Evans proof of smoothness of solutions to the heat equation

Question

This is Theorem 8 in Section 2.3 of Evans:

Theorem. Suppose $u \in C_{1}^{2}(U_{T})$ solves the heat equation in $U_{T}$. Then $u \in C^{\infty}(U_{T})$.

Here, $U_{T} := U \times (0,T]$. In his proof, he obtains a representation

$$u(x,t) = \int\int_{C} \Phi(x-y,t-s)[(\zeta_{s}(y,s) - \Delta \zeta(y,s))u(y,s) - 2D\zeta(y,s)\cdot Du(y,s)]\;dyds$$

where $\Phi$ is the fundamental solution, $C$ is a cylinder of radius $r$ and height $r^2$ whose top center point is some chosen $(x_{0},t_{0}) \in U_{T}$, $C'$ and $C''$ are similar cylinders with radii $3r/4$ and $r/2$, resp., and $(x,t) \in C''$. Evans integrates the last term by parts:

$$-\int\int_{C} 2\Phi(x-y,t-s) D\zeta(y,s) \cdot Du(y,s)\;dyds = \int\int_{C} [2D_{y}\Phi(x-y,t-s)\cdot D\zeta(y,s)]u(y,s)\;dyds$$

I really don't see how he obtains this. There are three factors in the integrand, so I always end up with some $\Delta \zeta(y,s)$ term which doesn't vanish.

Answer 1

Notice that D(ΦDζ∙u)=DΦDζ∙u+ΦΔζ∙u+ΦDζ∙Du, we get
$ \iint_{\partial C}ΦDζ\cdot u \cdot ν dS = \iint_{C}D_y ΦDζ\cdot u dyds + \iint_{C}Φ\Deltaζ\cdot u dyds + \iint_{C}ΦDζ\cdot Du dyds $,
and the LHS vanishes because that $Dζ=0$ for $ζ≡0$ near the parabolic boundary of C . So
$ \iint_{C}-2ΦDζ\cdot Du dyds=\iint_{C}2D_y ΦDζ\cdot u dyds + \iint_{C}2Φ\Deltaζ\cdot u dyds $,
and we get what exactly we want. :-)

Answer 2

Here's a slightly different approach. Note that

$$\langle \nabla \zeta, \nabla (\phi u) \rangle = u \langle \nabla \zeta, \nabla \phi \rangle + \phi \langle \nabla \zeta, \nabla u \rangle$$ so that $$- \phi \langle \nabla \xi, \nabla u \rangle = u \langle \nabla \zeta, \nabla \phi \rangle - \langle \nabla \zeta, \nabla (\phi u) \rangle.$$

Integrating over $C$, using integration by parts and the fact that $\nabla \zeta$ vanishes on $\partial C$, we obtain

\begin{align} \iint_C - \phi \langle \nabla \xi, \nabla u \rangle dyds &= \iint_C u \langle \nabla \zeta, \nabla \phi \rangle - \langle \nabla \zeta, \nabla (\phi u) \rangle dyds \\ &= \iint_C u \langle \nabla \zeta, \nabla \phi \rangle+ 2 \phi u \Delta \zeta dyds. \end{align}

Plugging this in to the expression for $u(x, t)$ gives the desired result.