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What can be said and done about the "SIGN-Gordon equation"? $$\varphi_{tt}- \varphi_{xx} + \text{sgn}(\varphi) = 0.$$ It came up here.

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Rescale $x$, then rescale $t$ identically so the derivatives both get the same constant multipliers. The constants that pop out don't affect the signum, so you can rescale $\varphi$. So this is a kind of scaling limit of sinh-Gordon. – Steve Huntsman Apr 25 '10 at 16:20
I thought about that but sinh would scale like O(\phi) in the small limit while sgn(\phi) scales like O(1). Thanks for the idea of scaling though. – Kaveh Khodjasteh Apr 25 '10 at 17:10
Jim, Kaveh is talking about the sgn function, not the sin function. (That is, it's a pun on the sine-Gordon equation.) – Qiaochu Yuan Apr 25 '10 at 18:28
Sorry, I was distracted by the header and didn't look far enough into the question. Some of my colleagues have been fond of sine-Gordon, but I'm an outsider. – Jim Humphreys Apr 25 '10 at 19:23
up vote 10 down vote accepted

Turns out it appears in literature as the "signum-Gordon equation". For example the paper Signum-Gordon wave equation and its self-similar solutions

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Thanks for the link. I think it passes for an answer! – Kaveh Khodjasteh Apr 26 '10 at 15:42

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