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

Question

I want to solve this PDE. I seems like a classic problem but I don't know what to do when in second order PDE coefficients are polynomial themselves.

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

Answer 1

Let us translate the given PDE into polar coordinates. Using $x = r \cos \theta$ and $y = r \sin \theta$, we get: $$ \frac{\partial z}{\partial r} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial r} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial r} = \cos \theta \frac{\partial z}{\partial x} + \sin \theta \frac{\partial z}{\partial y}.$$ Iterating this, $$ \frac{\partial^2 z}{\partial r^2} = \cos \theta \left( \cos \theta \frac{\partial^2 z} {\partial x^2} + \sin \theta \frac{\partial^2 z} {\partial y \partial x}\right) + \sin \theta \left( \cos\theta \frac{\partial^2 z}{\partial x \partial y} + \sin\theta \frac{\partial^2 z}{\partial y^2}\right) = \\ \cos^2 \theta \frac{\partial^2 z}{\partial x^2} + 2 \cos \theta \sin \theta \frac{\partial^2 z}{\partial x \partial y} + \sin^2 \theta \frac{\partial^2 z}{\partial y^2} = \\ \frac{1}{r^2} \left( x^2 \frac{\partial^2 z}{\partial x^2} + 2xy \frac{\partial^2 z}{\partial x \partial y} + y^2 \frac{\partial^2 z}{\partial y^2} \right).$$ Therefore, the given PDE is equivalent to $r^2 \frac{\partial^2 z}{\partial r^2} = 0$ which has general solution of $z = f(\theta) + r g(\theta)$ (on any domain which is an annular sector not including the origin).

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To relate this to the general solution from rafa11111's answer; or, to get a nice expression in terms of $x,y$: suppose our domain is contained either within the right half-plane or within the left half-plane. Then $\theta = \tan^{-1}(y/x)$, so we have: $$f(\theta) + r g(\theta) = f(\theta) + (r \cos \theta) (\sec \theta g(\theta)) = f(\tan^{-1}(y/x)) + x \left(\pm\sqrt{1 + (y/x)^2} g(\tan^{-1}(y/x)) \right)$$ (with the sign depending on the specific half-plane). Therefore, if $F(m) = f(\tan^{-1}(m))$ and $G(m) = \pm \sqrt{1+m^2} g(\tan^{-1}(m))$ (with the chosen branch of $\tan^{-1}$ also depending on the half-plane) then $z = F(y/x) + x G(y/x)$.

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As for how I came up with this approach: I first observed that the given PDE looks like an analogue of an Euler equation from ordinary differential equations. This suggested that some solution of the form $z = x^\alpha y^\beta$ might work. Plugging in this trial solution, we get: $$\alpha (\alpha-1) x^\alpha y^\beta + 2 \alpha \beta x^\alpha y^\beta + \beta (\beta-1) x^\alpha y^\beta = [(\alpha + \beta)^2 - (\alpha + \beta)] x^\alpha y^\beta = 0.$$ Therefore, this does indeed give a solution whenever $\alpha + \beta \in \{ 0, 1 \}$. Thus, we find particular solutions of the form $z = x^{-\beta} y^\beta = (y/x)^\beta$ and $z = x^{1-\beta} y^\beta = x (y/x)^\beta$.

Once we see this, it is reasonable to conjecture that in general, $f(y/x)$ and $x g(y/x)$ might be solutions - which is then straightforward to check. Furthermore, by linearity of the equation, $z = f(y/x) + x g(y/x)$ is also a solution. At this point, we heuristically have "enough degrees of freedom" that we think it could possibly be the general solution. Also, from the appearance of $y/x$ in the solution, this suggests that the approach of converting the PDE into polar coordinates could be fruitful. (Another approach would be to observe that the hypothesized general solution is linear on any ray from the origin, which would suggest examining the behavior of a solution on such curves $x = t \cos \alpha$, $y = t \sin \alpha$ for fixed $\alpha$. The calculation for this approach would end up looking much the same.)

Answer 2

Do you have any additional conditions? I tried and ansatz of the form $z = \log(Ax+By)$ and the left hand side yielded $-1$ so, since the equation is linear, any difference of logarithms of that form would solve the equation. Linear polynomials in $x$ and $y$ also make the left hand side vanish, so the following is a possible solution: $$ z(x,y) = \log(Ax+By) - \log(Cx+Dy) + Ex +Fy + G \,, $$ Where $A,\dots,G$ are constants. Not sure if there are more general solutions, but I hope this helps.

Answer 3

Let's introduce the variables $\xi(x,y)$ and $\eta(x,y)$. From sucessive application of chain rule one has $$ z_{xx} = z_{\xi \xi} \xi_x^2 + 2z_{\xi \eta} \xi_x \eta_x + z_{\eta \eta} \eta_x^2 + z_\xi \xi_{xx} + z_\eta \eta_{xx}, $$ $$ z_{yy} = z_{\xi \xi} \xi_y^2 + 2z_{\xi \eta} \xi_y \eta_y + z_{\eta \eta} \eta_y^2 + z_\xi \xi_{yy} + z_\eta \eta_{yy}, $$ $$ z_{xy} = z_{\xi \xi} \xi_x \xi_y + z_{\xi \eta} (\xi_x \eta_y+\xi_y \eta_x) + z_{\eta \eta} \eta_x \eta_y + z_\xi \xi_{xy} + z_\eta \eta_{xy}. $$ Substituting in the original PDE and colecting terms, $$ x^2 \left[z_{\xi \xi} \xi_x^2 + 2z_{\xi \eta} \xi_x \eta_x + z_{\eta \eta} \eta_x^2 + z_\xi \xi_{xx} + z_\eta \eta_{xx} \right] + 2xy \left[z_{\xi \xi} \xi_x \xi_y + z_{\xi \eta} (\xi_x \eta_y+\xi_y \eta_x) + z_{\eta \eta} \eta_x \eta_y + z_\xi \xi_{xy} + z_\eta \eta_{xy}\right] + y^2 \left[ z_{\xi \xi} \xi_y^2 + 2z_{\xi \eta} \xi_y \eta_y + z_{\eta \eta} \eta_y^2 + z_\xi \xi_{yy} + z_\eta \eta_{yy}\right] = 0 $$ $$ \left[ x^2 \xi_x^2 + 2xy \xi_x\xi_y +y^2\xi_y^2\right] z_{\xi \xi} + 2\left[x^2 \xi_x\eta_x+ xy(\xi_x\eta_y+\xi_y\eta_x)+y^2 \xi_y\eta_y\right] z_{\xi \eta} + \left[ x^2 \eta_x^2+2xy\eta_x\eta_y+y^2 \eta_y^2\right] z_{\eta \eta} = \phi, $$ in which $\phi$ represents the remaining terms. Setting the terms between the first pair of brackets to $0$ leads to $$ \left(\frac{x\xi_x}{y\xi_y}\right)^2 + 2 \left(\frac{x\xi_x}{y\xi_y}\right) + 1 = 0, $$ which is a quadratic equation on $x\xi_x/y\xi_y$, then $$ \frac{\xi_x}{\xi_y} = - \frac{y}{x}. $$ Since $d\xi=\xi_x dx + \xi_y dy$, for $\xi(x,y)=\mathrm{const}$ we have $d\xi=0$, leading to $$ \frac{dy}{dx} = - \frac{\xi_x}{\xi_y} , $$ then $$ \frac{dy}{dx}=\frac{y}{x}, $$ or $y=cx$. Therefore, $\xi=\mathrm{const}$ corresponds to $\xi=y/x$.

Setting the terms between the second pair of brackets to $0$, $$ x^2 \frac{\xi_x}{\xi_y}\eta_x+ xy\left(\frac{\xi_x}{\xi_y}\eta_y+\eta_x\right)+y^2 \eta_y=0. $$ Using what we found for $\xi_x/\xi_y$ we see that the expression vanishes identicaly to $0$, therefore the choice of $\eta_y$ is arbitrary. Defining $\eta=x$ leads to $$ z_{\eta \eta} = 0, $$ which is the canonical form of the parabolic equation. Integrating the equation, $$ z(\xi,\eta) = f(\xi) \eta + g(\xi), $$ or $$ z(x,y) = x f\left(\frac{y}{x}\right) + g\left(\frac{y}{x}\right). $$ That is the same solution presented by Daniel Schepler in the comments.

This answer was based on this text about canonical forms.