Problem Formulation

(Adapted from Katz and Plotkin's Low Speed Aerodynamics, Sec 3.2)

By Helmholtz' decomposition theorem, any velocity field $\mathbf{u}$ can be decomposed into a uniform component, an irrotational component, and a solenoidal component as

\[\begin{align*} \mathbf{u} = \underbrace{\mathbf{u}_\infty}_\text{uniform} + \underbrace{\nabla\phi}_\text{irrotational} + \underbrace{\nabla\times\boldsymbol{\psi}}_\text{solenoidal} ,\end{align*}\]

where $\mathbf{u}_\infty(t)$ is the freestream, $\phi (\mathbf{x}, t)$ is a scalar-potential field, and $\boldsymbol\psi (\mathbf{x}, t)$ is a vector-potential field.

NOTE: Even though here we show $\mathbf{u}_\infty$ as its own component, the freestream can be rolled into the potential field as a linearly-varying potential, giving a uniform velocity field.

Assuming incompressible flow, the continuity equation poses a Laplace equation for the scalar-potential as

\[\begin{align*} \nabla \cdot \mathbf{u} = 0 \quad\Rightarrow\quad \nabla^2 \phi = 0 .\end{align*}\]

Before we continue, let us derive a useful integral identity: Green's second identity. Let $f$ and $g$ be two differentiable functions, we have

\[\begin{align*} \nabla \cdot \left( f \nabla g - g \nabla f \right) = f \nabla^2 g - g \nabla^2 f .\end{align*}\]

The divergence theorem then leads to

\[\begin{align*} \int\limits_{\partial V} \left( f \nabla g - g \nabla f \right) \cdot \hat{\mathbf{n}} \,\mathrm{d}S & = \int\limits_{V} \left( f \nabla^2 g - g \nabla^2 f \right) \,\mathrm{d}V ,\end{align*}\]

over any volume $V$. Here we have defined the normal $\hat{\mathbf{n}}$ as pointing outward from $V$.

Using Green's second identity, we now define $f(\mathbf{x}) \equiv \frac{1}{\Vert \mathbf{x} - \mathbf{x}_p \Vert^2} = \frac{1}{r}$ and $g(\mathbf{x}) \equiv \phi(\mathbf{x})$ where $r$ is the distance to a point $\mathbf{x}_p$ defining an arbitrary fixed center, and the identity gives

\[\begin{align*} \int\limits_{\partial V} \left[ \frac{1}{r} \nabla \phi - \phi \nabla \left(\frac{1}{r}\right) \right] \cdot \hat{\mathbf{n}} \,\mathrm{d}S & = \int\limits_{V} \left[ \frac{1}{r} \cancel{\nabla^2 \phi}^{\,\,0} - \phi \cancel{\nabla^2 \left(\frac{1}{r}\right)}^{\,\,0} \right] \,\mathrm{d}V \end{align*}\]

\[\begin{align*} \Rightarrow \boxed{ \int\limits_{\partial V} \left[ \frac{1}{r} \nabla \phi - \phi\nabla \left(\frac{1}{r}\right) \right] \cdot \hat{\mathbf{n}} \,\mathrm{d}S = % 0 } .\end{align*}\]

The integrand becomes singular when evaluated at $\mathbf{x}_p$ since $\lim \limits_{r\rightarrow 0 }\frac{1}{r} = \infty$, which we will use to obtain some expressions depending on whether $\mathbf{x}_p$ lays inside, outside, or at the boundary of $V$.

In the case that $\mathbf{x}_p \cancel{\in} V$, the integral equation is automatically satisfied. In the case that $\mathbf{x}_p \in V$, we introduce a hole in $V$ in the form of a sphere of radius $\epsilon$ surrounding $\mathbf{x}_p$, and the integral equation is then automatically satisfied over the domain $V \backslash V_\epsilon$:

\[\begin{align*} & \int\limits_{\partial (V \backslash V_\epsilon)} \left[ \frac{1}{r} \nabla \phi - \phi\nabla \left(\frac{1}{r}\right) \right] \cdot \hat{\mathbf{n}} \,\mathrm{d}S = % 0 \\ \Leftrightarrow & \int\limits_{\partial V} \left[ \frac{1}{r} \nabla \phi - \phi\nabla \left(\frac{1}{r}\right) \right] \cdot \hat{\mathbf{n}} \,\mathrm{d}S - \int\limits_{\partial V_\epsilon} \left[ \frac{1}{r} \nabla \phi - \phi\nabla \left(\frac{1}{r}\right) \right] \cdot \hat{\mathbf{n}} \,\mathrm{d}S = % 0 .\end{align*}\]

Since $r$ is centered at $\mathbf{x}_p$ and the normal $\hat{\mathbf{n}}$ points radially inwards the sphere (since all normals point outside of the volume $V$), the second integral becomes

\[\begin{align*} \int\limits_{\partial V_\epsilon} \left[ \frac{1}{r} \nabla \phi - \phi\nabla \left(\frac{1}{r}\right) \right] \cdot \hat{\mathbf{n}} \,\mathrm{d}S = % - \int\limits_{\partial V_\epsilon} \left( \frac{1}{r} \frac{\partial \phi}{\partial r} + \frac{\phi}{r^2} \right) \,\mathrm{d}S .\end{align*}\]

In the limit that $\epsilon$ is infinitesimally small, the integral can be approximated as

\[\begin{align*} \int\limits_{\partial V_\epsilon} \left( \frac{1}{r} \frac{\partial \phi}{\partial r} - \frac{\phi}{r^2} \right) \,\mathrm{d}S \approx - \left( \frac{\partial \phi}{\partial r}(\mathbf{x}_p) \lim\limits_{r\rightarrow 0} \frac{1}{r} + \phi (\mathbf{x}_p) \lim\limits_{r\rightarrow 0} \frac{1}{r^2} \right) \lim\limits_{\epsilon\rightarrow 0} 4 \pi \epsilon^2 ,\end{align*}\]

where $4 \pi \epsilon^2$ is the surface area of the sphere.

Assuming $\phi(\mathbf{x}_p) \neq \pm\infty$ and $\frac{\partial \phi}{\partial r}(\mathbf{x}_p) \neq \pm\infty$, we have

\[\begin{align*} \int\limits_{\partial V_\epsilon} \left( \frac{1}{r} \frac{\partial \phi}{\partial r} - \frac{\phi}{r^2} \right) \,\mathrm{d}S & \approx - 4 \pi \frac{\partial \phi}{\partial r}(\mathbf{x}_p) \cancel{\lim\limits_{\rho\rightarrow 0} \frac{\rho ^2}{\rho}}^{\,\,{0}} - 4 \pi \phi(\mathbf{x}_p) \cancel{\lim\limits_{\rho\rightarrow 0} \frac{\rho^2}{\rho^2}}^{\,\,1} .\end{align*}\]

Thus,

Substituting this back into the original equation,

\[\begin{align*} \int\limits_{\partial V} \left[ \frac{1}{r} \nabla \phi - \phi\nabla \left(\frac{1}{r}\right) \right] \cdot \hat{\mathbf{n}} \,\mathrm{d}S - 4 \pi \phi(\mathbf{x}_p) = % 0 ,\end{align*}\]

and since $\mathbf{x}_p$ is any arbitrary point inside $V$, we conclude

\[\begin{align*} \boxed{ \phi(\mathbf{x}) = % \frac{1}{4 \pi } \int\limits_{\partial V} \left[ \frac{1}{r} \nabla \phi - \phi\nabla \left(\frac{1}{r}\right) \right] \cdot \hat{\mathbf{n}} \,\mathrm{d}S } , \quad \forall \mathbf{x} \in V .\end{align*}\]

If $\mathbf{x}_p$ lays in the boundary of $V$ (i.e., $\mathbf{x}_p \in \partial V$), we repeat the same derivation except that only half of the sphere $V_\epsilon$ is contained in the integration domain, and we conclude

\[\begin{align*} \boxed{ \phi(\mathbf{x}) = % \frac{1}{2 \pi } \int\limits_{\partial V} \left[ \frac{1}{r} \nabla \phi - \phi\nabla \left(\frac{1}{r}\right) \right] \cdot \hat{\mathbf{n}} \,\mathrm{d}S } , \quad \forall \mathbf{x} \in \partial V .\end{align*}\]

We merge the two equations to get

\[\begin{align*} \therefore \boxed{ \phi(\mathbf{x}) = % \frac{f_{\tiny \partial V}(\mathbf{x})}{\pi } \int\limits_{\partial V} \left[ \frac{1}{r} \nabla \phi - \phi\nabla \left(\frac{1}{r}\right) \right] \cdot \hat{\mathbf{n}} \,\mathrm{d}S } , \quad \forall \mathbf{x} \in V , \quad \text{where } f_{\tiny \partial V}(\mathbf{x}) = \begin{cases} 1/4 & \mathbf{x} \in V \backslash \partial V \\ 1/2 & \mathbf{x} \in \partial V \end{cases} .\end{align*}\]

Now, consider the case that $V$ is infinitely large in all directions (with bound $\partial V_\infty$) while also having a hole (bound $\partial V_b$, which represents a solid body immersed in the domain), leading to $\partial V = \partial V_b \cup \partial V_\infty$. In $V$, the potential is calculated as

\[\begin{align*} \phi(\mathbf{x}) = % \frac{f_{\tiny \partial V_b}(\mathbf{x})}{\pi } \int\limits_{\partial V_b} \left[ \frac{1}{r} \nabla \phi - \phi\nabla \left(\frac{1}{r}\right) \right] \cdot \hat{\mathbf{n}} \,\mathrm{d}S + \underbrace{ \frac{1}{4\pi } \int\limits_{\partial V_\infty} \left[ \frac{1}{r} \nabla \phi - \phi\nabla \left(\frac{1}{r}\right) \right] \cdot \hat{\mathbf{n}} \,\mathrm{d}S }_{ \equiv \phi_\infty(\mathbf{x}) } .\end{align*}\]

When the isolated body is considered as a volume of its own, $V_b$, we can define an internal potential (i.e., internal to the body) as $\phi_i$. When $\mathbf{x}_p$ is defined as being outside of $V_b$ (but still inside $V$), we have

\[\begin{align*} \frac{f_{\tiny \partial V_b}(\mathbf{x})}{\pi } \int\limits_{\partial V_b} \left[ \frac{1}{r} \nabla \phi_i - \phi_i\nabla \left(\frac{1}{r}\right) \right] \cdot \hat{\mathbf{n}} \,\mathrm{d}S = % 0 ,\end{align*}\]

from the Laplace equation.

Since the superposition of two solutions to the Laplace equation yields another valid solution, we can add this internal potential to the previous equation to obtain

\[\begin{align*} \phi(\mathbf{x}) = % \frac{f_{\tiny \partial V_b}(\mathbf{x})}{\pi } \int\limits_{\partial V_b} \left[ \frac{1}{r} \nabla \left(\phi - \phi_i\right) - \left(\phi - \phi_i\right) \nabla \left(\frac{1}{r}\right) \right] \cdot \hat{\mathbf{n}} \,\mathrm{d}S + \phi_\infty(\mathbf{x}) ,\end{align*}\]

where the negative sign accompanying $\phi_i$ comes from requiring $\hat{\mathbf{n}}$ to point outward from $V$.

This equation poses a boundary integral equation (BIE), with the potential $\phi$ being fully determined by its value at the boundaries. We can then define

\[\begin{align*} & \boxed{-\sigma \equiv \frac{\partial \phi}{\partial n} - \frac{\partial \phi_i}{\partial n}} \\ & \boxed{-\mu \equiv \phi - \phi_i} \end{align*}\]

resulting in the following BIE,

\[\begin{align*} \boxed{ \phi(\mathbf{x}) = % -\frac{f_{\tiny \partial V_b}(\mathbf{x})}{\pi } \int\limits_{\partial V_b} \left[ \sigma \frac{1}{r} - \mu \hat{\mathbf{n}} \cdot \nabla \left(\frac{1}{r}\right) \right] \,\mathrm{d}S + \phi_\infty(\mathbf{x}) } , \quad \text{where } f_{\tiny \partial V_b}(\mathbf{x}) = \begin{cases} 1/4 & \mathbf{x} \in V \backslash \partial V_b \\ 1/2 & \mathbf{x} \in \partial V_b \end{cases} ,\end{align*}\]

where $\mu$ and $\sigma$ are unknowns that we will later determine imposing boundary conditions.