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Unformatted text preview: Clarification about y = 0 in First Order ODEs
Consider the linear ODE y = 10  y with initial condition y(0) = 0. Then for all time t, y(t) < 10. The inequality is in fact strict. Why? Suppose for some time t1 , y(t1 ) 10. Then since y is continuous, there would exist some time t2 where y(t2 ) = 10. However, since we have a linear ODE, there is a unique solution that passes through the point (t2 , 10), namely y = 10. But y = 10 does not pass through (0, 0) so since that is the only solution where y(t2 ) = 10, we know the solution through (0, 0) does not pass through (t2 , 10) and hence must always have y < 10. Consider the nonlinear ODE 3 y = y 3 = f (t, y) 2 (1) (2) with initial condition y(0) = 0. f 9 = y 2 which is continuous everywhere (as is f ), so by a theorem from 2.4, we know there y 2 is a unique solution to this ODE. Note Consider by contrast the nonlinear ODE 3 y = y 1/3 2 with initial condition y(0) = 0. We have an infinite number of solutions to the initial value problem (3): y= 0 if t a; 3/2 if t > a (t  a) (3) where a 0 is arbitrary. (Can see that these are all solutions by plugging them in!) (1), (2) and (3) are all first degree autonomous initial value problems, but the linearity of (1) forces uniqueness (you can show this from the uniqueness theorem), the nonlinear problem (2) also has a unique solution by the theorem, but the nonlinear problem (3) has many solutions. 1 ...
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This note was uploaded on 07/17/2008 for the course MATH 415 taught by Professor Costin during the Fall '07 term at Ohio State.
 Fall '07
 COSTIN
 Differential Equations, Equations

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