Energy estimate for heat equation

Question

Consider the following heat equation in a bounded smooth domain $D \subset \mathbb{R}^d$: \begin{align*} u_t -\triangle u &= f, \qquad (x,t) \in D \times (0,T)\\ \partial_n u &=0, \qquad (x,t) \in \partial D \times (0,T)\\ u(x,0)&=0, \qquad x \in D. \end{align*} I am interested to some methods used to obtain an energy estimate of the form $$\|u\|_{H^1(0,T;L^2) \cap L^2(0,T;H^2)} \le C(T) \|f\|_{L^2(0,T;L^2)},$$ with better explicit constant $C(T)$. Note that Gronwall lemma gives an exponential constant in $T$.

Do you know any reference that covers such an estimate?

Answer 1

For a given function $g\in L^2(\Omega)$ let $\bar g:=\int_\Omega g\ dx$. Testing the equation with the constant function $1$, integrating by parts in space gives the ordinary differential equation $$ (\bar u)_t = \bar f, \ \bar u(0)=0. $$ So $\|\bar u\|_{L^\infty(0,T)} \le \|\bar f\|_{L^1(0,T)}$.

Testing the equation with $u-\bar u $, integration by parts in space, integration from $0$ to $t$ in time gives $$ \|u(t)-\bar u\|_{L^2}^2 - \| u(0)\|_{L^2}^2 + \|\nabla u\|_{L^2(0,t;L^2)}^2 = \int_0^t \int_\Omega f (u-\bar u) dt. $$ Now $$ \int_0^t\int_\Omega f (u-\bar u) dt =\int_0^t\int_\Omega (f-\bar f) (u-\bar u) dt \le \|f-\bar f\|_{L^2(0,t;L^2)} \|u-\bar u\|_{L^2(0,t;L^2)} , $$ the last factor is less than $c\|\nabla u\|_{L^2(0,t;L^2)} $ by Poincare inequality ($c$ depends on $\Omega$ only). And this expression can be compensated by the left-hand side.

Similarly, testing the equation with $-\Delta u$, integration by parts in space in the first term, integration from $0$ to $t$ in time gives $$ \|\nabla u(t)\|_{L^2}^2 - \|\nabla u(0)\|_{L^2}^2 + \|\Delta u\|_{L^2(0,t;L^2)}^2 = \int_0^t f (-\Delta u) dt. $$ Putting all these estimates together, we find: $$ \|\bar u\|_{L^\infty(0,T)}^2 + \|u-\bar u\|_{L^\infty(0,T;L^2)}^2 + \|\nabla u\|_{L^2(0,T;L^2)}^2 + \|\Delta u\|_{L^2(0,T;L^2)}^2 \le c\left( \|\bar f\|_{L^1(0,T)}^2 + \|f-\bar f\|_{L^2(0,T;L^2)}^2 + \|f\|_{L^2(0,T;L^2)}^2 \right), $$ where $c$ is independent of $T$. When estimating these $L^1$ and $L^\infty$ norms against $L^2$ norms, we would get additional factors depending on $T$. However, as we never used Gronwall inequality the constants are not exponential in $T$.

Answer 2

Here is another version of the previous answer.

You really only need an estimate for $\|\Delta u\|_{L^2(0,T;L^2)}$ and for $\|\bar u\|_{L^2(0,T)}$.

One the one hand, such an estimate implies an estimate for $u_t$ in $L^2(0,T;L^2)$, simply by using the equation and the triangle inequality.

On the other hand, an estimate for $\|u\|_{L^2(0,T;H^2)}$ also follows, since $$ \|u\|_{H^2}^2 \le const \cdot (\|\Delta u\|_{L^2}^2 + \bar u^2) $$ by standard elliptic regularity theory, where the constant depends on the geometry of $\Omega$. An estimate for $\Delta u$ now follows from testing the equation with $- \Delta u$, which gives after integration by parts $$ \int_0^T \int_\Omega \nabla u \cdot \nabla u_t + \int_0^T \int_\Omega |\Delta u|^2 = -\int_0^T \int_\Omega f \cdot \Delta u \, . $$
The first term on the left is non-negative (just integrate with respect to $t$), and the Cauchy-Schwarz inequality implies $$ \|\Delta u \|_{L^2(0,T;L^2)} \le 1 \cdot \|f \|_{L^2(0,T;L^2)} $$ As in the answer above, this can be sharpened by replacing $f$ with $f - \bar f$.