I am given the following; $u(x,y)$ satisfes,

$$2\frac{\partial ^ 2u}{\partial x^2} +3\frac{\partial ^ 2u}{\partial y^2}-7\frac{\partial ^ 2u}{\partial x \partial y}=0 $$

Use a change of variables, and suitable constants in the integers $\alpha$ and $\beta$ to transform the above into

$\frac{\partial ^ 2u}{\partial \eta \partial \xi} = 0$

where we have, $$\xi = x + \alpha y$$ $$\eta = \beta x + y$$

So here is my answer:

By writing $x$ and $y$ in terms of $\eta$ and $\xi$ we have, $$x = \frac{\alpha \eta - \xi}{\alpha \beta - 1}$$ $$y = \frac{\beta \xi - \eta}{\alpha \beta - 1}$$

This gives us that

$$\frac{\partial x}{\partial \eta} = \frac{\alpha}{\alpha \beta - 1}, \ \ \ \ \ \ \frac{\partial x}{\partial \xi} = \frac{-1}{\alpha \beta - 1}$$ $$\frac{\partial y}{\partial \eta} = \frac{-1}{\alpha \beta - 1}, \ \ \ \ \ \ \frac{\partial y}{\partial \xi} = \frac{\beta}{\alpha \beta - 1}$$

So using the chain rule, we get $$\frac{\partial }{\partial \xi} = \frac{\partial x}{\partial \xi}\frac{\partial }{\partial x} + \frac{\partial y }{\partial \xi}\frac{\partial}{\partial y} $$ $$\frac{\partial }{\partial \eta} = \frac{\partial x}{\partial \eta}\frac{\partial }{\partial x} + \frac{\partial y }{\partial \eta}\frac{\partial}{\partial y} $$

Putting these together, we get $$\frac{\partial ^2 }{\partial \xi \partial \eta} = (\frac{\partial x}{\partial \eta}\frac{\partial x}{\partial \xi })\frac{\partial ^2 }{\partial x^2} + (\frac{\partial y}{\partial \eta}\frac{\partial y }{\partial \xi})\frac{\partial ^2}{\partial y^2} + (\frac{\partial x}{\partial \xi}\frac{\partial y}{\partial \eta} + \frac{\partial x}{\partial \eta}\frac{\partial y}{\partial \xi})\frac{\partial ^2}{\partial x \partial y}$$ Plugging in our working for the partial derivatives of $x$ and $y$, we get

$$\frac{\partial ^2 }{\partial \xi \partial \eta} = \frac{-\alpha}{(\alpha \beta - 1)^2}\frac{\partial ^2 }{\partial x^2} + \frac{-\beta}{(\alpha \beta - 1)^2}\frac{\partial ^2}{\partial y^2} + \frac{\alpha \beta + 1}{(\alpha \beta - 1)^2}\frac{\partial ^2}{\partial x \partial y}$$

If we try to find a suitable $\alpha$ and $\beta$ we find that $\alpha$ and $\beta$ must both be negative but the $\alpha \beta + 1$ is positive which mean no such $\alpha$ , $ \beta$ exist Can someone find the mistake in my workings and tell me how to fix it?


Writing $x,y$ in terms of $\eta, \xi$ complicates things and we don't need to do that. Note that $$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial \eta}\frac{\partial \eta}{\partial x}+\frac{\partial u}{\partial \xi}\frac{\partial \xi}{\partial x} = \frac{\partial u}{\partial \eta}+\beta \frac{\partial u}{\partial \xi}$$ and so $$\frac{\partial^2 u}{\partial x^2} = \frac{\partial ^2 u}{\partial \eta^2}+\beta^2\frac{\partial ^2u}{\partial \xi^2}+2\beta \frac{\partial^2 u}{\partial \eta \partial \xi}$$ Similarly, $$\frac{\partial u}{\partial y} = \alpha \frac{\partial u}{\partial \eta}+\frac{\partial u}{\partial \xi}$$ $$\implies \frac{\partial^2 u}{\partial y^2} = \alpha^2 \frac{\partial^2 u}{\partial \eta^2}+\frac{\partial u^2}{\partial \xi^2}+2\alpha \frac{\partial^2 u}{\partial \eta\partial \xi}$$ Finally, $$\frac{\partial^2 u}{\partial x\partial y} = \alpha \frac{\partial^2 u}{\partial \eta^2}+\beta \frac{\partial^2 u}{\partial \xi^2}+(1+\alpha\beta)\frac{\partial^2 u}{\partial \eta\partial \xi}$$ and so $$0=2\frac{\partial^2u}{\partial x^2} + 3\frac{\partial^2 u}{\partial y^2}-7\frac{\partial^2u}{\partial x \partial y}$$ $$ = (2+3\alpha^2-7\alpha)\frac{\partial^2 u}{\partial \eta^2}+(2\beta^2+3-7\beta)\frac{\partial^2 u}{\partial \xi^2}+(6\beta+9\alpha-7-7\alpha\beta)\frac{\partial^2 u}{\partial \eta \partial \xi}$$ Now you can solve for $\alpha, \beta$ by setting the first two coefficients equal to zero.

  • $\begingroup$ Yes, I have solved the problem this way already which I agree is easier to do. However, I want to find out what is wrong with the answer above which in principle should work out $\endgroup$ – KingJ May 24 '17 at 17:26
  • $\begingroup$ Why do you suppose that $\alpha$ and $\beta$ must be negative? $\endgroup$ – florence May 24 '17 at 17:30
  • $\begingroup$ By equation coefficients we find that $\frac{- \alpha}{(\alpha \beta - 1)^2} = 2$ $\endgroup$ – KingJ May 24 '17 at 17:36
  • $\begingroup$ I think your problem is that you assumed that $\left(\frac{\partial x}{\partial \eta}\frac{\partial}{\partial x}\right)\left(\frac{\partial x}{\partial \xi}\frac{\partial}{\partial x}\right) = \frac{\partial x}{\partial \eta}\frac{\partial x}{\partial \xi}\frac{\partial^2}{\partial x^2}$, which is not the case, since $\frac{\partial}{\partial x}$ does not commute with $\frac{\partial x}{\partial \xi}$. $\endgroup$ – florence May 24 '17 at 18:00
  • $\begingroup$ Well yes, in the general case I would have to use the product rule - but given that $\frac{\partial x}{\partial \xi}$ is a constant - is it not correct in saying that the extra term will drop out? $\endgroup$ – KingJ May 24 '17 at 18:55

From V.S.Vladimirov, A Collection of Problems on the Equations of Mathematical Physics, Springer, 1986

enter image description here

The characteristic equation $$2(dy)^2+7dxdy+3(dx)^2=0$$ splits into two equations $$dx+2dy=0,$$ $$3dx+dy=0$$

with solutions $$x+2y=C_1$$ $$3x+y=C_2$$ Then change of variables $$\xi=x+2y,\\ \eta=3x+y$$ reduce equation $$2\frac{\partial ^ 2u}{\partial x^2} +3\frac{\partial ^ 2u}{\partial y^2}-7\frac{\partial ^ 2u}{\partial x \partial y}=0$$ into equation $$\frac{\partial ^ 2u}{\partial \eta \partial \xi} = 0.$$


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