Unformatted text preview: 2 MA 36600 LECTURE NOTES: MONDAY, FEBRUARY 9 Difference Equations
First Order Equations. Recall that every ﬁrst order diﬀerential equation is in the form
= G(t, y ),
dt y (t0 ) = y0 ; for some function G = G(t, y ). As discussed in the previous lecture, often it is too diﬃcult to ﬁnd an exact
solution, so we seek approximate solutions. We review the ideas.
Choose a step size h, and consider the equations
∆ tn = h
= G(tn , yn ).
dt (tn ,yn )
Equivalently, we have the recursive relations
tn = t0 + n h and yn+1 = yn + G(tn , yn ) h. We express the latter equation in the form
yn+1 = Γ(n, yn ) in terms of Γ(n, y ) = y + G(t0 + n h, y ) h. This is called a ﬁrst order diﬀerence equation. This course concerns the “Calculus of Diﬀerentials,” but in
the same way one can study the “Calculus of Diﬀerences.” While one can study diﬀerence equations in their
own right, we approach the subject through the motivation of diﬀerential equations.
Recall that a ﬁrst order diﬀerential equation is an autonomous equation if it is in the form
= f (y ),
dt y (t0 ) = y0 ; where f = f (y ) is a function which does not involve time t. As before, we approximate by substituting
= f (yn )
∆ tn =⇒ yn+1 = yn + f (yn ) h. We may express this equation in the form
yn+1 = φ(yn ) where φ(y ) = y + f (y ) h. We call such an equation an autonomous diﬀerence equation.
Linear Equations. We now consider a special class of diﬀerence equation. Consider the ﬁrst order linear
+ p(t) y = g (t),
y (t0 ) = y0 .
The associated diﬀerence equation is in the form
+ p(tn ) yn = g (tn )
∆ tn where ∆tn = h. We ﬁnd an expression for yn+1 in terms of yn :
yn+1 − yn
+ p(t0 + n h) yn = g (t0 + n h)
h =⇒ yn+1 = yn + g (t0 + n h) − p(t0 + n h) yn h. We may write this in the form
yn+1 = ρn yn + bn where ρn = 1 − p(t0 + n h) h, This is known as a ﬁrst order linear diﬀerence equation. bn = g (t0 + n h) h. ...
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- Spring '09
- Equations, Elementary algebra, yn, Tn, difference equation