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Simple Partial Differentiation problem

  1. Sep 13, 2013 #1
    1. The problem statement, all variables and given/known data
    If ##z=x\ln(x+r)-r## where ##r^2=x^2+y^2##, prove that
    $$\frac{∂^2z}{∂x^2}+\frac{∂^2z}{∂y^2}=\frac{1}{x+y}$$


    2. Relevant equations



    3. The attempt at a solution
    Since ##r^2=x^2+y^2##, ##∂r/∂x=x/r## and ##∂r/∂y=y/r##.
    Differentiating z w.r.t x partially,
    $$\frac{∂z}{∂x}=\ln(x+r)+x\cdot\left(\frac{1}{x+r}\right)\cdot \left(1+\frac{∂r}{∂x}\right)-\frac{∂r}{∂x}$$
    Using ##∂r/∂x=x/r##
    $$\frac{∂z}{∂x}=\ln(x+r)$$
    $$\frac{∂^2z}{∂x^2}=\frac{1}{x+r}\cdot\left(1+\frac{x}{r}\right)=\frac{1}{r}$$

    Differentiating z w.r.t y partially,
    $$\frac{∂z}{∂y}=\frac{x}{x+r}\cdot \frac{∂r}{∂y}-\frac{∂r}{∂y}=\frac{∂r}{∂y}\cdot \frac{-r}{x+r}$$
    Using ##∂r/∂y=y/r##
    $$\frac{∂z}{∂y}=\frac{-y}{x+r}$$
    $$\frac{∂^2z}{∂y^2}=-\frac{(x+r)-y(∂r/∂y)}{(x+r)^2}=-\frac{rx+r^2-y^2}{r(x+r)^2}$$
    Since ##r^2-y^2=x^2##
    $$\frac{∂^2z}{∂y^2}=-\frac{x}{r(x+r)}$$
    Adding the second order derivatives,
    $$\frac{∂^2z}{∂x^2}+\frac{∂^2z}{∂y^2}=\frac{1}{r}-\frac{x}{r(x+r)}=\frac{1}{x+r}$$
    Where did I go wrong? :confused:

    Any help is appreciated. Thanks!
     
  2. jcsd
  3. Sep 13, 2013 #2

    pasmith

    User Avatar
    Homework Helper

    I don't think you have; I also get the same answer.
     
  4. Sep 13, 2013 #3
    Thanks for the check pasmith! :smile:
     
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