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1 ODEs Cathal Ormond

2 2 1. Separable ODEs Contents 2. First Order ODEs 3. Linear ODEs

3 Chapter 1 Separable ODEs 1.1 Definition: An ODE An Ordinary Differential Equation (an ODE) is an equation whose unknown is a function of one variable and its derivatives. For example, given a function y(x), we may have y (x) = 2x y(x), which gives has the solution y(x) = Ce x2 for some C R. 1.2 Definition: The Order of an ODE The order of an ODE is the highest-order derivative in that ODE. In the above example, the ODE y (x) = 2x y(x) is a first order ODE. 1.3 Definition: An IVP An Initial Valued Problem (an IVP) is an ODE with initial conditions. In the above example, we may have y (x) = 2x y(x) and y(0) = 2, giving the solution of y(x) = 2e x Definition: A Separable ODE A first-order ODE is called separable if it can be written in the form y (x) = F (x)g(y) where F and G are some functions of x and y respectively Solving a Separable ODE Manipulating the above equation gives the following: dy G(y) = F (x)dx From this, we complete the integration and solve the above equation for G(y), and then isolate y to find a solution to the original ODE. Note In the above, we must firstly assume that G(y) 0 so that the division is defined. This may rule out some possibilities for the solutions to the ODE, so we must test the case that G(y) = 0 separately, when the above is complete.

4 4 1.5 Linearity An ODE is said to be linear if the coefficients of y, y, y, etc., are all functions of x alone. For example, x 2 y (x) + xy (x) + cos xy(x) = 0 is a second order, linear ODE, as all of the coefficients are functions of x. 1.6 Homogeneity A linear ODE is called homogeneous if it has y(x) = 0 as a solution. In the above example, x 2 y (x) + xy (x) + cos xy(x) = 0 is also a homogeneous ODE.

5 Chapter 2 First Order ODEs 2.1 First Order Linear, ODEs Sometimes, we may have a linear ODE which is easy to integrate. For example, the ODE xy (x) + y(x) = e x can be rewritten in the form (xy(x)) = e x. Thus, integrating both sides and rearranging gives y(x) = ex + c, for some constant c. This method works when one side may be rewritten due to x the product rule, but this is not always the case. 2.2 More on First Order, Linear ODEs Given a first order, linear ODE of the form y (x)+p (x)y(x) = Q(x) where P and Q are some functions of x, we may solve it using a method similar to the above method. We wish to multiply the ODE by some function, say µ(x), so that we may use the product rule and then integrate. Thus, the ODE becomes: µ(x)y (x) + µ(x)p (x)y(x) = µ(x)q(x) For some suitably chosen µ(x). If the product rule is to be applied, we must require that µ (x) = µ(x)p (x). Solving this last equation gives µ(x) = exp( P (x)dx). To summarize, given an ODE of the form y (x) + P (x)y(x) = Q(x), the solutions are given by µ(x)q(x)dx y(x) = µ(x) Where the function µ(x) is given by: 2.3 General First Order ODEs µ(x) = e P (x)dx Any first order ODE can be expressed as y (t) = f(t, y(t)) for some function f of t and y(t). In this section, we will only consider the solution to the general form of first order ODEs, as higher order ODEs can be reduced to a first order ODE. Heuristically speaking, there should exist a solution to the following IVP This IVP will be called the General IVP. y (t) = f(t, y(t)) y(t 0 ) = y 0

6 Reducing ODEs Higher ODEs can be treated similarly to first order ODEs using vectors instead of scalars. We will show this ( method ) by an example. ( ) Consider ( the ODE y (t) = t 2 y (t) + e t y(t), or y = t 2 y + e t y. y y Let y = y Thus, y y = y = ) ( ) ( ) 0 1 y y = t 2 y + e t = y e t t 2 y Letting the above matrix be denoted by A, we get y = A y which is a first order ODE. Thus, using matrix manipulation instead of scalar manipulation if necessary, we need only consider first order ODEs Lemma: Integral Formulation A function y = y(t) is a differentiable solution to the General IVP if and only if it is a continuous solution of: t y(t) = y 0 + f(s, y(s))ds t Theorem: Local Existence Given a general IVP, suppose that f and f y are continuous on a rectangle R centred about (t 0, y 0 ). Also, let R = {(t, y) : t t 0 a, y y 0 b, a, b R}. Then there exists a solution to the general IVP that is unique on some interval (t 0 ɛ, t 0 + ɛ), ɛ R. Note This theorem may be proven by proving the following statements: 1. There exists some L, M R such that for all (t, y), (t, y 1 ), (t, y 2 ) R: f(t, y) M f(t, y 1 ) f(t, y 2) L y 1 y 2 2. y 0 (t) = y 0 and y n+1 (t) = t 3. y n+1 (t) y n (t) MLn t t 0 (n + 1)! 4. y(t) = lim n y n(t) exists. t 0 f(s, y(s))ds for all n > 0. for all n 0 5. y(t) is a continuous solution to the integral equation Theorem: The Gronwall Inequality Let f, g be two non-negative and continuous functions with f(t) c + constants t 0, c.then f(t) c exp( all t Theorem t t 0 g(s) ds). Consequently, f(t) t t t 0 f(s)g(s) ds, for some t 0 f(s)g(s) ds f(t) = 0 for Let f and f y be continuous on some rectangle around (t 0, y 0 ). Then the IVP y (t) = f(t, y(t)), y(t 0 ) = y 0 can have at most one solution.

7 Theorem Let f and f y be continuous in a rectangle R = {(t, y) t t 0 a, y y 0 b} which is centred about (t 0, y 0 ). The general IVP has a unique solution when t t 0 ɛ, for ɛ > Theorem Let f and f y be continuous in an arbitrary subset A R 2 which contains the point (t 0, y 0 ). Thus the general IVP has a unique solution unless: g(t) becomes infinite in finite time (it blows up) The graph of y(t) meets the boundary of A Theorem Let f be a vector valued function depending on t R and y R n. If f and f yi are continuous on an arbitrary subset A R n containing the point (t 0, α). Then y = f(t, y), y(t 0 ) = α has a unique solution defined at all times unless: Note One of the components of y blows up The graph of y meets the boundary of A. The proof of this theorem is identical to the one for the scalar case, by adapting the notation and by n n n defining v = v i and A = a ij. Note also that A + B A + B and AB A B i=1 i=1 j=1 and define the integral of a vector to be componentwise.

8 Chapter 3 Linear ODEs 3.1 Definition: Linearity An ODE depending on y(t) is said to be linear if it is made up of derivatives of y only, where the coefficients of the derivatives are some functions of t. For example, the general system of linear ODEs is given by: y 1 (t) = a 11(t)y 1 (t) + a 12 (t)y 2 (t) + + a 1n (t)y n (t) + b 1 (t) y 2 (t) = a 21(t)y 1 (t) + a 22 (t)y 2 (t) + + a 2n (t)y n (t) + b 2 (t). y n(t) = a n1 (t)y 1 (t) + a n2 (t)y 2 (t) + + a nn (t)y n (t) + b b (t) Or more compactly, y (t) = A(t) y(t) + b(t) Definition: Homogeneous A linear ODE is said to be homogeneous if y(t) = 0 for all t is a solution. Note that the above system is homogeneous if b(t) = Theorem The set of all solutions of y (t) = A(t) y(t) forms a vector space over R. 3.2 Definition: Linear Independence The vector-valued functions y 1 (t), y 2 (t),..., y n (t) are said to be linearly independent if there exist a set of scalars c 1, c 2,..., c n not all equal to zero such that c 1 y 1 (t) + c 2 y 2 (t) + + c n y n (t) for all t Theorem Once n linearly independent solutions of y (t) = A(t) y(t) are found, all solutions may be determined. 3.3 Definition: The Wronskian We define the Wronskian of a set of n vectors to be: W (t) = det[ y 1 (t) y 2 (t)... y n (t)]

9 Theorem Given y (t) = A(t) y(t), where A(t) has continuous entries, the solutions y 1 (t), y 2 (t),..., y n (t) are linearly independent iff their Wronskian is non-zero, which is true iff the Wronskian is non-zero at some point Theorem Consider the n th order scalar ODE y (n) (t) + a n 1 y (n 1) (t) + + a 1 y(t) + a 0 = 0, and also consder its associated characteristic equation λ n + a n 1 λ n a 1 λ + a 0 = 0. If λ 1, λ 2,..., λ n C are the solutions to the characteristic equation, then every solution to the ODE has the form y(t) = c 1 e λ 1t + c 2 e λ 2t + + c n e λnt. 3.4 Definition: Matrix Exponentials Given an n n matrix A, we define its exponential to be: Where A 0 = I Theorem e A = k=0 The exponential is such that (e At ) is such that (e At ) = Ae At, and the columns of e At are linearly independent solutions of y (t) = A(t) y(t) Corollary The solution to the IVP y (t) = A(t) y(t) with y(0) = y 0 is given by y(t) = e At y Theorem: The Product Rule Given two square matrices A and B, then (AB) = A B + AB Theorem Given two commutative, square matrices A and B, then e A+B = e A + e B 3.5 Non-Homogeneous Systems The general form for a non-homogeneous system of equations is given by y (t) = A(t) y(t) + b(t), for some vector b(t). Here are some tequniques that may be used for solving linear non-homogeneous ODE s Theorem Suppose the entries of A(t) and b(t) are continuous. Then there exists a fundamental matrix Φ(t) which is defined at all times. A k k!

10 The General Method Given a system of equations, y (t) = A(t) y(t) + b(t), a solution is given by y(t) = Φ(t) Φ(s) 1 b(s) ds, where Φ(t) = e At. Note As Φ is a funcamental matrix, its columns are linearly independent, and thus det Φ 0, so Φ 1 exists Theorem If we know that the general solution of the scalar linear honogeneous section of our non-homogeneous ODE is given by y h (t), and any particular solution of the non-homogeneous solution is given by y p (t), then the general solution for the non-homogeneous ODE is given by y(t) = y h (t) + y p (t). Note There is no general formula for finding a particular solution to the non-homogeneous ODE, but there are some guidelines: If b(t) is a polynomial of degree m, then y p (t) = t k (b m t m + b m 1 t n b 0 ). If b(t) is e βt times some polynomial of degree m, let y p (t) = t k e βt (b m t m + b m 1 t n b 0 ). If b(t) = sin(βt) or cos(βt), let y(t) = t k (A sin(βt) + B cos(βt)). Note that in each case, k is the least positive integer such that y p does not contain terms already in y h Theorem: Reduction of Stability Suppose that y 1 is a non-zero solution of the ODE y (t) + P (t)y (t) + Q(t)y(t) = 0. Then y 2 = vy 1 is also a solution provided that ( ) 2y v P (t) v = 0 y 1

11 Chapter 4 Stability 4.1 Definition: Autonomous An Autonomous system is one of the form y (t) = f( y(t)). 4.2 Definition: Stability and Critical Points A solution y 0 (t) is said to be a critical point of an autonomous system if it is a constant solution, namely that y 0 (t) = f( y 0 (t)) = 0. A solution y 0 (t) is said to be stable if every solution that is near y 0 (t) initially remains near y 0 (t) at all times. Note For linear systems, assuming that A is invertible, this reduces to y (t) = A y(t) = Definition: Asymptotic Stability A critical point y 0 of an autonomous system is said to be asymptotically stable if it is stable, and all solutions which are near y 0 converge to it as t tends to infinity Theorem Given a second order, linear homogenous system, we may determine the stability (asymtotic or otherwise) of the zero solution by considering the eigenvalues and the following table. State of Eigenvalue Real and distinct, both positive Real and distinct, both negative Real and distinct, different signs Real and equal, both positive Real and equal, both negative Purely imaginary Complex with positive real part Complex with negative real part Stability Unstable Asymptotically stable Unstable Unstable Asymptotically stable Stable Unstable Asymptotically stable

12 Definition: Positive Definite We say that a mapping V = V (x, y) is positive definite iff: V (x, y) 0 for all x, y R and V (x, y) = 0 iff (x, y) = (0, 0) 4.5 Definition: Derivatives We define the derivative of V with respect to the autonomous system, to be V = V x x (t) + V y y (t). x (t) = f(x(t), y(t)) y (t) = g(x(t), y(t)) 4.6 Definition: Lyapunov Functions A function V = V (x, y) is Lyapunov function of the system, x (t) = f(x(t), y(t)) y (t) = g(x(t), y(t)) iff V is positive definite, and V 0 in some open region R about the origin Theorem Given a system, x (t) = f(x(t), y(t)) y (t) = g(x(t), y(t)) where f and g vanish at the origin, and first order partial derivatives exist, if there exists a Lyapunov function for this system, then the the zero solution is stable.

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