1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    [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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    [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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Need to check the answer of some simple PDE
  1. Answer need checking! (Replies: 0)

  2. Need an answer check (Replies: 7)

Loading...