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Feb 21 at 21:16 comment added Connor Mooney If you let $H(s) := s\sigma^2(s)$ then the equation takes the appealing form $\partial_t(H(u)) = H'(u)\Delta(H(u))$. If $H' > 0$ then standard uniformly parabolic theory (e.g. from the book of Lieberman) can be applied.
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Feb 13 at 10:04 history edited Akira CC BY-SA 4.0
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Feb 13 at 10:04 comment added Akira @Fiktor Thank you for your comment! It is a typo and yess it should be $\sigma \in C_b^{1,\alpha}(\mathbb{R})$. By $\sigma^2(u(t,x))$, I meant $|\sigma(u(t,x))|^2$.
Feb 11 at 20:54 comment added Fiktor Could you clarify? 1. Did you mean $\sigma \in C_b^{1,\alpha}(\mathbb{R})$? 2. By $\sigma^2(u(t,x))$, do you mean $\sigma(\sigma(u(t,x)))$ or $(\sigma(u(t,x)))^2$?
Feb 10 at 18:46 history edited Akira CC BY-SA 4.0
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Feb 3 at 15:01 history edited Akira CC BY-SA 4.0
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Feb 3 at 11:56 history edited Akira CC BY-SA 4.0
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Feb 3 at 11:49 history edited Akira CC BY-SA 4.0
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Jan 19 at 6:37 history edited Akira CC BY-SA 4.0
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Jan 19 at 6:22 history asked Akira CC BY-SA 4.0