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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!

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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 $t<T$ 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 $$ \int^T_0 \int_{\mathbb{R}^n} |\nabla_x u(x,\tau)|^2 dx d\tau \leq \frac{1}{2}\int_{\mathbb{R}^n} g(x)^2 dx $$

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Jon, thanks for you answer, you got my question precisely. You have mentioned that $\frac{d\vert\vert u(⋅,t)\vert\vert^2_{H^1}}{dt}$ is not well-defined for u either. What do you mean? Because I think I can show for the case of $u_t=(a(x)u_x)_x$ stability in $H^1$, however I am still working on showing that for $u_t=(a(x)u_x)_x+b(x)u_x+c(x)u+d(x)$. I can't find that in the book so I am in process of getting this result and your answer made me think it is impossible... –  Medan Apr 29 '12 at 2:41
    
@Medan You are correct, I was too indoctrinated by the weak formulation, and thought of the weak solution situation. If $\displaystyle \|\frac{d}{d t}u(\cdot,t)\|_{H^1(\mathbb{R})}^2$ exists, we are requiring the time derivative of the gradient is $L^2$-integrable too, this isn't necessarily true for weak solution though. However you could refer to L.C.Evans's PDE book Page 360 about the improved regularity for the linear evolution equation, and there is a bound for the second order time derivative which is related to the growth rate of $\nabla u$. –  Shuhao Cao Apr 29 '12 at 20:14
    
@Medan and by any chance you are working on the design of a stable numerical scheme for the variant coefficient parabolic equation? If you are, I am thinking Stig Larsson and Vidar Thomée's book Partial Differential Equations With Numerical Methods may have something that you need. –  Shuhao Cao Apr 29 '12 at 20:17
    
thanks for suggestion, I am working on proving stability of CN for variable coefficients advection-diffusion. So far have it in L^2 but would be nice to see in other norms. I will take a look at Evans book as well. I think I can show the stability in H^1 as well but that requires equation to be strong parabolic and a(x) degenerates in my equation. May be those books have something. thanks! –  Medan Apr 29 '12 at 23:28

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