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Unformatted text preview: Solutions for Homework 3 1. 3.4.16 a) The potential is V ( x ) = rx + x 3 3 . You could add a constant C , but that would not make any essential difference; see below. Here is the graph of V for r = 2, r = 0, and r = 2:42 2 45 5 r=242 2 45 5 r=042 2 45 5 r=2 The fixed points of the differential equation are the stationary points of V , that is, the values x for which V ( x ) = 0. Local minima are stable fixed points, since in a local minimum, V ( x ) changes from negative to positive, and therefore V ( x ) changes from positive to negative. Local maxima are unstable fixed points. Stationary points that are neither local maxima, nor local minima are semistable points. The potential V is defined to be the negative of an antiderivative of the righthand side of the differential equation. Why the minus sign? The answer is this: If we were to leave out the minus sign, simply writing the equation in the form x = dV / dx , then stable fixed points would be local maxima of V , and unstable ones would be local minima. This is unintuitive, if you think of a little ball rolling on the graph of V ! Back to our example. There are no stationary points of V for r &lt; 0, exactly one for r = (namely, 0), and two for r &gt; 0 (namely, r ). If we had added a constant C to V , the graph of V would have shifted upwards or downwards, but the stationary points would have remained the same. b) The potential is V ( x ) = r x 2 2 + x 3 3 . Here is the graph of V for r = 2, r = 0, and r = 2: 142 2 45 5 r=242 2 45 5 r=042 2 45 5 r=2 For any r negationslash = 0, there are two stationary points of V , and therefore two fixed points of the dif ferential equation, namely x = 0 and x = r . For r = 0, there is only one, namely x = 0. For r &lt; 0, x...
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This note was uploaded on 02/28/2011 for the course MATH 150 taught by Professor Dr.borgers during the Spring '09 term at Tufts.
 Spring '09
 Dr.Borgers

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