Reduce $u_{xx} + x^2u_{yy} = 0$ to canonical form

Question

Reduce $u_{xx} + x^2u_{yy} = 0$ to canonical form.

I got it here, which says

Why is it that

$$\eta = -iy + x^2/2, \ \xi = iy + x^2/2$$?

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What I tried:

We have $$A = 1, B = 0, C = x^2$$

$$\to B^2 - 4AC = -4x^2 < 0$$

Hence, $u_{xx} + x^2u_{yy} = 0$ is an ellptic-type PDE.

$$A\alpha^2 + B\alpha + C = 0$$

$$\to \alpha^2 + x^2 = 0$$

$$\to \alpha^2 = -x^2$$

$$\to \alpha = \pm ix$$

$$\to \lambda_1 = ix, \ \lambda_2 = -ix$$

$$\to \frac{dy}{dx} = \pm ix$$

$$\to dy = \pm ix dx$$

$$\to y = \pm i (x^2/2 + c)$$

$$\to y = i (x^2/2 + c_1), \ y = -i (x^2/2 + c_2)$$

$$\to y/i - x^2/2 = c_1, \ y/i + x^2/2 = c_2$$

So we choose

$$\to \eta = y/i - x^2/2, \ \xi = y/i + x^2/2$$

right?

Answer 1

Actually both approaches are valid as $$1/i = -i.$$ You can get the same result as in the example if you do the following:

In your approach consider

$$\xi = y/i + x^2/2= yi/i^2 + x^2/2=-iy + x^2/2$$ which is the same as $\eta$ in the example.

Now let's have a look at $\eta$ in your approach. As you mentioned

$$\to y = i (x^2/2 + c_1)$$

$$\to y/i - x^2/2 = c_1 $$ $$\to -y/i + x^2/2 = -c_1 $$ $$\to -y/i + x^2/2 = c$$ where $ c = -c_1$

from here take $\eta = -y/i + x^2/2 = -yi/i^2+x^2/2=iy+x^2/2 $ which is the same as $\xi$ in the example. Hope this helps.