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.

Consider the following integral,

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

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

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

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., ${\displaystyle \mu \nabla ^{2}u_{i}=\nabla p}$ , we can write

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

Again doing an Integration by parts gives

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

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

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

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

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

and after a rearrangement using above identity, we get

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

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

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

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

The Poiseuille flow theorem^[7] 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.