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For a smooth constant-density exterior potential flow satisfying the finite-energy condition for planar exterior potential flow, use $\mathbf{u}=\nabla \varphi$ and the planar streamfunction convention $u={\psi }_y$, $v=-{\psi }_x$. Since ${\nabla }^2\varphi =0$, $\nabla \cdot (\varphi \nabla \varphi )=|\mathbf{u}{|}^2$. Apply the divergence theorem to a truncated exterior region. The outer term vanishes; on a counterclockwise body boundary the body's outward normal satisfies ${\partial }_n\varphi ds=d\psi$, while the fluid's outward normal points the opposite way. This gives the displayed kinetic energy per unit span. The formula yields the added mass of a circular cylinder directly from its surface values of $\varphi$ and $\psi$.

Use instantaneous polar coordinates centered on the moving circular cylinder, but measure the fluid velocity in the laboratory frame. Impermeability equates the normal components of body and fluid velocity:Since ${\mathbf{e}}_x\cdot {\mathbf{e}}_r=\cos \theta$ and $u_r={\varphi }_r$, the no-penetration boundary condition isThe velocity potential obeys the exterior Laplace equation, by incompressibility. The boundary data have only the first cosine harmonic. A decaying harmonic potential of this form is $\varphi =C\cos \theta /r$. Matching its radial derivative gives $-C/a^2=U$, so $C=-U{a}^2$. The zero circulation condition rules out an angular potential term, and the integrated boundary condition gives zero net volume flux, ruling out a radial logarithmic source term. The remaining decaying harmonics have zero coefficients by the boundary data. HenceUsing the streamfunction in polar coordinates, $u_r=r^{-1}{\psi }_{\theta }$, $u_{\theta }=-{\psi }_r$, givesOne may also obtain both functions as the real and imaginary parts of $w(z)=-U{a}^2/z$, a two-dimensional potential doublet. The velocity decays as $r^{-2}$ and the source of this singular representation lies inside the excluded body. These are the fields used to calculate the added mass of a circular cylinder.

On the translating circular cylinder $r=a$, the preceding velocity potential and streamfunction giveUsing the boundary integral for kinetic energy of planar potential flow,ThereforeAs a direct check, $|\mathbf{u}{|}^2=U^2{a}^4/r^4$, and integration over $r>a$ gives the same result.

The fluid kinetic energy has the same dependence on body speed as the kinetic energy of an extra mass $M$ carried by the body. Thus $M$ is its added mass, here equal to the mass of displaced fluid per unit cylinder length. For a cylinder of physical mass per length $m$, the combined kinetic energy is $\frac{1}{2}(m+M)U^2$. Accelerating the body requires accelerating the surrounding fluid, producing the inertial reaction $-M\dot{U}$ along the direction of motion. This is the added mass of a circular cylinder.

For inviscid flow that is also incompressible flow and irrotational flow with velocity potential $\mathbf{u}=\nabla \varphi$, the Euler equations integrate to the Unsteady Bernoulli equationThe same $F(t)$ applies throughout a connected irrotational fluid region; an additive time-dependent potential can absorb it. With gravity absent, fluid at rest at infinity and $\varphi \to 0$, it is $p_{\infty }(t)/\rho$.

Take the cylinder centre at laboratory position $(X(t),0)$ with $\dot{X}=U(t)$, and define $r,\theta$ relative to this moving centre. The circulation-free potential flow satisfies Laplace's equation, decays at infinity, and has surface condition ${\varphi }_r(a,\theta )=U\cos \theta$. The $\cos \theta$ harmonic solution is consequentlyThe velocities are $u_r=U{a}^2\cos \theta /r^2$ and $u_{\theta }=U{a}^2\sin \theta /r^2$, hence $|\mathbf{u}{|}^2=U^2{a}^4/r^4$.

The time derivative must be taken at fixed laboratory coordinates, not fixed $r,\theta$. At a fixed fluid point $\dot{r}=-U\cos \theta$, $\dot{\theta }=U\sin \theta /r$, soSubstituting into the Unsteady Bernoulli equation gives the entire pressure fieldAt the cylinder surface this is $p_{\infty }+\rho a\dot{U}\cos \theta +\rho U^2(\cos 2\theta -1/2)$. Pressure acts against the outward radial normal of the cylinder, so the force per unit length isAll steady terms cancel by trigonometric orthogonality. ThereforeThis identifies the added mass of a circular cylinder as the mass of displaced fluid per unit length. The solution assumes the ambient fluid is initially at rest and has no separately prescribed circulation; these select the usual pure translating-cylinder flow.