10Ma2aAnNotes1

# 10Ma2aAnNotes1 - Ma2a (analytical) Fall 2010 PART I: FIRST...

This preview shows pages 1–4. Sign up to view the full content.

Ma2a (analytical) Fall 2010 PART I: FIRST ORDER ODEs 1. Introduction 1.1. Fundamental theorem of calculus. The simplest (”trivial”) d.e.: y ( x ) = f ( x ) . Examples: (i) Find all ( maximal ) solutions of the equation y ( x ) = 1 1 + x 2 . Answer: y ( x ) = dx 1 + x 2 = arctan x + C, ( x R ) . (ii) Solve the IVP y ( x ) = 1 1 + x 2 , y (1) = 0 . Answer: y ( x ) = dx 1 + x 2 = arctan x π 4 , ( x R ) . [Set x = 1 in the ”general solution” y ( x ) = arctan x + C .] (iii) Solve the IVP y ( x ) = 1 x , y ( 1) = 0 . Answer: the (maximal) solution is y ( x ) = log( x ) , ( x < 0) . Theorem. Suppose f is a continuous function on some interval I R . Let x 0 I and y 0 R . Then the IVP y = f ( x ) , y ( x 0 ) = y 0 , has ( i ) a unique ( ii ) solution on the whole interval ( iii ) I . The solution is given by the formula ( iv ) y ( x ) = y 0 + x x 0 f ( s ) ds, ( x I ) . Note. Several important concepts: (i) existence of a solution, (ii) uniqueness, (iii) interval/domain of a solution, (iv) solution formula. 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 1.2. First order ODEs. The most general form is F ( x,y,y ) = 0 , where F is a given function of 3 real variables; x is an independent variable; y is the unknown function y = y ( x ); y = dy/dx is the usual ( ordinary ) derivative. Any diﬀerentiable function y ( x ) that satisﬁes the equation (i.e. if we substitute y ( x ) for y , then the equation becomes an identity) is called a solution . Given two numbers x 0 ,y 0 , the corresponding initial value problem (”IVP”) is to ﬁnd a solution y ( x ) such that y ( x 0 ) = y 0 . Important : we always assume that a solution is deﬁned on some open interval of the real line; the interval can be inﬁnite, e.g. ( −∞ , + ). Example : we don’t consider the function y ( x ) = { log( x ) , x < 0 , log x + C, x > 0 , which is deﬁned on the union of two intervals ( −∞ , 0) (0 , + ), to be a solution of the equation y = 1 /x. Otherwise, the IVP y ( 1) = 0 would have inﬁnitely many solutions. A solution is maximal if it can not be extended to a solution on a larger interval. 1.3. Higher order ODEs. The general form of a 2-nd order ODE is F ( x,y,y ,y ,y ′′ ) = 0 , and solutions are twice diﬀerentiable functions (deﬁned on some intervals) satisfying the equation. Solving for the second derivative, we obtain Newton’s equation with 1 degree of freedom (the motion of a particle m = 1 on the real line): ¨ x = f ( t,x, ˙ x ) . [Here we have changed the notation: now time t is an independent variable, the position x = x ( t ) is the unknown function, and dot means time derivative, so ¨ x is acceleration. The RHS is the ”force”; e.g. f = 0 for a free particle, f = kx for a Hooke’s law spring, etc.] Example: the general solution of Eq. ¨ x = 0 is x ( t ) = C 1 + C 2 t, and the IVP x ( t 0 ) = x 0 , ˙ x ( t 0 ) = ˙ x 0 , has a unique solution on R .
3 The n -th order ODE is F

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 11/19/2010 for the course ANTH 122 taught by Professor 323 during the Spring '10 term at Centennial College.

### Page1 / 32

10Ma2aAnNotes1 - Ma2a (analytical) Fall 2010 PART I: FIRST...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online