1
$\begingroup$

In my notes there is the following:

Find the canonical form of the differential equation $$4u_{xx}-12u_{xy}+9u_{yy}+u_{y}=0$$

$$\Delta=(12)^2-4^2 \cdot 3^2=0$$

The canonical form will be of parabolic.

$$\begin{pmatrix}\frac{\partial{\xi}}{\partial{x}}-p_1 \frac{\partial{\xi}}{\partial{y}}=0 \Leftrightarrow (1,-p_1) \nabla \xi=0 \\ \xi=x-p_1 y\end{pmatrix}$$

$$\xi=ax+by, \eta =cx+dy$$

$$u(x,y)=U(\xi, \eta)$$

$$\xi=ax+by \\ \eta =cx+dy$$

$$u_x=aU_{\xi}+cU_{\eta} \Rightarrow u_{xx}=a(aU_{\xi \xi}+cU_{\xi \eta})+c(a U_{\eta \xi}+c U_{\eta \eta}=a^2U_{\xi \xi}+2acU_{\xi \eta}+c^2U_{\eta \eta} \\ u_y=bU_{\xi}+dU_{\eta} \Rightarrow u_{yy}=b^2U_{\xi \xi}+2bdU_{\xi \eta}+d^2U_{\eta \eta} \\u_{xy}a(bU_{\xi \xi}+dU_{\xi \eta}+c(bU_{\eta \xi}+dU_{\eta \eta})=abU_{\xi \xi}+(ad+bc)U_{\eta \xi}+cdU_{\eta \eta}$$

$$\dots \dots \dots \dots \dots$$

Could you explain to me the following??

$$\frac{\partial{\xi}}{\partial{x}}-p_1 \frac{\partial{\xi}}{\partial{y}}=0 \Leftrightarrow (1,-p_1) \nabla \xi=0 \\ \xi=x-p_1 y$$

What does this mean??

$\endgroup$
2
  • 1
    $\begingroup$ mary star, don't you remember the characteristics of the first order pde's. $\endgroup$
    – abel
    Apr 17, 2015 at 22:09
  • $\begingroup$ I remember them... But what is $p_1$ ?? @abel $\endgroup$
    – Mary Star
    Apr 17, 2015 at 22:48

1 Answer 1

2
$\begingroup$

I'm assuming you know the method of characteristics, so we want the solution $\xi$ to be constant in the direction $(1,-p_1)$ (obtain by looking at the coefficients of $\partial_x\xi$ and $\partial_y\xi$). So we want the derivative in the direction $(1,-p_1)$ to be zero, i.e., $$\nabla \xi\cdot(1,-p_1)=0.$$ (remember $\nabla\xi\cdot(1,-p_1)=(\partial_x\xi,\partial_y\xi)\cdot(1,-p_1)=\partial_x\xi-p_1\partial_y\xi$).

This holds for $\xi(x,y) = x-p_1y$.

We have that $\Delta=0$, this is the discriminant of the quadratic equation $4x^2-12x+9x^2=0$, which thus has a repeated root $x=\frac{3}{2}$, so we can express $$4\partial^2_{xx}u-12\partial^2_{xy}u+9\partial^2_{yy}u=4(\partial_x -\frac{3}{2}\partial_y)^2u.$$ Dividing our original equation by $4$ we obtain: $$(\partial_x -\frac{3}{2}\partial_y)^2u+\frac{1}{4}\partial_xu=0,$$ so $p_1=\frac{3}{2}.$

$\endgroup$
2
  • $\begingroup$ What is $p_1$ ?? @ellya $\endgroup$
    – Mary Star
    Apr 17, 2015 at 22:48
  • $\begingroup$ @MaryStar see my edit. $\endgroup$
    – Ellya
    Apr 17, 2015 at 23:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .