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Question regarding D'Alembert solution for one dimension wave equation

  1. Feb 13, 2010 #1
    In am studying PDE and I have question about D'Alembert solution for one dimension wave equation.

    I am going to reference Wolfram:

    http://mathworld.wolfram.com/dAlembertsSolution.html

    1) I want to verify the step of [tex]\frac{\partial y_0}{\partial t}[/tex] of step (14) of the page.

    [tex]\Rightarrow\; \frac{\partial y_0}{\partial t}=v_0\; =\; \frac{ \partial f(x-ct)}{\partial (x-ct) } \frac{ \partial (x-ct)}{\partial t }\; + \;\frac{ \partial g(x+ct)}{\partial (x+ct) } \frac{ \partial (x+ct)}{\partial t }[/tex]

    [tex] =\; c[\frac{ \partial f(x-ct)}{\partial (x-ct) }\;- \; \frac{ \partial g(x+ct)}{\partial (x+ct) }]|_{t=0} \;= \; -c\frac{ \partial f(x)}{\partial (x) }\;+ \; c\frac{ \partial g(x)}{\partial (x) } \;=\; -cf'(x)+cg'(x)\;\;\; (14)[/tex]

    Am I correct on the steps?



    2) I don't follow step (16)

    [tex] \int_{\alpha} ^x \; v_0(s)\; ds = -cf(x) +cg(x) \;\;\; (16)[/tex]

    a) Where is [tex]s[/tex] come from? Is it just a dummy variable for substitude for x and [tex]\alpha [/tex] later?

    Where is [tex]\alpha[/tex] come from?? Is it supposed to be 0 instead?


    b) [tex] \int_{\alpha} ^x \; v_0(s)\; ds = \int_{\alpha} ^x -cf'(x) +cg'(x) \; = [-cf(x)+cg(x)]_{\alpha}^x[/tex]

    This don't agree with [tex] \int_{\alpha} ^x \; v_0(s)\; ds = -cf(x) +cg(x) \;\;\; (16)[/tex]


    Please refer to (14) and (16) in Wolfram page.
     
    Last edited: Feb 13, 2010
  2. jcsd
  3. Feb 13, 2010 #2
    1) It is correct. I am guessing [tex]y_{0}[/tex] is the wave function at t=0? (you'll have to be more accurate with notations)

    2)Yes, s is just a dummy variable. [tex]\alpha[/tex] is actually a meaningless constant (could be zero as well). I agree with you that technically the integration is not correct (there should be another term with s=[tex]\alpha[/tex]), but if you look forward into the formula derivation you'll see that this extra term cancells out and has no part in the final formula, so it's not very critical.
     
  4. Feb 13, 2010 #3
    thanks for the response

    I was refering to the page in Wolfram provided, that's the reason I did not explain what is y0.

    Is D'Alembert only for one dimension only?
     
  5. Feb 14, 2010 #4
    New question on D'Alembert solution.

    Please refer to this:

    http://mathworld.wolfram.com/dAlembertsSolution.html

    [tex](17)\;\Rightarrow \; f(x)=\frac{1}{2}y_0\; -\; \frac{1}{2c} \int_{\alpha}^x \; v_0(s) ds [/tex]

    [tex](18)\;\Rightarrow \; g(x)=\frac{1}{2}y_0\; +\; \frac{1}{2c} \int_{\alpha}^x \; v_0(s) ds [/tex]

    [tex]Plug\;(17)\;&\;(18)\; into \; (13)\; \Rightarrow\; y_0(x) \;=\; f(x)\;+\; g(x)\; = \; y_0\;!!! [/tex]

    How does it become [tex] y(x,t) \; = \;\frac{1}{2} y_0(x\;-\;ct)\; +\; \frac{1}{2} y_0(x\;+\;ct)\; +\; \frac{1}{2c} \int_{x-ct}^{x+ct} \; v_0(s) ds \;\; ??[/tex]

    My eqestion is:

    1) If I pluging (17) and (18) into (13), I only get Y0(x) that cannot turn into (19) by putting t back in.






    The way I explain to myself, please tell me I am correct. Also I still have one more question below. Yes, I have no confidence on my calculus, I need some reassurance. I also added f(a) and g(a) terms from integration, please look and tell me if I am correct.

    [tex]y_0(x) \;= f(x) + g(x)\;,\; putting \;t \;back\; \Rightarrow\; y(x,t)\;= \;f(x-ct)\; +\; g(x+ct)[/tex]

    [tex](16)\Rightarrow\; \int_a^x \; v_0(s)\; ds \;=\; c[-f(s)\; +\;g(s)]_a^x \;=\; c[-f(x) \;+\; f(a) \;+\; g(x) \;-\;g(a)] \;\;where\; a\; is\; a\; dummy\; constant[/tex]

    [tex](17)\; \Rightarrow \; f(x-ct) \; =\;\frac{1}{2} y_0(x-ct)\; - \frac{1}{2c}\int_{\alpha}^{x-ct} \;v_0(s)\; ds \;-\frac{f(a)}{2c}\;+ \;\frac{g(a)}{2c}\;[/tex]

    [tex]=\; \frac{1}{2} y_0(x-ct)\; + \frac{1}{2c}\int _{x-ct}^{a} \;v_0(s)\; ds \;-\frac{f(a)}{2c}\;+ \;\frac{g(a)}{2c}\;[/tex]

    [tex](18)\; \Rightarrow \; g(x+ct) \; =\;\frac{1}{2} y_0(x+ct)\; - \frac{1}{2c}\int_{\alpha}^{x+ct} \;v_0(s)\; ds \;+\; \frac{f(a)}{2c}\;- \;\frac{g(a)}{2c}[/tex]

    [tex]Therefore\;\; y(x,t)\;= \;f(x-ct)\; +\; g(x+ct)\; = \;\frac{1}{2} y_0(x\;-\;ct)\; +\; \frac{1}{2} y_0(x\;+\;ct)\; +\; \frac{1}{2c} \int_{x-ct}^{x+ct} \; v_0(s) ds \;\;[/tex]

    [tex]What \;is \; y_0(x+ct) \; and \; y_0(x-ct)\; ?[/tex]

    Is it just take [tex]y_0(x)[/tex] and substitude every x with x+ct or x-ct?
     
    Last edited: Feb 15, 2010
  6. Feb 15, 2010 #5
    You explained it very well.

    And yes, y(x+ct) is making the change x->x+ct in y(x)
     
  7. Feb 15, 2010 #6
    I want to confilm that you mean [tex]y_0(x)[/tex] which is y(x) with t=0. When equating x=x+ct or x=x-ct on all the x in [tex]y_0(x)[/tex] will not equal back to the original y(x,t). Also [tex]y_0(x-ct)\; and\; y_0(x-ct)[/tex] are not equal.

    Thanks very much of your time. I don't go to school and just studying at home. I don't quite trust myself. I find going to school is very high pressure and I was very disappointed after looking at what the class cover. The class skip over the Laplace and Poisson equation. Only cover lightly on Bessel and Legendre differential equations. I am studying electromagnetics and antenna, not really studying math. Somehow both the ODE and PDE class spent a lot more time in heat equation than wave equation and Poisson.

    Anyway I don't want to bore you with all this. Thanks.

    Alan
     
    Last edited: Feb 15, 2010
  8. Feb 16, 2010 #7
    I meant a general function y(x), not necessarily related to the problem.

    Of course d'Alambert's formula does this with y0 and twice (once with x+ct and the other time with x-ct)
     
  9. Feb 16, 2010 #8
    Thanks.
     
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