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How to prove that the L2 norm is a non-increasing function of time?

  1. Sep 13, 2014 #1
    1. The problem statement, all variables and given/known data
    image.jpg


    2. Relevant equations
    How can I start the proof? Shall I use the Poincare inequality?

    3. The attempt at a solution
    Well, I know that this norm is defined by wmfWa.gif p4kUw.gif , but still I don't know how to start constructing the proof?
     
    Last edited: Sep 13, 2014
  2. jcsd
  3. Sep 13, 2014 #2

    Ray Vickson

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    Start by omitting the square root. Take ##\partial / \partial t## inside the integral sign (justify!), then use the DE to eliminate, or at least, modify ##\partial{u^2}/\partial t##.
     
  4. Sep 13, 2014 #3
    Thanks! I have done what you said (note that I just made a change in variables so I stick to the symbol convention of integration by parts.):

    [itex]\\2\int_{0}^{L}\frac{\partial^2 s}{\partial x^2}s dx
    \\
    \\
    \\u=s, dv=\frac{\partial^2 s}{\partial x^2}dx
    \\du=\frac{\partial s}{\partial x}dx,v=\frac{\partial s}{\partial x}

    \\
    \\
    \\\therefore 2\int_{0}^{L}\frac{\partial^2 s}{\partial x^2}s dx=2s\frac{\partial s}{\partial x}\mid-2\int_{0}^{L}\frac{\partial s}{\partial x}\frac{\partial s}{\partial x}dx=-2\int_{0}^{L}(\frac{\partial s}{\partial x})^{2}dx[/itex]

    Is that sufficient?
     
    Last edited: Sep 13, 2014
  5. Sep 14, 2014 #4
    Standard for me would be to start with ## \frac{\partial u}{\partial t}= \frac{\partial^2 u}{\partial x^2}## and multiply by ##u## then integrate over the spatial domain ##[0,L]##.
     
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