This is a question from my first class on multivariable calculus, where we learned partial derivatives.

given $u(x) = \ln({\sqrt{(x-x_0)^2 + (y-y_0)^2}})$

show that $u$ satisfies the laplace equation

$\frac {\partial^2u}{\partial x^2} + \frac {\partial^2u}{\partial y^2} = 0$

I have tried to compute these partial derivatives and have seen that subbing in values can produce 0 but computationally somehow fail to do so. I tried it by hand and got the same answer as in maple. In Maple, I got this using $x_0 = a$ and $y_0 = b$

$\frac {2}{(x-a)^2 + (y-b)^2} - \frac{1}{2}\frac {(2x-2a)^2}{((x-a)^2 + (y-b)^2)^2} + \frac{1}{2}\frac {(2y-2b)^2}{((x-a)^2 + (y-b)^2)^2}$

but I have no reason to understand why this is equal to zero. Maybe I am manipulating them wrong algabreically or something... could anyone explain how its satisfied?

  • $\begingroup$ If you heard about polar coordinates before, try to change it. en.wikipedia.org/wiki/… $\endgroup$ – Chee Han Sep 14 '17 at 2:22
  • $\begingroup$ I've heard of them but I don't think that's the exercise here. $\endgroup$ – Alex Van de Kleut Sep 14 '17 at 2:30

We don't need to worry about the square root, since $\log{a^n}=n\log{a}$ for positive $a$. We have $$ \frac{\partial u}{\partial x} = \frac{1}{2}\frac{2(x-x_0)}{(x-x_0)^2+(y-y_0)^2}, \\ \frac{\partial u}{\partial y} = \frac{1}{2}\frac{2(y-y_0)}{(x-x_0)^2+(y-y_0)^2}, $$ and differentiating again, $$ \frac{\partial^2 u}{\partial x^2} = \frac{[(x-x_0)^2+(y-y_0)^2] -2(x-x_0)^2}{[(x-x_0)^2+(y-y_0)^2]^2} = \frac{(y-y_0)^2 -(x-x_0)^2}{[(x-x_0)^2+(y-y_0)^2]^2} \\ \frac{\partial^2 u}{\partial x^2} = \frac{[(x-x_0)^2+(y-y_0)^2] -2(y-y_0)^2}{[(x-x_0)^2+(y-y_0)^2]^2} = \frac{(x-x_0)^2-(y-y_0)^2}{[(x-x_0)^2+(y-y_0)^2]^2} $$ by the quotient rule, and these sum to give zero.

  • $\begingroup$ Thank you, I completely forgot about logs and exponents. $\endgroup$ – Alex Van de Kleut Sep 14 '17 at 2:39

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