$H^{-1}$ norm and $H^{-1}$ gradient flow I know how to calculate the $L^2$ gradient flow. But I also noticed some books mentioned the $H^{-1}$ gradient flow. For instance, the Dirichlet Energy $$E(u) = \frac{1}{2} \int_{\Omega} |\nabla u|^2 dx$$
It is easy to get the $L^2$ gradient flow, which is $-\Delta u$. But how to calculate the $H^{-1}$ gradient flow? I know the answer is ${\Delta}^2 u$.
Just curious, what is the $H^{-1}$ norm on $H^{-1}$ space? I understand that $H^{-1}$ space is the dual space of $H^{1}$ space and naturally equipped with the norm $\displaystyle \lVert F {\rVert}_{H^{-1}} = \sup_{f\in H^1}{\frac{|F(f)|}{\lVert f \rVert_{H^1}}}$. But is there a way to calculate the $H^{-1}$ norm more explicitly?
 A: Since $H^{-1} := (H_0^1)'$ is the dual of a Hilbert space, we have a canonical isomorphism $H_0^1 \simeq H^{-1}$ given by $f^\flat = \langle f, \cdot \rangle_{H^1_0};$ i.e. a function $f \in H^1_0$ corresponds to the linear functional $$f^\flat(\phi)= \int_\Omega \nabla f \cdot \nabla \phi$$ on $H_0^1.$ From integration by parts we see that $f^\flat(\phi) = -\langle \Delta f, \phi\rangle_{L^2};$ so the isomorphism $\flat : H_0^1\to H^{-1}$ can be viewed as the distributional Laplacian $-\Delta.$ This isomorphism is important to us because it is how we define the inner product on the dual space $H^{-1}:$ by definition we have $$\langle f^\flat, g^\flat\rangle_{H^{-1}} = \langle f,g \rangle_{H^1_0},$$ or more explicitly $$\langle u,v \rangle_{H^{-1}} = \langle \Delta^{-1}u,\Delta^{-1}v\rangle_{H_0^1}.$$
The directional derivative of $E$ is $$DE|_u(v) = \frac{d}{ds}\Big|_{s=0} E(u+sv)= \int_\Omega \nabla u\cdot \nabla v=\langle u,v \rangle_{H^1_0}. $$
The $H^{-1}$ gradient of $E$ is defined by the relation $\langle\nabla E|_u, \phi\rangle_{H^{-1}}=DE|_u(\phi);$ i.e. $$\langle \Delta^{-1} \nabla E|_u, \Delta^{-1}\phi \rangle_{H^1_0}=\langle u, \phi \rangle_{H^1_0}.$$ Integrating by parts on the LHS gives  $$\langle\Delta^{-2}\nabla E|_u,\phi\rangle=\langle u,\phi\rangle$$ for all $\phi,$ and thus we conclude $\nabla E|_u = \Delta^2 u.$
