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I'm given the PDE $$2u_{xx} -6u_{xy} -8u_{yy}=0$$

Substituting $\xi = y-x$, $\eta=y+4x$, I turn it into:

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

Now I have the initial conditions:

$$u(x,0)=5x^2$$ $$u_y (x,0) =0$$

I don't know how to use the initial conditions. Should I express $x$ in terms of the new variables? We haven't done this at all in class (using initial conditions) but it is required on the exam.

Here's my attempt:

$$u_{\xi \eta}=0 \Rightarrow u(\xi,\eta)=f(\xi)+g(\eta) \Rightarrow u(x,y)=f(y-x)+g(y+4x)$$ $$u_{y}(x,y)=\frac{\partial}{\partial y}f(y-x)+ \frac{\partial}{\partial y} g(y+4x)$$

Now for the initial conditions:

$$u(x,0)=f(-x)+g(4x)=5x^2$$ $$u_y (x,0)=f_y(-x)+g_y(4x)=0$$

Integrating the second one gives:

$$-f(-x)+\frac{1}{4}g(4x)=C$$

So now we have a system:

$$ \begin{cases} f(-x)+g(4x)=5x^2 \\ -f(-x)+\frac{1}{4}g(4x)=C \end{cases}$$

This produces the solutions

$$\begin{cases} f(-x)=x^2 - \frac{4}{5}C \\ g(4x)=4x^2 +\frac{4}{5}C \end{cases}$$

Which is equivalent to

$$\begin{cases} f(x)=x^2 - \frac{4}{5}C \\ g(x)=\frac{1}{4}x^2 +\frac{4}{5}C \end{cases}$$

Adding $f(y-x)+g(y+4x) = (y-x)^2 + \frac{1}{4}(y+4x)^2 = 5x^2 + \frac{5}{4}y^2 = u(x,y)$

Is this correct?

Spine Feast
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  • Your PDE is not an equation at all, just a differential expression. – Ron Gordon Jun 24 '13 at 13:25
  • @DepeHb:do you mean ?$$2u_{xx} -6u_{xy} -8u_{yy}=0$$ – M.H Jun 24 '13 at 13:31
  • Yeah, of course, corrected this now. – Spine Feast Jun 24 '13 at 13:48
  • I have checked your result and it is correct. [Side note: You can do such checks quickly but just plugging the result into the equation and into initial conditions and see if it fits. Here symbolic computation packages (Mathematica, Maple, for example) prove to be very useful. ] – Giuseppe Negro Jun 24 '13 at 15:07

1 Answers1

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I think that what you are planning to do is too complicated. It is easier to just solve $u_{\xi\eta}=0$, which means that $u(\xi, \eta)=f(\xi)+g(\eta)$ for arbitrary (regular enough) functions $f$ and $g$. Then you go back to the original coordinates $x, y$, impose initial conditions and so determine $f, g$.

P.S.: I had not seen your edit. Looks like you did everything well.

Giuseppe Negro
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