Answer to Question #197607 in Differential Equations for Esha Asad

Question

Reduce to canonical form :

𝑢𝑥𝑥 − 𝑢𝑥𝑦 + 𝑢𝑦𝑦 + 𝑢𝑥 = 0

Answer 1

$$\Delta = a_{12}^2 - {a_{11}}{a_{22}} = {(-\frac{1}{2})^2} - 1 \cdot 1 = - \frac{3}{4} < 0$$

 - we have an equation of elliptic type.

Characteristic equation:

$$\frac{{dy}}{{dx}} = \frac{{{a_{12}} + \sqrt \Delta }}{{{a_{11}}}} = \frac{{-\frac{1}{2} + \frac{{\sqrt 3 }}{2}i}}{1} \Rightarrow y =- \frac{x}{2} + \frac{{\sqrt 3 }}{2}ix + {C_1}$$

or

$$\frac{{dy}}{{dx}} = \frac{{{a_{12}} - \sqrt \Delta }}{{{a_{11}}}} = \frac{{-\frac{1}{2} - \frac{{\sqrt 3 }}{2}i}}{1} \Rightarrow y = -\frac{x}{2} - \frac{{\sqrt 3 }}{2}ix + {C_2}$$

Let

$$\xi (x,y) = \frac{{{C_1} + {C_2}}}{2}$$ $$= \frac{1}{2}\left( {y + \frac{x}{2} - \frac{{\sqrt 3 }}{2}ix + y+\frac{x}{2} + \frac{{\sqrt 3 }}{2}ix} \right) = y +\frac{x}{2}$$

$$\eta \left( {x,y} \right) = \frac{{{C_2} - {C_1}}}{{2i}}$$ $$= \frac{1}{{2i}}\left( {y +\frac{x}{2} + \frac{{\sqrt 3 }}{2}ix - y - \frac{x}{2} + \frac{{\sqrt 3 }}{2}ix} \right) = \frac{{\sqrt 3 }}{2}x$$

$$u(x,y) = v\left( {\xi ;\,\eta } \right)$$

Then

$${u_x} = {v_\xi }{\xi _x} + {v_\eta }{\eta _x} = \frac{1}{2}{v_\xi } + \frac{{\sqrt 3 }}{2}{v_\eta }$$

$${u_y} = {v_\xi }{\xi _y} + {v_\eta }{\eta _y} = {v_\xi }$$

$${u_{xx}} = \frac{1}{2}\left( {{v_{\xi \xi }}{\xi _x} + {v_{\xi \eta }}{\eta _x}} \right) + \frac{{\sqrt 3 }}{2}\left( {{v_{\eta \xi }}{\xi _x} + {v_{\eta \eta }}{\eta _x}} \right)$$

$$= \frac{1}{2}\left( { \frac{1}{2}{v_{\xi \xi }} + \frac{{\sqrt 3 }}{2}{v_{\xi \eta }}} \right) + \frac{{\sqrt 3 }}{2}\left( { \frac{1}{2}{v_{\eta \xi }} + \frac{{\sqrt 3 }}{2}{v_{\eta \eta }}} \right)$$

$$= \frac{1}{4}{v_{\xi \xi }} +\frac{{\sqrt 3 }}{2}{v_{\xi \eta }} + \frac{3}{4}{v_{\eta \eta }}$$

$${u_{yy}} = {v_{\xi \xi }}{\xi _y} + {v_{\xi \eta }}{\eta _y} = {v_{\xi \xi }}$$

$${u_{yx}} = {v_{\xi \xi }}{\xi _x} + {v_{\xi \eta }}{\eta _x} = \frac{1}{2}{v_{\xi \xi }} + \frac{{\sqrt 3 }}{2}{v_{\xi \eta }}$$

Substitute the found values ​​into the original equation:

$$\frac{1}{4}{v_{\xi \xi }} +\frac{{\sqrt 3 }}{2}{v_{\xi \eta }} + \frac{3}{4}{v_{\eta \eta }}- \frac{1}{2}{v_{\xi \xi }} -\frac{{\sqrt 3 }}{2}{v_{\xi \eta }}$$

$$+{v_{\xi \xi }}+\frac{1}{2}{v_\xi } + \frac{{\sqrt 3 }}{2}{v_\eta }=0$$

$$\frac{3}{4}{v_{\xi \xi }} + \frac{3}{4}{v_{\eta \eta }}=-\frac{1}{2}{v_\xi } - \frac{{\sqrt 3 }}{2}{v_\eta }$$

$${v_{\xi \xi }} + {v_{\eta \eta }} =- \frac{2}{3}{v_\xi } - \frac{{2\sqrt 3 }}{3}{v_\eta }$$

Answer: 

$${v_{\xi \xi }} + {v_{\eta \eta }} = -\frac{2}{3}{v_\xi } - \frac{{2\sqrt 3 }}{3}{v_\eta },$$

where 

$$\xi (x,y) = \ y + \frac{x}{2},\ \eta \left( {x,y} \right) = \frac{{\sqrt 3 }}{2}x,$$

$$u(x,y) = v\left( {\xi ;\,\eta } \right)$$