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Solution to homogeneous wave equation

  1. Feb 16, 2014 #1
    1. The problem statement, all variables and given/known data
    Prove by direct substitution that any twice differentiable function of (t-R[itex]\sqrt{με}[/itex]) or of (t+R[itex]\sqrt{με}[/itex]) is a solution of the homogeneous wave equation.


    2. Relevant equations
    Homogeneous wave equation = ∂2U/ ∂R2 - με ∂2U/∂t2 = 0
    3. The attempt at a solution

    Could you guys give me a direction or a way I should think. I have no idea on how I should approach this problem.
     
    Last edited: Feb 16, 2014
  2. jcsd
  3. Feb 16, 2014 #2

    PeroK

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    Why not start by differentiating a general function and see what you get?
     
  4. Feb 16, 2014 #3
    By differentiating a general function you mean differentiating t-R√με in terms of t?
     
  5. Feb 16, 2014 #4

    PeroK

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    No, that's not a general function. What do you think this means:

    Any twice differentiable function of (t-R√με) or of (t+R√με)
     
  6. Feb 16, 2014 #5
    I think it means f"(t-R√με)
     
  7. Feb 16, 2014 #6

    PeroK

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    Not quite. That's the second derivative of a general function (f) of t - R√με.

    You also need to relate partial derivatives (wrt t and R) to ordinary derivatives.
     
  8. Feb 16, 2014 #7
    Well isn't a twice differentiable function one that has a second derivative?
    Could you elaborate or demonstrate another example about linking partial derivatives to the ordinary ones
     
  9. Feb 16, 2014 #8

    PeroK

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    There's a difference between a twice differentiable function and the second derivative of that function. E.g:

    [tex]f(x) = x^3[/tex]
    Is a twice differentiable function. And:
    [tex]f''(x) = 6x[/tex] is its second derivative.

    To get you started, you would say something like:

    [tex]Let \ U(t, R) = f(t - R\sqrt{μ ε})[/tex] where f is a twice differentiable function.

    Now you need to calculate the partial derivatives of U. Do you know how to do that?
     
  10. Feb 16, 2014 #9
    ∂U/∂t = 1

    ∂U/∂R = √με

    So that is the partial derivative of U(t,R)

    I can't take the second derivative of this because it is constants.
     
  11. Feb 16, 2014 #10

    PeroK

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    That's not right. You will need to use the chain rule for the partial derivatives of U.
     
  12. Feb 16, 2014 #11
    I know how to solve via chain rule
    but there is no given value of t & R so...

    ∂U/∂f = ∂U/∂t * dt/dU + ∂U/∂R * dR/dU
    = (1) * (?) + (√με) * (?)
     
  13. Feb 16, 2014 #12

    PeroK

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    The chain rule works like this:

    [tex]\frac{\partial{U}}{\partial t} = f'(t - R\sqrt{με}).1 \ \ and \ \ \frac{\partial{U}}{\partial R} = f'(t - R\sqrt{με}).(-\sqrt{με})[/tex]

    What you actually have here is:

    [tex]U(t, R) = f(g(t,R)) \ where \ g(t, R) = t - R\sqrt{με}[/tex]So:

    [tex]\frac{\partial{U(t, R)}}{\partial t} = f'(g(t, R)).\frac{\partial{g(t, R)}}{\partial t}[/tex]
     
  14. Feb 16, 2014 #13
    What I am confused about is what are u deriving f'(g(t,R) in terms of?
    So for example the answer to
    ∂U(t,R)/∂t = f'g(t,R) * 1
    I don't understand where we are going with this.
    I am sorry for testing your patience but I am confused and I want to learn this
     
  15. Feb 16, 2014 #14

    PeroK

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    Okay. I find it really important with multi-variable calculus to key an eye on what everything is is function of. Functions can be functions of one variable, two variables or vectors and be real or vector valued. So:

    In this case, U is real-valued function of two variables (R and t). So, it has partial derivatives wrt R and t.

    f, on the other hand, is an ordinary real valued function of a single variable. So, f' and f'' are also ordinary real-valued functions. These are well-defined functions (if f is twice differentiable).

    And, g is the specific function of two variables: g(t, R) = t - R√με

    Now, one form of the chain rule is:

    [tex]\frac{\partial{U(t, R)}}{\partial t} = f'(g(t, R)).\frac{\partial{g(t, R)}}{\partial t}[/tex]

    That's the one that's relevant here. It's probably worth taking time to fully understand this.

    For example, if f(x) = sin(x), then f'(x) = cos(x) etc.

    And, if [tex]U(r, R) = sin(t^2 - R^2) \ with \ g(t, R) = t^2 - R^2[/tex], then:

    [tex]\frac{\partial{U(t, R)}}{\partial t} = f'(g(t, R)).\frac{\partial{g(t, R)}}{\partial t} = cos(t^2 - R^2).(2t)[/tex]
     
  16. Feb 16, 2014 #15

    PeroK

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    ... alternatively, you could let x = t - R√με, then:

    U = f(x) and

    [tex]\frac{\partial{U}}{\partial{t}} = \frac{df}{dx}\frac{\partial{x}}{\partial{t}}[/tex]

    Maybe you prefer to look at it that way?

    You can think of these things in terms of functions or variables.
     
  17. Feb 16, 2014 #16
    Thanks the example helped
    So according to this ∂U(t,R)/∂t = t -R√με * (1)
     
  18. Feb 16, 2014 #17

    PeroK

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    What happened to f?
     
  19. Feb 16, 2014 #18
    Actually if x = t-R√με
    & U = f(x)

    ∂U/ ∂t = df/dx * ∂x/∂t
    = 1 * 1
    = 1
     
  20. Feb 16, 2014 #19
    This is how I interpreted your example
    x=t2 + R2
    ∂U(t,R)/∂t = f'(x) * ∂x/∂t
    = cos(x) * (2t)
    = cos(t2 + R2) * 2t
     
  21. Feb 16, 2014 #20

    PeroK

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    So, the derivative of any function is 1?

    What if f(x) = sin(x)? Then df/dx = cos(x). It's certainly not 1.
     
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