This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: 18.03 Lecture #2 Sept. 11, 2009: notes First topic is the mathematical theorem (more or less as it appears on page 24 of EP) saying that the general problem we’re studying dy dx = f ( x,y ) , ODE y ( x ) = y IC can be solved. Theorem Suppose that the function f of two variables x and y is continuous on a rectangle a ≤ x ≤ b, c ≤ y ≤ d in the plane. Suppose also that ( x ,y ) lies strictly inside this rectangle: a < x < b, c < y < d. Then there is some interval I containing x so that (ODE), (IC) has a solution y ( x ) defined on the interval I . Suppose in addition that the partial derivative ∂f/∂y exists and is continuous. Then the solution y ( x ) is unique. A proof of this theorem is beyond the scope of 18.03. The uniqueness assertion at the end is not very difficult. The existence assertion is harder; in difficulty it’s a bit like the proof of the fact that a continuous function on an interval with endpoints must have a maximum value. The point of putting the theorem here is the same as the point of telling you the theorem about maxima of continuous functions before you start looking for maxima in calculus: so that the professor will sleep more comfortably. The goal of 18.03 is to learn how to find the solutions y ( x ) whose existence is guaranteed by the theorem. Most of the course will be about finding explicit formulas for solutions. It’s also important to understand how to calculate approximate solutions numerically, for at least two reasons. First,to understand how to calculate approximate solutions numerically, for at least two reasons....
View Full Document
- Fall '09
- Derivative, Euler, interval I. Suppose