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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 dy = 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 ￿ ∆yn dy ￿ ￿ ∆ tn = h and = = G(tn , yn ). ∆ tn 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 dy = 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 ∆ yn = 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 differential equation dy + p(t) y = g (t), y (t0 ) = y0 . dt The associated difference equation is in the form ∆ yn + 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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