Solutions of Ordinary Differential Equations

Transcription

1 Solutions of Ordinary Differential Equations Consider the n-th order ODE F(x, y, y!, y!!,...,y ( n) ) = 0 (1). Definition: Let g(x) be a real-valued function defined on a interval I, having the n-th derivative for all x in I. g(x) is called an explicit solution of the equation (1) on the interval I, if: 1) F(x,g(x), g!(x), g!!(x),...,g ( n) (x)) is defined for all x in I ) F(x,g(x), g!(x), g!!(x),...,g ( n) (x)) = 0 for all x in I 1) Verify that the function g(x) = e x is an explicit solution of the equation F(x,y,y,y ) = y + y 6y = 0 We have g (x) = e x and g (x) = 4e x, when we substitute in the equation, we get F(x, g(x), g (x), g (x))= 4e x + e x 6 e x = 0 and defined for all x real. ) Verify that the function h(x) = x 3 defined on the interval (0, 3) is an explicit solution of the equation F(x,y,y ) = 3xy y = 0. Since g (x) = 3 x! 1 3 is defined on the interval (0, 3) and F(x, g(x), g (x)) = 3x 1 3 x! 3! x 3 = x 3! x 3 = 0 and defined for all x in (0, 3) Definition: A relation H(x,y) = 0 is called an implicit solution of the ODE (1) if this relation produces at least one real-valued function g(x) defined on the interval I, such that g(x) is an explicit solution of (1) on I. 1) The relation x + y 4 = 0 is an implicit solution of the equation F(x,y,y ) = x + yy = 0 on the interval (-, ). The relation produces the functions h(x) = y = 4! x and also the function k(x) = y = - 4! x both defined on (-, ). "x Since h!(x) =, then 4 " x F(x,h(x),h (x)) = x + 4! x!x = 0 and defined on (-, ). 4! x ) The relation x + y + 4 = 0 is NOT an implicit solution of the equation F(x,y,y ) = x + yy = 0 on the interval (-,).

2 The relation does not produce a real-valued function on the interval, solving for y we get p(x) = y =!4! x and this function is undefined since - 4 x < 0. If we differentiate the relation implicitly, we get x + yy = 0 or x + yy = 0, that is the equation we want to solve. Then, we say that the relation x + y + 4 = 0 is a formal solution of the differential equation. Remark: A first order ODE can be given by the expression dy dx = f(x,y) Geometric interpretation of the first order ODE A first order ODE dy dx = f(x,y) associates with each point (x 0, y 0 ) in a region D!! a direction m = dy dx = f(x 0,y 0 ). The direction at each point (x 0, y 0 ) is the slope of the tangent line to a curve, with equation g(x,y) = c, passing through the point. The region D with the direction at each point creates what is called a direction field. Solving the differential equation means to find curves whose tangent lines at the point (x 0, y 0 ) has slope m = dy dx = f(x 0,y 0 ). The ODE dy dx = x creates a directional field in!. Every curve y = x + c where c is an arbitrary constant, has a tangent line at the point (x,y) with slope m = x. Remark: We saw in the above example that functions of the form h c (x) = x + c where c is any constant, are solutions of the ODE dy = x. The constant c is called a parameter. dx Then, h c (x) = x + c is a one-parameter family of solutions of the ODE. Every first order ODE has a one-parameter family of solutions. Initial Value Problem Let s consider the following problem: Find a function k(x) that is a solution of the ODE dy = x, such that k(1) = 4. dx This means that: 1- k(x) must satisfy the ODE, so k (x) = x for all x - k(1) = 4 This is called an initial value problem (I.V.P.) and it is denoted by: " dx = x y(1) = 4

3 Since the ODE dy dx = x has a one-parameter family of solutions h c(x) = x + c where c is an arbitrary constant, imposing the condition x =1 then y = 1, we get 4 = (1) + c or c = 3. Therefore, k(x) = x + 3 satisfies the I.V.P.. Remark: For any first order ODE, we denote an I.V.P. by " dx = f(x,y) y(x 0 ) = y 0 The I.V.P. can be extended to ODE of higher order. Consider the problem " d y dx + y = 0 y(0) =!1 y'(0) = 0 % We have to find a solution h(x) such that satisfies the ODE and moreover h(0) = -1 and h (0) = 0. For ODE of order or higher there is another type of problem called boundary-value problem and it is given by " d y dx + y = 0 y(0) = 1 y(!) = 0 % The conditions relate to the two different values of x, 1 and π. Theorem: Basic Existence and Uniqueness Theorem Given the I.V.P. " dx = f(x,y) y(x 0 ) = y 0 If the function f(x,y) and the partial derivative!f (x,y) are continuous in a rectangle R centered at (x 0,y 0 ), R = {(x,y): x 0 - h < x < x 0 + h, y 0 - k < y < y 0 + k}

4 Then, there is a function p(x) defined on the interval (x 0 h, x 0 + h) that is the unique solution of the I.V.P. Examples: 1) Consider the I.V.P. " dy dx =! x y % y(3) = 4 The relation x + y = c is a one-parameter family of solution of the ODE dy dx =! x y. The condition y(3) = 4 implies = 5 = c Solving for y we get to functions: h(x) = 5! x and k(x) = - 5! x both defined on the interval [-5,5]. Since h(3) = 4 and k(3) = -4, the only solution of the I.V.P. is h(x) = 5! x. Notice that f(x,y) =! x!f and y = x are continuous on a rectangle centered at (3,4) and y does not include the x-axis. ) Consider the I.V.P. " dy dx =! x y % y(1) = 0

5 Since f(x,y) =! x "f!!and!! y "y = x y create a rectangle centered at (1,0) where f and!f are NOT continuous along the x-axis, then we cannot the I.V.P. has two solutions h(x) = 1! x and k(x) = - 1! x are continuous. The theorem fails and 3) Consider the I.V.P. Since!f = y " dx = y y(0) = 0 is NOT defined at (0,0), then we cannot create a rectangle centered at (0,0) where both f and!f are continuous, therefore the theorem fails and the I.V.P. has two different solutions, h(x) = x and k(x) = 0.