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    Q: If w = f(x,y), where x = e^r \cos \theta and y=e^r \sin \theta. Show that w_{xx} + w_{yy} = e^{-2r} (w_{rr} + w_{\theta \theta}).

    Using the chain rule \displaystyle \frac{\partial w}{\partial r} = \displaystyle \frac{\partial w}{\partial x} \displaystyle \frac{\partial x}{\partial r} + \displaystyle \frac{\partial w}{\partial y} \displaystyle \frac{\partial y }{\partial r}, I get w_r = f_x (x,y).e^r \cos \theta + f_y (x,y). e^r \sin \theta.

    But now I don't know how to find w_{rr}, when I have the function w_r in terms of f_x (x,y), f_y (x,y), \theta and r. If it would've been in terms of theta and r only, then it'd have been easy. How do I evaluate it further? I presume I'd get a form of f_{xy} (x,y) maybe.

    Please help!
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    (Original post by Zishi)

    Please help!
    As no one else has replied.

    Yes, you'll get terms in f_{xy} (x,y),f_{yx} (x,y),f_{yy} (x,y),f_{xx} (x,y) as well as f_{y} (x,y),f_{x} (x,y) via chain rule and product rule.
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    (Original post by ghostwalker)
    As no one else has replied.

    Yes, you'll get terms in f_{xy} (x,y),f_{yx} (x,y),f_{yy} (x,y),f_{xx} (x,y) as well as f_{y} (x,y),f_{x} (x,y) via chain rule and product rule.
    Sorry for late reply. I'll try doing it via chain rule and product rule. Just wanted to ask one more thing. I learned that putting value of \cos \theta and \sin \theta i.e \dfrac{x}{e^r} and \dfrac{y}{e^r} in \sin^2 \theta + \cos^2 \theta = 1 and making the right hand side of that equation equal to zero will gives us the left hand side as the function of x and y, i.e f(x,y). I used that function directly(i.e not working in terms of f(x,y)), and I was able to prove the equation.

    My question is how does that way work? That doesn't make any sense - I mean that function f(x,y) could have any equation, so how can we get its equation by just using an identity?
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    (Original post by Zishi)
    Sorry for late reply. I'll try doing it via chain rule and product rule. Just wanted to ask one more thing. I learned that putting value of \cos \theta and \sin \theta i.e \dfrac{x}{e^r} and \dfrac{y}{e^r} in \sin^2 \theta + \cos^2 \theta = 1 and making the right hand side of that equation equal to zero will gives us the left hand side as the function of x and y, i.e f(x,y). I used that function directly(i.e not working in terms of f(x,y)), and I was able to prove the equation.

    My question is how does that way work? That doesn't make any sense - I mean that function f(x,y) could have any equation, so how can we get its equation by just using an identity?
    You'll have to elaborate as I don't follow. Also the bit in bold.
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    (Original post by ghostwalker)
    You'll have to elaborate as I don't follow. Also the bit in bold.
    I meant to say that I have been told that putting \cos \theta and \sin \theta i.e \dfrac{x}{e^r} and \dfrac{y}{e^r} in the identity \sin^2 \theta + \cos^2 \theta = 1 gives  \left(\dfrac{x}{e^r}  \right)^2 + \left(\dfrac{y}{e^r} \right)^2 = 1 \Rightarrow x^2 + y^2 = e^{2r} \Rightarrow x^2 + y^2 - e^{2r} = 0. And then taking f(x,y) = x^2 + y^2 - e^{2r} and putting f(x,y) = x^2 + y^2 - e^{2r} in w_r = f_x (x,y).e^r \cos \theta + f_y (x,y). e^r \sin \theta(from post #1) and then solving the equation(which is easy to do when we've got an equation of f(x,y)) proves the question in the first post.

    I just wanted to ask that why does that work? How is an identity useful in finding the equation of a function when we're given it's variables in terms of other variables(x, y in r and theta)?
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    (Original post by Zishi)
    I meant to say that I have been told that putting \cos \theta and \sin \theta i.e \dfrac{x}{e^r} and \dfrac{y}{e^r} in the identity \sin^2 \theta + \cos^2 \theta = 1 gives  \left(\dfrac{x}{e^r}  \right)^2 + \left(\dfrac{y}{e^r} \right)^2 = 1 \Rightarrow x^2 + y^2 = e^{2r} \Rightarrow x^2 + y^2 - e^{2r} = 0. And then taking f(x,y) = x^2 + y^2 - e^{2r} and putting f(x,y) = x^2 + y^2 - e^{2r} in w_r = f_x (x,y).e^r \cos \theta + f_y (x,y). e^r \sin \theta(from post #1) and then solving the equation(which is easy to do when we've got an equation of f(x,y)) proves the question in the first post.

    I just wanted to ask that why does that work? How is an identity useful in finding the equation of a function when we're given it's variables in terms of other variables(x, y in r and theta)?
    If I read this right, you've not actually proved the original question for a general f(x,y) with this method, but rather taken a specific f(x,y), i.e. x^2 + y^2 - e^{2r} which actually involves an r as well, and shown that it's true for that.
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    (Original post by ghostwalker)
    If I read this right, you've not actually proved the original question for a general f(x,y) with this method, but rather taken a specific f(x,y), i.e. x^2 + y^2 - e^{2r} which actually involves an r as well, and shown that it's true for that.
    Yes, that r has to be taken as a constant. Although I'd always like to prove for a general f(x,y), but I don't understand this method of finding a specific f(x,y).
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    (Original post by Zishi)
    Yes, that r has to be taken as a constant. Although I'd always like to prove for a general f(x,y), but I don't understand this method of finding a specific f(x,y).
    You say "method". Is this something you've come up with/noticed?

    PS: I'm not the most knowledgeable person on partial derivatives, so anyone more up to speed, feel free to chip in.
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    (Original post by ghostwalker)
    You say "method". Is this something you've come up with/noticed?

    PS: I'm not the most knowledgeable person on partial derivatives, so anyone more up to speed, feel free to chip in.
    Nah, I was having problem with evaluating the derivatives(chain rules, etc) in terms of f(x,y), f_{xy} (x,y),f_{yx} (x,y),f_{yy} (x,y),f_{xx} (x,y). So I asked a friend to solve this question and he came up with this "method". The thing is that he was unable to explain how it works.
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    (Original post by Zishi)
    Nah, I was having problem with evaluating the derivatives(chain rules, etc) in terms of f(x,y), f_{xy} (x,y),f_{yx} (x,y),f_{yy} (x,y),f_{xx} (x,y). So I asked a friend to solve this question and he came up with this "method". The thing is that he was unable to explain how it works.
    If you write practically any function of x and y and have w=f(x,y), x=e^r cos t and y =e^r sin t then it will satisfy that pde. That's the whole point. You just need the various derivatives to exist.

    Did you prove it in the end?
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    (Original post by BabyMaths)
    If you write practically any function of x and y and have w=f(x,y), x=e^r cos t and y =e^r sin t then it will satisfy that pde. That's the whole point. You just need the various derivatives to exist.

    Did you prove it in the end?
    So I need to write any function of f(x,y) which will NOT give f_{xy} (x,y),f_{yx} (x,y),f_{yy} (x,y),f_{xx} (x,y) equal to zero, right? And btw now I understand how to do it generally, thanks to your solution. (PRSOM)
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    (Original post by Zishi)
    So I need to write any function of f(x,y) which will NOT give f_{xy} (x,y),f_{yx} (x,y),f_{yy} (x,y),f_{xx} (x,y) equal to zero, right? And btw now I understand how to do it generally, thanks to your solution. (PRSOM)
    For the question you gave you don't have to find f(x,y) at all.

    The derivatives being = 0 is not a problem.

    f(x,y)=0 works.

    f(x,y)=x+y^2 works.

    I think, in fact, that we just need w_{xx} etc to be continuous.
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    (Original post by BabyMaths)
    For the question you gave you don't have to find f(x,y) at all.

    The derivatives being = 0 is not a problem.

    f(x,y)=0 works.

    f(x,y)=x+y^2 works.

    I think, in fact, that we just need w_{xx} etc to be continuous.
    Alright. Thanks again.
 
 
 
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