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Prove that f(x,y)=U(x+y)+V(x-y)

  1. Nov 5, 2016 #1
    Member advised that the homework template is not optional
    Let ##h(u,v)=f(u+v,u-v)## and ##f_{xx}=f_{yy}## for every ##(x,y)\in\mathbb{R}^2##. In addition, ##f\in{C^2}.## Show that ##f(x,y)=U(x+y)+V(x-y)##.

    I think applying the Taylor theorem would be useful.

    ##f(x,y)=f(x+h_1,y+h_2)-\left(\frac{\partial{f(x,y)}}{\partial{x}}h_1+\frac{\partial{f(x,y)}}{\partial{y}}h_2\right)-\frac 1 2\left(\frac{\partial^2{f(x,y)}}{\partial^2{x}}h_1^2+\frac{\partial^2{f(x,y)}}{\partial{x}\partial{y}}h_1h_2+\frac{\partial^2{f(x,y)}}{\partial^2{y}}h_2^2\right)-R(h_1,h_2)##

    How would one proceed?
    An alternative, maybe more elegant way would be to denote ##x+y=\epsilon##, ##x-y=\mu## and solve the hyperbolic partial differential equation ##\frac{\partial^2{f}}{\partial{\epsilon}\partial{\mu}}=0##, but we haven't actually covered them in class yet.

    ##\frac{\partial^2{f}}{\partial{\epsilon}\partial{\mu}}=0## ⇔ ##\frac{\partial{f}}{\partial{\epsilon}}=C(\epsilon)## etc.
     
    Last edited by a moderator: Nov 5, 2016
  2. jcsd
  3. Nov 5, 2016 #2

    Orodruin

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    Have you tried following through on your last alternative?
     
  4. Nov 5, 2016 #3
    Yes,

    ##\frac{\partial{f}}{\partial{\epsilon}}=C(\epsilon)##⇔##\int{C(\epsilon)d\epsilon}+D(\mu)=E(\epsilon)+D(\mu)=E(x+y)+D(x-y).##
    That's pretty straightforward, isn't it? However, we are not yet taught how to solve partial differential equations, so I am not sure if that's what the question setter intended. I am not completely sure how to derive the situation ##\frac{\partial^2{f}}{\partial{\epsilon}\partial{\mu}}=0## either. First we take partial derivatives of ##f## with respect to ##x## and ##y## using the chain rule, but what is the next step?



    ##
    \frac{\partial f}{\partial x} =
    \left(
    \frac{\partial f}{\partial \xi}
    \right)
    \frac{\partial \xi}{\partial x}
    +
    \left(
    \frac{\partial f}{\partial \eta}
    \right)
    \frac{\partial \eta}{\partial x}
    =
    \frac{\partial f}{\partial \xi}
    +
    \frac{\partial f}{\partial \eta}
    \\
    \frac{\partial f}{\partial y} =
    \left(
    \frac{\partial f}{\partial \xi}
    \right)
    \frac{\partial \xi}{\partial y}
    +
    \left(
    \frac{\partial f}{\partial \eta}
    \right)
    \frac{\partial \eta}{\partial y}
    =
    -\frac{\partial f}{\partial \xi}
    +
    \frac{\partial f}{\partial \eta}

    ##
     
    Last edited: Nov 5, 2016
  5. Nov 5, 2016 #4

    LCKurtz

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    I haven't worked through this problem, but I have a couple of suggestions. First, I don't think Taylor series is what you want. Second, you haven't used anywhere that ##f_{xx} = f_{yy}##. I'm thinking you might want to take partial derivatives and use the fact that ##f_{xx} = f_{yy}## somehow.
     
  6. Nov 5, 2016 #5
    That's true, please take a look at my post above.
     
  7. Nov 5, 2016 #6

    Orodruin

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    So these are the first derivatives, what about the second derivatives that are in ##f_{xx} = f_{yy}##?
     
  8. Nov 5, 2016 #7
    ##\frac{\partial^2{f}}{\partial{x^2}}=\frac{\partial{f}}{\partial{x}}(\frac{\partial{f}}{\partial{\epsilon}}+\frac{\partial{f}}{\partial{\eta}})##
    and
    ##\frac{\partial^2{f}}{\partial{y^2}}=\frac{\partial{f}}{\partial{x}}(-\frac{\partial{f}}{\partial{\epsilon}}+\frac{\partial{f}}{\partial{\eta}})##

    ##\frac{\partial{f}}{\partial{x}}(\frac{\partial{f}}{\partial{\epsilon}}+\frac{\partial{f}}{\partial{\eta}})-\frac{\partial{f}}{\partial{x}}(-\frac{\partial{f}}{\partial{\epsilon}}+\frac{\partial{f}}{\partial{\eta}})=0##

    ##\frac{\partial^2{f}}{\partial{x}\partial{\epsilon}}+\frac{\partial^2{f}}{\partial{x}\partial{\eta}}+\frac{\partial^2{f}}{\partial{x}\partial{\epsilon}}-\frac{\partial^2{f}}{\partial{x}\partial{\eta}}=0##

    ##2\frac{\partial^2{f}}{\partial{x}\partial{\epsilon}}=0##

    ##\frac{\partial^2{f}}{\partial{x}\partial{\epsilon}}=0##

    But there should be ##\eta## instead of ##x##. Maybe it requires coordinate transformation or something? I am afraid that's what I can't do. Since ##f\in{C^2}##, the fact that ##f_{xy}=f_{yx}## may be useful?
     
    Last edited: Nov 5, 2016
  9. Nov 5, 2016 #8

    Ray Vickson

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    You don't need to use much information about partial differential equation. Just use the fact that ##\partial^2 f/\partial \epsilon \partial \mu = 0## implies that ##\partial f/ \partial \epsilon## is not a function of ##\mu## (i.e., is a constant with respect to varying ##\mu##), and that ##\partial f/\partial \mu## is not a function of ##\epsilon##. That ought to be enough.
     
  10. Nov 5, 2016 #9
    I suppose the partial differential equation needs to be derived somehow first, which is giving me trouble.
     
  11. Nov 5, 2016 #10
    Hint: Calculate $$\frac{\partial ^2h}{\partial u \partial v}$$ and use information you are given.
     
  12. Nov 5, 2016 #11

    Orodruin

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    He is already doing that, his problem is deriving that differential equation from the wave equation.

    This is not correct, it is the second derivative of ##f## you are after, not the product of two first derivatives.
     
  13. Nov 5, 2016 #12

    PeroK

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    It seems to me you got off on the wrong foot. What you know about ##f## is that ##f_{xx} = f_{yy}##.

    Then, it seems to me you were given a hint to define:

    ##h(u, v) = f(u+v, u-v)##

    With the implication that you analyse the function ##h## and in particular the pd's ##h_{uv}## and ##h_{vu}##.
     
  14. Nov 6, 2016 #13
    ##\frac{\partial{h}}{\partial{u}}=\frac{\partial{f}}{\partial{\epsilon}}\frac{\partial{(u+v)}}{\partial{u}}+\frac{\partial{f}}{\partial{\eta}}\frac{\partial{(u-v)}}{\partial{u}}=\frac{\partial{f}}{\partial{\epsilon}}+\frac{\partial{f}}{\partial{\eta}}##
    How to derivate that with respect to ##v##?I am confused. Anyway, I have seen derivative factorization like this:
    $$
    \left(
    \frac{\partial}{\partial x}\frac{\partial}{\partial x}
    \right) f
    =
    \left(
    \frac{\partial}{\partial \xi}
    +
    \frac{\partial}{\partial \eta}
    \right)
    \left(
    \frac{\partial}{\partial \xi}
    +
    \frac{\partial}{\partial \eta}
    \right) f
    =
    \left(
    \frac{\partial}{\partial \xi}
    \right)^2 f
    +
    2
    \left(
    \frac{\partial}{\partial \xi}
    \frac{\partial}{\partial \eta}
    \right) f
    +
    \left(
    \frac{\partial}{\partial \eta}
    \right)^2 f \iff \\
    f_{xx} = f_{\xi\xi} + 2 f_{\xi\eta} + f_{\eta\eta}
    \\
    \left(
    \frac{\partial}{\partial y}\frac{\partial}{\partial y}
    \right) f
    =
    \left(
    -\frac{\partial}{\partial \xi}
    +
    \frac{\partial}{\partial \eta}
    \right)
    \left(
    -\frac{\partial}{\partial \xi}
    +
    \frac{\partial}{\partial \eta}
    \right) f
    =
    \left(
    \frac{\partial}{\partial \xi}
    \right)^2 f
    -
    2
    \left(
    \frac{\partial}{\partial \xi}
    \frac{\partial}{\partial \eta}
    \right) f
    +
    \left(
    \frac{\partial}{\partial \eta}
    \right)^2 f \iff \\
    f_{yy} = f_{\xi\xi} - 2 f_{\xi\eta} + f_{\eta\eta},$$

    thus, $$f_{xx}-f_{yy}=0⇔$$ $$\frac{\partial^2{f}}{\partial{\epsilon}\partial{\eta}}=0$$What is that factorization based on? At a first glance it would look like squaring the partials, but it's not. Are there alternative ways or approaches? Does the problem require solving the PDE? I mean, I don't want to write down anything without fully understanding what's going on.
     
    Last edited: Nov 6, 2016
  15. Nov 6, 2016 #14

    PeroK

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    This problem highlights that you need to understand what functions and partial derivatives are.

    If you have a function of two variables, the you can take the partial derivative with respect to the first and the second variable. One confusion arises from how to denote those partial derivatives. For a single-variable function you can write ##f'## and that does not imply any particular variable. But, for a multi-variable functions, we usually write ##f_x## and ##f_y##. So, what exactly are ##x## and ##y## here?

    ##f_x## is the well-defined function of two variables obtained by taking the pd of ##f## with respect to its first variable and similarly for ##f_y##

    In this case, therefore, ##x## and ##y## are dummy variables with ##x## standing for "the first variable" and ##y## "the second variable".

    You could, if you wanted, change the terminology and use ##f_1, f_2##, say, instead of ##f_x, f_y## and in many ways this would be simpler:

    If you had something like:

    ##g(x, y) = h(x^2, xy)## then the chain rule would be:

    ##g_1(x, y) = h_1(x^2, xy)(2x) + h_2(x^2, xy)y##

    But, it's usual to write this as:

    ##g_x(x, y) = h_x(x^2, xy)(2x) + h_y(x^2, xy)y##

    (But, I personally still read this as: the pd of ##g## wrt its first variable equals the pd of ##h## wrt its first variable times the derivative of ##h##'s first variable wrt ##g##'s first variable etc. And that's the chain rule.

    In summary, one way to understand the chain rule properly is to understand what these functions ##g_x, g_y, h_x, h_y## really are.

    If we go back to the problem, you were originally given:

    ##h(u, v) = f(u+v, u-v)##

    There is no need to introduce additional variables ##\epsilon, \mu## here. Can you apply the chain rule to this equation?
     
  16. Nov 6, 2016 #15
    Well, ##h_u(u,v)=f_u(u+v)(1)+f_v(u-v)(-1)=f_u(u+v)-f_v(u-v)##?
    ##h_{uv}(u,v)=##
     
  17. Nov 6, 2016 #16

    PeroK

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    The pd's are also functions of two variables. As long as you use ##x## as a dummy variable, then you can use ##x## for all the pd's but, given that ##h## was defined in terms of ##(u, v)## it's more conventional to write:

    ##h_u(u, v) = f_x(u+v, u-v) + f_y(u+v, u-v)##

    How I see it personally (to keep it all straight) is that:

    We define some ##f## in terms of ##x## and ##y##. We then calculate the pd's of ##f##, which we label as ##f_x, f_y##.

    Then we define ##h## in terms of ##u, v##, hence we ought to label the pd's as ##h_u, h_v##

    But, in a sense, how we label the pd's doesn't matter as long as we know what they mean mathematically.
     
    Last edited: Nov 6, 2016
  18. Nov 6, 2016 #17
    So now ##h_u=f_1(u+v,u-v)-f_2(u+v,u-v)## and ##h_{uv}=f_{12}(u+v,u-v)(1)-f_{22}(u+v,u-v)(-1)##?
     
  19. Nov 6, 2016 #18

    PeroK

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    No, you need to take more care. Let's stick to the conventional notation:

    ##h_u(u, v) = f_x(u+v, u-v) + f_y(u+v, u-v)##

    To get ##h_{uv}(u, v)## you must partially differentiate both ##f_x## and ##f_y## with respect to ##v##. These are just functions, like ##f## is. So, you're going to get two terms for each.
     
    Last edited: Nov 6, 2016
  20. Nov 6, 2016 #19

    PeroK

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    It's also a good idea to have an example to differentiate to check your formulas. You could try:

    ##f(x, y) = x^2y + 2xy^2##

    You can now define ##h(u, v) = f(u+v, u-v)## and calculate ##f_x, f_y, h_u, h_v## etc. explicitly.
     
  21. Nov 6, 2016 #20
    This is hopeless.
    ##\frac{\partial{h}}{\partial{u}}(x,y)=\frac{\partial{f}}{\partial{x}}(x,y)+\frac{\partial{f}}{\partial{y}}(x,y)## and ##\frac{\partial{h}}{\partial{v}}(x,y)=\frac{\partial{f}}{\partial{x}}(x,y)-\frac{\partial{f}}{\partial{y}}(x,y)##

    ##\frac{\partial^2{h}}{\partial{u}}(x,y)=\frac{\partial^2{h}}{\partial{x^2}}(x,y)+\frac{\partial^2{h}}{\partial{y^2}}(x,y)+2\frac{\partial^2{h}}{\partial{x}\partial{y}}##

    ##\frac{\partial^2{h}}{\partial{v}}(x,y)=\frac{\partial^2{h}}{\partial{x^2}}(x,y)+\frac{\partial^2{h}}{\partial{y^2}}(x,y)-2\frac{\partial^2{h}}{\partial{x}\partial{y}}##

    ##\frac{\partial^2{f}}{\partial{x^2}}(u+v,u-v)-\frac{\partial^2{f}}{\partial{y^2}}(u+v,u-v)=(\frac{\partial^2{f}}{\partial{x^2}}(u+v,u-v)+\frac{\partial^2{f}}{\partial{y^2}}(u+v,u-v)+2\frac{\partial^2{f}}{\partial{x}\partial{y}}(u+v,u-v))-(\frac{\partial^2{f}}{\partial{x^2}}(u+v,u-v)+\frac{\partial^2{f}}{\partial{y^2}}(u+v,u-v)-2\frac{\partial^2{f}}{\partial{x}\partial{y}}(u+v,u-v))=0##

    Why would you consider ##h_{uv}## and ##h_{vu}## instead of ##f_{xx}## and ##f_{yy}##?
     
    Last edited: Nov 6, 2016
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