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?
Best Answer
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.