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How can I prove that I can write the wave equation $\partial_x^2 u - \partial_y^2 u = 0$ in factored form $(\partial_x - \partial_y)(\partial_x + \partial_y) = 0$?

Attempt 1:

$u_{xx} - u_{yy} = u_{xx} - u_{yy} + u_{xy} - u_{yx} = $ (by symmetry of second derivative - is this same as smoothness assumption?) $(u_x - u_y)(u_x + u_y)$

Attempt 2: $$\partial_x^2 u - \partial_y^2 u = \partial_x \partial_x u - \partial_y \partial_y u = (\partial_x \partial_x - \partial_y \partial_y)u = (\partial_x \partial_x - \partial_y \partial_y + \partial_x \partial_y - \partial_y \partial_x)u = (\partial_x - \partial_y)(\partial_x + \partial_y)u$$

1 Answers 1

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Sure there is, and it is exactly as you have stated it, i.e.: $$(\partial_x - \partial_y)(\partial_x + \partial_y)u = 0$$ It's nothing but a consequence of Schwarz Theorem (I presume you assume the solution to be $C^2$)

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    @boonheT . I apologize. I should have been more clear. How can I prove that the wave equation can be written in factored form. And, yes, I'm assuming that the solution is $C^2$. Edited the question with an attempt, thanks to your link! – 2017-01-30
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    Your attempt is incorrect. You don't have to apply the differential operators to $u$ and then multiply these together. Instead you have to compose these operators, i.e apply them, one after the other, on $u$. To make an easy example, note that $$\partial_x^2 u=\partial_x\partial_x u=\partial_{xx}u\neq \partial_x u \partial_x u,$$ and you are doing exactly the thing on the RHS. Moreover, no $C^\infty$ is of course a much stronger assumption, $C^2$ is sufficient. – 2017-01-30
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    I think I got it now on my second attempt above. Thank you! – 2017-01-30
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    yes. that's correct – 2017-01-30