Notes about ordinary differential equations. Master BME, Math Level - PowerPoint PPT Presentation

Notes about ordinary differential equations. Master BME, Math Level 2 October 10, 2019 1/33

Differential equations and systems of differential equations Basic notions and examples ◮ Ordinary Differential Equation (ODE) : equation involving an unknown function x ( t ), its derivative x ′ ( t ), and possibly its higher-order derivatives x ′′ ( t ), x (3) ( t ), etc. up to a given order x ( n ) ( t ), that is supposed to be satisfied for each t . ◮ examples: ◮ x ′ ( t ) + x ( t ) = 0 is an ODE of order 1, ◮ x ′ ( t ) = tx ( t ) + t 2 is an ODE of order 1, ◮ x ′′ ( t ) = cos( x ′ ( t )) x ( t ) is an ODE of order 2. ◮ ODE are called ”ordinary” to distinguish them from Partial Differential Equations (PDE) which involve partial derivatives of functions of several variables. 2/33

Differential equations and systems of differential equations Basic notions and examples ◮ About notations : dt ( t ), d 2 x ◮ derivatives are sometimes written dx dt 2 ( t ) (the standard notation), ˙ x ( t ), ¨ x ( t ), instead of x ′ ( t ), x ′′ ( t ). ◮ here we use the notation x ( t ), meaning that x is the unknown function and t is the variable, but many other notations can be found in books : f ( t ), f ( x ), y ( t ), y ( x )... ◮ sometimes we drop out the ”( t )” in the equation to simplify the notation. For example the 3 previous ODEs could be written also x ′ + x = 0, x ′ = tx + t 2 , x ′′ = cos( x ′ ) x . 3/33

Differential equations and systems of differential equations Two easy things with ODEs 1) Knowing an ODE and a candidate solution, it is easy to check that it is actually a solution. Examples: ◮ x ( t ) = e − t is solution of x ′ + x = 0, because indeed x ′ ( t ) = − e − t = − x ( t ), so x ′ ( t ) + x ( t ) = 0. ◮ x ( t ) = cos( t ) is solution of x ′′ + x = 0, because x ′ ( t ) = − sin( t ) and x ′′ ( t ) = − cos( t ) = − x ( t ). 4/33

Differential equations and systems of differential equations Two easy things with ODEs 2) From a function x ( t ), it is easy to find an ODE which it satisfies. Examples: ◮ if x ( t ) = t 2 + 2, then x ′ ( t ) = 2 t , so 1 2 tx ′ ( t ) − x ( t ) + 2 = 0. Hence 2 tx ′ − x + 2 = 0 is an ODE which admits x ( t ) = t 2 + 2 as solution. 1 ◮ if x ( t ) = sin( t ), then x ′ ( t ) = cos( t ), so we have sin( t ) x ′ ( t ) − cos( t ) x ( t ) = 0. Hence sin( t ) x ′ − cos( t ) x = 0 is an ODE which admits x ( t ) = sin( t ) as solution. But in this case we can also notice that x ′′ ( t ) = − sin( t ) so we have that x ′′ ( t ) + x ( t ) = 0. Hence, x ′′ + x = 0 is another ODE which admits x ( t ) = sin( t ) as solution. 5/33

Differential equations and systems of differential equations Terminlogy of ODEs ◮ Autonomous ODE : An ODE is autonomous when the variable t does not appear in the equation, except through the function x ( t ) and its derivatives. For example x ′′ ( t ) 2 x ′ ( t ) + x ( t ) = 0 is an autonomous ODE, while x ′′ ( t ) = t + cos( x ( t )) is not autonomous. Remark : if x ( t ) is the solution of an autonomous ODE, then any x ( t + λ ) where λ is a constant number, is also a solution. For example, x ( t ) = cos( t ), x ( t ) = cos( t + 2) or x ( t ) = cos( t + π/ 2) = − sin( t ) are all solutions of the ODE x ′′ + x = 0. Graphically, this means that by translating the graph of a solution along the horizontal axis, we get the grpahs of other solutions of the equations. 6/33

Differential equations and systems of differential equations Terminlogy of ODEs ◮ Explicit ODE : An ODE is explicit if it is of the form x ( n ) ( t ) = ... , where ... contains terms involving derivatives of x ( t ) of order lower than n . For example x ′ = e x + t 2 , x ′′ = x ′ + x 2 − sin( t ) are explicit ODEs, while x ′′ 2 cos( x ) = x ′ is not (so it is implicit ). Remark : Most of the times an implicit ODE can be put in explicit form, but this may involve making some assumptions about the solution. For example x ′′ 2 cos( x ) = x ′ is equivalent to the explicit form x ′′ = � x ′ / cos( x ), but only when cos( x ( t )) � = 0 and x ′ ( t ) / cos( x ( t )) ≥ 0. 7/33

Differential equations and systems of differential equations Linear equations ◮ A linear ODE is an ODE of the form a 0 ( t ) x ( t ) + a 1 ( t ) x ′ ( t ) + · · · + a n ( t ) x ( n ) ( t ) = b ( t ) , where the a i ( t ) and b ( t ) are known functions. ◮ A linear ODE is homogeneous if b ( t ) = 0. ◮ A linear ODE is with constant coefficients if the a i ( t ) are just constants. ◮ examples: ◮ tx + e t x ′ = 2 t is a linear ODE, ◮ x ′′ − x ′ + 4 x = 5 cos( t ) is a linear ODE with constant coefficients, ◮ x ′ − tx = 0 is a homogeneous linear ODE. 8/33

Differential equations and systems of differential equations Solutions of linear ODEs ◮ A classical way to solve a linear ODE is first to consider the associated homogeneous ODE , obtained by replacing b ( t ) by 0. This is because of the following important result: the solutions of a linear ODE form an affine subspace, whose direction (i.e. its unique parallel linear subspace) is the space of solutions of the associated homogeneous equation. ◮ Consequently, one gets the general solutions of a linear ODE by adding any particular solution of this ODE to the general solutions of the associated homogeneous ODE . 9/33

Differential equations and systems of differential equations Solutions of first order linear ODEs with constant coefficients It can be put in the form : x ′ = ax + b ( t ). ◮ The associated homogeneous ODE is x ′ = ax and has general solution x ( t ) = λ e at , λ ∈ R . ◮ To get a particular solution of the original ODE, it is possible to use the method of variation of constant : we look at a solution of the form x ( t ) = λ ( t ) e at . Then writing the ODE, one gets after simplification that λ ′ ( t ) e at = b ( t ), so we are left to compute an integral of the function e − at b ( t ). Hence if F ( t ) is such an integral, one gets the particular solution as F ( t ) e at , and thus the general solution of the original ODE is x ( t ) = F ( t ) e at + λ e at = ( F ( t ) + λ ) e at , λ ∈ R . 10/33

Differential equations and systems of differential equations Domain of a solution and maximal domains ◮ When considering an ODE, one may look at solutions defined only on a specific domain , i.e. a specific range of t values: ◮ either because the ODE itself assumes some constraints over t . For example, the ODE x ′ + x = √ t − 1 is defined only when t ≥ 1, so we cannot look at solution in a domain which is larger than [0 , + ∞ ). ◮ or because for some other reason we want to restrict to a specific domain, e.g. quite arbitrary we may wish to look at solutions only over the domain (1 , 5) for the previous equation. ◮ A domain is maximal for a given solution of an ODE when we cannot get the same solution over a larger domain of R . For example, x ( t ) = e t is a solution of x ′ = x in the domain (0 , + ∞ ) but this domain is not maximal because x ( t ) = e t can be defined over all R , and it is still a solution of x ′ = x over R . Hence R is the maximal domain for this solution. See exercise 1 for a less trivial example. 11/33

Differential equations and systems of differential equations Initial Value Problems (IVP) ◮ Most of the times, an ODE has an infinite number of solutions (for example x ′ = x admits all functions x ( t ) = λ e t with λ ∈ R a solutions. ◮ But most of the times also, one gets a unique solution when fixing the values and derivatives of x at a specific t 0 . The ODE, together with these constraints, form an initial value problem (IVP). Usually the number of constraints one has to fix must correspond to the order of the ODE. ◮ examples (to check as an exercise) � x ′ = x ◮ The IVP has unique solution x ( t ) = e t − 1 . x (1) = 1 x ′′ = 2 x ′ − x   ◮ The IVP has unique solution x ( t ) = (1 − 2 t ) e t . x (0) = 1 x ′ (0) = − 1  12/33

Differential equations and systems of differential equations Numerical solutions of ODEs ◮ Euler scheme is the simplest and most well-known method to solve an IVP numerically. Consider any first order ODE in explicit form, with an initial condition: � x ′ = f ( t , x ) x ( t 0 ) = x 0 The Euler scheme simply approximates x ′ ( t ) by a difference quotient: x ′ ( t ) ≈ x ( t + h ) − x ( t ) , h which yields x ( t + h ) ≈ x ( t ) + hf ( t , x ) . Thus, starting from position x 0 at t 0 , and fixing a stepsize h > 0, one can compute approximations x 1 ≈ x ( t 0 + h ), x 2 ≈ x ( t 0 + 2 h ), etc. iteratively using the rule x k +1 = x k + h f ( t 0 + kh , x k ) . 13/33

Differential equations and systems of differential equations Numerical solutions of ODEs ◮ example : Consider the ODE (1 + x 2 ) x ′ = sin( t 2 ). It can be put directly in explicit form x ′ = f ( t , x ) with f ( t , x ) = sin( t 2 ) 1+ x 2 . Here is a plot of 100 iterations of the Euler scheme for t 0 = 0, x 0 = 1, with stepsize h = 0 . 05: ◮ The Euler method is very simple, but not very accurate. One has to decrease the stepsize h , hence to perform more iterations, in order to get precise results. Numerical solvers such as the function ode45 of Matlab usually implement more complex schemes called Runge-Kutta methods that allow to get accurate results with less iterations (see exercise 4). 14/33