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I'm having trouble reducing this parabolic equation to canonical form.

$$\frac{\partial^2 u}{\partial x^2} + 2\frac{\partial^2 u}{\partial x \partial y} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial u}{\partial x} - \frac{\partial u}{\partial y} = 0$$

I know it's parabolic because I checked: $B^2 - AC0$, $$\begin{align} A = 1,\\ B = 1,\\ C = 1,\\ B^2 - AC = 1 - (1)(1) = 0\end{align}$$ so it's parabolic

I'm really not sure where to go from here. I know a change of variables is involved but I'm not sure how to reduce this to canonical form. I appreciate any help. Thanks in advance!

user114014
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2 Answers2

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Set $$ \frac{\partial}{\partial t}=\frac{\partial}{\partial y}-\frac{\partial}{\partial x}\quad \quad\text{and}\quad \frac{\partial}{\partial z}=\frac{\partial}{\partial x}+\frac{\partial}{\partial y}, $$ and you have that $$ \frac{\partial^2 u}{\partial x^2} + 2\frac{\partial^2 u}{\partial x \partial y} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial u}{\partial x} - \frac{\partial u}{\partial y}=\frac{\partial}{\partial z^2}-\frac{\partial}{\partial t}. $$

Yiorgos S. Smyrlis
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  • Sorry, typo! It is parabolic, there is a 2 in front of the xy term. – user114014 Jan 26 '14 at 18:06
  • I'm really not sure where to go from there - could you please clarify what I am supposed to do next? Why should I use this substitution? – user114014 Jan 26 '14 at 19:11
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    @user114014: You should just say that with the transformation $$x=z-t,,y=z+t,$$ we obtain the heat equation $u_t=u_{zz}$. – Yiorgos S. Smyrlis Jan 26 '14 at 19:24
  • Indeed, you have: $x-y = -2t = \xi, \ x+y = 2z = \eta$, where $\xi$ is the characteristic obtained in my comment so $2z$ stands for $f(x,y)$. The term $u_x - u_y$ gives you a hint for choosing $f$. Cheers! – Dmoreno Jan 26 '14 at 19:52
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Edit:

The PDE is parabolic and the characteristics are to be found from the equation:

$$\xi_x^2 + 2 \xi_x \xi_y + \xi_y^2 = (\xi_x + \xi_y)^2 = 0.$$

You can solve for $\xi_x/\xi_y$ as follows:

$$\frac{\xi_x}{\xi_y} + 1 = 0 \Rightarrow \frac{\xi_x}{\xi_y} = -1 =-\frac{dy}{dx},$$

and hence you have information of only one characteristic since the solution of the equation above is double:

$$\begin{align} \xi = & x-y, \\ \eta = & f(x,y). \end{align}$$

Beeing $f(x,y)$ any function so $\eta$ is not a linear combination of $\xi$. Use now the chain rule, remembering that:

$$u_x = u_\xi \xi_x + u_\eta \eta_x, \quad u_y = u_\xi \xi_y + u_\eta \eta_y, \quad u_{xx} = \ldots$$

I'm sure you can work from here. Let us know about your progress.

I hope this is useful to you.

Cheers!

Dmoreno
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  • I just realized I missed the 2 when I was typing it out. That's why it's parabolic, there is in fact a 2 in front of the xy term. How would this change the solution? – user114014 Jan 26 '14 at 18:06
  • Ok! Let me edit my post. – Dmoreno Jan 26 '14 at 18:15
  • Thanks! A few questions: Did you get ξ=x−y from solving the separable ODE -1 = -dy/dx?

    Also I'm not actually sure where to go from there at the end. Why exactly are you using chain rule? Isn't ξ=x−y the solution and we're done?

     – user114014 Jan 26 '14 at 18:54
  • What exactly is the canonical form we're aiming for here? – user114014 Jan 26 '14 at 19:12
  • Hi there! Yes I obtained the characteristics from solving the ODE-1 you refer to. The canonical form of the PDE is the equation you obtain after the mapping $(x,y) \to (\xi,\eta)$, which hopefully yields to $u_{\xi\xi} + u_{\eta \eta} = 0$ (canonical form of an elliptic PDE), after an appropriate choice of the function $f$. To do this you need to re-write the equation in terms of $\xi$ and $\eta$ solely by using the chain rule. Cheers! – Dmoreno Jan 26 '14 at 19:46
  • Sorry to keep asking questions, I'm just having trouble figuring out what to do. So when I'm re-writing the original equation in terms of ξ and η using the chain rule, I end up with a very long string of terms (especially from the 2nd order terms). This much longer equation hardly seems like a reduction to me. When you said "which hopefully yields to uξξ+uηη=0", did you mean that once I re-write the equation, things will cancel / be factored out and eventually that's what I should end up with? Or is there something I'm missing? – user114014 Jan 26 '14 at 19:59
  • Also, via the chain rule for the rewritten equation you just mean plug in ux=uξξx+uηηx, uy=uξξy+uηηy, uxx=… etc like you said above? Because nothing is cancelling for me and I have no idea how I'm supposed to reduce this massive equation. Thanks in advance. – user114014 Jan 26 '14 at 20:00
  • If you make use of chain rule, you should have: $u_{xx} = u_{\xi\xi} + 2 u_{\xi \eta} + u_{\eta\eta}$, $u_{xy} = -u_{\xi \xi} + u_{\eta\eta} $, $u_{yy} = u_{\xi \xi} -2u_{\xi \eta} u_{\eta \eta}$, and of course: $u_x = u_\xi + u_\eta,$ $u_y = -u_\xi + u_\eta$. Substitute this back into the original PDE and you will have: $$2u_{\eta\eta} + u_\xi = 0,$$ which is the canonical form of a parabolic equation, also known as the (non dimensional) heat equation (without heat addition in an isotropic medium). Cheers! – Dmoreno Jan 26 '14 at 20:54
  • Of course, using $\eta = x+y$. If you set $\xi \to -\xi$, you obtain the heat equation in a more useful engineering meaning: $2u_{\eta\eta} = u_\xi$. – Dmoreno Jan 26 '14 at 21:00