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.

  • 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 – KingJ May 24 '17 at 17:26
  • Why do you suppose that $\alpha$ and $\beta$ must be negative? – florence May 24 '17 at 17:30
  • By equation coefficients we find that $\frac{- \alpha}{(\alpha \beta - 1)^2} = 2$ – KingJ May 24 '17 at 17:36
  • 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 $\frac{\partial x}{\partial \xi}$. – florence May 24 '17 at 18:00
  • 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? – 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 $$\frac{\partial ^ 2u}{\partial \eta \partial \xi} = 0.$$

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.