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    Hi there,

    I am given that u = F(x - ct), where F() is ANY function.

    I have to calculate partial du/dt and partial du/dx [for the adjective eqn.)

    So i started off with letting s = x - ct

    So then u = F(s),

    and partial ds/dt = -c, and partial ds/dx = 1

    I think I have to work out partial du/dF, and partial du/ds,

    so then partial du/dt = [partial ds/dt] . [partial du/ds]

    and partial du/dx = [partial ds/dx] . [partial du/ds]

    but im unsure how to go about doing these, as the F being ANY function is really confusing me.

    Thank you very much
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    (Original post by Jaswarbrick)
    Hi there,

    I am given that u = F(x - ct), where F() is ANY function.

    I have to calculate partial du/dt and partial du/dx [for the adjective eqn.)

    So i started off with letting s = x - ct

    So then u = F(s),

    and partial ds/dt = -c, and partial ds/dx = 1

    I think I have to work out partial du/dF, and partial du/ds,

    so then partial du/dt = [partial ds/dt] . [partial du/ds]

    and partial du/dx = [partial ds/dx] . [partial du/ds]

    but im unsure how to go about doing these, as the F being ANY function is really confusing me.

    Thank you very much
    Are you sure it doesn't ask for \displaystyle \frac{\partial u}{\partial t} in terms of \displaystyle \frac{\partial u}{\partial x}?

    \displaystyle \frac{\partial u}{\partial t} and \displaystyle \frac{\partial u}{\partial x} are dependent on F.
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    (Original post by notnek)
    Are you sure it doesn't ask for \displaystyle \frac{\partial u}{\partial t} in terms of \displaystyle \frac{\partial u}{\partial x}?

    \displaystyle \frac{\partial u}{\partial t} and \displaystyle \frac{\partial u}{\partial x} are dependent on F.
    The question is:

    "Verify (not prove) that u = F(x - ct) is the general solution of the adjective equation 'partial du/dt + c. partial du/dx = 0 where F() is ANY function."

    Thank you
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    (Original post by Jaswarbrick)
    The question is:

    "Verify (not prove) that u = F(x - ct) is the general solution of the adjective equation 'partial du/dt + c. partial du/dx = 0 where F() is ANY function."

    Thank you
    You've almost done it:

    ...and partial ds/dt = -c, and partial ds/dx = 1...
    ...so then partial du/dt = [partial ds/dt] . [partial du/ds]

    and partial du/dx = [partial ds/dx] . [partial du/ds]...
    Substituting ds/dt and ds/dx gives

    \displaystyle \frac{\partial u}{\partial t}=-c\frac{\partial u}{\partial s}

    \displaystyle \frac{\partial u}{\partial x}=\frac{\partial u}{\partial s}

    Can you finish it?
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    (Original post by notnek)
    You've almost done it:


    Substituting ds/dt and ds/dx gives

    \displaystyle \frac{\partial u}{\partial t}=-c\frac{\partial u}{\partial s}

    \displaystyle \frac{\partial u}{\partial x}=\frac{\partial u}{\partial s}

    Can you finish it?
    Well i'm getting:

    partial du/ds = s. partial dF/ds + F.1 (using product rule, but I don't know if this is right), and

    partial du/dF = s. partial dF/dF (which doesn't make sense) + F. partial dF/ds

    cheers
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    (Original post by Jaswarbrick)
    Well i'm getting:

    partial du/ds = s. partial dF/ds + F.1 (using product rule, but I don't know if this is right), and

    partial du/dF = s. partial dF/dF (which doesn't make sense) + F. partial dF/ds

    cheers
    Actually I might not need partial du/dF after all, I thought I did?
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    (Original post by Jaswarbrick)
    Well i'm getting:

    partial du/ds = s. partial dF/ds + F.1 (using product rule, but I don't know if this is right), and

    partial du/dF = s. partial dF/dF (which doesn't make sense) + F. partial dF/ds

    cheers
    It's easier than you're making it - you don't need to work out du/ds.

    Look again at the equations I gave in my last post and then look at what you're trying to prove...
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    (Original post by notnek)
    It's easier than you're making it - you don't need to work out du/ds.

    Look again at the equations I gave in my last post and then look at what you're trying to prove...
    Ah I see, so because:

    partial du/ds = partial du/dx,

    I can substitute that into the first of your equations to give

    partial du/dt = -c. [partial du/ds]

    => partial du/dt = -c. [partial du/dx]

    so it's just a substitution? Thank you I think I have it now,

    also part 2 of the question is:

    2. Assume periodic boundary conditions u(0,t) = u(1,t) and following initial conditions:

    I1: u(x,0) = exp[-100(x-1/4)^2] (Gaussian)

    I2: u(x,0) = 1 when 1/4 <= x <= 1/2 (0 otherwise, Rectangular)

    Describe how an initial shape evolves and write down the exact solution for the initial conditions I1 and I2.
    [domain 0 <= x < 1, t>= 0]

    Thank you very much, or should this be in a new thread?
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    Any help with the above please?

    Cheers
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    Anyone?

    Cheers
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    (Original post by notnek)
    It's easier than you're making it - you don't need to work out du/ds.

    Look again at the equations I gave in my last post and then look at what you're trying to prove...
    Any help for the above mate?

    Cheers
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    (Original post by Jaswarbrick)
    Any help for the above mate?

    Cheers
    I won't be around much to help you with this (plus it's been a while..).

    Maybe start a new thread?
 
 
 
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