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Unformatted text preview: MAE294B/SIO203B: Methods in Applied Mechanics Winter Quarter 2010 http://maecourses.ucsd.edu/mae294b Solution I 1 The function f ( x ) = x ( 1- log x ) has zeros at x = e (where log x = 1) and also tends to zero as x → 0. Hence there are fixed points at 0 and e. The derivative of f ( x ) is f ( x ) =- log x , which is positive for x < 1, zero for x = 1 and negative for x < 0. Hence the fixed point at x = e is stable. Near the origin f ( x ) > 0 so the origin is unstable. At the origin f ( x ) is unbounded but the stability argument is concerned with positive values near x = 0. To solve the equation exactly, separate variables: d x x ( 1- log x ) = d t . Now- x- 1 is the derivative of g ( x ) = 1- log x , so the fraction takes the form- g ( x ) / g ( x ) , which is the derivative of log | g ( x ) | . Hence log | 1- log x | =- t + log | 1- log x | , where x is the value of x at t = 0. (The solution with x = 0 is x = 0, which can be viewed as the limit of this equation.) Hence | 1- log x | = e- t | 1- log x | . For large times, the right-hand side tends to 0 and log x → 1, so x ( t ) → e, which is consistent with the phase line argument. In fact one can drop the absolute values : the explicit solution for x ( t ) is x ( t ) = exp 1- e- t ( 1- log x ) ....
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- Derivative, λt