Lecture 2. Introduction to Differential Equations. Roman Kitsela. October 1, Roman Kitsela Lecture 2 October 1, / 25

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1 Lecture 2 Introduction to Differential Equations Roman Kitsela October 1, 2018 Roman Kitsela Lecture 2 October 1, / 25

2 Quick announcements URL for the class website: URL for MATLAB website: This has been updated recently with the correct dates for Fall The MATLAB TAs should be your main point of contact for MATLAB related questions. URL to sign up to Piazza: piazza.com/ucsd/fall2018/math20dfall2018dsections Roman Kitsela Lecture 2 October 1, / 25

3 Today s lecture Section Background First examples of differential equations Basic terminology: Dependent vs Independent variables Ordinary D.E. vs partial D.E. Order of a D.E. Linear vs non-linear D.E. Section Solutions and IVPs Explicit vs Implicit solutions Initial value problems Existence and uniqueness theorem: Formal statement Examples (how to use the theorem). Roman Kitsela Lecture 2 October 1, / 25

4 Section Modelling and differential equations Scientists and engineers use mathematical models to attempt to understand (and predict) physical phenomena. These models will often involve rates of change which are expressed as derivates of some (possibly unknown) function. Equations involving derivatives are called differential equations. Roman Kitsela Lecture 2 October 1, / 25

5 Basic examples Example (Body falling under the force of gravity) Object released from some height falls under the force of gravity. Model: Newton s 2nd Law Differential Equation: F = m a m d 2 h dt 2 = mg Roman Kitsela Lecture 2 October 1, / 25

6 Basic examples Example (Radioactive decay) A radioactive substance experiences radioactive decay. Model: Rate of decay is proportional to the amount of radioactive substance present. Differential Equation: da dt = ka, k > 0 Roman Kitsela Lecture 2 October 1, / 25

7 What these examples tell us about differential equations 1 Solution to differential equations are functions. 2 Integration is an important tool in solving differential equations. 3 We cannot expect to get a unique solution unless given more information. This information will be called the initial conditions of the problem (such as the initial position, velocity, acceleration etc...) Roman Kitsela Lecture 2 October 1, / 25

8 More examples Example (Compound interest) dp dt = r P, t in years 100 Example (Charge in an electric circuit) L d 2 q dt 2 + R dq dt + 1 C q = E(t) Example (Modelling rate of learning) Example (Wave propagation) dy/dt 2p y 3/2 = (1 y) 3/2 n 2 u t 2 c2 2 u x 2 = 0 Roman Kitsela Lecture 2 October 1, / 25

9 Basic Terminology To facilitate the study of differential equations we need to know some common words used to describe differential equations: Dependent vs Independent variables Ordinary D.E. vs partial D.E. Order of a D.E. Linear vs non-linear D.E. Roman Kitsela Lecture 2 October 1, / 25

10 Dependent vs independent variables Basic example: y(x) (i.e. y is a function of x) Here y depends on x and so y is the dependent variable, x is the independent variable. Differential equations involve derivatives such as dy dx dq dt dp dt 2 u x t Dependent variables: y, q, P, u Independent variables: x, t, t, x and t Conclusion: Top of the derivative = dependent variable Bottom of the derivative = independent variable(s) Roman Kitsela Lecture 2 October 1, / 25

11 Ordinary vs partial differential equations A differential equation involving only ordinary derivatives with respect to a single (independent) variable is called an ordinary differential equation. Example: da dt = ka A differential equation involving partial derivatives with respect to more than one (independent) variables is called an partial differential equation. Example: 2 u x u y 2 = 0 Roman Kitsela Lecture 2 October 1, / 25

12 Order of a differential equation The order of a differential equation is the order of the highest-order derivatives present in the equation. Example: da dt = ka and 2 u x u y 2 = 0 First-order and Second-order differential equations respectively. Roman Kitsela Lecture 2 October 1, / 25

13 Linear vs nonlinear differential equaions Definition A differential equation is called linear if it has the format a n (x) d n y dx n + a n 1(x) d n 1 y dx n a 1(x) dy dx + a 0(x)y = F (x) Linear dy dx + y = x 2 1 dy dt = yt 1 t cos(x) dy dx sin(x)y = e2x cos(x) dr dθ = θ2 Non-linear dy dx y = x 2 1 dy dt = yt y t dy dx sin(y)x = e2x ( dr dθ ) 2 = θ Roman Kitsela Lecture 2 October 1, / 25

14 Classifying differential equations Equation Dep. var? Indep. var? ODE or PDE? Order? Linear? m d2 h dt 2 = mg h t ODE 2 da dt = ka A t ODE 1 d 3 y dx 3 + y 3 = 0 y x ODE 3 dp dt = r 100 P P t ODE 1 L d2 q + R dq dt 2 dt + 1 C q = E(t) q t ODE 2 dy/dt = 2p y 3/2 (1 y) 3/2 n y t ODE 1 2 u c 2 2 u = 0 u x, t PDE 2 t 2 x 2 Roman Kitsela Lecture 2 October 1, / 25

15 Section Explicit solutions A general nth-order differential equation (with x independent and y dependent) can be expressed in the following form: F ( x, y, dy dx, d 2 y dx 2,, d n ) y dx n, = 0 (1) We assume that this equation holds for all x in some open interval I. Sometimes we can isolate the highest order derivative and rewrite (4) as: d n ( y dx n = f x, y, dy dx, d 2 y dx 2,, d n 1 ) y dx n 1 (2) Definition A function φ(x) that satisfies (4) (or (2)) for all x in I when substituted for y is called an explicit solution to (4) (or (2)). Roman Kitsela Lecture 2 October 1, / 25

16 Reminder: Open intervals Quick reminder: Open interval are usually written as (a, b) and contain all x such that: a < x < b (Note: a and b are allowed to be and respectively) Examples: ( 3, π) contains all x such that 3 < x < π (0, ) contains all x > 0 (i.e. all positive x) (, 0) contains all negative x (, 1) contains all negative x < 1 (, ) contains all (real) x Roman Kitsela Lecture 2 October 1, / 25

17 Explicit solutions - An example Example (Example 1 from textbook) Show that φ(x) = x 2 x 1 is an explicit solution to the linear equation but ψ(x) = x 3 is not. Important point: d 2 y dx 2 2 x 2 y = 0 (3) Explicit solutions need to be defined on an open interval, not just a single value of x. Roman Kitsela Lecture 2 October 1, / 25

18 Implicit solutions We are still discussing solutions to a general nth-order differential equation ( F x, y, dy dx, d 2 y dx 2,, d n ) y dx n, = 0 (4) Definition A relation G(x, y) = 0 is said to be an implicit solution to (4) on the interval I if it defines one or more explicit solutions on I. Important point: We will not always be able to rewrite implicit solutions as y = φ(x) for some φ(x). In these cases we use implicit differentiation. Roman Kitsela Lecture 2 October 1, / 25

19 Implicit solutions - examples Example (Example 3 from textbook) Show that the relation y 2 x = 0 implicitly defines a solution to the nonlinear equation on the interval (2, ). dy dx = 3x 2 2y Roman Kitsela Lecture 2 October 1, / 25

20 Implicit solutions - examples Example (Example 4 from textbook) Show that the relation x + y + e xy = 0 is an implicit solution to the nonlinear equation (1 + xe xy ) dy dx yexy = 0 Roman Kitsela Lecture 2 October 1, / 25

21 Initial value problems In general solving an nth-order differential equation introduces n unknowns (often as constants of integration). So a differential equation with no initial conditions will not have a unique solution. A differential equation together with (sufficiently many) initial conditions is called an initial value problem. Example (From practice final) Solve the initial value problem: dy dx y x = xex, y(1) = e 1 Roman Kitsela Lecture 2 October 1, / 25

22 The higher the order of the differential equation, the more initial conditions we need to determine an exact solution... Example (Example 6 from textbook) Show that φ(x) = sin(x) cos(x) is a solution to the initial value problem d 2 y dx 2 + y = 0; y(0) = 1, dy dx (0) = 1 Note: In chapter 4 we will learn how to actually solve these types of problems. Without the initial conditions y(0) = 1, y (0) = 1 we would only be able to get the general solution: φ(x) = A cos(x) + B sin(x) where A and B are unknown constants. Roman Kitsela Lecture 2 October 1, / 25

23 Existence and uniqueness theorem for first-order IVPs Theorem Consider the initial value problem If f and f y dy dx = f (x, y), y(x 0) = y 0 (5) are continuous functions in some rectangle R = {(x, y) : a < x < b, c < y < d} that contains (x 0, y 0 ), then the initial value problem (5) has a unique solution φ(x) in some interval x 0 δ < x < x 0 + δ. Translation: If f and f y behave nicely at the initial condition (x 0, y 0 ) then you can guarantee a unique solution to the IVP close to that initial condition (otherwise you cannot apply the theorem). Roman Kitsela Lecture 2 October 1, / 25

24 How to use the existence and uniqueness theorem Example For the initial value problem 3 dy dx = x 2 xy 3, y(1) = 6 does the existence and uniqueness theorem imply the existence of a unique solution? Solution. Calculate: f (x, y) = x 2 xy 3 3 = f y = xy 2 Check continuity: In this case f and f y are both continuous at (1, 6) so we can apply theorem (and deduce the existence of a unique solution). Roman Kitsela Lecture 2 October 1, / 25

25 How to use the existence and uniqueness theorem Example For the initial value problem dy dx = 3y 2/3, y(2) = 0 does the existence and uniqueness theorem imply the existence of a unique solution? Solution. Calculate: f (x, y) = 3y 2/3 = f y = 2y 1/3 Check continuity: In this case f y is not continuous at (2, 0) so we cannot apply the Theorem. Roman Kitsela Lecture 2 October 1, / 25