sol1 - MAE294B/SIO203B Methods in Applied Mechanics Winter...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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 0 | , where x 0 is the value of x at t = 0. (The solution with x 0 = 0 is x = 0, which can be viewed as the limit of this equation.) Hence | 1 - log x | = e - t | 1 - log x 0 | . 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 0 ) . 2 For the case with two distinct eigenvalues, we see that Λ n = λ n 1 0 0 λ n 2 .
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern