# Linear coupled parabolic PDE system with Holder continuous coefficients

I am interested in proving existence/uniqueness to: find $u(x,t)$, $v(x,t)$ such that $$u_t - a_1u_{xx} - a_2u_x - a_3u -a_4v = f$$ $$v_t - a_5u_{xx} - a_6u_x - a_7u - a_8v_{xx} - a_9v_x - a_{10}v = g$$ where the coefficients $a_i(x,t)$ are in a parabolic version of Holder space (say $C^{k, \alpha}([0,1]\times[0,t])$.

Is there already literature where this is treated? I tried Ladyschenkaja but there this type of system is not present (I believe in that book, they require $a_1 \equiv a_5$). I would appreciate if anyone had any references to this problem.

I believe though that I can probably apply a fixed-point argument to this -- again I would appreciate if someone gave a pointer as where such things are discussed. Thanks.

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## 2 Answers

There is a classical counterexample due to Plis of an elliptic differential operator with Hölder continuous coefficients without Cauchy uniqueness. This was refined with a counterexample in divergence form by Miller in a 1974 Arch. Rat. Mech.(vol. 54) article for the elliptic and parabolic case.

Hölder continuity is not enough to get uniqueness results for parabolic or elliptic equations.

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Thanks. How about weak solutions? The space doesn't really matter to me. – Bloop Aug 15 '12 at 11:00
What is Cauchy uniqueness? Is it relevant to OP's question? The system as stated looks completely fine to me. – timur Aug 23 '12 at 1:07

A good reference is Krylov's Lectures on elliptic and parabolic equations in Hölder spaces. On another note, if $a_1\not\equiv a_5$ is the only reason you are stuck, then it should be possible to adapt the arguments you have for $a_1\equiv a_5$ to $a_1\not\equiv a_5$, by using uniform ellipticity.

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Thank you, can you expand a little on your last sentence? – Bloop Aug 14 '12 at 18:21