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1 Math 220, Differential Equations Professor Charles S.C. Lin Office: 528 SEO, Phone: Office Hours: MWF 2:00 p.m. & by appointments address:

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2 Teaching Assistant Mr. Diego Dominici Office: 607 SEO Office Hours: ? Phone:

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3 Please check: ml for Syllabus, assignments, etc... for interactive CD

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4 Differential Equation: Classifications Ordinary differential equations, order! Partial differential equations Linear equations: i.e. linear in the dependent variable(s). Nonlinear differential equations: not linear For example:

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5 Explicit solution and Implicit solution If a function satisfies a differential equation, for example: Such a function, defined explicitly as a function of independent variable x is called an explicit solution. On the other hand, the equation

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6 Given by the following: The equation is said to defined an implicit solution of the equation (*) above.

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7 In fact, there are many solutions to a D.E such as (*) above. To find a solution passing through a specific point in xy-plane, we need to impose a condition, known as : initial value, i.e. y(x 0 ) = y 0. This is known as the initial value problem. We shall assume that the function f(x,y) is sufficiently smooth, that a solution always exists. Namely: Theorem (Existence and uniqueness): The I.V.P. always has a unique solution in a rectangle containing the point (x 0, y 0 ), if f and f x are continuous there.

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8 Consider the first order D. E. the equation specifies a slope at each point in the xy-plane where f is defined. It gives the direction that a solution to the equation must have at each point. Direction Fields

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9 A plot of short line segments drawn at various points in the xy- plane showing the slope of the solution curve this is called a “ direction field ” for the differential equation. The direction field gives us the “ flow of solutions ”.

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10 Example For the equation Using Maple with(DEtools); eq:=diff(y(t),t)=t^2-y(t); DEplot(eq,y(t),t=-5..5,y=-5..5,arrows=slim)

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11 Using Maple program, we have the following graph:

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12 Another example: Consider the logistic equation for the population of a certain species: using maple, we write in commands: eq:=diff(p(t),t)=3*p(t)-2*p(t)^2; DEplot(eq,p(t), t=0..5,p=0..5,arrows=slim); and get

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13 like this: Its direction field

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14 Consider the differential equation (*) dy/dx = f(x,y). The set of points in the xy-plane where all the solutions have the same slope dy/dx; i.e. the level curves for the function f(x,y) are called the isoclines for the D. E. (*). This is the family of curves f(x,y) = C. This gives us a way to draw direction field. The Method of Isoclines

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15 Example For the differential equation f(x,y) = x + y, and the set of points where: x + y = c, are straight lines with slope (-1). We can now draw the isoclines for the D. E. and the solution passing through a given initial point can also be drawn.

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16 Let us graph the isoclines of f(x,y) = x + y. and compare it to the direction field of it, we see

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17 Maple example Let us consider the IVP for y’ = x^2 - y, with three sets of initial points: [0,-1], [0,0] and [0,2]. What will be the corresponding solutions?

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18 Separable Equation Given a differential equation If the function f(x,y) can be written as a product of two functions g(x) and h(y), i.e. f(x,y) = g(x) h(y), then the differential eq. is called separable.

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19 Example The equation is separable, since

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20 Method for solving separable equation Separable equation can be solved easily, Rewrite the equation:

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21 Example Consider the initial value problem

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22 Using Maple: we can solve the IVP with the following Maple commands. ODE:=diff(y(x),x)=(y(x)-1)/(x-3); IC:=y(-1)=0; IVP:={ODE,IC}; GSOLN:=dsolve(ODE,y(x)); Then use the IC to find the arbitrary constant.

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23 Linear Equations We shall study how one can solve a first order linear differential equation of the form: We first rewrite the above equation in the so called “standard form”:

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24 Integration Factor Suppose we multiply a function (x) to the above equation, we get: Is it possible for us to find (x) such that the left hand side ?

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25 Since We see that this can be done, if

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26 In this case, we can solve it by integration. Note that:

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27 Examples: Consider the D.E. Solution: Another example: solve the following initial value problem:

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28 Consider a large tank holding 1000 L of water into which a brine solution of salt begins to flow at a constant rate of 6 L/min. The solution inside the tank is kept well stirred and is flowing out of the tank at a rate of 6 L/min. If the concentration of salt in the brine entering the tank is 1 kg/L, determine when the concentration of salt in the tank will reach 0.5 kg/L Application: Mixing Problems (Compartmental Analysis)

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29 Let x(t) be the mass of salt in the tank at time t. The rate at which salt enters the tank is equal to “input rate - output rate”. Thus

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30 The equation is separable We can solve it easily, using the initial condition, we get

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31 Suppose P(x) and Q(x) are continuous on the interval (a,b) that contains the point x 0. Then the initial value problem: y + P(x)y = Q(x), y(x 0 )=y 0 for any given y 0. has a unique solution on (a,b). Existence and Uniqueness Theorem

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32 Application to Population Growth If we assume that the growth rate of a population is proportional to the population present, then it leads to a D.E.: Let p(t) be the population at time t. Let k > 0 be the proportionality constant for the growth rate and let p 0 be the population at time t = 0. Then a mathematical model for a population could be:

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33 This can be solved easily. Example: In 1790 the population of the United States was 3.93 million, and in 1890 it was million. Estimate the U.S. population as a function of time.

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34 The study of motion of objects and the effect of forces acting on those objects is called Mechanics. A model for Newtonian mechanics is based on Newton’s laws of motion: Let us consider an example: An object of mass m is given an initial velocity of v 0 and allowed to fall under the influence of gravity. Assuming the gravitational force is constant and the force due to air resistance is proportional to the velocity of the object. Determine the equation of motion for this object. Application to Newtonian Mechanics

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35 Since the total force acting on the object is F = F G - F A = mg - k v(t). And according to Newton’s 2nd law of motion, F = m a, we see that m a = mg - k v. Let x(t) be the position function of the object at time t, and v(t) = dx/dt, a = dv/dt. Solution

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36 Equation of motion can be rewritten as: The following separable initial value problem. We can solve the equation easily, and obtain:

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37 Suppose that at t = 0, the object is x 0 units above the ground, i.e. x(0) = x 0. Then for the position function x(t), we have the following I.V.P. This can be solved easily. Now, to find the position function x(t)

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38 We obtain: The equation of motion:

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39 We shall now consider linear 2nd order equations of the form: Linear Differential Operators (4.2)

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40 is the equation Homogeneous equation associated with ( )

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41 We have L[ y 1 + y 2 ]= L[ y 1 ]+ L[y 2 ], for any constants and , and any twice differentiable functions y 1 and y 2. Theroem1. If y 1 and y 2 are solutions of the homogeneous equation (HE): y +py+qy=0, then any linear combination y 1 + y 2 of y 1 and y 2 is also a solution of (HE). Remark on linearity of the operator L, and linear combinations of solutions to homogeneous equation.

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42 L[y] = y + 4y + 3y, We use the convention Dy = y, D 2 y = y , D n y = y (n), and rewrite L[y] = D 2 y + 4Dy +3y, or symbolically, L = D 2 + 4D +3. Since formerly D 2 + 4D +3 = (D + 3)(D + 1), we see that: L[y] = (D + 3)(D + 1)[y]. The solutions for L[y] = 0 are y 1 = e -x, and y 2 = e -3x. Consider an example

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43 Theorem 2. Let p(x), q(x) and g(x) be continuous on an interval (a,b), and x 0 (a,b). Then the I.V.P. Existence and Uniqueness of 2nd order equation

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44 Let us first define the notion of the Wronskian of two differentiable functions y 1 and y 2. The function Fundamental Solutions of Homogeneous Equations

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45 Fundamental solution set A pair of solutions [y 1, y 2 ] of L[y] = 0, on (a,b) where L[y] = y +py+qy is called a fundamental solution set, if W[y 1, y 2 ](x 0 ) 0 for some x 0 (a,b). A simple example: Consider L[y] = y +9y. It is easily checked that y 1 = cos 3x and y 2 = sin 3x are solutions of L[y] = 0. Since the corresponding Wronskian W[y 1, y 2 ](x) = 3 0, thus {cos 3x, sin 3x} forms a fundamental solution set to the homogenenous eq: y + 9y = 0. We see that any linear combination c 1 y 1 + c 2 y 2 also satisfies L[y] = 0. This is known as a general solution

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46 Linear Independence, Fundamental set and Wronskian Theorem. Let y 1 and y 2 be solutions to the equation y + py + qy = 0 on (a,b). Then the following statements are equivalent: (A) {y 1, y 2 } is a fundamental solution set on (a,b). (B) y 1 and y 2 are linearly independent on (a,b). (C) The W[y 1, y 2 ](x) is never zero on (a,b). For the proof, we need some linear algebra, i.e. Linearly dependent vectors,uniqueness theorem etc...

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47 Reminder First Hour Exam: Date: June 15 (Friday) Room: TBA

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48 Homogeneous Linear Equations With Constant Coefficients Recall: For equations of the form ay + by + cy = 0, by subsituting y = e r x, we obtain the auxiliary eq: ar 2 + br + c = 0. If r 1 and r 2 are two distinct roots, then a general solution is of the form y = c 1 exp(r 1 x)+ c 2 exp(r 2 x), where c 1 and c 2 are arbitrary constants.

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49 Repeated Roots If in the above equation, r 1 = r 2 = r, then a general solution is of the form y = c 1 exp(rx) + c 2 x exp(rx), Example: consider the D.E. : y + 4y´ + 4 = 0. Its auxiliary equation is: r 2 + 4r + 4 = 0, hence r = -2 is a double root, the general solution is y = c 1 e -2x + c 2 x e -2x,

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50 Cauchy-Euler Equations If an equation is of the form: ax 2 y + bxy´ + cy = h(x), a, b, c are constants, then by letting x = e t, we transform the original equation into:(with t as the independent variable), ay + (b-a)y´ + cy = h(e t ). An equation with constant coefficients. Hence can be solved by the method of constant coefficients. The equation above is known as a Cauchy-Euler Equation.

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51 Reduction of order We know,in general, a second order linear differential equation has two linearly independent solutions. If we already have one solution, how can we find the other one? Let f(x) be a solution to y + p(x)y´ + q(x)y = 0. We will try to find another solution of the form y(x) = v(x)f(x), with v(x) a non-constant function. Formerly, we have y ´ = v´f + vf ´, and y = … set w = v ´, etc…, we obtain a separable eq. in w. Finally find v from w by integration.

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52 Example Given f(x) = x -1 is a solution to x 2 y - 2xy´ -4y = 0, x > 0; find a second linearly independent solution. First write the D.E. in standard form. Next compute v. Finally, 2nd independent solution is y = v f.

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53 Auxiliary Eq. With Complex Roots If the auxiliary equation of a linear 2nd order D. E. with constant coefficents: ar 2 + br + c = 0, has complex roots, (when b 2 - 4ac < 0 ). i.e. r 1 = + i and r 2 = - i , where and are real numbers, then the solutions are y 1 = e ( + i )x, and y 2 = e ( - i )x. Since we know that e i x = cos x + i sin x, we simply take y 1 = e x cos x, and y 2 = e x sin x as the two linearly independent solutions.

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54 And the general solution is of the form y(x) = c 1 e x cos x + c 2 e x sin x, where c 1 and c 2 are arbitrary constants. Remark about complex solution: z(x) = u(x) + iv(x) to L[z] = 0 and the fact that in this case, we also have L[u] = 0 and L[v] = 0. Thus the real part and the imaginary part of a complex solution to L[y] = 0 are also solutions of L[y] = 0.

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55 Example Consider the D.E.

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56 Nonhomogeneous Equation And the the method of Superposition Let L be a linear operator of 2nd order, i.e. L= D 2 + pD + q, and g 0. The equation: L[y] = g, is called a Nonhomogeneous eq. We wish to solve the equation L[y] = g, using a particular solution to L[y] = g, and a fundamental solution set to L[y] = 0. First let me introduce the concept of the method of superposition.

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57 Theorem: Let y 1 be a solution to the equation L[y] = g 1, and let y 2 be a solution to the equation L[y] = g 2, where g 1 and g 2 are two functions. Then for any two constants c 1 and c 2, the linear combination c 1 y 1 + c 2 y 2 is a solution to the equation L[y] = c 1 g 1 + c 2 g 2. (This is known as the Superposition principle).

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58 Proof

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59 Representation Theorem of L[y] = g. Theorem: Let y p (x) be a particular solution to the nonhomogeneous equation (*) L[y] = g(x), where L[y]= y + p(x) y + q(x) y, on the interval (a,b) and let y 1 (x) and y 2 (x) be a fundamental solution set of L[y] = 0 on the interval (a,b). Then every solution of (*) can be written in the form (**) y(x) = y p (x) + c 1 y 1 (x) +c 2 y 2 (x). This is known as the general solution to (*).

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60 Example Given that y p (x) = x 2 is a particular solution of the equation: (*) y - y = 2 - x 2, find a general solution of (*). Note the auxiliary equation is r = 0. It follows that a general solution of (*) is of the form y = x 2 + c 1 e -x + c 2 e x.

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61 Superposition Principle & the Method of Undetermined coefficients. Example: Find a general solution to the D.E. Step 1: We first consider the associated homogenous equation:

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62 Step 2: Find particular solution to the Non-homogenous equation using the Superposition Principle There are “2” equations:

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63 To find a particular solution to each of the above equations We use the method of undetermined coefficients, that is: for the first equation, we try y p = ax 2 + bx + c, and For the second equation, we try y p = Ae x. If any term in the trial expression for y p is a solution to the corresponding homogeneous equation, then we replace y p by x y p, etc…. See table 4.1 on Page 208 of your book.

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64 Next we present a more general method, known as: The method of variation of parameters,

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65 Consider the non-homogenous linear second order differential equation : The Method of Variation of Parameters

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66 two equations (by avoiding 2nd order derivatives for the unknows and from L[y p ]=g): Where v 1 and v 2 are functions to be determined. We obtain:

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67 two equations (by avoiding 2nd order derivatives for the unknows and from L[y p ]=g): Where v 1 and v 2 are functions to be determined. We obtain:

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68 Finally, solution is found by integration. Example:

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