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My final Diff Eq question for now

  1. Sep 22, 2005 #1
    Here u(x,t) refers to D'ALembert's solution for the one dimensional wave. L represnets the legnth of the string and c, the speed of the wave.

    Find u(3/4,2) where l=c=1 f(x) = x(1-x), g(x) = x^2 (1-x)
    Ok i know that from earlier that i ahve to extend g as an even periodic function over this interval

    But what about f? Does f have to be extended ?

    For teh extension of g... Do i simply move that function to right?
    so [tex] g_{1} (r) = r^2 (1-r) [/tex]

    for [tex] g_{2} (r) = (-r-\frac{1}{2})^2)[1-(-r-\frac{1}{2})] [/tex] over the interval [5/4,7/4]

    and finally [tex] g_{3} (r) = (r-1)^2)[1-(r-1)] [/tex] over the interval [7/4,2]

    am i going in the right direction??

    Simialr question i have is
    If [itex] f(x) = x^2 (1-x)^2, g(x) = 1, \partial u/ \partial x (0,t)=0, c=1, find u(3/4,7/2)[/itex]

    does f(x) have to be extended
    if so would it be extended by a polynomial of degree 4? in which case
    for [1,2] (x-1.5)^4
    for [2,3.5] (2.5-x)^4 [/tex]

    your help would be greatly appreciated!
    p.s. the other question was not required to be answered tahts why you dont see that one, thanks!
    Last edited: Sep 22, 2005
  2. jcsd
  3. Sep 22, 2005 #2


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    Stunner, the extensions depend on the boundary conditions: if the boundary conditions are of a form u(0,t)=0 or u(L,t)=0 then odd-extend about those points. If the boundary conditions are of the form [itex]u_x(0,t)=0[/itex] or [itex]u_x(L,t)=0[/itex], then even-extend about the points. If the left boundary is u(0,t) then odd-extend about that point and if the right boundary is [itex]u_x(L,t)=0[/itex], then even-extend about that point.

    You don't state the boundary conditions above.
  4. Sep 22, 2005 #3
    the first one is just like this question from earlier https://www.physicsforums.com/showthread.php?t=90027

    the second one has the following conditions
    [itex] f(x) = x^2 (1-x)^2, g(x) = 1, \partial u/ \partial x (0,t)=\partial u/ \partial x (1,t)0, c=1 [/itex] find u(3/4,7/2)

    ok to odd extend is for a function to follow this rules f(-x) = -f(x)
    and to even extend is fora function to follow f(-x) = f(x)

    i nthe first case i need to odd extend - but am right in what i have done?

    in the second case i need to even extend

    now heres the stupid part - how do i even or odd extend?
    Last edited: Sep 22, 2005
  5. Sep 22, 2005 #4


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    Ok, the first one has boundary conditions:

    u(0,t)=0 and u(L,0)=0

    So then both f(x) and g(x) would be odd-extended about 0 and L. The second one has derivative boundary conditions so f(x) and g(x) for that one would be even-extended. Can you draw plots of both of these extensions?
  6. Sep 22, 2005 #5
    the one on the left applies to the first one

    te one of hte right applies to the second one

    Attached Files:

  7. Sep 22, 2005 #6


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    Not exactly Stunner. You need to extend both f(x) and g(x) for each problem. Plots below are the odd-extensions for the first problem. First is odd-extension of f(x)=x(1-x), second one is the odd-extension for g(x)=x^2(1-x). Both are extended in the range of -5/4 to 11/4 which corresponds to the intervals needed in order to calculate the value for u(3/4, 2). Can you come up with the equations for each interval for each plot in order to plug it into D'Alembert's formula?

    Attached Files:

  8. Sep 22, 2005 #7
    ok so then for the first one
    For f(x)
    x(1-x) [0,1]
    (x-1)(x-2) [1,2]
    (2-x)(x-3) [2,3]

    x(x+1) [-1,0]
    (x+1)(x+2) [-2,-1]

    for g(x)
    x^2 (1-x) [0,1]
    (x-2)^2 (-1-x) [1,2]
    -x(x+1)^2 [-1,-2]
  9. Sep 23, 2005 #8


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    This is what I get:

    For f(x):

    [tex](-1,0)\Rightarrow \quad x(1+x)[/tex]
    [tex](0,1)\Rightarrow \quad x(1-x)[/tex]
    [tex](1,2)\Rightarrow \quad -(x-1)+(x-1)^2[/tex]
    [tex](2,3)\Rightarrow \quad (x-2)-(x-2)^2 [/tex]

    for g(x):

    [tex](-1,0)\Rightarrow \quad -x^2(1+x)[/tex]
    [tex](0,1)\Rightarrow \quad x^2(1-x)[/tex]
    [tex](1,2)\Rightarrow \quad -(2-x)^2[1-(2-x)][/tex]
    [tex](2,3)\Rightarrow \quad (x-2)^2[1-(x-2)][/tex]
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