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Homework Help: Need to check the answer of some simple PDE

  1. Feb 6, 2010 #1
    I don't have the answer of these question. Can someone take a look at a) and tell me am I correct? I don't even know how to solve b)

    a)1. The problem statement, all variables and given/known data

    a) Show [tex]u(x,t)=F(x+ct) + G(x-ct)[/tex] is solution of [tex]\frac{\partial^2 u}{\partial t^2}= c^2\frac{\partial^2 u}{\partial x^2}[/tex]



    3. The attempt at a solution

    let [tex]\alpha = x+ct,\beta = x-ct \Rightarrow \frac{\partial \alpha}{\partial x}= 1, \frac{\partial \alpha}{\partial t}= c,[/tex] and also [tex] \frac{\partial \beta}{\partial x}= 1, \frac{\partial \beta}{\partial t}= -c[/tex]

    [tex]\frac{\partial u}{\partial x} = \frac{\partial F(\alpha)}{\partial \alpha} \frac{\partial \alpha}{\partial x} + \frac{\partial G(\beta)}{\partial \beta}\frac{\partial \beta}{\partial x} = \frac{\partial F(\alpha)}{\partial \alpha} + \frac{\partial G(\beta)}{\partial \beta}[/tex]

    [tex]\frac{\partial^2 u}{\partial x^2} = \frac {\partial (\frac{\partial F(\alpha)} {\partial \alpha}) }{\partial \alpha} \frac{\partial \alpha}{\partial x} +

    \frac {\partial (\frac{\partial G(\beta)}{\partial \beta })} {\partial \beta} \frac{\partial \beta}{\partial x}

    = \frac{\partial^2 F(\alpha)}{\partial \alpha^2} + \frac{\partial^2 G(\beta)}{\partial \beta^2}[/tex]



    [tex]\frac{\partial u}{\partial t} = \frac{\partial F(\alpha)}{\partial \alpha}\frac{\partial \alpha}{\partial t} + \frac{\partial G(\beta)}{\partial \beta}\frac{\partial \beta}{\partial t} = c[\frac{\partial F(\alpha)}{\partial \alpha} - \frac{\partial G(\beta)}{\partial \beta}][/tex]

    [tex]\frac{\partial^2 u}{\partial t^2} = c\frac {\partial (\frac{\partial F(\alpha)} {\partial \alpha}) }{\partial \alpha} \frac{\partial \alpha}{\partial t} -

    c\frac {\partial (\frac{\partial G(\beta)}{\partial \beta })} {\partial \beta} \frac{\partial \beta}{\partial t}

    = c^2[\frac{\partial^2 F(\alpha)}{\partial \alpha^2} + \frac{\partial^2 G(\beta)}{\partial \beta^2}][/tex]


    [tex]\Rightarrow \frac{\partial^2 u}{\partial t^2}= c^2\frac{\partial^2 u}{\partial x^2}[/tex]

    Therefore:
    [tex]u(x,t)=F(x+ct) + G(x-ct)[/tex] is solution of [tex]\frac{\partial^2 u}{\partial t^2}= c^2\frac{\partial^2 u}{\partial x^2}[/tex]



    b) Transform [tex]\frac{\partial^2 u}{\partial t^2}= c^2\frac{\partial^2 u}{\partial x^2}[/tex] in to [tex]\frac{\partial^2 u}{\partial \alpha \partial \beta}=0[/tex]

    where [tex]u(x,t)=F(x+ct) + G(x-ct)[/tex]

    and [tex]\alpha = x+ct,\beta = x-ct[/tex]

    Can someone give me a hint how to go by this?
     
    Last edited: Feb 6, 2010
  2. jcsd
  3. Feb 6, 2010 #2

    vela

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    Your solution to (a) is fine.

    For part (b), try start by solving for x and t in terms of [itex]\alpha[/itex] and [itex]\beta[/itex].
     
  4. Feb 6, 2010 #3
    Thanks for checking on a).

    [tex]\alpha = x+ct,\beta = x-ct \\Rightarrow x=\frac{\alpha + \beta}{2}, t=\frac{\alpha - \beta}{2c}[/tex]

    [tex]\frac{\partial u}{\partial \alpha}= \frac{\partial u}{\partial x} \frac{\partial x}{\partial \alpha}=\frac{\partial F(\alpha}{\partial \alpha}+ \frac{\partial G(\beta)}{\partial x}\frac{\partial x}{\partial \beta}[/tex]
     
  5. Feb 6, 2010 #4
    Thanks for checking on a).

    [tex]\alpha = x+ct,\beta = x-ct \Rightarrow x=\frac{\alpha + \beta}{2}, t=\frac{\alpha - \beta}{2c}[/tex]

    [tex]\frac{\partial u}{\partial \alpha}= \frac{\partial u}{\partial x} \frac{\partial x}{\partial \alpha} + \frac{\partial u}{\partial t} \frac{\partial t}{\partial \alpha} = \frac{\partial F(\alpha)}{\partial \alpha}+ \frac{\partial G(\beta)}{\partial x}\frac{\partial x}{\partial \beta} + \frac{\partial G(\beta)}{\partial t}\frac{\partial t}{\partial \beta}[/tex]

    But how do I find [tex]\frac{\partial x}{\partial \alpha},\frac{\partial x}{\partial \beta}?[/tex]
     
  6. Feb 6, 2010 #5

    vela

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    [tex]x=\frac{\alpha+\beta}{2} \Rightarrow \frac{\partial x}{\partial \alpha} = \frac{1}{2}[/tex]

    and so on. You do pretty much the same thing you did in part (a) just with different variables.
     
  7. Feb 6, 2010 #6
    I try this part and I got a different answer:

    [tex]\frac{\partial x}{\partial \alpha} = \frac{\partial (\alpha + \beta)}{2} = \frac{1}{2} [\frac{\partial \alpha}{\partial \alpha} + \frac{\partial \beta}{\partial x} \frac{\partial x}{\partial \alpha}] = \frac{1}{2}[ 1 + \frac{\partial x}{\partial \alpha}][/tex]

    [tex]\Rightarrow \frac{\partial x}{\partial \alpha} = 1 [/tex]

    Which is the same as [tex]\frac {1}{(\frac{\partial \alpha}{\partial x})} = 1[/tex]

    What did I do wrong?
     
  8. Feb 6, 2010 #7

    vela

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    [itex]x=x(\alpha,\beta)[/itex] is a function of [itex]\alpha[/itex] and [itex]\beta[/itex]. When you take the partial derivative wrt [itex]\alpha[/itex], you hold [itex]\beta[/itex] constant.
     
  9. Feb 6, 2010 #8
    The thing that confuse me is both [tex] \alpha, \beta[/tex] have x and t. [tex] \alpha, \beta[/tex] are not independent variables like x and y in the normal case. [tex] \alpha = 2x-\beta[/tex].

    That is the thing that really throw me off all this time. That is the reason I did it in post #6. Why it is wrong in post #6.
     
  10. Feb 6, 2010 #9

    vela

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    You can say the same thing about x and t. They both depend on [itex]\alpha[/itex] and [itex]\beta[/itex]. It's like a change of basis from x and t to [itex]\alpha[/itex] and [itex]\beta[/itex].
     
  11. Feb 6, 2010 #10
    I still don't get it. Let say x is position, t is time. We equate [tex]\alpha = x+ct, \beta=x-ct[/tex]. This imply only [tex]\alpha, \beta[/tex] is depending on both position and time. That cannot reverse to imply x and t have a relation. x and t are completely indenpend variable. I just cannot turn this around in my mind.

    Also in post #6, I don't assume anything there, I just step by step derive the formula. I look at it again and again, I just don't see what I did wrong. Can you comment about post #6 where exactly I did wrong that it won't work even if [tex]\alpha, \beta[/tex] are being treated as independent variables.

    Thanks for your patient. I feel I am stuck in a spot.
     
  12. Feb 6, 2010 #11

    vela

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    Just as x and t are independent when you write [itex]\alpha=\alpha(x,t)[/itex] and [itex]\beta=\beta(x,t)[/itex], you consider [itex]\alpha[/itex] and [itex]\beta[/itex] to be independent when you write [itex]x=x(\alpha,\beta)[/itex] and [itex]y=y(\alpha,\beta)[/itex].

    In #6, you're taking the partial derivative with respect to [itex]\alpha[/itex]. That means [itex]\beta[/itex] is held constant so any derivative of [itex]\beta[/itex] will equal 0.
     
  13. Feb 6, 2010 #12
    Please bear with me, I am still stuck!!! What is wrong with this:

    [tex]\frac{\partial \beta}{\partial \alpha} = \frac{\partial \beta}{\partial x} \frac{\partial x}{\partial \alpha} = \frac{\partial \beta}{\partial x}[/tex]
     
  14. Feb 6, 2010 #13
    Think about it in terms of linear algebra: (1,0) and (0,1) are two independent vectors that span R^2. Spanning R^2 means that given any point p in R^2, you can find coordinates x and y such that p = x (1,0) + y (0,1). (1,0) and (1,1) are also independent and form a basis for R^2, but they are not orthogonal.

    The same is true for alpha and beta. They are two independent coordinates (in the linear algebra sense) and form a basis for R^2, but they are not orthogonal.
     
  15. Feb 6, 2010 #14
    Consider another example: take standard x-y coordinates and define a function t(x,y) = x+y. By the same argument,

    [tex]\frac{\partial y}{\partial x} = \frac{\partial y}{\partial t} \frac{\partial t}{\partial x} = 1 \cdot 1 = 1[/tex]

    Do you see what happened?
     
  16. Feb 7, 2010 #15
    I got out the multi variable book and read up on the chain rule, I am still reading and thinking about it. I thing I see wrong with my assumption is [tex]\frac{\partial x}{\partial \alpha} [/tex]is not legal in the chain rule because it is travelling "up" the diagram.

    I can see

    [tex]\frac{\partial y}{\partial x} = \frac{\partial y}{\partial t} \frac{\partial t}{\partial x} = 1 \cdot 1 = 1[/tex]

    Is wrong because t is function of y, y is not a function of t.

    [tex]\frac{\partial y}{\partial t}[/tex] is not allowed.

    Am I correct?

    Also is [tex] \alpha,\beta[/tex] independent variable that [tex]\frac{\partial \alpha}{\partial \beta}=0[/tex]?
     
    Last edited: Feb 7, 2010
  17. Feb 7, 2010 #16
    The diagrams are a good way to take partial derivatives. When you try to differentiate alpha with respect to beta, you are either moving up in the diagram or you need to build a circular diagram (beta is a function of x and t, x is a function of alpha and beta, etc).

    Have you tried drawing a picture of what the alpha and beta coordinate curves look like? Once you draw a picture, you can see that you can freely vary alpha without affecting beta. You can also do this algebraically: Suppose beta = x-ct = const or t = (x - const)/c. If you move along the line t = (x - const)/c, you will vary alpha while keeping beta fixed.

    It's the analog of moving parallel to the x axis (keeping the y coordinate fixed) in the x-y world. Alpha and beta are two independent coordinates, and the equations alpha = x + ct, beta = x - ct tell you how to convert between alpha/beta and x/y coordinates.
     
  18. Feb 8, 2010 #17
    Can anyone show me how to prove this part?

    Transform [tex]\frac{\partial^2 u}{\partial t^2}= c^2\frac{\partial^2 u}{\partial x^2}[/tex] in to [tex]\frac{\partial^2 u}{\partial \alpha \partial \beta}=0[/tex]

    where [tex]u(x,t)=F(x+ct) + G(x-ct)[/tex]

    and [tex]\alpha = x+ct,\beta = x-ct[/tex]

    This is not a school work and I am still struggling on the Chain rule. I am hopping if I get the steps to solve this problem, I can reverse it and try to understand this. I think I spent enough time on this already.

    Thanks
     
  19. Feb 8, 2010 #18
    I think you and vela have established that

    du/d(alpha) = du/dx dx/d(alpha) + du/dt dt/d(alpha) = 1/2 du/dx + 1/2 du/dt
    du/d(beta) = du/dx dx/d(beta) + du/dt dt/d(beta) = 1/(2c) du/dx - 1/(2c) du/dt

    Can you say anything about the second partial derivatives of u with respect to alpha and beta?
     
  20. Feb 8, 2010 #19
    Or if you already know this part:
    [tex]
    u(x,t)=F(x+ct) + G(x-ct) = F(\alpha) + G(\beta)
    [/tex]
    it's even easier:

    The derivative of F(alpha) with respect to beta is zero. Differentiating G'(beta) with respect to alpha makes that zero. QED. Alpha and beta are independent variables!
     
  21. Feb 8, 2010 #20
    I gave this a lot of thoughts. I still cannot accept [tex]\frac{\partial G(\beta)}{\partial \alpha}=0 [/tex]

    Please bear with me. Let's take a look at this example:

    Let [tex]G(\beta)=-\beta = \beta - 2\beta = x-ct-2x+2ct=(x+ct)-2x=\alpha -2x[/tex]

    [tex]\Rightarrow\frac{\partial G(\beta)}{\partial \alpha}= \frac{\partial \alpha}{\partial \alpha} + \frac{\partial 2x}{\partial \alpha}= 1+0=1 [/tex]

    Yes I know I am playing around and the argument is very thin, but never the less, it is valid. My whole point is just because [tex] \alpha[/tex] and [tex]\beta[/tex] are independent variable, [tex]\frac{\partial G(\beta)}{\partial \alpha}[/tex] not necessary equal 0.

    [tex]\frac{\partial \beta}{\partial \alpha}=0 [/tex] do not imply at all [tex]\frac{\partial G(\beta)}{\partial \alpha}=0 [/tex]

    I don't think the question is good to say [tex]\frac{\partial^2 u}{\partial \alpha \partial \beta}}=0[/tex] I have been struggling on this very point for two days!!!!

    Please tell me if I am correct.

    Thanks a million.

    Alan
     
    Last edited: Feb 9, 2010
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