Ordinary Differential Equations

Transcription

1 Ordinary Differential Equations (MA102 Mathematics II) Shyamashree Upadhyay IIT Guwahati Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 1 / 1

2 Books Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 2 / 1

3 Differential Equations : Formal definition An equation involving derivatives of one or more dependent variables with respect to one or more independent variables is called a differential equation. Examples (i) dy = x. dx (ii) d2 y + dy = sin x. dx 2 dx (iii) 2 u + 2 u = 0. x 2 y 2 Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 3 / 1

4 Classification by type: Ordinary and Partial If a differential equation contains only ordinary derivatives of one or more dependent variables with respect to a single independent variable, then it is called an ordinary differential equation (ODE). Examples (i) dy = x dx, (ii) d2 y + dy = sin x. dx 2 dx If a differential equation contains partial derivatives of one or more dependent variables with respect to two or more independent variables, then it is called a partial differential equation (PDE). Examples (i) 2 u + 2 u = 0 x 2 y 2 (ii) u + u + v = x y y xy2. Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 4 / 1

5 Order and Degree of an ODE In this course, we will deal with ODE s only. The order of a differential equation (ODE or PDE) is the order of the highest derivative in the equation. Example The order of d3 y dx 3 + 5x dy dx = yex is 3. The degree of a differential equation is the power of the highest order derivative occuring in the differential equation (after rationalizing the differential equation as far as the derivatives are concerned). Example = dy dx 1 + ( d2 y ) 1 dy dx 2 2. After rationalizing, we get ( dx 1)2 = d2 y. dx 2 This equation has order 2 and degree 1. Note that the degree is not 2! Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 5 / 1

6 General form of ODEs The general form of a n-th order ODE is given by F(x, y, y, y,..., y (n) ) = 0. where F is a real valued function of the n + 2 many variables x, y, y,..., y (n). If the equation F(x, y, y, y,..., y (n) ) = 0 can be expressed in the form d n y dx n = f (x, y, y,..., y (n 1) ) then this new equation is called the normal form of the previous equation. Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 6 / 1

7 Linear and non-linear ODEs The differential equation F(x, y, y, y,..., y (n) ) = 0 is said to be linear if F is linear in y, y,..., y (n). In other words, an n-th order linear differential equation can be written as a 0 (x) dn y dx n + a 1(x) dn 1 y dx n a n(x)y = g(x). In a linear differential equation F(x, y, y, y,..., y (n) ) = 0, The dependent variable and its derivatives occur in first degree. For instance, expressions like y 2, xy 2, ( dy dx )2 should not appear in the differentail equation. No product of the dependent variable and its derivatives should occur in the differential equation. For instance, expressions like y dy should not appear in the dx differentail equation. No transcendental functions of the dependent variable and its derivatives should occur in the differential equation. For instance, expressions like sin y, e y, log y should not appear in the equation. Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 7 / 1

8 Solution of an ODE A function φ defined on an interval I and possessing at least n derivatives, which when substituted into an n-th order ODE reduces the equation to an identity, is said to be a solution of the ODE on the interval I. In other words, a solution of an n-th order ODE F(x, y, y, y,..., y (n) ) = 0 is a function φ that possesses at least n derivatives and for which for all x I. F(x, φ(x),..., φ (n) (x)) = 0 Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 8 / 1

9 Particular solution and Singular Solution Any solution to an n-th order ODE involving n arbitrary constants is called a general solution of the ODE. For example, y = x 2 + c is a general solution of the ODE dy dx = 2x. If we assign some particular values to these n arbitrary constants, then the resulting solution is called a particular solution of the ODE. For example, y = x 2 is a particular solution of the ODE dy = 2x. This solution is obtained dx from the general solution y = x 2 + c by assigning the value 0 to the constant c. If a solution to an ODE cannot be obtained from a general solution, then it is called a singular solution. Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 9 / 1

10 Explicit and Implicit solutions Consider the n-th order ODE F(x, y, y,..., y (n) ) = 0 A solution of the above ODE in which the dependent variable y is expressed solely in terms of the independent variable x and constants, is said to be an explicit solution of the ODE. The relation G(x, y) = 0 is said to be an implicit solution of the above ODE if it gives rise to at least one explicit solution of the above ODE. Example y = x 2 + c is an explicit solution of the ODE dy dx = 2x. And x2 + y 2 = 1 is an implicit solution of the ODE y dy dx + x = 0 on the interval [ 1, 1] because the equation x2 + y 2 = 1 gives rise to 2 explicit solutions of the ODE y dy dx + x = 0 on the interval [ 1, 1], namely, y = 1 x 2 and y = 1 x 2. Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 10 / 1

11 Initial value problem (IVP) A problem to the type On some interval I containing x 0 Find a solution y(x) of the ODE F(x, y, y,..., y (n) ) = 0 sub ject to the conditions y(x 0 ) = a 0, y (x 0 ) = a 1,..., y (n 1) (x 0 ) = a n 1 where a 0, a 1,..., a n 1 are arbitrarily specified real constants, is called an Initial value problem or IVP. The conditions y(x 0 ) = a 0, y (x 0 ) = a 1,..., y (n 1) (x 0 ) = a n 1 are called initial conditions. The above problem is also called an n-th order initial value problem. Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 11 / 1

12 A well-posed IVP Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 12 / 1

13 Peano Theorem Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 13 / 1

14 Does every IVP has a solution? Consider the IVP x dy = 4y sub ject to y(0) = 1 dx It can be easily verified that y = cx 4 is a general solution of the ODE x dy = 4y, but this dx general solution does not satisfy the initial condition y(0) = 1. Hence the above IVP has no solution. Question: If a IVP has a solution, is it unique? Answer: No! Consider the following IVP: Solve dy dx = y subject to y(0) = 0. Observe that y 0 and y = { different solutions of the same IVP! x 2 4 i f x 0 0 i f x < 0 are two Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 14 / 1

15 An IVP having a unique solution Consier the IVP dy = 2x sub ject to y(0) = 1 dx The general solution of the ODE dy dx = 2x is given by y = x2 + c. Putting, the initial condition y(0) = 1, we get c = 1. Hence y = x is a unique solution of the above IVP. Two fundamental questions: (i) When does an IVP has a solution? (ii) If an IVP has a solution, is it unique or when is it unique? These 2 questions are answered by the existence and uniqueness theorem, which we will state now. Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 15 / 1

16 Lipschitz Condition Let f (x, y) be a real valued function defined on a bounded domain D. We say that f satisfies Lipschitz condition in the (second) variable y with a Lipschitz constant K if f (x, y 1 ) f (x, y 2 ) K y 1 y 2 for any (x, y 1 ) and (x, y 2 ) in D. Example: Let D = {(x, y) R 2 : 0 < x < 2 and 10 < y < 10}. Let f (x, y) = x 2 sin(y) for (x, y) D. Then, for any (x, y 1 ) and (x, y 2 ) in D we have f (x, y 1 ) f (x, y 2 ) = x 2 sin(y 1 ) sin(y 2 ) x 2 y 1 y 2 4 y 1 y 2. Therefore f satisfies Lipschitz condition on the bounded domain D in the variable y. Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 16 / 1

17 Existence and Uniqueness of Solutions to IVPs: Picard Theorem Let R = {(x, y) R 2 : x x 0 a, y y 0 b, a > 0, b > 0} be a closed and bounded rectangle containing the point (x 0, y 0 ). The initial value problem ( ) is given by ODE y = f (x, y) on R, Initial Condition: y(x 0 ) = y 0. Theorem Picard Theorem: If f (x, y) is continuous and satisfies Lipschitz condition in the variable y with Lipschitz constant K on the closed rectangle R, then there exists a unique solution y(x) to the IVP ( ) on the interval x x 0 h where h = min(a, b/m) and f (x, y) M for all (x, y) R. Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 17 / 1

18 Example: IVP with a unique solution ODE: y = 4y for x R, Initial Condition: y(0) = 1. The solution to this IVP is y(x) = e 4x for x R. Further it is a unique solution to this IVP. Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 18 / 1

19 Example: IVP having infinitely many solutions The solutions to this IVP are: ODE: y = 3 y 2/3 for x R, Initial Condition: y(0) = 0. y c (x) = { 0 if x c (x c) 3 if x c where c 0. For each real number c 0, we have a solution y c (x) to the IVP. Therefore, this IVP has infinitely many solutions. Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 19 / 1

20 Theorem Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 20 / 1

21 Theorem Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 21 / 1

22 Method to find the solution to the IVP provided by Picard Theorem (Method of Successive Approximations) Step 1: Set y 0 (x) = y 0 for all x R. Step 2: Iterative Step Compute x y n (x) = y 0 (x) + f (s, y n 1 (s)) ds for n = 1, 2, x 0 This y n (x) is called the Picard Successive Approximation or Picard Iterate. Step 2: Limit of Iterates y(x) := lim n y n (x) for x I = [x 0 h, x 0 + h]. Under the hypothesis of Picard s theorem, {y n (x)} converges uniformly on the interval I and the limit function y(x) is the unique solution of the given IVP ( ) in I. Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 22 / 1

23 Worked out Example Solve y = 2y with y(0) = 1 by the method of successive approximations. Here f (x, y) = 2y for (x, y) R 2. Step 1: Initial Approximation Set y 0 (x) = y 0 = 1 for all x R. Step 2: Computing Successive Approximations y 1 (x) = y 0 (x) + x y 2 (x) = y 0 (x) + x x 0 f (s, y 0 (s)) ds = 1 + x 0 f (s, y 1 (s)) ds = 1 + x y 3 (x) = y 0 (x)+ f (s, y 2 (s)) ds = 1+ x 0 x 0 x 0 x 0 2 ds = 1 + 2x. 2(1 + 2s) ds = 1 + 2x + 2x 2. 2(1+2s+2s 2 ) ds = 1+2x+2x 2 + 4x3 3. Step 3: Limit of Successive Approximations (if possible compute) y n (x) = n (2x) k e 2x =: y(x) for x R as n. k! k=0 Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 23 / 1

24 Corollary to Picard Theorem Let R = {(x, y) R 2 : x x 0 a, y y 0 b, a > 0, b > 0} be a closed and bounded rectangle containing the point (x 0, y 0 ). The initial value problem ( ) is given by ODE y = f (x, y) on R, Initial Condition: y(x 0 ) = y 0. Corollary to Picard Theorem: If f (x, y) is continuous and f is continuous y on the closed rectangle R, then there exists a unique solution y(x) to the IVP ( ) on the interval x x 0 h where h = min(a, b/m) and f (x, y) M for all (x, y) R. If f is continuous on R then by applying the MVT for the variable y, it y follows that f satisfies Lipschitz condition in the variable y with Lipschitz constant K on R and then Picard theorem completes the remaining proof. Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 24 / 1

25 Direction field Suppose we have the differential equation dy = f (x, y). dx Then the direction field corresponding to this differential equation will be a collection of line segments drawn at various points (x, y) of a rectangular grid (xy-plane) such that at each point (x, y), the slope of the line segment equals the value f (x, y). To see how to plot an approximate solution to a differential equation of the form dy = f (x, y) dx passing through a point (x 0, y 0 ), please look at the black board. Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 25 / 1

26 Interpretation Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 26 / 1

27 Interpretation contd... Shyamashree Upadhyay ( IIT Guwahati ) Ordinary Differential Equations 27 / 1