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Why is differential equation equal to sum of partials?

  1. Feb 8, 2016 #1
    My current understanding of differential equations is extremely shaky, and my vocabulary is probably very incorrect, but I'm curious about something I've recently seen in some Khan Academy videos (specifically this one) and in other situations with differential equations. It seems that the following is true:
    [tex]\frac{d}{dx} \psi (x, y) = \frac{∂\psi}{∂x}\frac{dx}{dx} + \frac{∂\psi}{∂y}\frac{dy}{dx} [/tex]
    Why is this? The video on Khan Academy shows a "proof" by which he assumes ##\psi## can be represented by a sum of products of functions of x and y (##\psi = f_1(x)g_1(y) + ... + f_n(x)g_n(y)##). What is the basis of this "proof"? Why could some function of x and y, psi, be displayed as products of separate functions of x or y?

    I feel as though following the flow of videos on Khan Academy isn't explaining differential equations well-enough for me. Can anyone here recommend where else I might look to learn about differential equations?

    Thanks for any help!
     
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  3. Feb 8, 2016 #2

    Orodruin

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    I suggest you go back to the basic definition of a derivative instead and take the actual limit.
     
  4. Feb 8, 2016 #3

    HallsofIvy

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    The "chain rule" for functions of x and y, each of which is a function of t, says that
    [tex]\frac{d\phi}{dt}= \frac{\partial \phi}{\partial x}\frac{dx}{dt}+ \frac{\partial \phi}{\partial y}\frac{dy}{dt}[/tex]

    For your result, think of y as a function of x and replace each "t" with "x".
     
  5. Feb 8, 2016 #4
    Okay, so what I'm looking at is the chain rule for PDEs? In that case, could you help me figure out how this is derived? I've looked over some things after a quick Google search (Paul's Online Notes, Imperial College of London [pdf], and Harvey Mudd College [pdf]), but none of these resources show any derivation of this or why it is so, unless I'm misreading them.

    The first link shows the typical chain rule for calculus (##y = f(x), x = g(t), \frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt}##), and then goes straight into "Case 1" (the only case I'm considering at the moment), showing your example. I don't see how the chain rule for PDEs relates to the chain rule for basic differentiation based on this; where does the addition come from?

    The second link again only defines this chain rule with no derivation.

    The third link references a "formal proof" that I have yet to find, and continues to use some intuition based on a diagram that doesn't seem at all to be mathematical proof.
    So the basic definition of a derivative would be an infinitesimal change in something with respect to an infinitesimal change in something else, right? By take the actual limit, do you mean use the First Principle on my function Psi with respect to x?
    [tex]\frac{d}{dx}\psi (x, y) = \lim_{h\to0} \frac{\psi (x + h, y) - \psi (x, y)}{h}[/tex]
    I'm not sure I'm following what your hinting at, or at least I don't know how I would use the above limit to deduce the chain rule for PDEs.

    I'm sorry, I appear to be missing some extremely fundamental and basic intuition or concept. Am I getting any closer to an understanding, or is it best for me at this point to start off back at the basics of calculus and work my way back to this point?
     
  6. Feb 8, 2016 #5

    HallsofIvy

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    This not the "chain rule for PDEs", it is the chain rule for any function of several variables. You should have learned it in Calculus before you started studying Differential Equations and well before studying Partial Differential Equations".
     
  7. Feb 8, 2016 #6
    I see. In that case, I'm guessing that most of what I've asked/said in this thread has been completely nonsensical. At any rate, it looks like I'll have to start over at the beginning of some calculus textbook.

    Thanks for your help up until now, @Orodruin and @HallsofIvy!
     
  8. Feb 9, 2016 #7

    Orodruin

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    Yes, but you have to remember that y is also a function of x and you are taking a total derivative. You should replace x with x+h everywhere in the first term, including in y(x).
     
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