# Find $P(x^{2}-y^{2})$ in terms of $x^{2}-y^{2}$ [closed]

Let $$f(z)=P(x^{2}-y^{2})+iQ(x,y)$$ be a holomorphic function where P and Q are of class $$C^{2}$$

I tried to use the fact that $$P$$ is harmonic , but couldn't continue

## closed as off-topic by cmk, Lee David Chung Lin, Xander Henderson, nmasanta, LeucippusJul 30 at 5:05

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Let $$u = \Re(h)$$ where $$h$$ is analytic and non-constant. More generally, $$P(u)$$ is harmonic iff $$P$$ is linear.

Proof: The if direction is clear. Assuming $$P(u)$$ is harmonic, we have $$(u_x^2+u_y^2)P''(u) = 0$$ by direct calculation of the Laplacian. If $$u_x(x_0, y_0) = u_y(x_0, y_0) = 0,$$ we obtain $$h'(z_0) = 0$$ where $$z_0 = x_0+iy_0.$$

Let $$K = \text{Ker}(h').$$ If $$K$$ is dense in $$\mathbb{C},$$ we have $$h' \equiv 0$$ on all of $$\mathbb{C}$$ by any number of arguments, contradiction. Otherwise, there must be an open neighborhood $$S \subseteq \mathbb{C}$$ on which $$h'$$ is non-zero. This corresponds to an open neighborhood $$S' \subseteq \mathbb{R}^2$$ on which $$R(u) \equiv 0,$$ where $$R = P''.$$ By the open mapping theorem for harmonic functions, this corresponds to an open set $$S''=u(S') \subseteq \mathbb{R}$$ on which $$R \equiv 0.$$ We wish to show that $$R \equiv 0$$ on all of $$\mathbb{R}.$$

Unfortunately, we are thwarted at this point by the existence of bump functions. We must strengthen the hypothesis to $$P$$ being analytic (here, having a convergent Taylor series) to get rid of them. Then $$R$$ is analytic too, and hence $$R \equiv 0.$$

I do not claim to have come up with the general case immediately. Here is the original draft:

Note that $$P(x^2-y^2) = P(\Re((x+iy)^2)) = P(\Re(z^2)).$$ With that in mind, we should aim to prove that $$P$$ is constant or linear so that $$f(z) = az^2+b$$ or $$f(z) \equiv c$$ are the only solutions (by applying the Cauchy-Riemann equations).

It turns out that your abandoned line of reasoning works wonders: We have $$0 = \Delta P(x^2-y^2) = 4(x^2+y^2)P''(x^2-y^2).$$ Choosing $$(x,y) = (\sqrt{t}, 0), (0, \sqrt{t}), (1,1)$$ as $$t$$ traverses the positive reals shows that $$P'' \equiv 0,$$ which implies the result.