F1.9MO2. F1.9MO2 Ordinary Differential Equations. Lecture notes Andrew Lacey

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1 F.9MO2 i F.9MO2 Ordinary Differential Equations Lecture notes 29 - Andrew Lacey Contents An Introduction to Differential Equations. General remarks First-order ODEs Separable equations Linear equations Change of variables Exact equations Existence and uniqueness of solution Direction fields General remarks about solvable second-order ODEs Reduction of order Equations which do not depend on y Equations which do not depend on x The general linear homogeneous equation The general linear inhomogeneous equation Linear with constant coefficients Euler equations (L. Euler, ) Method of undetermined coefficients Systems of ODEs 8 2. General remarks Existence and uniqueness of solutions of systems of ODEs Linear systems of ODEs Homogeneous linear systems Matrix methods for finding the FSS The Fundamental Matrix Inhomogeneous linear systems Higher-order ODEs Picard s Theorem The FSS and the Wronskian The FSS for equations with constant coefficients Inhomogeneous linear nth-order ODES: Variation of parameters 34

2 F.9MO2 ii 3 Laplace transforms Definition and basic properties Examples Simple properties of Laplace transforms Solution of initial-value problems Inverting Laplace transforms ODEs involving discontinuous functions The Step Function The Dirac delta function The convolution integral Linear Systems of ODEs Boundary-Value Problems 5 4. General Comments Green s functions Sturm-Liouville Problems General comments Inner product spaces Eigenvalues, eigenfunctions and eigenfunction expansions Phase Planes Introduction Phase Planes for Linear Systems Types of Equilibrium Points Stability A Revision of Vector Spaces 86 B Revision of Matrices 88 C Table of Standard Laplace Transforms 89

3 F.9MO2 An Introduction to Differential Equations. General remarks The study of differential equations occupies a central place in both applied and pure mathematics. It is fair to say that the subject is the most important part of mathematics for understanding the physical sciences. Applications also abound in other fields, such as engineering, biology and economics. The questions encountered in studying differential equations have in turn inspired many key ideas and concepts in pure mathematics, in particular in analysis. Examples include the theory of Fourier series and Hilbert spaces. A differential equation is an equation involving a function and its derivatives. If the function depends on one variable only we call it an ordinary differential equation. If it depends on several variables we call it a partial differential equation. In both cases, the order of the differential equation is the order of the highest derivative occurring in it. Examples: dy dx xy = d 2 y dx 2 ( dy dx u t = 2 u x 2 is a first-order ordinary differential equation. ) 3 = 3 is a second-order ordinary differential equation. is a second-order partial differential equation. In this course we are only concerned with ordinary differential equations, which will be referred to as ODEs for brevity..2 First-order ODEs The most general first-order ODE is of the form ( F x, y, dy ) =. dx (.) ode The function y is a solution of (.) if ( F x, y(x), dy ) dx (x). (.2) We call a first-order ODE explicit if it is of the form dy dx = f(x, y). (.3) explicit Otherwise the ODE is called implicit. Before we try to understand the general case, we consider some examples where solutions can be found by elementary methods.

4 F.9MO Separable equations The simplest example of a separable differential equation is of the elementary form dy dx = f(x). (.4) elem The solution is (by definition) by integration y(x) = f(x) dx. (.5) Note that the right-hand side is an indefinite integral - determined only up to a constant. As we shall see, solutions of first-order differential equations are only determined up to one arbitrary constant. In general, the solution of an ODE containing the maximum number of arbitrary constants is called the general solution, and a solution with no arbitrary constants is call a particular solution. Let us compute the general solution of the following slightly more complicated equation: dy dx = 2xy2. (.6) sep It can be solved by separating the variables, i.e. by bringing all x dependence to one side and all y dependence to the other: Integrating once yields dy = 2x dx. y2 y = x2 + c, where c is an arbitrary real constant. Hence the general solution of (.6) is y =. (.7) x 2 solfam + c The constant c is determined if we give the value of y at one point, e.g. at x =. Note: In dividing by y we have lost one solution: y also solves (.6), but it is not included in the general family (.7). Any equation of the form dy dx = f(x)g(y) (.8) can be solved by separating the variables - at least in principle. The solution y(x) is determined implicitly by g(y) dy = f(x) dx. (.9) In practice it may not be possible to express the integrals in terms of elementary functions and to solve explicitly for y.

5 F.9MO Linear equations A first-order explicit equation that can be written in the form dy + a(x)y = b(x) (.) dx is called a linear equation. Consider the following example where, for a change, we have denoted the independent variable by t: dy dt + 2ty = t. (.) First-order linear equations can always be solved by using an integrating factor. In the above example, we multiply both sides by exp t 2 to obtain Now the LHS has become a derivative: dy t2 e dt + 2tet2 y = te t2. (.2) d dt (et2 y) = te t2. (.3) This is of the elementary form (.4). Integrating once yields so that the general solution is More generally, equations of the form e t2 y = 2 et2 + c y(t) = 2 + ce t2. (.4) dy dt + a(t)y = b(t), (.5) linode where a and b are arbitrary functions of t, can be solved using the integrating factor ( ) I(t) = exp a(t) dt. (.6) Since the indefinite integral a(t) dt is only determined up to an additive constant, the integrating factor is only determined up to a multiplicative constant: if I(t) is an integrating factor, so is CI(t). Multiplying (.5) by I(t) we obtain d (I(t)y(t)) = I(t)b(t). (.7) dt Now integrate and solve for y(t) to find the general solution.

6 F.9MO Change of variables Sometimes ODEs can be simplified and solved by changing variables. We illustrate how this works by considering two important classes of ODEs. a)homogeneous equations These are equations of the form dy ( y ) dx = f, (.8) homog x for example, dy dx = 2xy x 2 + y 2 = If we define u = y/x, so y = xu, and use the product rule we obtain Hence the equation (.8) can be rewritten as 2(y/x). (.9) homex + (y/x) 2 dy dx = u + xdu dx. (.2) x du dx = f(u) u (.2) homexx which is separable and hence solvable. In our example (.9) we obtain the following equation for u: x du dx = 2u + u 2 u (.22) which becomes, after separating variables, ( + u 2 ) du u( u)( + u) = dx. (.23) x Using partial fractions, + u 2 u( u)( + u) = u + u + u, (.24) we deduce that u is determined as a function of x by u = cx (.25) u2 where c is an arbitrary constant. The general solution is thus given by b) Bernoulli equations (Jakob Bernoulli, ) These are equations of the form y = c(x 2 y 2 ). (.26) dy dx + a(x)y = b(x)yα, (.27) bernoulli

7 F.9MO2 5 where α is a real number not equal to. Let u = y α. Then du dx = du dy dy dx Therefore, (.27) becomes, after multiplying by ( α)y α, = ( α)y α dy dx. (.28) du + ( α)a(x)u = ( α)b(x). (.29) dx This is linear and can be solved by finding an integrating factor For example the equation becomes a linear equation for u = y 3 : which we can solve using the integrating factor exp( 3x). dy dx + y = y4 (.3) du 3u = 3 (.3) dx.2.4 Exact equations Consider the following equation If we define (ln x 2) dy dx + y + 6x =. (.32) exact x ψ(x, y) = y ln x 2y + 3x 2 (.33) then (.32) can be written and thus solved by d ψ(x, y(x)) = (.34) total dx ψ(x, y(x)) = c (.35) for some constant c. Equations which can be written in the form (.34) for some function ψ are called exact. It is possible to determine whether a general equation of the form a(x, y) dy + b(x, y) = (.36) genab dx is exact as follows. Suppose the above equation were exact. Then we should be able to write it in the form (.34) for some ψ : R 2 R. However, (.34) is equivalent to ψ dy y dx + ψ x =. (.37)

8 F.9MO2 6 Thus (.36) is exact if a(x, y) = ψ y ψ (x, y) and b(x, y) = (x, y) (.38) x for some function ψ. therefore A necessary condition for the existence of such a function is a x = b y. (.39) Thus for example is exact because (x + cos y) dy dx + y = (.4) y (x + cos y) = x y. (.4) Can you find the corresponding function ψ? Note that it is sometimes possible to make a non-exact equation exact by multiplying with a suitable integrating factor. However, it is only possible to give a recipe for computing the integrating factor in the linear case. In general one has to rely on inspired guesswork..2.5 Existence and uniqueness of solution Most of the differential equations one encounters in applications cannot be solved by the elementary methods described so far. However, even in the most general case one would still like to know whether a solution exists and whether it is unique. Our experience so far suggest that solutions exist and are unique once we specify the value of the unknown function at one point. We could therefore ask whether the equations dy dx = f(x, y), y(x ) = α (.42) init have a unique solution. The two equations together are called an initial-value problem. An important theorem by C.E. Picard (856-94) says that, under fairly mild assumption on f, initial-value problems have unique solutions, at least locally. Thm:Picard Theorem.. Picard s theorem. Suppose f : R 2 R is continuous in some rectangle x x a, y α b and that the partial derivative f is also continuous y there. Then there is an interval x x h a in which the initial-value problem (.42) has a unique solution.

9 F.9MO2 7 Remarks. Although Picard s theorem guarantees the existence of a solution for a large class of equations, it is not generally possible to find an explicit form of the solution in terms of elementary functions. For example dy dx = sin(xy), y() = satisfies the condition of Picard s theorem but no explicit formula is known for the solution. 2. If f is not continuous then (.42) may not have a solution. For example { if x f(x, y) = and y() =. if x < This would imply that y(x) = x for x and y(x) = for x <, which is not differentiable at x =. 3. If f does not have a continuous first-order partial derivative (.42) may have more than one solution. For example is solved by y(x) = dy dx = y 3, y() = { ( 2 3 (x c)) 3 2 for x c for c < for any c. y() = gives another solution. 4. Picard s theorem guarantees a solution for x close to x, but this solution may not exist for all x R: dy dx = 2xy2, y() = has the solution y(x) = x 2 which tends to infinity as x ±. 5. Very straightforward-looking equations may have no solution: x dy dx + y = x, y() =. d Writing the equation as (yx) = x we deduce y(x) = x/2 + c/x, which is dx incompatible with y() =. Why does Picard s theorem not apply?

10 F.9MO Direction fields For explicit first-order equations of the form dy dx = f(x, y) (.43) explicitt a geometrical viewpoint is helpful for understanding general properties of solutions. Suppose we have a solution curve y(x) and draw its graph in the (x, y)-plane. Then equation (.43) tells us that the gradient at a point (x, y) on the solution curve is given by f(x, y). To understand the general nature of all solutions it is therefore helpful to draw a short line segment through a number of points (x, y) with slope f(x, y). The collection of all such line segments is called the direction field of the differential equation (.43). Any solution curve must be tangential to the direction field at every point. Example.2. Sketch the direction field for the differential equation dy dx = y(y ) (.44) df and hence sketch the solutions with initial conditions (i) y() =.5, (ii) y() =., (iii) y() =.5. The simplest way to start is to find the values of x and y for which dy/dx =, dy/dx >, dy/dx <. In this question these are respectively: y =, ; y > or y < ; and < y < y x Figure : Direction field for (.44) Example.3. Sketch the direction field for the differential equation and hence sketch the solutions with initial condition y() =. dy dx = x2 + y 2 (.45) df2

11 F.9MO y x Figure 2: Direction field for (.45) ss:2nd.3 General remarks about solvable second-order ODEs In this section we shall consider specifically second-order ODEs, i.e., equations of the form y = f(x, y, y ). (.46) gensecond The theory of second-order ODEs is an extensive and highly developed part of mathematics, not least because of its numerous applications in the physical sciences. For example, Newton told us, with impressive foresight, that the radial motion of the space shuttle is governed by the equation d 2 r dt 2 = MG r 2, (.47) where M is the mass of the earth and G is the gravitational constant. As another sample, a vibrating spring of mass m obeys the equation m d2 x dt + γ dx + kx = F (t), 2 dt where γ is a damping coefficient, k the spring constant, and F (t) the applied force. In Chapter 2, we develop an arsenal of techniques, and show how some of them specialise to second-order equations; most importantly, we show how the method of variation of parameters gives a practical method of solving linear inhomogeneous 2nd-order equations. In this section, we shall try not to repeat this, but consider instead a range of new techniques that work best specifically for 2nd-order ODEs. First of all though let us indeed repeat ourselves once, and consider the specialisation of Picard s Theorem to the initial-value problem d 2 y dx 2 = f(x, y, y ), y(x ) = α, y (x ) = β. (.48) initii Theorem.4. Suppose f : R 3 R is continuous in some cube x x a, y α b, y β c and that the partial derivatives f f and are also continuous there. y y Then there is some interval x x h a in which the initial-value problem (.48) has a unique solution.

12 F.9MO2 An obvious consequence of this is that a general solution to a second-order ODE should have two arbitrary constants in it; these correspond to the two initial conditions in (.48). As very elementary example, consider the simple harmonic oscillator problem y = y. (.49) favex The function f(x, y, y ) = y is clearly continuous in x, y, y and the conditions of Picard s Theorem are met. You might recall (or simply check) that both sin x and cos x solve this equation. More generally, the superposition y(x) = c cos x + c 2 sin x, (.5) where c and c 2 are two arbitrary constants, also solves (.49). The constants, and hence the solution, are determined uniquely if we specify y(x ) = α, y (x ) = β. (.5) inittwo Now we shall go on to consider the range of promised new techniques that are most useful for second-order systems..4 Reduction of order Sometimes a second-order equation can be reduced to a first-order equation and can then hopefully be solved:.4. Equations which do not depend on y For equations of the form y = f(x, y ) (.52) we define z = y. Then y = z and the equations becomes first order: Example.5. Find the general solution of y + 2y = e x. z = f(x, z). (.53) Solution: Letting z = y, we have z + 2z = e x, which is linear and can be solved by the integrating factor e 2x, to give z = e x + Ae 2x. Integrating once more gives the general solution y = e x + Be 2x + C.

13 F.9MO2.4.2 Equations which do not depend on x Consider equations of the form y = f(y, y ). (.54) silly In such a case, let z = y and regard z as a function of y. Then y = d ( ) dy = dz dx dx dx = dz dy dy dx = z dz dy. (.55) Hence equation (.54) becomes which is again of first order. z dz dy Example.6. Find the general solution of Solution: Letting z = y gives which is separable and solved by = f(y, z) (.56) yy + (y ) 2 = (.57) silex yz dz dy + z2 =. (.58) z = A/y (.59) for some constant A. Recalling that z = dy dx general solution of (.57): and integrating once again we obtain the y 2 = Bx + C. (.6).4.3 The general linear homogeneous equation Consider the equation y + a(x)y + b(x)y =. Suppose we know one solution u(x) of this equation. Then letting y = uw, we obtain y = u w + uw, y = u w + 2u w + uw. Substituting into the equation, we obtain u w + 2u w + uw + a(x)(u w + uw ) + b(x)uw =.

14 F.9MO2 2 Rearranging gives (u + a(x)u + b(x)u)w + uw + (a(x)u + 2u )w =. The first term is zero and the other terms give uw + (a(x)u + 2u )w =. Writing z = w, as above, we then have a first-order ODE z + (a(x) + 2u /u)z =, which is linear and can be solved by an integrating factor. So, in principle, we can find the second solution y(x) = u(x)w(x). ex:2ndlsp Example.7. One solution of y 2αy + αy = is y(x) = e αx. Use the method of reduction of order to show that the general solution is given by y(x) = Ae αx + Bxe αx Solution: Letting y(x) = e αx w(x), we have y = u w + uw, y = u w + 2u w + uw and so substituting into the equation gives 2u w + uw 2αuw =, which gives w = (since u = αu). Integrating twice (we don t really need the z = w substitution for this trivial case) gives w = C + Dx. Hence, the second solution is of the form y(x) = (C + Dx)e αx, and the general solution is as specified. Example.8. One solution of x 2 y (x 2 + 2x)y + (x + 2)y = is y(x) = x. Use the method of reduction of order to find a second solution and hence write down the general solution. Solution: Letting y = xw, we get y = w + xw, y = (2w + xw ). Substituting into the equation gives w w =, which we can solve by letting z = w, such that z z =. The solution of this equation is z = Ae x. Integrating again gives w = Ae x + B. Thus a second solution is y = x(ae x + B). The general solution is therefore y(x) = Cx + Dxe x. All these examples illustrate the result that for a homogeneous linear second-order ODE, if we have two independent solutions, say y () (x) and y (2) (x), linearity implies that the general solution is of the form y(x) = Ay () (x) + By (2) (x). (.6) eq:gen2h

15 F.9MO The general linear inhomogeneous equation Consider now the equation y + a(x)y + b(x)y = f(x). Suppose we know one particular solution y p (x) of this equation. Then letting y = y p + y c, we obtain y c + a(x)y c + b(x)y c =. This ODE is the homogeneous differential equation corresponding to the original homogeneous one and has a general solution, as (.6), of the form y c (x) = Ay () (x) + By (2) (x). We call this the complementary function. The general solution of our inhomogeneous ODE is then of the form complementary function + particular solution: y(x) = y c (x) + y p (x) = Ay () (x) + By (2) (x) + y p (x). LCC.5 Linear with constant coefficients Motivated by a first-order ODE of the form y + ay =, which has the general solution y = Ae x, when confronted with a linear, second-order, constant-coefficient, homogeneous ODE y + ay + by =, (.62) eq:2ndlcc we can look for a solution y(x) = Ae λx, with λ to be determined and A an arbitrary but non-zero constant. Substituting this into (.62) leads to the auxiliary equation, or characteristic equation, λ 2 + aλ + by =. (.63) eq:aux Provided that a 2 > 4b there are two distinct real solutions to (.63), say λ = 2 ( a + (a 2 4b) ) and λ 2 = 2 ( a (a 2 4b) ), so we get two independent solutions to the ODE, namely y () (x) = e λ x, y (2) (x) = e λ 2x. These form a fundamental system for (.62); we say more about such systems in Chap. 2. Because of the linearity of (.62), any linear combination of y () and y (2) is also a solution so the general solution can now be written as y(x) = A y () + A 2 y (2) = A e λ x + A 2 e λ 2x, (.64) eq:2ndla for arbitrary constants A and A 2. With a 2 = 4b, we have a case as in Example.7, and a single root λ = a/2 of (.63). There are then two independent solutions of the ODE e λx and xe λx, and the general solution is y(x) = A e λx + A 2 xe λx. (.65) eq:2ndlb

16 F.9MO2 4 Finally, for a 2 < 4b, (.63) has complex roots, say λ = p ± iq so that e (p+iq)x solves (.62). Taking real and imaginary parts, independent solutions for the ODE are y () (x) = e px cos qx, y (2) (x) = e px sin qx. A linear combination now gives the general solution as y(x) = e px (A cos qx + A 2 sin qx). (.66) eq:2ndlc.6 Euler equations (L. Euler, ) These are equations of the form x 2 d2 y dx 2 + a x dy dx + a y = (.67) Euler where a and a are constants. Remarkably, this equation can be transformed into a linear ODE with constant coefficients by writing it in terms of the variable u = ln x. Then and Hence we have Equation (.67) becomes d 2 y = d ( ) dy dx 2 dx x du = dy x 2 du + x dy dx = dy du du dx = dy x du x dy dx = dy du, which has constant coefficients, as promised. (.68) = dy x 2 du + d dy x dx du (.69) du d 2 y dx du = dy 2 x 2 du + d 2 y x 2 du. 2 (.7) dy x2 dx = d2 y du dy 2 du. Example.9. Find two independent solutions of d 2 y du + (a 2 ) dy du + a y =, (.7) x 2 d2 y dx + 2xdy + y =. (.72) 2 dx Solution: In terms of the variable u = ln x the equation becomes d 2 y du + dy + y =, (.73) 2 transform du

17 F.9MO2 5 which we can solve with the techniques of the previous section. equation is The characteristic λ 2 + λ + = (.74) with solutions λ = + 3 i and λ = 3i. Hence two solutions of (.73) are 2 2 ( ) ( ) 3 3 y () = e u/2 cos 2 u, y (2) = e u/2 sin 2 u. (.75) Transforming back to x we obtain solutions for (.67): ( ) ( ) y () (x) = x 3 2 cos 2 ln x, y (2) (x) = x 3 2 sin 2 ln x. (.76).7 Method of undetermined coefficients In Chapter 2, we consider a quite general technique, the method of variation of parameters which works in principle for any linear second-order ODE of the form d 2 y dx + a 2(x) dy 2 dx + a (x)y = b(x) (.77) as well as for higher-order ODEs and general linear inhomogeneous systems. When a and a 2 are constants, and b(x) is a polynomial, exp, sin or cos, there is a simpler method to obtain a particular solution. This you have met in previous courses, and which we here call the method of undetermined coefficients. The idea is to choose a particular solution that has a similar form to b(x). The following table gives recipes involving unknown coefficients which one can determine by substituting into the equation. There is no deep reason for these recipes other than that they work. In the table homogeneous equation has been abbreviated by HE. To illustrate the recipes given in the table, we consider some examples, beginning with the polynomial case. We try and find, by inserting (.79) into (.78), that Comparing coefficients yields y + y = x 2. (.78) poly x p (t) = b + b x + b 2 x 2, (.79) guess (2b 2 + b ) + b x + b 2 x 2 = x 2. (.8) b 2 =, b =, b = 2 (.8) Regarding the first (polynomial) case, if is a solution of HE, try instead c x + + c n+ x n+ ; if and x are solutions of HE, try instead c 2 x c n+2 x n+2.

18 F.9MO2 6 b(t) particular solution b + b x b n x n c + c x + + c n x n e λx e λx is not a solution of HE try ce λx e λx is a solution of HE try cx exp(λx) e λx and xe λx solutions of HE try cx 2 exp(λx) sin(ωx) or cos(ωx) sin(ωx), cos(ωx) are not solutions of the HE try c sin(ωx) + c 2 cos(ωx) sin(ωx), cos(ωx) are solutions of the HE try c x sin(ωx) + c 2 x cos(ωx) tab:undcoef Table : Forms of particular solutions. (These can be extended to cover products of polynomials, exponentials, and either sines or cosines.) so that Continuing with the exponential case, consider y p (t) = x 2 2. (.82) d 2 y dx dy 2 dx 2y = 2e x (.83) lininhom The roots of the characteristic equation are λ = and λ = 2. Thus the table tells us to try Substituting into the equation, we find y p (x) = cxe x. c = 2/3. Thus, a particular solution is y p (t) = 2 3 xe x. Finally we turn to the recipe for the oscillatory case f(t) = cos ωt given in the last row of Table. Suppose we want to solve y + a y + a y = cos(ωx). (.84) This equation is that of the damped, driven, simple harmonic oscillator. We could just use the recipe, but it perhaps interesting to consider how it arise from the exponential case. Since cos ωx is the real part of exp(iωx) we could try to solve y + a y + a y = e iωx (.85) complex

19 F.9MO2 7 first and then take the real part of the solution we obtain. This turns out to be an efficient method. Suppose that iω is not a solution of the characteristic equation. Then try y(x) = C exp(iωx). Inserting into (.85) yields Dividing by exp(iωx) and solving for C we find C = Ce iωx ( ω 2 + ia ω + a ) = e iωx. (.86) ω 2 + ia ω + a = e iφ (a ω 2 ) 2 + a 2 ω 2 (.87) where tan φ = a ω a ω 2. (.88) Taking the real part of y(x) = C exp(iωx) we therefore find the solution y p (x) = cos(ωx + φ). (.89) mess (a ω 2 ) 2 + a 2 2 ω Since cos(ωx + φ) = cos φ cos(ωx) sin φ sin(ωx) this solution can also be written as y p (x) = cos φ (a ω 2 ) 2 + a 2 ω cos(ωx) sin φ sin(ωx) (.9) 2 (a ω 2 ) 2 + a 2 2 ω which is of the form given in the last row of Table. You can read off the coefficients c and c 2 - note that they depend on the frequency ω! One can repeat this exercise in the case that iω is a solution of the characteristic equation (try it!). In summary, in the cases when the coefficients a 2 (x), a (x) in (.83) are independent of x, and b(x) is one of the forms given in Table, it usually relatively easy to use the method of undetermined coefficients. Otherwise we must resort to the more general method of variation of parameters described in the next chapter.

20 F.9MO2 8 chap:syst 2 Systems of ODEs 2. General remarks In many applications we need several real-valued functions of one real variable to describe a system. Examples are the temperature and pressure as a function of time, or the concentrations of two chemicals in a solution as a function of time, or the three position coordinates of a particle as a function of time. In this section we shall use the notation that the independent variable will be called t, with the dependent variables x,..., x n. Each of these functions obeys a differential equation, which may in turn depend on the other functions, e.g. dx dt dx 2 dt = tx + x 2, More generally, we might have (writing for d/dt): = x 2 4t 2 x 2. (2.) systex ẋ (t) = F (x (t), x 2 (t),, x n (t), t), ẋ 2 (t) = F 2 (x (t), x 2 (t),, x n (t), t),. ẋ n (t) = F n (x (t), x 2 (t),, x n (t), t). (2.2) Such a set of coupled equations is called a system of differential equations. Clearly it makes sense to use vector notation. Defining x : R R n, x(t) = F : R n+ R n, F (x, t) = x (t)., (2.3) xvec x n (t) F (x,, x n, t)., (2.4) F n (x,, x n, t) we can write the system of differential equations as one equation For example, equations (2.) can be written dx (t) = F (x(t), t). (2.5) genform dt with x = ( x x 2 ) dx (t) = F (x(t), t), (2.6) dt, F (x, t) = ( tx + x 2 x 2 4t 2 x 2 ). (2.7)

21 F.9MO2 9 Although the general form (2.5) seems to involve only a first derivative, it includes differential equations of higher order as a special case. An nth-order differential equation of the form d n x dt n = f ( x, dx dt,..., dn x dt n, t can be written as a first-order equation for the vector-valued function x in (2.3) as x x n x 2 ) (2.8) nthorder d x 2 dt. = x 3.. (2.9) rewrite f(x,..., x n, t) We recover the differential equation (2.8) by identifying x = x. Then the first n components of (2.9) tell us that x 2 = dx dt, x 3 = dx 2 dt = d2 x dt 2,..., x n = dn x dt n (2.) so that the last component is precisely the equation (2.8). eqlist Example 2.. Write the following differential equations as first-order systems: (a) d2 x dt 2 + dx dt + 4x = t2 ; (b) xẍ + (ẋ) 2 = ; (c) d2 x dt 2 = x + y, d 2 y dt 2 = x + y. For (a) we need a two-component vector x = ( x x 2 ). We identify x with x and want our equation to ensure that x 2 = ẋ = x. The required equation is ( ) ( ) ẋ x 2 =. (2.) 4x x 2 + t 2 exa ẋ 2 (b) can be written as ( ẋ ẋ 2 ) ( ) x 2 =. (2.2) x 2 2/x For (c) we need a four-component vector. We identify x with x and x 3 with y. Our equation should include ẋ = x = x 2 and ẏ = x 3 = x 4. The required equation is d dt x x 2 x 3 x 2 = x + x 3 x 4 (2.3) x 4 x + x 3

22 F.9MO2 2 ss:thersys When F (x, t) is a linear function of x, x 2,, x n, we can write the right-hand side of (2.5) in the form A(t)x(t) + b(t), where A(t) is an n n matrix valued function and b(t) : R R n (i.e. a vector-valued function). For example, we can write (2.) as d dt ( x x 2 ) = ( 4 ) ( x x 2 ) ( ) +. (2.4) t 2.2 Existence and uniqueness of solutions of systems of ODEs Picard s theorem generalises to systems dx(t) = F (x, t) as follows dt Theorem 2.2. Suppose F : R n+ R n is continuous in some region U I, where I = (t, t 2 ) is an open interval and U R n is an open set such that the partial derivatives in the derivative matrix F F x... x n DF =.. (2.5) derivativ F n F x... n x n are also continuous there. Then, for every t I and x U, the initial-value problem dx dt (t) = F (x(t), t), x(t ) = x (2.6) geninit has a unique solution in some open interval containing t. Example 2.3. Compute the derivate matrix, and hence show the existence and uniqueness of a solution ( ) for the initial-value problem consisting of (2.) and the initial condition x() =. The derivative matrix is given by ( ) ( ) F F x DF = x 2 t F 2 F 2 = (2.7) x x 2 2x 4t 2 ( ) which is continuous in an open set in R 3 that contains t =, x =. We will not prove this version of Picard s theorem either, but note a corollary for linear systems of the form dx (t) = A(t)x(t) + b(t). (2.8) genlin dt The derivative matrix is now simple DF (t) = A(t), and so if A and b are continuous functions of t, the conditions of Picard s theorem are satisfied for all initial conditions x(t ) = x. Hence we have linexist Corollary 2.4. If A : R R n2 is a continuous matrix-valued function and and b : R R n is a continuous vector-valued function then the linear differential equation (2.8) has a unique solution for all initial conditions x(t ) = x and this solution exists for all t R.

23 F.9MO Linear systems of ODEs In this section we develop a general theory of linear systems of the form dx (t) = A(t)x(t) + b(t), (2.9) genlinn dt where A is an n n matrix-valued function and b is R n -valued function of t. Both A and b are assumed to be continuous functions on all of R, so that, according to Corollary 2.4, there is unique solution of (2.9) satisfying a given initial condition x(t ) = x Homogeneous linear systems For a given continuous n n matrix-valued function A of t we define the differential operator and note that L is a linear operator: The equation L[x] = dx dt Ax (2.2) linop L(αx + βy) = α dx dt + β dy αax βay = αl[x] + βl[y]. (2.2) dt L[x] = dx dt Ax = (2.22) genhomo is called a homogeneous linear equation. The space of solutions of (2.22) is characterised by the following theorem. solspace Theorem 2.5. (Solution space for homogeneous linear systems.) The space S = {x C (R, R n ) L[x] = } (2.23) solutions of solutions of the homogeneous equation (2.22) is a vector space of dimension n since the evaluation map is a vector-space isomorphism. Ev t : S R n, x x(t ) (2.24) Proof : The fact that S is a vector space follows from the linearity of L (i.e. x, y S = αx + βy S). To show that Ev t is a vector-space isomorphism, we first note that it is a linear map: Ev t [αx + βy] = αx(t ) + βy(t ) = αev t [x] + βev t [y]. Picard s theorem implies that it is bijective: It is surjective since, for any given x R n, there exists solution x S so that x(t ) = x. It is injective since this solution is unique. Hence Ev t is a bijective linear map, i.e. a vector-space isomorphism. Since the space containing x is R n, and S is isomorphic to it, S is n dimensional.

24 F.9MO2 22 Definition 2.6. (Fundamental sets solutions for homogeneous systems.) A basis {y (),..., y (n) } of the set S defined in (2.23) is called a fundamental set of solutions of (2.22). The following lemma is an immediate consequence of Theorem 2.5. determinant Lemma 2.7. (Determinant for fundamental sets.) A set of n solutions y (),..., y (n) of the homogeneous linear equation (2.22) forms a fundamental set of solutions if and only if the matrix y () (t)... y (n) (t) Y (t) =.. (2.25) ymat y n () (t)... y n (n) (t) has a non-zero determinant det(y (t)) at one value t = t (and hence for all values of t). Note that if the determinant is non-zero at one value t = t, then using Ev t, this establishes that y (),, y (n) are a basis of S. But we can then use Ev t, to obtain another set of basis vectors of R n with determinant non-zero. Hence, if the determinant is non-zero for one value t, it is also non-zero for all values of t. Example 2.8. Find a fundamental set of solutions for the homogeneous equation ( ) ( ) ( ) d x x =. (2.26) eigex dt To solve (2.26) we try a solution of the form x 2 x 2 x(t) = e λt v. (2.27) guessagai Inserting into (2.26) we find that this is a solution if ( ) λv = v, (2.28) so that ((2.27) ) is a solution if λ is an eigenvalue of and v an eigenvector of the matrix. A short calculation shows that the eigenvalues are and, with ( ) ( ) eigenvectors v () = and v (2) =. Hence we obtain the solutions ( ) ( ) y () (t) = e t e t, y (2) (t) = e t e t Clearly, y () () and y (2) () are linearly independent since ( ) det = 2.. (2.29)

25 F.9MO2 23 Hence, {y (), y (2) } form a fundamental set. Note that fundamental sets are not unique: we can take linear combinations to obtain other fundamental sets, e.g. ( ) ( ) 2 (y() (t) + y (2) cosh t (t)) =, sinh t 2 (y() (t) y (2) sinh t (t)) =. (2.3) cosh t As we have seen, homogeneous linear n-th order equations d n x dt + a d n x n n dt a x = (2.3) n genn are a special case of linear homogeneous systems (2.22). Here we define { } s = x C n (R, R) d n x dt + a d n x n n dt a x =. n (2.32) solutions In this case, the evaluation map takes the form Ev t : s R n, x x(t ) dx (t dt ). d n x(t dt n ) (2.33) which assigns to each solution the vector made out of the values of the function x and its first n derivatives at t. Theorem 2.5 then states that this map is a vector-space isomorphism. This is sometimes useful for checking linear independence of solutions of an nth-order homogeneous linear equation: two solutions are independent if and only if their images under Ev t are independent in R n. For the two solutions y () (t) = cos t and y (2) (t) = sin t of ẍ + x =, the evaluation map at t = gives ( ) ( ) Ev (y () ) =, Ev (y (2) ) =. (2.34) The FSS {y (),, y (n) } of a homogeneous linear system dx (t) = A(t)x(t) (2.35) hlinsys dt has three main uses. Firstly, we can write the general solution of (2.35) as y(t) = c y () (t) + + c n y (n) (t) for constants c,, c n. Secondly, we can find the particular values of c,, c n that give the unique solution of the initial-value problem specified by (2.35) and the condition x(t ) = x. Thirdly, as we shall see in Section 2.3.4, we can use the FSS in order to construct a particular solution of the inhomogeneous problem dx (t) = A(t)x(t) + b(t). dt Example 2.9. Find the general solution of ( ) ( ) ( d x = dt x 2 x x 2 ), (2.36) eigex2

26 F.9MO2 24 and hence find the( solution ) of the initial-value problem specified by (2.36) and the initial condition x() =. We have already found the FSS for this problem; hence the general solution is ( ) ( ) ( ) e t e t c e t + c 2 e t x(t) = c + c e t 2 =. (2.37) e t c e t c 2 e t The solution of the initial-value problem is given by the choice ( ) ( ) c + c 2 =, (2.38) c c 2 with solution c = /2, c 2 =. Hence the solution is 2 ( ) sinh(t) x(t) =. (2.39) cosh(t) matrixmeth Matrix methods for finding the FSS In practice, the FSS can only be found explicitly in rare happy cases. An example is the case where the matrix A is constant, and this is the case we study in some detail this section. Consider the system of linear equations ẋ(t) = Ax(t), (2.4) constcoef where x : R R n and A is a constant, real, n n matrix. To find solutions we make the ansatz (=inspired guess), suggested by what happens for a single ODE, Inserting into (2.4) we obtain x(t) = e λt v. (2.4) expansatz Av = λv (2.42) so that (2.4) is a solution iff v is an eigenvector. If A has n real linearly independent eigenvectors v (),..., v (n) with real eigenvalues λ,..., λ n (which need not be distinct), a fundamental set of solutions is given by y () (t) = e λ t v (),..., y (n) (t) = e λnt v (n). (2.43) FSS However, as discussed in Appendix B, not all real n n matrices have a basis of n real linearly independent eigenvectors. Let us consider the different situations which can arise: (i) n distinct real eigenvalues In this case, we do get n linearly independent eigenvectors and the FSS is given by (2.43)

27 F.9MO2 25 Example 2.. Consider the equation (2.4) for n = 2 and ( ) 4 A =. (2.44) This matrix has eigenvalue ( ) λ if det(a λi) = ( λ) 2 4 = i.e. if λ = or λ = 3. The vector v () = v v 2 is an eigenvector for eigenvalue if v + 4v 2 = v, and v + v 2 = v 2. (2.45) ( ) Hence v () 2 = is an eigenvector for eigenvalue. Similarly we find that ( ) v (2) 2 = is an eigenvector for eigenvalue 3. The solutions ( ) ( ) y () (t) = e t 2, y (2) (t) = e 3t 2 (2.46) are independent and form a fundamental set. (ii) Repeated real eigenvalues. If a given eigenvalue has algebraic multiplicity m (i.e. it is repeated m times) it will have q m linearly independent eigenvalues. Let us consider the cases q = m, and q < m separately. q = m Example 2.. Find the FSS of (2.4) for ( ) 2 A = (2.47) 2 ( ) ( ) A has v () = and v (2) = as eigenvectors for the repeated eigenvalue 2. In this case we find a fundamental set of solutions: ( ) ( ) y () (t) = e 2t =, y (2) (t) = e 2t. (2.48) q < m. As we have said, a matrix might fail to have a basis of eigenvectors if an eigenvalue is repeated. Suppose for simplicity that a real eigenvalue λ has multiplicity m = 2 (i.e. the characteristic polynomial contains a factor (λ λ ) 2 ) and that v is the only eigenvector for this eigenvalue (i.e. q = ). Then y () (t) = e λ t v is a solution of (2.4). To find a second solution we are motivated by Sec..5 and try y (2) (t) = te λ t u + e λ t w. (2.49) repguess

28 F.9MO2 26 Inserting into (2.4) we find that this is a solution if i.e. if e λt (λ tu + u + λ w) = e λ A(tu + w) (2.5) Au = λu and (A λ I)w = u. (2.5) geneig The first of (2.5) indicates that u = v and then one can show that the second equation can always be solved for w. Since (A λ I)w but (A λ I) 2 w =, w is sometimes called a generalised eigenvector of A. Example 2.2. ( ) 9 A =. (2.52) 5 We find that λ is an eigenvalue if (λ + 2) 2 =. Hence 2 is a repeated eigenvalue. The only eigenvector is ( ) 3 v = (2.53) so that one solution of (2.4) is ( ) y () (t) = e 2t 3. (2.54) To find the second solution we need to solve ( ) ( ) (A + 2I)w = v w =. (2.55) 3 With elementary row operations we find ( ) w = (2.56) so that a second solution is given by ( ) ( ) ( ) y (2) (t) = e 2t 3 t + e 2t = e 2t 3t. (2.57) t (iii) Complex eigenvalues If the real matrix A in (2.4) has a complex eigenvalue λ = α + iβ with corresponding complex eigenvector v = v () + iv (2), then y(t) = e λt v is a complex solution (note that y (t) = e λ t v is another solution). However, since A is real we have ( ) d dy Re (y) = Re = Re (Ay) = A Re (y), dt d Im (y) dt = Im dt ) = Im (Ay) = A Im (y), ( dy dt

29 F.9MO2 27 so Re (y(t)) and Im (y(t)) are both real solutions. Thus we obtain two real solutions by taking the real and imaginary parts of y(t). Example 2.3. Find the FSS of (2.4) with ( ) 5 A =. (2.58) 3 The eigenvalues are ± i, with eigenvectors solution ( ) 2 ± i. Thus we have the complex ( ) y(t) = e (+i)t 2 + i ( ) = e t 2 + i (cos t + i sin t) ( ) ( ) = e t 2 cos t sin t + ie t cos t 2 sin t cos t sin t (2.59) so that a fundamental set of solutions is ( ) ( ) y () (t) = e t 2 cos t sin t, y (2) (t) = e t cos t 2 sin t. (2.6) cos t sin t Note that the other complex solution is ( ) y (t) = e ( i)t 2 i and we have Re (y (t)) = y () (t), Im (y (t)) = y (2) (t). Thus, in order to obtain two real solutions, we only need to deal with one of the pair of complex conjugate solutions, and it doesn t matter which one The Fundamental Matrix If {y (),, y (n) } is a FSS, the non-singular matrix Y (t) = y () y (2) y (n)... y n () y n (2) y n (n) is called the fundamental matrix. The general solution of the homogeneous linear ODE ẋ(t) = A(t)x(t) is then x(t) = c y () (t) + c 2 y (2) (t) + c n y (n) (t), (2.6) eq4

30 F.9MO2 28 where c,, c n are constants. Eqn. (2.6) can be written in the form c c x(t) = Y (t) c, where c = 2. (2.62) eq5. The solution to the initial-value problem consisting of ẋ(t) = A(t)x(t) and the initial condition x() = x is given by (2.62) with c specified by Thus we can write c n x = Y (t )c, such that c = Y (t ) x. (2.63) eq6 x(t) = Y (t)y (t ) x. (2.64) eq7 Note that Y (t) exists since det(y (t)). Eqn. (2.64) is not of much practical use as a way to compute the solution - it is usually easier just to find c from x = Y (t )c by row reduction of Y (t ). It is however conceptually important and of interest in the next section. Another useful property of Y (t) is that it satisfies This follows since the ith column of both sides is simply ẏ (i) y. (i) = A(t).. ẏ (i) y (i) Ẏ (t) = A(t)Y (t). (2.65) Ydot inhomlin Inhomogeneous linear systems We return to the more general form dx (t) = A(t)x(t) + b(t), (2.66) genlinnn dt of linear systems with b. Such systems are called inhomogeneous linear systems. Using again the linear operator L defined in (2.2) we have the following analogue of Theorem 2.5. Theorem 2.4. (Solution space for inhomogeneous linear systems.) Let {y (),..., y (n) } be a fundamental set of solution for the homogeneous equation ẋ = Ax. Then any solution of the inhomogeneous equation (2.66) is of the form x(t) = n c i y (i) + x p, (2.67) geninsol i= where x p is a particular solution of (2.66) and c,..., c n are real constants.

31 F.9MO2 29 Proof : It is easy to check that (2.67) satisfies the inhomogeneous equation (2.66), using the fact that each of the y (i) satisfy ẏ (i) = Ay (i). To show that every solution can be written in this way, suppose that x is a solution of (2.66). Then x x p satisfies the homogeneous equation d dt (x x p) = A(x x p ) and therefore can be expanded x x p = n c i y (i) (2.68) i= for some real constants c,..., c n. The most systematic way of finding a particular solution of an inhomogeneous linear equation is called the method of variation of the parameters. The idea is to look for a particular solution of the form x p (t) = n c i (t)y (i) (t), (2.69) ansatz i= where {y (),..., y (n) } is a fundamental set of the homogeneous equations as before, but, crucially, the c i are now functions of t. As above, we can write (2.69) in terms of the fundamental matrix Y as x p = Y c. (2.7) Using the product rule and the fact that Ẏ = AY we deduce Hence x p = Ẏ c + Y ċ = AY c + Y ċ. (2.7) x p = Ax p + b AY c + Y ċ = AY c + b ċ = Y b. We can now compute c by integration - at least in principle. We summarise this result as follows. varpthm Theorem 2.5. Method of variation of the parameters. Let {y (),..., y (n) } be a fundamental set of solutions of ẋ = Ax and let Y be the matrix constructed from {y (),..., y (n) } according to (2.25). Then x p (t) = Y (t) is a particular solution of the inhomogeneous equation ẋ = Ax + b. t t Y (τ) b(τ)dτ (2.72) varpar The formula (2.72) is very general, and therefore very powerful. To get a feeling for how it works we need to study examples - see Problem Sheet 5! Let us consider the n = case in more detail and show that equation (2.72) reduces to the formula for the solution of a linear first-order equation in terms of the integrating factor. With the conventions of this section, we consider the first-order differential equation ẋ(t) = a(t)x(t) + b(t) ẋ(t) a(t)x(t) = b(t) (2.73) firstagai

32 F.9MO2 3 and note that the homogeneous equation ẋ(t) = a(t)x(t) (2.74) has the solution ( t ) y(t) = exp a(τ) dτ, (2.75) t which is the inverse of the integrating factor ( t ) I(t) = exp a(τ) dτ t (2.76) for (2.73). Thus the formula t b(τ) x p (t) = y(t) dτ (2.77) t y(τ) for the particular solution is precisely the solution we would obtain using the integrating factor /y(t).

33 F.9MO Higher-order ODEs Recall that an nth-order differential equation of the form ( d n x dt = f x, dx ) n dt,..., dn x dt, t n (2.78) nthi can be written as a system of systems of ODEs by identifying Then we have x = x, x 2 = dx dt, x 3 = dx 2 dt = d2 x dt,..., x 2 n = dn x. (2.79) dtn x x n x 2 d x 2 dt. = x 3.. (2.8) nlin f(x,..., x n, t) where the last component is precisely the equation (2.78). The initial-value problem corresponds to (2.8) and a condition x(t ) = x. For the equation (2.78) this is equivalent to specifying x(t ), dx dt (t ), d2 x dt 2 (t ),..., dn x dt n (t ), (2.8) i.e. the function and its first (n ) derivatives. All of the results of Sec. 2.2 specialise to the case of nth-order ODES. In this section we consider some of these specialisations Picard s Theorem The derivative matrix for (2.8) is DF = f x 3 f x f x 2 f x n f x n. Thus (2.8) satisfies the conditions of Picard s theorem if the function f and the partial derivatives f,..., f (2.82) x x n are continuous. In that case there will be a unique solution, at least locally, if we specify x(t ), i.e., the function x(t ) and its first (n ) derivatives.

34 F.9MO2 32 Example 2.6. Show that x(t) = t 2 cannot be a solution of the initial-value problem d 2 x dt 2 = ( ) 2 dx + x 3, x() =, dt dx () =. dt The function f(t, x, dx f ) on the right-hand side and the two partial derivatives = dt x 3x2, f = 2ẋ, are continuous at t =, x =, ẋ =. The initial-value problem therefore ẋ has a unique solution from Picard s Theorem. One solution is x(t) = (and ẋ(t) = ). Therefore there can not be another solution of the form given The FSS and the Wronskian A linear nth-order ODE is one of the form system d n x dt + a n(t) dn x n dt + + a (t)x = b(t). (2.83) n genni The equation is said to be homogeneous when b(t) =. Any n linearly independent y () (t), y (2) (t),..., y (n) (t) solutions of the homogeneous linear equation constitute a Fundamental Solution Set. The counterpart of Lemma 2.7 has a special name for nth-order equations: wronskian Lemma 2.7. Wronskian for n-th order linear homogeneous equations. A set of n solutions y (),..., y (n) of the n-th order homogeneous linear equation (2.3) forms a fundamental set of solutions if and only if the matrix y ()... y (n) dy () dy... (n) Y = dt dt (2.84) fundmat.. d n y () d... n y () dt n dt n has a non-zero determinant W =det(y ) at one value t = t (and hence for all values of t). The determinant W of Y is called the Wronskian of y (),..., y (n). Example 2.8. The Wronskian of two solutions y (), y (2) of the second-order equation ẍ + a 2 ẋ + a x = is W = y () ẏ (2) y (2) ẏ (). If we have n solutions of a homogeneous, linear nth-order ODE, and we want to check for linearly independence, we simply have to evaluate the Wronskian at any value of t (or whatever the variable is called). Example 2.9. Two solutions of the equation d2 y + 4y = are cos 2x and sin 2x. dx2 Show that these two solutions constitute a FSS. The Wronskian is W = cos(2x) cos(2x) + sin(2x) sin(2x) =, hence the solutions are linearly independent and are an FSS.

35 F.9MO The FSS for equations with constant coefficients Consider the linear homogeneous equation dx n dt + a n d n x dt n a x = (2.85) eq47 in the case when a n,, a are constant coefficients. Then we can write the equation in matrix form (after identifying x = x ) as... dx. = A x(t), where A = dt. (2.86) eq57... a a 2... a n a n As above, we can press on and solve this equation by finding the eigenvalues λ that satisfy det(a λi) =. It is simple exercise (see Problem Sheet 4) to show that this equation reduces to the characteristic equation λ n + a n λ n + a 2 λ + a =. (2.87) char For n real, distinct solutions λ,, λ n, n linearly independent solutions of (2.86) are given by e λ t e λ 2t e λ nt λ e λ t λ 2 e λ 2t λ n e λ nt λ n e λ t λ n n e λ 2t λ n e λnt The corresponding solutions of (2.85) are just e λ t,, e λ nt. Of course there is a far easier way of deriving these results for the (2.85). We just try a solution of the form e λt and substitute into (2.85) to get the equation (2.87) as we would do for the n = 2 case. The solutions are then e λ it for distinct eigenvalues. However, it is hopefully illuminating to see how these results can be found by specialising from the matrix methods discussed earlier. Example 2.2. Find the general solution of d 4 x dt + d3 x 4 dt x 3 7d2 dt dx + 6x =. 2 dt Hence find the solution that satisfies the initial conditions x() =, dx dt () =, n d 2 x d () = 2, dt2 3 x () =. dt3 The characteristic equation is λ 4 + λ 3 7λ =, with roots λ =,, 2, 3. Hence the general solution is x(t) = c e t + c 2 e t + c 3 e 2t + c 4 e 3t.