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# lecture_11 (dragged) - MA 36600 LECTURE NOTES FRIDAY...

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Unformatted text preview: MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 6 Numerical Solutions Euler’s Method. Consider an initial value problem dy = G(t, y ), dt y (t0 ) = y0 ; in terms of some function G(t, y ). In general it is too diﬃcult to solve such an equation. Indeed, we have seen that we can solve this equation only if we have some type of trick – such as using integrating factors or studying separable equations. Instead we will discuss how to compute an approximate solution. Say that we have a solution y = y (t) to the initial value problem above. Choose a value t1 which is “near” t0 . We explain how to choose a value y1 which is “near” y (t1 ). Consider a secant line through the two points (t0 , y0 ) and (t1 , y1 ). Also consider the line tangent to the curve y = y (t) at the point (t0 , y0 ). We want the slope of the secant line to be equal to the slope of the tangent line: ￿ ∆y0 dy ￿ y1 − y 0 ￿ = that is, = G(t0 , y0 ). ￿ ∆ t0 dt (t0 ,y0 ) t1 − t 0 Solving for y1 gives the expression y1 = y0 + G(t0 , y0 ) · (t1 − t0 ) ≈ y (t1 ). We may proceed in a recursive fashion. Choose a value tn+1 which is “near” tn . Consider the secant line through two points (tn , yn ) and (tn+1 , yn+1 ); as well as the tangent line to y = y (t) at (tn , yn ). We deﬁne ￿ ∆ yn dy ￿ yn+1 − yn ￿ = that is, = G(tn , yn ). ∆ tn dt ￿(tn ,yn ) tn+1 − tn Solving for yn+1 gives the expression yn+1 = yn + G(tn , yn ) · (tn+1 − tn ) ≈ y (tn ). We explain how to use these ideas to ﬁnd an approximate solution to the initial value problem. Consider the following steps: #1: Choose a sequence of numbers t0 < t1 < t2 < · · · < tn < · · · . #2: Deﬁne a sequence of numbers recursively by yn+1 = yn + G(tn , yn ) · (tn+1 − tn ) , y0 = y (t0 ). #3: Plot the points {. . . , (tn , yn ), . . . }. The plot will be an approximate plot of the graph of the actual solution y = y (t). This technique for ﬁnding an approximate solution is known as Euler’s Method. There is a standard way to choose the sequence t0 < t1 < t2 < · · · < tn < · · · . Fix a positive number h; this is called the step size. We may choose ∆ tn = h that is, tn = t0 + n h. Then the y -coordinated can be expressed in the form yn+1 = yn + Gn h in terms of 1 Gn = G(tn , yn ). ...
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