Answer to Question #137955 in Differential Equations for Ramalingam

Question

Classify the equation zxx + xzyy = 0 for x > 0 and reduce it to canonical form.

Answer 1

 

$$z_{xx}+xz_{yy}=0$$

Comparing the given equation with the equation

$$Az_{xx}+Bz_{xy}+Cz_{yy}+Dz_x+Ez_y+Fz=G,$$

we have

$A=1, B=0, C=x, D=0, E=0, F=0, G=0$

The discriminant is

$$B^2-4AC=0-4(1)(x)=-4x<0, x>0$$

The given equation is elliptic for all $x>0.$

 The characteristic equations are given by 

$$\dfrac{dy}{dx}=\dfrac{B-\sqrt{B^2-4AC}}{2A}=\dfrac{-\sqrt{-4x}}{2}=-i\sqrt{x}$$

$$\dfrac{dy}{dx}=\dfrac{B+\sqrt{B^2-4AC}}{2A}=\dfrac{\sqrt{-4x}}{2}=i\sqrt{x}$$

Integrating these two equations, we get 

$iy+\dfrac{2}{3}x\sqrt{x}=c_1$ and $-iy+\dfrac{2}{3}x\sqrt{x}=c_2$

Assume that

$$\xi=iy+\dfrac{2}{3}x\sqrt{x}, \eta=-iy+\dfrac{2}{3}x\sqrt{x}$$

Introducing the transformations $\alpha=\dfrac{1}{2}(\xi+\eta)$ and $\beta=\dfrac{1}{2i}(\xi-\eta),$ we obtain

$$\alpha=\dfrac{1}{2}(iy+\dfrac{2}{3}x\sqrt{x}-iy+\dfrac{2}{3}x\sqrt{x})=\dfrac{2}{3}x\sqrt{x}$$

$$\beta=\dfrac{1}{2i}(iy+\dfrac{2}{3}x\sqrt{x}+iy-\dfrac{2}{3}x\sqrt{x})=y$$

The transformed equation is obtained by first computing

$$\bar{A}=A\alpha_{x}^2+B\alpha_{x}\alpha_{y}+C\alpha_{y}^2=x$$

$$\bar{B}=3A\alpha_{x}\beta_{x}+B(\alpha_{x}\beta_{y}+\alpha_{y}\beta_{x})+2C\alpha_{y}\beta_{y}=0$$

$$\bar{C}=A\beta_{x}^2+B\beta_{x}\beta_{y}+C\beta_{y}^2=x$$

$$\bar{D}=A\alpha_{xx}+B\alpha_{xy}+C\alpha_{yy}+D\alpha_{x}+E\alpha_{y}=\dfrac{1}{2\sqrt{x}}$$

$$\bar{E}=A\beta_{xx}+B\beta_{xy}+C\beta_{yy}+D\beta_{x}+E\beta_{y}=0$$

$$\bar{F}=F=0\ and\ \bar{G}=G=0$$

Substitute in the following equation

$$\bar{A}z_{\alpha\alpha}+\bar{B}z_{\alpha\beta}+\bar{C}z_{\beta\beta}+\bar{D}z_\alpha+\bar{E}z_\beta+\bar{F}z=\bar{G}$$

Thus, canonical form of the given partial differential equation is

$$xz_{\alpha\alpha}+xz_{\beta\beta}+\dfrac{1}{2\sqrt{x}}z_\alpha=0$$

Or

$$z_{\alpha\alpha}+z_{\beta\beta}=-\dfrac{1}{2x\sqrt{x}}z_\alpha$$

$$z_{\alpha\alpha}+z_{\beta\beta}=-\dfrac{1}{3\alpha}z_\alpha$$