intuition behind stability in different norms I want to understand connection between property of solution and energy estimates. Say I have a pde $u_t=Lu$ and I proved that there exist a unique solution that is in $C^2$. In pde analysis using the energy methods we can estimate the norm of the solution by the initial data. For example $\frac{\partial \vert\vert^2 u(x,t) \vert\vert}{\partial t}\leq 0$. Now the norm that is typically used in the book is $L^2$ norm. Can I show similar estimates using $H^1$ norm? And if I do so, what would that imply? That's is how do I decide which norm to choose in order to get the estimates for the norm of the solution? thanks! 
 A: I think you meant to say that there exists a unique solution $u\in C^{2,1}(\mathbb{R}^n \times (0,T))\bigcap C(\mathbb{R}^n \times [0,T])$(? or other reasonably regularity for $u$), it has $C^2$ and $C^1$ differentiability in space and time respectively, also continuous up to the boundary, due to the regularity you gave, I am guessing $L$ should be a second order linear differential operator, for example, $L = \Delta$ then you have a homogeneous heat equation on the whole space, and I will use this as an example to illustrate the intuition behind the "energy estimate", so the language may be a little bit informal.
The estimate you gave was:
$$
\frac{d}{d t} \|u(\cdot,t)\|_{L^2(\mathbb{R})}^2 \leq 0
$$
If we go back to the derivation of the heat equation, the solution $u$ represents concentration of a certain quantity(heat, chemical substance, probability density), the estimate above implies two things:


*

*The concentration decays as we marching in time without any "input", otherwise we should have $u_t = Lu + f$ and the estimate would be different.

*The $L^2$-integrability implies that there is control of this concentration outside a ball of certain radius, just think of the normal distribution density function, the total concentration far from the origin is rather small.
Now back to your question, why the $L^2$-norm for energy estimate? Because it is most natural estimate originated from the weak formulation of this equation. Why not the $H^1$-norm for the energy estimate? By $H^1$-norm, I am guess you would like to get the same estimate for $\displaystyle \frac{d}{d t}\|u(\cdot,t)\|_{H^1(\mathbb{R})}^2$, the meaning of this would be that you have decaying property on the flux density $\boldsymbol{F} = -\nabla u$! Unfortunately for general $L$ and even for $\Delta$, this is not true, physically speaking, the flux $-\nabla u$ describes the flow across $\partial \Omega$ from the higher concentration region to lower concentration region in the following sense:
$$
\frac{d}{dt}\int_{\Omega} u = - \int_{\partial \Omega} \boldsymbol{F}\cdot \boldsymbol{n} \,dS
$$
And that the concentration is decaying over time doesn't imply the flux has the temporal decaying property too. Secondly $\displaystyle \frac{d}{d t}\|u(\cdot,t)\|_{H^1(\mathbb{R})}^2$ is not well-defined for $u$ either. 
However, we could have another type of estimate for $\|u(\cdot,t)\|_{H^1(\mathbb{R})}^2$, let's say for the heat equation:
$$
\left\{
\begin{aligned}
u_t &= \Delta u & \text{ for }x\in \mathbb{R}^n
\\
u(0,x) &= g(x)
\end{aligned}
\right.
$$
Apply the same technique that gets us to the energy decaying estimate:
$$
\frac{1}{2}\frac{d}{dt}\int_{\mathbb{R}^n} u^2(x,t) dx + \int_{\mathbb{R}^n} |\nabla_x u(x,t)|^2 dx = 0 
$$
then we integrate w.r.t time on $[0,t]$ for any $t&ltT$ we have:
$$
\frac{1}{2}\int_{\mathbb{R}^n} u^2(x,t) dx + \int^t_0 \int_{\mathbb{R}^n} |\nabla_x u(x,\tau)|^2 dx d\tau = \frac{1}{2}\int_{\mathbb{R}^n} g(x)^2 dx 
$$
hence we have a rather primitive estimate
$$
\int^T_0 \int_{\mathbb{R}^n} |\nabla_x u(x,\tau)|^2 dx d\tau \leq \frac{1}{2}\int_{\mathbb{R}^n} g(x)^2 dx 
$$
