Answer to Question #197611 in Differential Equations for Esha

Question

Reduce to canonical form

u_xx - u_xy + u_yy + u_x = 0

Answer 1

$$A=1, B=-1, C=1$$

$$B^2-4AC=1-4=-3<0$$

The given partial differential equation is elliptic.

 The characteristic equations are given by 

$$\dfrac{dy}{dx}=\dfrac{B-\sqrt{B^2-4AC}}{2A}=\dfrac{-1-\sqrt{-3}}{2(1)}$$

$$=\dfrac{-1-i\sqrt{3}}{2}$$

and

$$\dfrac{dy}{dx}=\dfrac{B+\sqrt{B^2-4AC}}{2A}=\dfrac{-1+\sqrt{-3}}{2(1)}$$

$$=\dfrac{-1+i\sqrt{3}}{2}$$

Integrating these two equations, we get 

$$y=x(\dfrac{-1-i\sqrt{3}}{2})+c_1$$

and

$$y=x(\dfrac{-1+i\sqrt{3}}{2})+c_2$$

Assume that

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

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

The transformation

$$\alpha=\dfrac{1}{2}(\xi+\eta)$$

$$\beta=\dfrac{1}{2i}(\xi-\eta)$$

$$\alpha=y+x$$

$$\beta=\dfrac{\sqrt{3}}{2}x$$

$$\alpha_x=1, \alpha_y=1$$

$$\alpha_{xx}=0, \alpha_{xy}=0, \alpha_{yy}=0$$

$$\beta_x=\dfrac{\sqrt{3}}{2}, \beta_y=0$$

$$\beta_{xx}=0, \beta_{xy}=0, \beta_{yy}=0$$

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

$$\bar{A}=A\alpha_x^2+B\alpha_x\alpha_y+C\alpha_y^2=1-1+1=1$$

$$\bar{B}=2A\alpha_x\beta_x+B(\alpha_x\beta_y+\alpha_y\beta_x)+2C\alpha_y\beta_y$$

$$=2(1)(\dfrac{\sqrt{3}}{2})-1(0+(1)\dfrac{\sqrt{3}}{2})+2(1)(0)=\dfrac{\sqrt{3}}{2}$$

$$\bar{C}=A\beta_x^2+B\beta_x\beta_y+2C\alpha_y\beta_y=$$

$$=\dfrac{3}{4}+0+0=\dfrac{3}{4}$$

$$\bar{D}=A\alpha_{xx}+B\alpha_{xy}+C\alpha_{yy}+D\alpha_x+E\alpha_y=1$$

$$\bar{E}=A\beta_{xx}+B\beta_{xy}+C\beta_{yy}+D\beta_x+E\beta_y=\dfrac{\sqrt{3}}{2}$$

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

$$u_{\alpha\alpha}+\dfrac{\sqrt{3}}{2}u_{\alpha\beta}+\dfrac{\sqrt{3}}{2}u_{\beta\beta}+u_{\alpha}+\dfrac{\sqrt{3}}{2}u_\beta=0$$