Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Find all solutions to $u_{xx}+u_{yy}=0$, where $u=\log p(x,y)$, with $p(x,y)$ a quadratic polynomial.

Assume $p(x,y)=ax^2+bxy+cy^2+dx+ey+f$, I computed $u_{xx}+u_{yy}$, then all coefficients in it have to be zero. But I failed to solve the correspnding system of algebraic equations...

I guess there would be a trickier way to find the solutions. Looking forward to your suggestions. Thanks!

share|cite|improve this question
(I had a wrong answer below, there were a bunch of comments on it which may be orphans now.) – user29743 Sep 8 '12 at 21:37
@countinghaus Thank you, though. :) – Vladimir Sep 8 '12 at 22:44

The important point to be taken here is that we do not need to work in the coordinates we have been given from the exercise.

Observe that if $\log p(x,y)$ is a solution then $\log p(x-X, y-Y)$ is also a solution for arbitrary constants $X, Y$. This corresponds to moving the origin. But note also that transformation that rotates the $xy$ plane around the origin also creates a new solution. In other words, we can assume that the polynomial is in the form $$p(x,y) = Ax^2 + By^2 + C,$$ linear terms being destroyed by moving the origin appropriately and the crossterm by a suitable rotation. Finding the solutions is now simple since $$ 0 = u_{xx} + u_{yy} = {p(p_{xx} + p_{yy}) - p_x^2 - p_y^2 \over p^2}$$ and so $$ (Ax^2 + By^2 + C)(2A + 2B) = 4A^2x^2 + 4B^2y^2.$$ If $A = -B$ then the LHS vanishes and $A = B = 0$ giving us a constant solution with $C$ arbitrary. Otherwise we must have $A = B$. Therefore the only nonconstant solutions in these coordinates are $u = \log(A(x^2 + y^2)) = \log(x^2 + y^2) + \log A$. Moreover, since this solution is invariant to rotations (being only a function of the radial distance $r^2 = x^2 + y^2$) the full set of solutions is recovered from this by translations. Therefore, the only nonconstant solutions must be of the form $$u = \log ((x-X)^2 + (y-Y)^2) + Z$$ with $X, Y, Z$ arbitrary. You can check that all of these are indeed solutions.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.