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Help me understand wave equation ?

  1. Sep 27, 2012 #1
    hi,

    It is a bit confusing, because when I say "wave equation" the connotation it invokes in my mind is if I apply the equation I will find the function of the wave I'm exploring.
    Of course I know now this is wrong.. it is the other way around, I have to know my wave-function in advance and substitute it into the "wave equation".
    Tell me if so far I'm on the right track ?

    My next question is what do I get back ?
    I can see there second-partial-derivative of position OR second-partial-derivative of time, but what is their meaning.. what is the idea ?

    For example in the Newton law I can understand the second-derivative of position against time is acceleration, but in the wave equation we have second-partial-derivative of position against "frozen"-time , what does this mean ?
    And then similar for time ?

    Help me make the connection, thank you ?
     
  2. jcsd
  3. Sep 27, 2012 #2

    Simon Bridge

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    Sounds like you are utterly confused to me.

    I take it you are trying to understand: $$\frac{\partial^2}{\partial t^2}y(x,t) = \frac{1}{c^2}\frac{\partial^2}{\partial x^2}y(x,t)$$

    Nothing. It's nonsense.

    probably easiest to think in terms of a transverse wave on a string.

    All the parts of the string are moving up and down with time. If I take a snapshot of the string at some particular time I call t=0 I will get a function y(x) which tells me how far each point x has been displaced in the y direction. dy/dx is just the gradient of the string at point x, and d2y/dx2 is the curvature of y at x.

    At a later time t, this will have changed.

    Similarly, if I just watch one point on the string, I'll see it move up and down in the y direction over time.

    The second time derivative of y is the y-acceleration of the point x with time.
    The second space derivative of y is the curvature.
    The wave equation says that the acceleration in the y direction is related to the curvature in the x direction. A function that has this particular relation will be a wave.

    y does not have to be a physical displacement however. If can be anything that could have different values in different places - pressure, say, or the magnitude of the electric field.
     
    Last edited: Sep 27, 2012
  4. Sep 28, 2012 #3
    What do you mean by "curvature in x direction" ?

    If you know the "wave function" of the wave, why do you need to use "wave equation" , couldn't you calculate any parameter of the wave just from the "wave function" ? this is probably where my confusion lies !
     
    Last edited: Sep 28, 2012
  5. Sep 28, 2012 #4

    Simon Bridge

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    Exactly what you think I do: how much the function curves at the point.

    You'll be more used to thinking in terms of critical points of a function:
    If you have a function y=f(x), then f'(x)=0 tells you the location of the turning points, and f''(x) at the turning points tells you their concavity (concave-up, or down, or a point of inflection.)

    You don't of course ... the trouble is that you don't always have the wave function.

    You may have a function, but you don't know if it is a wave function.
    You may just have a bunch of material properties and conditions (say, a hammer banging on a sheet of metal) and you want to know the waves that get set up ... you have to solve the wave equation.
    Often you have some differential equations which describe the physics of the situation and being able to rearrange them into the form of the wave equation tells you that your solutions are waves.

    Example: You have a long line of masses on a frictionless surface connected by springs of constant k and unstretched length Δx. If we displace one of the masses a small distance y, and let it go, it sends a pulse along the line. How fast does the pulse travel?
     
  6. Sep 28, 2012 #5
    thank you, could you point me to a place which describes more "Examples".
    At the moment I'm more interested of figuring out what is the idea behind the "wave equation" i.e. what do you normally put "in" and what do you get as "out" with connection with real life examples.

    From your description it seems that sometimes you know the wave-function and you solve the wave-equation to see if this wave-function could be used for particular problem you want to solve.
    And sometimes you don't have even the wave-function, but just some partial information about the problem you want to solve and you put this partial-info into the wave-equation to get back some form of wave-equation..
     
  7. Sep 28, 2012 #6
    Hello mraptor

    The 'wave equation' is a partial differential equation.

    You are probably familiar with the integration (=solution) of ordinary differential equations where you need to add an arbitrary constant (2 for second order).

    With partial differential equations you need to introduce arbitrary functions not constants.

    Just as with ODEs the actual values of the constants are determined by the boundary conditions not the ODE itself, so with partial differential equations the arbitrary functions are determined by the boundary conditions, not the PDE.

    So you need to use the boundary conditions to determine the appropriate arbitrary functions, since they are your wave functions.
     
  8. Sep 29, 2012 #7

    Simon Bridge

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    The wave equation is used to:
    1. find wavefunctions from the parameters of a physical system
    2. demonstrate that a particular function is a wavefunction
    3. demonstrate that a particular set of physical relations have wave solutions

    you've seen examples 1 and 2 ... for 3, Maxwel's equations can be shown rearranged to have the form of the wave equation ... showing that electromagnetic waves can exist. Solving the EM form of the wave equation gets you EM wave functions.

    You can come up with your own examples easily enough
    ... start with a rope lying along the floor, lift one end, then bring that end down sharply.
    ... start with a string stretched between two supports - lift the middle so the string forms a triangle between the supports - let go.
    ... the B field of a wire carrying a current that changes direction periodically.

    These have wave-equation solutions. The first will be a damped travelling wave and the second should give stationary waves. The last one gives EM waves. Solving the wave equation for each of these is left as an exercise :)
     
  9. Oct 1, 2012 #8
    thanks guys, I studied calculus (not differential eq, just simple calculus) 10++ years ago..seems I have to freshen up on those things.
    It seems my main misunderstanding stems from there...

    PS> There seems to be no good tutorial to grasp the ideas behind the application of differential equations unless you plunge deep into using them.
    Looked at couple of books, they also seem to require pretty advanced knowledge before you start to get basic idea.
     
  10. Oct 1, 2012 #9
    OK, you may well have been taught something along the lines of "Integrate the following"


    [tex]I = \int {3xdx} [/tex]

    and come up with


    [tex]I = \frac{{3{x^2}}}{2} + C[/tex]

    Now this can be rewritten


    [tex]dy = \int {3xdx} [/tex]

    or



    [tex]\frac{{dy}}{{dx}} = 3x[/tex]

    Which is a simple differential equation. When we say solve this we mean find a function y =f(x) that satisfies this equation.

    The conventional way to write this is


    [tex]\frac{{dy}}{{dx}} - 3x = 0[/tex]

    And the solution is


    [tex]y = \frac{{3{x^2}}}{2} + C[/tex]

    Which you can see is what happens if we integrate both sides of the equation. dy on one side and the expression on the other.

    If we are told that at x=0, y=0 we can calculate a value for C (=0).

    This is known as a boundary condition or (in this case) an initial condition.

    Pretty well all the theory of differential equations boils down to more complicated versions of the above.

    A good place to start is the equation for constant acceleration motion in a straight line.

    Perhaps you would like to try these? You can derive all the standard equations from here quite simply.


    [tex]\frac{{dv}}{{dt}} = \frac{{{d^2}s}}{{d{t^2}}} = v\frac{{dv}}{{ds}} = f = {\rm{a}}\;{\rm{constant}}[/tex]
     
  11. Oct 1, 2012 #10
    Starts to make some sense .. So in sense you have a differential equation , you integrate to solve it.. and then in your original experiment you would have had some initial (boundary) condition depending on how do you setup the experiment.
    You fill those conditions and get your final results ?

    What about the case like in the current example you have the wave-equation to solve and you pick this y(x,t) function from "somewhere" .. this case seems abit different..
    First is it possible you don't know this y(x,t) function in advance ?
    Second if you know it, does the procedure you described apply i.e. substitute solve(integrate), populate initial conditions..get final result.
     
  12. Oct 1, 2012 #11
    The wave equation has an enormous number of possible solutions.

    However we can immediately discard most of them.

    We don't want an arbitrary function to goes off to infinity ie we want one that remains bounded.

    So that lets out ex, x2, x3 and so on.

    It even lets out some trigonometric functions eg tan(x).

    However cos(x) and sin(x) <= 1 so are always bounded.

    Again cos(0) ≠ 0 so that is no use if we want y=0 when x=0

    That leaves sin(x) since sin(0)=0.

    If we wanted to consider a wave with an initial push or acceleration we might want to use cos(x).
     
  13. Oct 1, 2012 #12

    Simon Bridge

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    Well ... some wave problems do blow up :)
    However, studiot's main point is that we let the physics select what sort of equations to guess for y(x,t)... and that is what solving differential equations is all about: making the right guess.

    We usually make some sort of guess about the general shape of the equation, leaving important parameters variable, then use the rest of the boundary conditions to tell us what those parameters have to be in order for that shape to satisfy the wave equation.

    If nothing will then we've made the wrong guess or the physics does not produce wave motion.

    sine and cosine functions are very useful because we can prove that any other periodic function can be built up by summing sine and cosine functions.

    But you are right - to understand how the wave equation is used, properly, you will have to learn about how to solve differential equations in general. Usually students won't meet the wave equation without having some experience at this. It may well be that this is the main source of your confusion.
     
  14. Oct 2, 2012 #13
    Whole encyclopedias have been written about differential equations.

    Please remember that post#9 was a one post introduction to them.
     
  15. Oct 2, 2012 #14
    Found couple of things which I'm reading now... when I'm bit more knowledgeable will revisit the question again ..
    http://tutorial.math.lamar.edu/Classes/DE/DirectionFields.aspx


    Setting up the problem initial conditions is the bigger part of it... mentioning it is a wave-equation is just sort of qualification that group a real-world phenomena in sort of categories, so that when people find problems which may have similar solutions you can have some sort of routine methods to handle them..
    Is that part of it ?
     
    Last edited by a moderator: Sep 25, 2014
  16. Oct 2, 2012 #15

    Simon Bridge

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    That's part of it ... you are on your way.
    Have fun.
     
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