Kelvin's minimum energy theorem

Let ${\displaystyle \mathbf {u} }$ be the velocity field of an incompressible irrotational fluid and ${\displaystyle \mathbf {u_{1}} }$ be that of any other incompressible fluid motion with same normal component velocity ${\displaystyle \mathbf {u} \cdot \mathbf {n} =\mathbf {u_{1}} \cdot \mathbf {n} }$ at the boundary of the domain, where ${\displaystyle \mathbf {n} }$ is the unit vector of the bounding surface (and, if the domain extends to infinity, ${\displaystyle \mathbf {u} \cdot \mathbf {n} =\mathbf {u_{1}} \cdot \mathbf {n} =0}$ there). Then the difference between the kinetic energy is given by

${\displaystyle T_{1}-T={\frac {1}{2}}\rho \int (\mathbf {u} _{1}^{2}-\mathbf {u} ^{2})\ dV}$

can be rearranged to give

${\displaystyle T_{1}-T={\frac {1}{2}}\rho \int (\mathbf {u} _{1}-\mathbf {u} )^{2}\ dV+\rho \int (\mathbf {u} _{1}-\mathbf {u} )\cdot \mathbf {u} \ dV.}$

Since ${\displaystyle \mathbf {u} }$ is irrotational and the domain is simply-connected, a single-valued velocity potential exists, i.e., ${\displaystyle \mathbf {u} =\nabla \phi }$. Using this, the second integral in the above equation can be written as

${\displaystyle \int (\mathbf {u} _{1}-\mathbf {u} )\cdot \nabla \phi \ dV=\int \nabla \cdot [(\mathbf {u} _{1}-\mathbf {u} )\phi ]\ dV-\int \phi \nabla \cdot (\mathbf {u} _{1}-\mathbf {u} )\ dV.}$

The second integral is identically zero for steady incompressible fluid, i.e., ${\displaystyle \nabla \cdot \mathbf {u} =\nabla \cdot \mathbf {u} _{1}=0}$. Applying the Gauss theorem for the first integral we find

${\displaystyle \int (\mathbf {u} _{1}-\mathbf {u} )\cdot \nabla \phi \ dV=\int \phi (\mathbf {u} _{1}-\mathbf {u} )\cdot \mathbf {n} \ dA}$

where the surface integral is zero since normal component of velocities are equal there. Thus, one concludes

${\displaystyle T_{1}-T={\frac {1}{2}}\rho \int (\mathbf {u} _{1}-\mathbf {u} )^{2}\ dV\geq 0}$

or in other words, ${\displaystyle T_{1}\geq T}$, where the equality holds only if ${\displaystyle \mathbf {u} _{1}=\mathbf {u} }$, thereby proving the theorem.