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The wave equation

  1. Jul 27, 2004 #1
    can anyone give me the most simple for someone with little diffrential equations back ground, on the proof for where the 1D wave equation was developed, i mean whats the proof for it. The 1D wave equation is a partial diffrential equation [tex]\frac{\partial^2\psi}{\partial x^2} = \frac{1}{c^2}\frac{\partial^2\psi}{\partial t^2}[/tex]
     
  2. jcsd
  3. Jul 27, 2004 #2
    What do u mean ?
    U need the solution of the equation
    /?
     
  4. Jul 27, 2004 #3

    Galileo

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    Consider a function with arbitrary shape representing some measurable physical quantity. It is a wave (per definition) if it travels through space with some velocity c. So that the function has the form f(x-ct).

    You can show that the differential equation that such a function obeys is the wave equation. Conversely, one can show that the general solution of the wave equation are any solution of the form F(x+ct)+G(x-ct) for arbitrary (twice differentiable) functions F and G.
     
  5. Jul 27, 2004 #4
    Can you give me an example?
     
  6. Jul 27, 2004 #5

    robphy

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  7. Jul 28, 2004 #6
    Well, here is a simple one dimensional acoustic wave eq. for solid isotropic material.
    Gives me a chance to practice my latex.
    Consider a long thin bar which will have only longitudinal plane waves,
    we take a small (infintesimal) section and call it delta x. NOW : the inertial force from Newton's Second Law :

    [itex]
    \[
    \,\frac{{\partial (mv)}}{{\partial {\kern 1pt} t}} = \Delta x \cdot \frac{{\partial (\rho A\dot u)}}{{\partial {\kern 1pt} t}} = \Delta x\frac{\partial }{{\partial {\kern 1pt} t}}\left( {\rho A\frac{{\partial {\kern 1pt} u}}{{\partial {\kern 1pt} t}}} \right)
    \]
    [/itex]

    Where u is the element displacement from it's rest position, rho is the mass density and A is the cross sectional area. The stress at any point in the bar is obtained from Hook's Law, vis ;

    [itex]
    \[
    T = YS = Y\frac{{\partial {\kern 1pt} u}}{{\partial {\kern 1pt} x}}\,
    \]
    [/itex]

    Where Y is the material Young's modulus, and positive strains correspond to positive stresses correspond to tension. The restoring force on the element is the change in force across the element, and may be written :

    [itex]
    \[
    \Delta x\frac{{\partial F}}{{\partial {\kern 1pt} x}} = \Delta x\frac{{\partial ( - AT)}}{{\partial {\kern 1pt} x}} = \Delta x\frac{\partial }{{\partial {\kern 1pt} x}}\left( { - AY\frac{{\partial {\kern 1pt} u}}{{\partial {\kern 1pt} x}}} \right)
    \]
    [/itex]

    Where the force is defined as the negitive of the stress times the area. Conservation of energy requires that the sum of the forces is zero, or ;

    [itex]
    \[
    \frac{\partial }{{\partial {\kern 1pt} t}}\left( {\rho A\frac{{\partial {\kern 1pt} u}}{{\partial {\kern 1pt} t}}} \right) = \frac{\partial }{{\partial {\kern 1pt} x}}\left( {AY\frac{{\partial {\kern 1pt} u}}{{\partial {\kern 1pt} x}}} \right)
    \]
    [/itex]

    Now if we specify that the density and area are independant of time and that the Young's modulus is independant of position :

    [itex]
    \[
    \frac{{\partial ^2 {\kern 1pt} u}}{{\partial {\kern 1pt} x^2 }} + \frac{1}{A}\frac{{\partial A}}{{\partial {\kern 1pt} x}}\frac{{\partial {\kern 1pt} u}}{{\partial {\kern 1pt} x}} - \frac{1}{{c^2 }}\frac{{\partial ^2 {\kern 1pt} u}}{{\partial {\kern 1pt} t^2 }} = 0
    \]
    [/itex]

    Which is known as Webster's Horn Equation. If the area is not a function of position the second term drops out and we have the one dimensional wave equation from Newton's Second law combined with Hook's law and conservation of energy.

    Best
     
    Last edited: Jul 29, 2004
  8. Jul 28, 2004 #7

    Galileo

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    I think what you would physically want to define as a wave is (as before) is a function with fixed shape moving with uniform velocity. i.e. [itex]\psi(x,t)=f(x\pm vt)[/itex] where v is the speed of the wave.

    Letting [itex]x'=x \pm vt[/itex], using the chain rule we have:
    [tex]\frac{\partial \psi}{\partial x}=\frac{\partial f}{\partial x'}[/tex]
    [tex]\frac{\partial^2 \psi}{\partial x^2}=\frac{\partial^2 f}{\partial x'^2}[/tex]
    Similarly:
    [tex]\frac{\partial \psi}{\partial t}=\pm v \frac{\partial f}{\partial x'}[/tex]
    [tex]\frac{\partial^2 \psi}{\partial t^2}=v^2 \frac{\partial^2 f}{\partial x'^2}[/tex]

    Combining the results:
    [tex]\frac{\partial^2 \psi}{\partial x^2}=\frac{1}{v^2}\frac{\partial^2 \psi}{\partial t^2}[/tex]
    Thus any function of the form [itex]f(x\pm vt)[/itex] obeys the wave equation.

    Conversely, write the wave equation as:
    [tex]\psi_{tt}-v^2\psi_{xx}=0[/tex]
    Make a change of variables: [itex]\xi=x+vt[/itex], [itex]\nu=x-vt[/itex]
    and let [itex]\psi(x,t)=h(\xi,\nu)[/itex].
    Then you can write the equation as:
    [tex]4v^2h_{\xi\nu}=0[/tex]
    Following your nose and integrating with respect to [itex]\nu[/itex] and then with respect to [itex]\xi[/itex]:
    [tex]h_{\xi}=f(\xi)[/tex] and
    [tex]h(\xi,\nu)=F(\xi)+G(\nu)[/tex]
    So the general solution is:

    [tex]\psi(x,t)=F(x+vt)+G(x-vt)[/tex]
     
  9. Oct 10, 2004 #8
    Can someone show me that [tex]f(x, t) = A\cos(K(x-vt) + \phi)[/tex] is in fact a solution of the wave equation?
    I tried to go backward but doesn't seem to work very well (probably just my weak calculus)...

    Thank you very much!
     
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