Math 225 Differential Equations Notes Chapter 1 Michael Muscedere September 9, 2004 1 Introduction 1.1 Background In science and engineering models are used to describe physical phenomena. Often these models yield equations involving the derivatives of unknown functions. The equations are known as differential equations. As an example consider an object in free fall. The following describes it s motion. m d2 h dt = mg Let s try to solve this for the function h(t). d 2 h dt = g dh dt = gt + c h = h(t) = gt2 + c 1 t + c 2 2 If we add additional information we could solve for the constants of integration. If we know (for example) the initial height and initial velocity the h(t) can be uniquely determined. 1
Here is another example of a model of radioactive decay. da dt = ka; k > 0 Here A is that amount of the radioactive material at time t and k is the proportionality constant to the rate of change. We can solve this equation for the unknown function by: Integrate both sides. 1 da = kdt A 1 A da = kdt ln A + C 1 = kt + C 2 A = A(t) = e ln A = e kt e C 1 C 2 = Ce kt Differential equations occur in many subject areas (careers) economics medicine psychology operations research electrical engineering mechanical engineering biology and many more 1.1.1 Electrical Circuits Using Kirchhoff s voltage laws around an electrical circuit gives: L d2 q dt 2 + Rdq dt + 1 C q = E(t) 2
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1.1.2 Gravitational Equilibrium of a Star 1.1.3 Learning 1 r ( d r 2 dr ρ dp dr ) == 4πρG dy dt 2p = y 3/2 (1 y) 3/2 n y represents the learner s skill level as a function of time. 1.1.4 Vibrating String (Wave Equation) d 2 u dt u 2 c2d2 dx = 0 2 where t is time, x is the location on the string, c is the speed of the wave and u is the displacement of the string as a function of time. Definitions Ordinary Differential Equation: A differential equation (DE)containing only ordinary derivatives of the dependent variable with respect to a single independent variable. Partial Differential Equation: A DE involving the partial derivative of the dependent variable w.r.t the independent variable. Order: The order of the highest-order derivative involved in the DE. 4
Definition A Linear DE is one with involves and additive combination of derivatives of the independent variable. a n (x) dn y dx n + a n 1(x) dn 1 y dx n 1 +... + a 1(x) dy dx + a 0(x)y = F (x) Example 1: t 2 d 2 y dt 2 = 0 is linear. Example 2: x 2 d 2 y + y 2 = 0 is not linear since y 2 term is not present in the dx 2 definition. Example 3: d 2 y dx 2 + y = tan(x) is linear Example 4: d 2 y + y dy dy dx 2 dx = x is not linear because of the y dx term. 5
1.2 Solutions and Initial Value Problems The general form of a nth-order differential equation is F (x, y, dy dx dn y d n x ) = 0 we assume the equation holds for all x in an interval I = (a < x < b) were a and/or b may be infinite Definition A function φ(x) that when substituted in the equations above satisfy the equation for all values of the x (independent variable) in the interval I is called and explicit solution to the equation on I Example 5: Show φ(x) = x 2 x 1 is an explicit solution to the linear equation: d 2 y dx 2 2 x 2y = 0 Solution: The function φ(x), φ (x) = 2x+x 2,and φ (x) = 2 2x 3 are all defined for all x 0. Substituting back into the DE we get (2 2x 3 ) 2 x 2(x2 x 1 ) = (2 2x 3 ) (2 2x 3 ) = 0 Question: For what interval is this solution valid? 6
Example 6: Show that for any choice of the constants c 1 and c 2 the function φ(x) = c 1 e x + c 2 e 2x is and explicit solution to the linear equation y y 2y = 0 Solution: Computing φ (x) = c 1 e x +2c 2 e 2x and φ (x) = c 1 e x + 4c 2 e 2x. We substitute back into the DE and get (c 1 e x + 4c 2 e 2x ) ( c 1 e x + 2c 2 e 2x ) 2(c 1 e x + c 2 e 2x ) = (c 1 + c 1 2c 1 )e x + (4c 2 2c 2 2c 2 )e 2x = 0 Question: For what interval is this solution valid? Definition A relation G(x,y)=0 is said to be an implicit solution to DE (on the interval I) if it defines one or more explicit solution on I. Example 7: Show that x + y + e xy = 0 Is an implicit solution to the nonlinear DE equation (1 + xe xy ) dy dx + 1 + yexy = 0 Can we solve for y directly? How do we know a solution exist for any interval I? We apply the Implicit Function Theorem. 7
Theorem 1.1 Implicit Function Theorem Let F be a continuously differentiable, real-valued function defined on in a rectangle R = {(x, y) : a < x < b, c < y < d} and let (x 0, y 0 ) be a point in R for which F (x 0, y 0 ) = 0 and such that F x (x 0,y 0 ) 0 Then there exists an open interval I = (x 0 δ, x 0 + δ) containing y 0 and a unique function f : I R which is continuously differentiable and such that f(y 0 ) = x 0 and for all y I F (f(y), y) = 0 Example 7 continued. Now that we know y is differentiable function of x we can use implicit differentiation and chain rule. or d dx (x + y + exy ) = 1 + dy dx + exy (y + x dy dx ) = 0 (1 + xe xy ) dy dx + 1 + yexy = 0 8
Definition Initial Value Problem Let F (x, y, dy dx dn y d n x ) = 0 be defined as an nth-order differential equation then the IVP is: Find a solution to the nth-order DE on an interval I containing x 0 which also satisfies the n initial conditions y(x 0 ) = y 0 dy dx (x 0) = y 1. dn 1 y d n 1 x = y n 1 where x 0 I and y 0, y 1,..., y n 1 are given constants. 9
Example 8: We know from a previous example that φ(x) = c 1 e x + c 2 e 2x is and explicit solution to the linear equation y y 2y = 0 Determine c 1 and c 2 if the following initial conditions are given: y(0) = 2 y (0) = 3 Solution so φ(0) = c 1 e 0 + c 2 e 0 = 2 φ (0) = c 1 e 0 + 2c 2 e 0 = 3 c 1 + c 2 = 2 c 1 + 2c 2 = 3 adding the two equations we get 3c 2 = 1 which implies c 2 = 1/3 and since c 1 + c 2 = 2 we have c 1 = 7/3. So the solution to the IVP is: φ(x) = (7/3)e x (1/3)e 2x 10
Theorem 1.2 Existence and Uniqueness Theorem Given the IVP dy dx = f(x, y) y(x 0 ) = y 0 assume that f and f y are continuous functions in the rectangle R = {(x, y) : a < x < b, c < y < d} that contains the point (x 0, y 0 ). Then the IVP has a unique solution φ(x) in some interval x 0 δ < x < x 0 + δ where δ > 0. Class Exercise: Everyone draw what that this theorem says using the rectangle R. 11
d y 0 c a x 0 x 0 x 0 + b Example 9: Does the theorem apply for the following IVP? and y(1)=6. dy dx = x2 xy 3 Solution: The function f(x, y) = x 2 xy 3 and f/ y = 3xy 2. Are these continuous functions in a rectangle about the point (1, 6)? Example 10: Does the theorem apply for the following IVP? dy dx = 3y2/3 and y(2)=0. Relax; Solution: The function f(x, y) = 3y 2/3 and f/ y = 2y 1/3. Are these continuous functions in a rectangle about the point (2, 0)? 12
1.3 Directional Fields One technique that is useful in visualizing the solution to a first order differential equation is to sketch the direction field to the equation. Example 11: Let s sketch the direction field for the following DE. dy dx = x2 y 13
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Example 12: Let s sketch the direction field for the following DE. dy dx = 2y Example 13: Let s sketch the direction field for the following DE. dy dx = 3y2/3 Example 14: Can we answer the following questions by sketching the direction field for the logistic ode which models population growth. If the initial population is 3000 what can we say about the limiting population as t? Can a population of 1000 ever decrease to 500? Can a population of 1000 ever increase to 3000? dp dt = p(2 p) Definition Isoclines are a set of points on the xy-plane where all the solutions of a given DE have the same slope. Example 15: Let s sketch the isoclines for the following DE. y = x + y Solution: The isoclines are the curves that satisfy x + y = C or y = x + C Class Exercise: Draw the isoclines for the above ode. 15
Example 16 Let s sketch the isoclines for the following DE. y = x 2 y Solution: The isoclines are the curves that satisfy x 2 y = C or y = x 2 + C 1.4 Method of Euler Consider the following first order IVP. y = f(x, y) y(x 0 ) = y 0 (x 1,y 1 ) (x 2,y 2 ) (x 0,y 0 ) m2 = f(x 1,y 1 ) m3 = f(x 2,y 2 ) m1 = f(x 0,y 0 ) (x 3,y 3 ) 0 x 0 x 1 x 2 x 3 Let h be a fixed positive number called the step size and consider the equally spaced points. x n = x 0 + nh At the point (x 0, y 0 ) the slope of the solution is given by dy/dx = f(x 0, y 0 ). Therefore the tangent line to the solution curve at the initial point (x 0, y 0 ) is: 16
so y y 0 x x 0 = f(x 0, y 0 ) y = y 0 + (x x 0 )f(x 0, y 0 ) Using this tangent line to approximate φ(x), we find for the point x 1 = x 0 + h φ(x 1 ) y 1 := y 0 + (x 0 + h x 0 )f(x 0, y 0 ) = y 0 + hf(x 0, y 0 ) Next, starting at (x 1, y 1 ) we construct the line with the slope given by the direction field at the point (x 1, y 1 ) that is f(x 1, y 1 ) so stepping to x 2 = x 1 + h we get: φ(x 1 ) y 2 := y 1 + (x 1 + h x 1 )f(x 1, y 1 ) = y 1 + hf(x 1, y 1 ) Repeating the process for all x n is called Euler s method which is summarized by: 1. x n+1 = x n + h 2. y n+1 = y n + hf(x n, y n ), n = 0, 1, 2,... Example 17 Use Euler s Method to approximate e. Solution: Consider the IVP: y = y y(0) = 1 We observe the solution to this IVP is e x. so let s apply Euler s method discretizing the interval x (0, 1). We note f(x n, y n ) = y n,x 0 = 0 and y 0 = 1. 17
1. x n+1 = x n + h 2. y n+1 = y n + hy n = (1 + h)y n, n = 0, 1, 2,... N h Approx for φ(1) = e 1 1.0 2.0 2 0.5 2.25 4 0.25 2.44141 8 0.125 2.56578 16 0.0625 2.63793 18