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Prove time invariance

  1. Oct 5, 2016 #1
    1. The problem statement, all variables and given/known data
    let y(x, t) be a solution to the quasi-linear PDE
    [tex]\frac{\partial y}{\partial t} + y\frac{\partial y}{\partial x} = 0[/tex]
    with the boundary condition
    [tex]y(0, t) = y(1, t) = 0[/tex]
    show that
    [tex]f_n(t) = \int_0^1 y^n\,\mathrm{d}x[/tex]
    is time invariant for all n = 1, 2, 3,...
    2. Relevant equations


    3. The attempt at a solution
    By differentiation under the integral sign i obtain
    [tex]\frac{\mathrm{d}}{\mathrm{d}t}f_n(t) = \int_0^1ny^{n-1}\frac{\partial y}{\partial t}\,\mathrm{d}x \\
    = -n\int_0^1y^n\frac{\partial y}{\partial x}\,\mathrm{d}x \\
    = -n\underbrace{[y^{n+1}]_0^1}_{=0} + n^2\int_0^1y^n\frac{\partial y}{\partial x}\,\mathrm{d}x[/tex]
    where the last equality in valid by integration by parts.

    My goal was to obtain a differential equation for f_n.
     
  2. jcsd
  3. Oct 5, 2016 #2

    Ssnow

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    Hi, I don't understand why at the end it is ## n^2\int_{0}^{1}y^n\frac{\partial y}{\partial x} dx## instead of ##n^2\int_{0}^{1}y^n dx=n^2f_{n}##
     
  4. Oct 5, 2016 #3
    Because,
    [tex] \int_0^1 y^n\frac{\partial y}{\partial x}\mathrm{d}x = [y^{n+1}]_0^1 - \int_0^1 ny^{n-1}\frac{\partial y}{\partial x}\,y\,\mathrm{d}x[/tex]
    as
    [tex]\int fg^\prime = fg - \int f^\prime g[/tex]
     
  5. Oct 5, 2016 #4

    Ssnow

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    ok sorry I lost a term deriving... yes it is no easy to find the form ##f_{n}## for a possible equation. The fact is that ##f_{n}## depends only by the time and also the quantity ##F=\int_{0}^{1}y^{n}(x,t)\frac{\partial y(x,t)}{\partial x}dx ## is a function only of the time, so you equation will be

    ## \frac{df_{n}}{dt}=n^2F(t)##
     
  6. Oct 5, 2016 #5

    Ssnow

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    Hi, but if you can think ##y^n\frac{\partial y}{\partial x}=\frac{1}{n+1}\frac{\partial y^{n+1}}{\partial x}## you have that the integral will be:

    ##\frac{n^2}{n+1}[y(1,t)^{n+1}-y(0,t)^{n+1}]=0##

    so your equation is ##\frac{d f_{n}}{dt}=0##
     
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