# Does smoothness of solution to parabolic equation require smoothness of coefficients?

I have a function that solves a parabolic partial differential equation $$\partial_tu - Lu = 0$$ with a linear second order uniformly elliptic-in-space differential operator $$L$$, whose coefficients are $$C^1$$. Is it true that $$u$$ is $$C^{\infty}$$? (If it helps, I might also be able to argue that the second-order coefficients are smooth.)

Theorem 4.5 on page 19 of Lectures on second order elliptic and parabolic partial differential equations proves that if $$L$$ is uniformly elliptic in space and all coefficients of $$L$$ are $$k$$-times differentiable, then any solution is $$k+2$$ times differentiable (I'm ignoring some other assumptions here that I believe aren't important to the question).

On the other hand, the Wikipedia entry on Hoermander's condition states that if "the vector fields [...] have bounded derivative [...] [and satisfy] the parabolic Hoermander condition, then its solutions admit a smooth density with respect to Lebesgue measure". (And I'm only interested in parabolic equations where the Hoermander condition is trivially satisfied.)

In particular, the Wikipedia article only requires coefficients to be once differentiable to deduce that the solution is smooth. (Hoermander's article only discusses smooth coefficients, but I could believe that that's just for convenience of presentation.)

My question: Is Wikipedia right, and regularity of the coefficients is not as important as one might think from the lecture notes? Is there maybe a crucial difference in the setups that I'm missing? Or does Wikipedia simply mean "differentiable" by "smooth"?

In general, it is easy to see that for PDE with non-zero RHS $$f$$, smoothness results must impose regularity of $$f$$, because any non-smooth $$u$$ satisfies its own parabolic PDE $$\partial_t - Lu = f$$ with $$f:= \partial_t - Lu$$.

Unfortunately, I have no intuition or idea how to argue even informally about the necessity of regularity of the other coefficients for smoothness results.

Yes, smoothness of solutions to parabolic equation depend quite directly on the smoothness of coefficients. E.g. consider a uniformly parabolic operator in non divergence form $$Lu=\partial_tu -\sum_{i,j=1}^na_{ij}(x,t)\partial_{ij}u.$$ If the coefficients belong to anisotropic Holder space $$C^{k,\alpha}(\Omega)$$, $$k\ge0$$, $$0<\alpha<1$$, for domain $$\Omega\subset \mathbb R^{n+1}$$ then it can be proven that any classical solution $$u$$ of homogeneous equation $$Lu=0$$ in $$\Omega$$ belongs to $$C^{k+2,\alpha}_{loc}(\Omega)$$.

And, in general, no more smoothness is guarantied. To see that, take one dimensional case $$\partial_tu-a(x,t)\partial_{x}^2u=0.$$ Then $$\partial_tu-\partial_{x}^2u=(a(x,t)-1)\partial_{x}^2u.$$

Let $$a\in C^{0,\alpha}(\bar \Omega)$$ and $$a(x,t)\ne1$$ in $$\Omega$$. Then $$u$$ belongs to $$\in C^{2,\alpha}_{loc}(\Omega)$$, but not, say, $$C^{3,\alpha}_{loc}(\Omega)$$. Assuming the opposite the reasoning is reduced to your argument for an equation with rhs. Namely, if $$u\in C^{3,\alpha}_{loc}(\Omega)$$ then the rhs $$(a(x,t)-1)\partial_{x}^2u$$ generally belongs to $$C^{0,\alpha}_{loc}(\Omega)$$ and, therefore, either $$\partial_tu$$ or $$\partial_{x}^2u$$ doesn't belong to $$C^{1,\alpha}_{loc}(\Omega)$$.

• Sorry, I don't follow your counterexample / counterargument. Why can't $(a-1)\partial_x^2u$ be more smooth than $C^{0,\alpha}$? Commented Jun 10 at 20:47
• @Bananach by assumption $u\in C^{3,\alpha}_{loc}(\Omega)$, so $\partial_{x}^2u\in C^{1,\alpha}_{loc}(\Omega)$. Take $a \in C^{0,\alpha} (\Omega)$ s.t. $a \not\in C^{0,\beta }_{loc}(\Omega)$ for any $\beta>\alpha$. Then $a-1 \in C^{0,\alpha}(\Omega)$ and $(a(x,t)-1)\partial_{x}^2u$ is generally no more smooth than $C^{0,\alpha}_{loc}(\Omega)$. Commented Jun 11 at 3:26
• But $u$ solves a PDE that involves $a$. Can't that imply that $a\partial_x^2u$ is more smooth? Commented Jun 11 at 6:34
• @Bananach By assumption $\partial_{x}^2u\in C^{1,\alpha}_{loc}(\Omega)$. If this function doesn't turn to zero and $a \in C^{0,\alpha} (\Omega)$ and not more then the product $a\partial_{x}^2$ is not smoother than $C^{0,\alpha}_{loc}(\Omega)$. o see that consider function of one variable $f(x)=|x|^\alpha g(x)$ , where $g\in C^{1,\alpha}([-1,1])$, $g(0)\ne0$. Then as can be easily checked $f\in C^{0,\alpha}([-1,1])$. Commented Jun 11 at 14:57
• @Bananach But $f\not\in C^{0,\beta}([-1,1])$ for any $\beta>\alpha$, since $$\sup_{x,x+\Delta x\in[-1,1],\Delta x\ne0}\frac{| f(x+\Delta x)-f(x)|}{|\Delta x|^{\beta}}\ge \sup_{\Delta x\in[-1,1],\Delta x\ne0}\frac{| f(\Delta x)-f(0)|}{|\Delta x|^{\beta}}=\sup_{\Delta x\in[-1,1],\Delta x\ne0}\frac{| \Delta x|^\alpha}{|\Delta x|^{\beta}}|g(\Delta x)|=\infty.$$ Commented Jun 11 at 14:57

If $$\partial_t u + a \partial_x^2u =0$$ for non-smooth $$a$$, then in any neighborhood where $$\partial_x^2u\not=0$$, the ratio $$\partial_t u / \partial_x^2 u = a$$ is not smooth, and therefore either $$\partial_t u$$ or $$\partial_x^2u$$ must be non-smooth. In particular $$u$$ is not smooth unless it's trivial.

This trivially extends to all PDE with suitable modifications.