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Simple PDE question

  1. Dec 7, 2005 #1
    Find u(3/4,2) when l=c=1, f(x) = x(1-x), [itex] g(x) = x^2 (1-x) [/itex]
    all i need to do is find the value using d'Alembert's solution of the one dimensional wave.

    now it is easy for me to extend f(x)
    for f(x)
    [tex](-1,0)\Rightarrow \quad x(1+x)[/tex]
    [tex](0,1)\Rightarrow \quad x(1-x)[/tex]
    [tex](1,2)\Rightarrow \quad -(x-1)+(x-1)^2[/tex]
    [tex](2,3)\Rightarrow \quad (x-2)-(x-2)^2 [/tex]

    but for g(x)
    for extnesion into the (1,2) interval i get
    [tex] (x-1)^2 (x-2) [/tex]
    but i was told that the answer is [tex] -(2-x)^2 (x-1)][/tex]
    which is switched from my answer. WHy is it the oppsoite? WHo is correct? WHat i did is isketched this little piece of function for the (0,1) interval and then reflected it on the X axis. I then moved it right by one place to theh right by -1 factor.

    Also is it ok the solve the one dimensional wave equation using separation of variables rather than using d'Alembert's solution?

    Please advise!
    Thank you for your help!
     
  2. jcsd
  3. Dec 8, 2005 #2
    can anyone help i need to figure this out!

    also could u have a look at this thread, on a similar question
    https://www.physicsforums.com/showthread.php?t=103134

    my exam is on saturday i need to understand this once and for all, or at least have the doubt about how to extend the function odd or even clarified.

    PLease help!

    Thank you!
     
  4. Dec 8, 2005 #3
    You generally use D'Alembert's solution when you are given the 1-D wave equation with intial conditions (t=0), and use separation of variables when you are given both initial condtions and boundary conditions. For the latter method you simply suppose the the solution u(x,t)=X(x)Y(t) and then substitute this value for u into the wave equation. You will get 2 separate ODE's which you then solve, based on your initial conditions. I may be wrong but the problem of extending a function even/odd to a periodic one even/odd in some interval is a separate issue from solving the 1-D wave equation. Substitution of your initial conditions into the D'Alembert equation leads to a quick solution for which u(3/4,2) can be found.
     
  5. Dec 8, 2005 #4
    i should have stated earlier that the second question about separation is not related to the first question. ALso i know that i need to extend the functions as even or odd functions depending on the initial conditions and simple substitituion doesnt work since such an answer (which i have tried to use) is not correct.

    All i want to know is if my extensions of f and g are correct or not?
     
  6. Dec 8, 2005 #5
    Ok, here is the deal.

    You use the d'Alembert's formula when you are given a wave equation, but your domain is the entire real line. You use seperation of variables if your domain is bounded, say from 0 to L.

    Now, I am a bit confused about your problem. I can help you figure it out, but can you state it in better words. You are given certain initial conditions, is that it? Whats the domain for your problem?

    - harsh
     
  7. Dec 8, 2005 #6
    ok before you ask me for initial conditions and all please look at this thread and see how saltydog helped me solve it
    https://www.physicsforums.com/showthread.php?t=90027

    not initial conditions were needed to be considered for it.

    also i belive the string is assuemd to be fixed at its end points so that all i need to do is even extend f and g
    THat was the assumption made in that thread cited above
     
  8. Dec 9, 2005 #7

    saltydog

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    Stunner . . . first precisely state the problem:

    [tex]\text{DE:}\quad \frac{\partial^2 u}{\partial t^2}=\frac{\partial^2 u}{\partial x^2};\quad 0\le x\le 1,\quad -\infty<t<\infty[/tex]

    [tex]\text{BC:}\quad u(0,t)=0,\quad u(1,t)=0[/tex]

    [tex]\text{IC:}\quad u(x,0)=f(x),\quad u_t(x,0)=g(x)[/tex]

    See, that's for a finite string, same dif for infinite or semi-infinite: Use the method of images (odd and even extensions) and D'Alembert's method. Note for the above boundary conditions, you odd-extend the initial conditions but for derivative boundary conditions you even-extend the initial conditions. As far as how to extend them, well first try extending them across the origin, that's an easy way to see how it's being extended. What you did above, reflect it, is NOT correct which I think you'd see if you first extended it across the origin and then continued to extend it so that it looks kinda like a sine-wave. The odd extension of g(x) in [1,2] is indeed:

    [tex]g_{(1,2)}(x)=-(2-x)^2(x-1)[/tex]

    Plot it and see.

    Also remember when you're integrating that part in D'Alembert's solution, you need to remember to integrate it piece-wise, that is, integrate each extension over it's interval of definition.
     
  9. Dec 9, 2005 #8
    ok i see what u mean.. but is there a more systematic method of obtaining the result liek that? ALso if in this case the ends of the string were not kept fixed how would i even extend the functions?
     
  10. Dec 9, 2005 #9
    Ok, this is what even and odd extensions mean: Even extensions are symmetric over the y-axis, while odd extensions are symmetric over the origin. I dont have a way to show you how those work, but you can plot a few functions from say 0<x<1, and extend then in an even or odd manner.

    To Saltydog, thats a new way to solve PDE's that I didnt know before. In our class, we only used d'Alembert's if the domain was -infinity< x <infinity, t > 0. Does t> - infinity even make sense?

    - harsh
     
  11. Dec 9, 2005 #10
    Another thing is, you can easily solve the problem using seperation of variables, but saltydo's method seems fairly straightforward as well, but you must be careful with what you are integrating over what limits.

    - harsh
     
  12. Dec 9, 2005 #11

    saltydog

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    Well that does look a little odd. That's just the standard notation used in my favorite PDE text book, "Basic PDEs" by D. Bleecker and G. Csordas. Perhaps one could say that as long as t is finite, then it's greater than -infinity.
     
  13. Dec 9, 2005 #12

    saltydog

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    You know what, need to make sure this is all correct. What is u(3/4,2) using the method of images discussed above and is it the answer that is published?

    Also, with regards to a systematic approach: suppose have f(x) defined on [0,1] and f(0)=0 and f(1)=0. Then to odd-extend it, can you see the logic with the following sequence:

    [tex]f_{(1,2)}(x)=-f_{(0,1)}(2-x)[/tex]

    [tex]f_{(2,3)}(x)=f_{(0,1)}(x-2)[/tex]

    [tex]f_{(3,4)}(x)=-f_{(0,1)}(4-x)[/tex]

    [tex]f_{(4,5)}(x)=f_{(0,1)}(x-4)[/tex]

    The first plot is the odd extension. Can you come up with the sequence for the second plot which is the even-extension?

    Edit: Corrected f(4,5)
     

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    Last edited: Dec 9, 2005
  14. Dec 9, 2005 #13
    but inyour answer F(1,2) is the same thing as F(2,3) ? How so?
    The answer for this question is
    905/256
     
  15. Dec 10, 2005 #14

    saltydog

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    Don't understand you Stunner. In the sequence above for

    [tex]f_{(0,1)}(x)=x(1-x)[/tex]

    I obtained:

    [tex]f_{(1,2)}=-(2-x)(x-1)[/tex]

    [tex]f_{(2,3)}=(x-2)(3-x)[/tex]

    Another question: I solved for u(3/4,2) for the IBVP I stated above and obtained 3/16. Now, that could be wrong but I think I'm pretty good at these. Is your IBVP exactly the one I stated above?
     
    Last edited: Dec 10, 2005
  16. Dec 10, 2005 #15

    saltydog

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    This is mine:

    [tex]u(3/4,2)=\frac{1}{2}\left(\widetilde{f_0}(11/4)+
    \widetilde{f_0}(-5/4)\right)+
    \frac{1}{2}\int_{-5/4}^{11/4}\widetilde{g_0}(r)dr\right)[/tex]

    where:

    [tex]\widetilde{f_0},\quad\widetilde{g_0}[/tex]

    are the odd-extensions of the initial conditions.

    Thus:

    [tex]
    \begin{align*}
    u(3/4,2)&=\frac{1}{2}\left(f_{(2,3)}(11/4)+f_{(-2,-1)}(-5/4)\right) \\
    &+\frac{1}{2}\left(\int_{-5/4}^{-1} g_{(-2,-1)}(r)dr+\int_1^2 g_{(1,2)}(r)dr+\int_2^{11/4} g_{(2,3)}(r)dr\right) \\
    &=1/2\left[(11/4-2)(3-11/4)-(2-5/4)(1-5/4)\right]+0 \\
    &=3/16
    \end{align*}
    [/tex]
     
    Last edited: Dec 10, 2005
  17. Dec 11, 2005 #16

    saltydog

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    Hey Stunner, just for the record I checked 3/16 against both back-substitution into the IBVP and against numerical calculations both of which agreed with this answer. I suggest either your result is not correct or we are working on two different IBVPs. Do you understand the formula for u(3/4,2) that I gave above?

    Cheers,
    Salty
     
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