21.04.2015 Views

Lecture Notes in Differential Equations - Bruce E. Shapiro

Lecture Notes in Differential Equations - Bruce E. Shapiro

Lecture Notes in Differential Equations - Bruce E. Shapiro

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

7<br />

Figure 1.1 illustrates what this means. The <strong>in</strong>itial condition is given by<br />

the po<strong>in</strong>t (t 0 , y 0 ) (the horizontal axis is the t-axis; the vertical axis is y). If<br />

there is some number M such that |∂f/∂y| < M everywhere <strong>in</strong> the box R,<br />

then there is some region N where we can draw the curve through (t 0 , y 0 ).<br />

This curve is the solution of the IVP. 3<br />

Figure 1.1: Illustration of the fundamental existence theorem. If f is<br />

bounded, cont<strong>in</strong>uous and differentiable <strong>in</strong> some neighborhood R of (t 0 , R 0 ),<br />

and the partial derivative ∂f∂y is also bounded, then there is some (possibly<br />

smaller) neighborhood of (t 0 , R 0 ) through which a unique solution to<br />

the <strong>in</strong>itial value problem, with the solution pass<strong>in</strong>g through (t 0 , y 0 ), exists.<br />

This does not mean we are able to f<strong>in</strong>d a formula for the solution.<br />

A solution may be either implicit or explicit. A solution y = φ(t) is said<br />

to be explicit if the dependent variable (y <strong>in</strong> this case) can be written<br />

explicitly as a function of the <strong>in</strong>dependent variable (t, <strong>in</strong> this case). A<br />

relationship F (t, y) = 0 is said to represent and implicit solution of the<br />

differential equation on some <strong>in</strong>terval I if there some function φ(t) such<br />

that F (t, φ(t)) = 0 and the relationship F (t, y) = 0 satisfies the differential<br />

equation. For example, equation 1.34 represents the solution of {dy/dt =<br />

t/y, y(1) = 2} implicitly on the <strong>in</strong>terval I = [− √ 3, √ 3] which (1.35) is an<br />

explicit solution of the same <strong>in</strong>itial value problem problem.<br />

3 We also require that |f| < M everywhere on R and that f be cont<strong>in</strong>uous and<br />

differentiable.

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!