G(π/2, z) = 0 implies C(z) = 0. This reduces our solution to the form { A(z) sin x if x < z; G(x, z) = (104) D(z) cos x if x > z. Now, apply the constraints for continuity of G, and the discontinuity in its derivatives, at x = z: A(z) sin z D(z) cos z = 0 (105) D(z) sin z A(z) cos z = 1. (106) This set of equations is easily solved for A and D, giving A(z) = cos z and D(z) = sin z, and thus the final expression for the Green s function is G(x, z) = { cos z sin x if x < z; sin z cos x if x > z. (107) Eq.(91) now gives us the solution to the original DE, recalling that we had R(x) = 1/ sin x, y(x) = = π/2 0 x 0 1 G(x, z) dz (108) sin z ( cos x) sin z sin z dz + π/2 x ( sin x) cos z sin z dz (109) = x cos x sin x [ln sin z ] π/2 x (110) = x cos x + sin x ln sin x. (111) As expected, this solution is not valid for x = 0. Note that this procedure has given us the full solution, homogeneous and particular, and that the solution is fairly non-trivial. Moreover, once we have the Green s function, we get the solution for any other R(x) for free, provided the LHS and boundary conditions are the same. If, on the other hand, we were to solve the same DE, but with different BC s, we would have to start over and find a different Green s function. The following example illustrates this: 13

14 Example: Let us find the Green s function for the same DE as in the previous example, y + y = R(x), (112) but now with different boundary conditions, y(0) = y (0) = 0. Up until eq.(103) the derivation is the same as previously, so we have G(x, z) = { A(z) sin x + B(z) cos x if x < z; C(z) sin x + D(z) cos x if x > z. (113) The difference comes when we implement the boundary conditions. Here we have both BCs given at x = 0, which corresponds to x < z. The first one says G(0, z) = 0 and implies B(z) = 0. The second one, G (0, z) = 0 (where the derivative is with respect to x) implies A(z) = 0. This leaves us with G(x, z) = { 0 if x < z; C(z) sin x + D(z) cos x if x > z. (114) Thus, the condition of continuity at x = z implies while the discontinuity of G (x, z) at x = z implies C(z) sin z + D(z) cos z = 0, (115) C(z) cos z D(z) sin z = 1. (116) Solving these two equations for the unknown C(z) and D(z) one finds C(z) = cos z and D(z) = sin z. Also note that cos z sin x sin z cos x = sin(x z). Thus, the final expression for the Green s function is { 0 if x < z; G(x, z) = (117) sin(x z) if x > z. Note that this is rather different from the Green s function (107) which was for the same DE but different BCs. The solution to our original DE (112) with the present BCs and an arbitrary R(x) is thus y(x) = (Note the integration limits.) 3.4 Series expansion methods 0 G(x, z) R(z) dz = x 0 sin(x z) R(z) dz. (118) In this final section we will present two methods applicable to second order homogeneous DEs in cases when the DE is difficult, in the sense that its solution cannot be expressed as elementary functions. In physics we meet several such DEs, important enough to carry the names of their discoverers, such as the Legendre, Bessel or Hermite equation. The basic idea is to represent the solutions as series expansions with unknown coefficients, and then use the DE (and BCs, if applicable) to determine those coefficients. In the simplest case, this series expansion will be a power series, y(x) = a n x n. (119) n=0 However, a series of integer powers may be too restrictive, and so we will also address the more general Fröbenius method where the solution is written as y(x) = x s n=0 a n x n (120) with s real or complex. This latter approach applies to a rather broad class of DEs, as we will see. 14

15 3.4.1 Power series Once again, we consider a homogeneous DE of the form y + P (x)y + Q(x)y = 0. (121) Unless P (x) and Q(x) are already polynomials (as they often are), we want to represent them by power series, and assume a solution of the form and thus y(x) = y (x) = y (x) = a n x n = a 0 + a 1 x + a 2 x (122) n=0 na n x n 1 = a 1 + 2a 2 x +... (123) n=0 n(n 1)a n x n 2 = 2a 2 + 6a 3 x +... (124) n=0 Inserting all this into the original DE and collecting terms with equal powers of x, one can solve the equation power by power, thus finding the unknown coefficients a k. Note that if no boundary conditions are specified, the DE involves two unknown coefficients as usual, i.e. two of the coefficients (say, a 0 and a 1 ) will remain undetermined, and the others expressed in terms of these. For examples, see lectures and Boas. In particular, an example important in physics, which will be discussed in the lectures, is the Legendre equation, and a class of its solutions called Legendre polynomials. This is also covered in Boas. It is worth pointing out that it is of course possible to expand the solution around other points x 0 0. It can be shown that if P (x) and Q(x) are both analytic at x = x 0, then all solutions of the DE are analytic at x 0 and can thus be represented by a power series in (x x 0 ), with some radius of convergence, R Fröbenius method When P (x) and Q(x) are not analytic at x 0, the simplest power series method discussed above, does not apply. We then say that x 0 is a singular point of the DE. In the following we will stick to x 0 = 0. The following theorem tells us what kind of singularities we can handle with the Fröbenius method: Any DE of the form y + b(x) x y + c(x) x 2 y = 0 (125) where b(x) and c(x) are analytic at x = 0, has at least one solution that can be represented in the form y(x) = x s n=0 a n x n (a 0 0) (126) where s may be real or complex. Of course, the DE generally has two linearly independent solutions, and it depends on the DE whether or not both are on the Fröbenius form. Again the idea will be to collect terms of the same order in x and equate order by order. To this end, write the DE as x 2 y + xb(x)y + c(x)y = 0, (127) expand b(x) and c(x), b(x) = b 0 + b 1 x + b 2 x (128) c(x) = c 0 + c 1 x + c 2 x (129) 15

16 and assume the announced series solution for y(x), y(x) = a 0 x s + a 1 x s+1 + a 2 x s (130) xy (x) = sa 0 x s + (s + 1)a 1 x s+1 + (s + 2)a 2 x s (131) x 2 y (x) = s(s 1)a 0 x s + (s + 1)sa 1 x s (132) Let us first look at the lowest power, x s. Inserting the above into the DE and collecting all terms of order x s, we find s(s 1) + b 0 s + c 0 = 0. (133) This is called the indicial equation; it is a quadratic equation that determines s. Now, there are three scenarios for possible solutions: The indicial equation has two distinct roots (s 1, s 2 ), and s 1 s 2 is not an integer. In this case we find two linearly independent solutions y i = x si a n x n. Two distinct roots but s 1 s 2 = integer (let s 1 > s 2 ). Often s 2 gives a complete solution (which includes the one corresponding to s 1 ), so it is wise to always try the smallest root first. However, it may happen that s 2 does not give a solution at all, in which case one starts from the solution y 1 corresponding to s 1, and finds the other by variation of constants, y 2 (x) = C(x)y 1 (x). Double root s 1 = s 2 = s. In this case we only find one solution of the Fröbenius form, y 1 (x) = a n x s+n, and the other must be found by variation of constants, y 2 (x) = C(x)y 1 (x). In the lectures we will use this method to solve the Hermite equation (see also Boas), which is one of the important equations in physics. We will also go through problem in detail. 16