Helmholtz minimum dissipation theorem

Let ${\displaystyle \mathbf {u} ,\ p}$ and ${\displaystyle E={\frac {1}{2}}(\nabla \mathbf {u} +(\nabla \mathbf {u} )^{T})}$ be the velocity, pressure and strain rate tensor of the Stokes flow and ${\displaystyle \mathbf {u} ',\ p'}$ and ${\displaystyle E'={\frac {1}{2}}(\nabla \mathbf {u} '+(\nabla \mathbf {u} ')^{T})}$ be the velocity, pressure and strain rate tensor of any other incompressible motion with ${\displaystyle \mathbf {u} =\mathbf {u} '}$ on the boundary. Let ${\displaystyle u_{i}}$ and ${\displaystyle e_{ij}}$ be the representation of velocity and strain tensor in index notation, where the index runs from one to three. Let ${\displaystyle \Omega \subset \mathbb {R} ^{3}}$ be a bounded domain with boundary ${\displaystyle \Gamma }$ of class ${\displaystyle C^{1}}$.^[7]

Consider the following integral,

${\displaystyle {\begin{aligned}\int _{\Omega }(e_{ij}'-e_{ij})e_{ij}\ dV&=\int _{\Omega }{\frac {\partial (u_{i}'-u_{i})}{\partial x_{j}}}e_{ij}\ dV\end{aligned}}}$

where in the above integral, only symmetrical part of the deformation tensor remains, because the contraction of symmetrical and antisymmetrical tensor is identically zero. Integration by parts gives

${\displaystyle \int _{\Omega }(e_{ij}'-e_{ij})e_{ij}\ dV=\int _{\Gamma }(u_{i}'-u_{i})e_{ij}n_{j}\ dA-{\frac {1}{2}}\int _{\Omega }(u_{i}'-u_{i})(\nabla ^{2}u_{i})\ dV.}$

The first integral is zero because velocity at the boundaries of the two fields are equal. Now, for the second integral, since ${\displaystyle u_{i}}$ satisfies the Stokes flow equation, i.e., ${\displaystyle \mu \nabla ^{2}u_{i}={\frac {\partial p}{\partial x_{i}}}}$, we can write

${\displaystyle \int _{\Omega }(e_{ij}'-e_{ij})e_{ij}\ dV=-{\frac {1}{2\mu }}\int _{\Omega }(u_{i}'-u_{i}){\frac {\partial p}{\partial x_{i}}}\ dV.}$

Again doing an Integration by parts gives

${\displaystyle \int _{\Omega }(e_{ij}'-e_{ij})e_{ij}\ dV=-{\frac {1}{2\mu }}\int _{\Gamma }p(u_{i}'-u_{i})n_{i}\ dA+{\frac {1}{2\mu }}\int _{\Omega }p{\frac {\partial (u_{i}'-u_{i})}{\partial x_{i}}}\ dV.}$

The first integral is zero because velocities are equal and the second integral is zero because the flow is incompressible, i.e., ${\displaystyle \nabla \cdot \mathbf {u} =\nabla \cdot \mathbf {u} '=0}$. Therefore we have the identity which says,

${\displaystyle \int _{\Omega }(e_{ij}'-e_{ij})e_{ij}\ dV=0.}$

The total rate of viscous dissipation energy over the whole volume ${\displaystyle \Omega }$ of the field ${\displaystyle \mathbf {u} '}$ is given by

${\displaystyle D'=\int _{\Omega }\Phi 'dV=2\mu \int _{\Omega }e_{ij}'e_{ij}'\ dV=2\mu \int _{\Omega }[e_{ij}e_{ij}+e_{ij}'e_{ij}'-e_{ij}e_{ij}]\ dV}$

and after a rearrangement using above identity, we get

${\displaystyle D'=2\mu \int _{\Omega }[e_{ij}e_{ij}+(e_{ij}'-e_{ij})(e_{ij}'-e_{ij})]\ dV}$

If ${\displaystyle D}$ is the total rate of viscous dissipation energy over the whole volume of the field ${\displaystyle \mathbf {u} }$, then we have

${\displaystyle D'=D+2\mu \int _{\Omega }(e_{ij}'-e_{ij})(e_{ij}'-e_{ij})\ dV}$.

The second integral is non-negative and zero only if ${\displaystyle e_{ij}=e_{ij}'}$, thus proving the theorem (${\displaystyle D'\geq D}$).

The Poiseuille flow theorem^[8] is a consequence of the Helmholtz theorem states that The steady laminar flow of an incompressible viscous fluid down a straight pipe of arbitrary cross-section is characterized by the property that its energy dissipation is least among all laminar (or spatially periodic) flows down the pipe which have the same total flux.