Canonical form of parabolic equations

Theorem 5   Suppose that the equation

$$au_{xx} + b u_{xy} + c u_{yy} + d u_x + e u_y + fu = g.$$ (2.14)

is parabolic in a domain $D.$ Then there exists a coordinate system $(\xi, \eta)$ in which this equation has the canonical form

$$w_{\xi \xi} + l_{1}[w] = G(\xi, \eta),$$

where $w(\xi, \eta) = u(x(\xi, \eta), y(\xi, \eta)),$ $l_{1}$ is a first order linear differential operator, and $G$ is a function which depends on given equation.

Proof: Under a generic change of coordinate system $(x,y) \rightarrow (\xi, \eta),$ the given PDE () changes to the PDE

$$Aw_{\xi \xi} + B w_{\xi \eta} + C w_{\eta \eta} + D w_{\xi} + E w_{\eta} + Fw = G.$$ (2.15)

where the new coefficients are given by ().

Given PDE is parabolic, and by the invariance of the type of PDE, we have the discriminant $B^2 - 4AC = 0$. This is true, when $B=0$ and $A$ or $C$ is equal to zero. It is enough to set $A$ or $C$ equal to zero, because then by the parabolicity of equation, we will automatically get $B \equiv 0$ and let $C=0,$

$$C(\xi, \eta) = a \eta_{x}^{2} + b \eta_{x} \eta_{y} + c \eta_{y}^{2},$$

$$= \frac{1}{a} \left ( a^2 \eta_{x}^{2} + ab \eta_{x} \eta_{y} + ac \eta_{y}^{2} \right ),$$

$$= \frac{1}{a} \left ( a^2 \eta_{x}^{2} + ab \eta_{x} \eta_{y} + \frac{b^2}{4} \eta_{y}^{2} \right ) \hspace{1cm} (\because b^2 - 4ac = 0),$$

$$= \frac{1}{a} \left ( a \eta_{x} + \frac{b}{2} \eta_{y} \right )^2 =0.$$

Thus $\eta$ is a solution of the first order linear PDE

$$a \eta_{x} + \frac{b}{2} \eta_{y} = 0.$$ (2.16)

The auxiliary equation is $$\frac{dx}{a} = \frac{dy}{b/2} = \frac{d \eta}{0}.$$ The solution $\eta$ is constant on the characteristic curve given by

$$\frac{dy}{dx} = \frac{b}{2a}.$$ (2.17)

Solving this equation, we get implicitly

$$\eta(x,y) = c_1, \hspace{1cm} c_1 =$$constant.

The second independent variable $\xi$ can be chosen as any function which is not linearly dependent on $\eta$, which will make the Non-zero Jacobian of transformation on $D$.

Note 1: Check that the parabolic PDEs have only one family of characteristics (Compare it with hyperbolic PEDs).

Note 2: Check that with the above choice, we automatically get $B \equiv 0$ ! (Check it yourself.)

Example 2.4.2   Show that the equation

$$x^2 u_{xx} - 2xy u_{xy} + y^2 u_{yy} = e^x,$$

is parabolic, and reduce it to its canonical form.

Solution Comparing the given equation with the standard form, one gets

$$a=x^2,~b=-2xy~c=y^2,$$

and discriminant is $\Delta=0$. Therefore, the given equation is parabolic.

The characteristic equation is

$$\frac{dy}{dx}=-\frac{y}{x},$$

$$\implies~xy=c_1,$$

where $c_1$ is constant.Take $\eta=xy~$and$~\xi=x^2.$ Now on changing the coordinates, one gets

$$u_x =u_{\xi}+u_{\eta}(y),$$

$$u_y =(0)u_{\xi}+u_{\eta}(x),$$

$$u_{xx} =4x^2u_{\xi \xi}+y^2u_{\eta \eta}+4xy u_{\xi \eta}+2u_{\xi},$$

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

$$u_{yy} =u_{\eta \eta}x^2$$

Putting it in given equation, one gets

$$4x^4u_{\xi \xi}+2x^2u_{\xi}-2xyu_{\eta} =e^{\sqrt{\xi}},$$

$$4\xi^2u_{\xi \xi}+2\xi u_{\xi}-2\eta u_{\eta} =e^{\sqrt{\xi}},$$

$$u_{\xi \xi} =-\frac{1}{2\xi}u_{\xi}-\frac{\eta}{4{\xi}^2}u_{\eta}+\frac{1}{4{\xi}^2}e^{\sqrt{\xi}}.$$

If one takes $\eta=y$ and $\xi=xy$, one will gets

$$u_{\eta \eta}=\frac{2\xi}{\eta^2}u_{\xi}+\frac{1}{\eta^2}e^{\frac{\xi}{\eta}}.$$