Canonical form of elliptic equations

For elliptic PDEs, the discriminant $B^2 - 4AC < 0$ as $b^2 - 4ac < 0,$ and the characteristic equations

$$\frac{dy}{dx} = \frac{- (-b - \sqrt{b^2 - 4ac})}{2a}$$ (2.18)

$$\frac{dy}{dx} = \frac{- (-b + \sqrt{b^2 - 4ac})}{2a}$$ (2.19)

gives complex conjugate coordinates, say $\xi$ and $\eta.$ Make another transformation from $(\xi, \eta)$ to $(\alpha, \beta)$ such that

$$\alpha = \frac{\xi + \eta}{2}, \hspace{2cm} \beta = \frac{\xi - \eta}{2i}$$

i.e.

$$\alpha =$$   Re$$(\xi), \hspace{1cm} \beta =$$   Im$${(\xi)},$$

using which one will get the canonical form

$$u _{\alpha \alpha} + u _{\beta \beta} = \phi(\alpha, \beta, u, u_{\alpha}, u_{\beta}).$$

Example 2.4.3   Find the canonical form of the Tricomi equation

$$u_{xx} + x u_{yy} = 0, \hspace{1cm} x >0.$$

Solution Compare the given equation with the standard form. One gets

$$a=1,~b=0~$$and$$~c=x.$$

$$\therefore~\Delta=b^2-4ac=-4x<0,$$

which implies that given equation is elliptic for $x>0.$ The corresponding characteristic equations are

$$\frac{dy}{dx} =\pm \frac{2\imath \sqrt{x}}{2},$$

$$\implies~\frac{3}{2}y \pm \imath x^{\frac{3}{2}}=c_1.$$

Therefore, the canonical variables are

$$\xi=\frac{3}{2}y,~\eta=x^{\frac{3}{2}},$$

$$u_x =\frac{3}{2}\sqrt{x}u_{\eta},$$

$$u_y =\frac{3}{2}u_{\xi},$$

$$u_{xx} =\frac{3}{4}\frac{1}{\sqrt{x}}u_{\eta}+\frac{9}{4}xu_{\eta \eta},$$

$$u_{yy} =\frac{9}{4}xu_{\xi \xi}.$$

Putting these in given equation, one gets

$$u_{\eta \eta}+u_{\xi \xi}=-\frac{1}{3x^{\frac{3}{2}}}u_{\eta}=-\frac{1}{3\eta}u_{\eta},$$

which means

$$u_{\eta \eta}+u_{\xi \xi}=-\frac{1}{3\eta}u_{\eta}.$$

Note One can explore further $2$nd linear PDEs with constant coefficient. For operator method read book written by Sankara Rao on PDEs.

Example 2.4.4   Find the canonical form of

$$u_{xx}+6u_{xy}-16u_{yy}=0.$$

Solution On comparing the given equation with the standard form, one gets

$$a=1,~b=6~$$and$$~c=-16.$$

$$\therefore~\Delta=b^2-4ac=100>0,$$

which implies that given equation is hyperbolic equation. The corresponding characteristic equations are

$$\frac{dy}{dx}=-2,~$$and$$~ \frac{dy}{dx}=8$$

$$\implies~y=-2x+c_1~$$and$$~y=8x+c_2.$$

Therefore, the canonical variables are

$$\xi=y+2x~$$and$$~=\eta=y-8x,$$

Therefore change of variables

$$u_x =2u_{\xi}-8u_{\eta},$$

$$u_y =u_{\xi}+u_{\eta},$$

$$u_{xx} =4u_{\xi \xi}-32u_{\xi \eta}+64u_{\eta \eta},$$

$$u_{yy} =u_{\xi \xi}+2u_{\xi \eta}+u_{\eta \eta}.$$

Putting these in given equation, one gets

$$u_{\xi \eta}=0.$$

This is the required canonical form.

Example 2.4.5   Find the canonical form of

$$u_{xx}-6u_{xy}+9u_{yy}=xy^2.$$

Solution On comparing the given equation with the standard form, one gets

$$a=1,~b=-6~$$and$$~c=9.$$

$$\therefore~\Delta=b^2-4ac=0,$$

which implies that given equation is parabolic equation. The corresponding characteristic equations are

$$\frac{dy}{dx}=-3$$

$$\implies~y=-3x+c_1.$$

Therefore, the canonical variables are

$$\xi=y+3x~$$and$$~=\eta=x^2,$$

Therefore change of variables

$$u_x =3u_{\xi}+2xu_{\eta},$$

$$u_y =u_{\eta},$$

$$u_{xx} =9u_{\xi \xi}+12xu_{\xi \eta}+4x^2u_{\eta \eta}+2u_{\eta},$$

$$u_{xy} =3u_{\xi \xi}+2x u_{\eta \xi},$$

$$u_{yy} =u_{\xi \xi}.$$

Putting these in given equation, one gets

$$4x^2u_{\eta \eta}+2u_{\eta}=xy^2.$$

$$\implies~u_{\eta \eta}=\frac{(\xi-3\sqrt{x})^2}{4\sqrt{\eta}}-\frac{u_{\eta}}{2\eta}.$$

This is the required canonical form.