Subsonic Potential Aerodynamics

A 3D Aerodynamic Potential-Flow Code

Theoretical Model

• A vector field $\underline{V}: \mathbb{U} \to \mathbb{R}^n$, where $\mathbb{U}$ is an open subset of $\mathbb{R}^n$, is said to be conservative if there exists a $\mathrm{C}^1$ (continuously differentiable) scalar field $\phi$ on $\mathbb{U}$ such that $\underline{V} = \nabla \phi$.

• According to Poincaré's Lemma, A continuously differentiable ($\mathrm{C}^1$) vector field $\underline{V}$ defined on a simply connected subset $\mathbb{U}$ of $\mathbb{R}^n$ ($\underline{V} \colon \mathbb{U} \subseteq \mathbb{R}^n \to \mathbb{R}^n$), is conservative if and only if it is irrotational throughout its domain ($\nabla \times \underline{V} = 0$, $\forall \underline{x} \in \mathbb{U}$).

• Circulation $\Gamma = \oint_{C} \underline{V} \cdot \mathrm{d} \underline{l} = \iint_S \nabla \times \underline{V} \cdot \mathrm{d}\underline{S}$.

• In a conservative vector field this integral evaluates to zero for every closed curve. $\Gamma = \oint_{C} \underline{V} \cdot \mathrm{d} \underline{l} = \iint_S \nabla \times \underline{V} \cdot \mathrm{d}\underline{S} = \iint_S \nabla \times \nabla \phi \cdot \mathrm{d}\underline{S} = 0$

Velocity field $\underline{V} \colon \mathbb{U} \subseteq \mathbb{R}^3 \to \mathbb{R}^3$

$\begin{array}{l} \bullet \text{ incompressible: } \nabla \cdot \underline{V} = 0 \\ \begin{drcases} \bullet \text{ irrotational: } \nabla \times \underline{V} = 0 \\ \bullet \text{ simply connected domain } \mathbb{U} \end{drcases} \implies \underline{V} = \nabla \phi \text{ (conservative)} \end{array}$

Laplace's Equation

$ \nabla \cdot \underline{V} = 0 \implies \nabla \cdot \nabla \phi = 0 \implies \nabla^2 \phi = 0$

Rotational Invariance of Laplace differential operator

A function defined on an inner product space is said to have rotational invariance if its value does not change when arbitrary rotations are applied to its argument. $f(\underline{x}') = f(\mathbf{R}\underline{x}) = f(\underline{x})$ This also applies for an operator that acts on a function $f : \mathbb{U} \subseteq \mathbb{R} \to \mathbb{U}$. In that case rotational invariance may also mean that the function commutes with rotations of elements in $\mathbb{U}$. An example is the Laplace differential operator $\nabla^2 = \frac{\partial^2()}{\partial x^2} + \frac{\partial^2()}{\partial y^2} + \frac{\partial^2()}{\partial z^2}$.

$\nabla_\mathbf{x}^2 u(\mathbf{x}) = \nabla_\mathbf{y}^2 v(\mathbf{y})$, where $\mathbf{y} = \mathbf{R} \mathbf{x}$ and $ u(\mathbf{x}) = v(\mathbf{y}) = v(\mathbf{R} \mathbf{x})$

Fundamental Solution of the Laplace operator

Since Laplace equation is invariant under rigid motions, it is natural to look for solutions to $\nabla^2 \psi = 0$ which have rotational symmetry and the form $\psi \colon \mathbb{R}^3 \to \mathbb{R} \colon \psi(\underline{r},\underline{r}_p) = \psi(\lVert \underline{r} - \underline{r}_p \rVert)$

Assuming that $\underline{r} \neq \underline{r}_p$ , it is true that

$$\begin{align*} &\frac{\partial \psi}{\partial x} = \psi'(\lVert \underline{r} - \underline{r}_p \rVert) \frac{x-x_p}{\lVert \underline{r} - \underline{r}_p \rVert} \, , \qquad \frac{\partial \psi}{\partial x} = \psi'(\lVert \underline{r} - \underline{r}_p \rVert) \frac{x-x_p}{\lVert \underline{r} - \underline{r}_p \rVert} \, , \qquad \frac{\partial \psi}{\partial x} = \psi'(\lVert \underline{r} - \underline{r}_p \rVert) \frac{x-x_p}{\lVert \underline{r} - \underline{r}_p \rVert} \\\ &\frac{\partial^2 \psi}{\partial x^2} = \frac{(x-x_p)^2}{\lVert \underline{r} - \underline{r}_p \rVert^2} \psi''(\lVert \underline{r} - \underline{r}_p \rVert) + \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert} \psi'(\lVert \underline{r} - \underline{r}_p \rVert) - \frac{(x-x_p)^2}{\lVert \underline{r} - \underline{r}_p \rVert^3} \psi'(\lVert \underline{r} - \underline{r}_p \rVert) \\\ &\frac{\partial^2 \psi}{\partial y^2} = \frac{(y-y_p)^2}{\lVert \underline{r} - \underline{r}_p \rVert^2} \psi''(\lVert \underline{r} - \underline{r}_p \rVert) + \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert} \psi'(\lVert \underline{r} - \underline{r}_p \rVert) - \frac{(y-y_p)^2}{\lVert \underline{r} - \underline{r}_p \rVert^3} \psi'(\lVert \underline{r} - \underline{r}_p \rVert) \\\ &\frac{\partial^2 \psi}{\partial z^2} = \frac{(z-z_p)^2}{\lVert \underline{r} - \underline{r}_p \rVert^2} \psi''(\lVert \underline{r} - \underline{r}_p \rVert) + \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert} \psi'(\lVert \underline{r} - \underline{r}_p \rVert) - \frac{(z-z_p)^2}{\lVert \underline{r} - \underline{r}_p \rVert^3} \psi'(\lVert \underline{r} - \underline{r}_p \rVert) \end{align*}$$ $$\nabla^2 \psi = \psi''(\lVert \underline{r} - \underline{r}_p \rVert) - \frac{2}{\lVert \underline{r} - \underline{r}_p \rVert} \psi'(\lVert \underline{r} - \underline{r}_p \rVert) = 0 \qquad \forall \underline{r} \in \mathbb{R}^3 - \{\underline{r}_p\}$$ $$\begin{align*} &\psi''(\lVert \underline{r} - \underline{r}_p \rVert) - \frac{2}{\lVert \underline{r} - \underline{r}_p \rVert} \psi'(\lVert \underline{r} - \underline{r}_p \rVert) = 0 \implies \\\ &\lVert \underline{r} - \underline{r}_p \rVert^2 \psi''(\lVert \underline{r} - \underline{r}_p \rVert) - 2 \lVert \underline{r} - \underline{r}_p \rVert \psi'(\lVert \underline{r} - \underline{r}_p \rVert) = 0 \implies \left[ \lVert \underline{r} - \underline{r}_p \rVert^2 \psi'(\lVert \underline{r} - \underline{r}_p \rVert) \right]' = 0 \implies \\\ &\psi'(\lVert \underline{r} - \underline{r}_p \rVert) = \frac{c}{\lVert \underline{r} - \underline{r}_p \rVert^2} \implies \psi(\lVert \underline{r} - \underline{r}_p \rVert) = - \frac{c}{\lVert \underline{r} - \underline{r}_p \rVert} + c' \, , \qquad c, c' \in \mathbb{R} \end{align*}$$

When $c = \frac{1}{4 \pi}$ and $c' = 0$ are chosen, so that $\psi(\lVert \underline{r} - \underline{r}_p \rVert) = - \frac{1}{4 \pi} \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert}$, it can be shown that

$$\nabla^2 \psi(\underline{r}, \underline{r}_p) = \delta(\underline{r} - \underline{r}_p) \qquad \text{and} \qquad \iiint_V f(\underline{r}) \delta(\underline{r} - \underline{r}_p) \mathrm{d}V = f(\underline{r}_p)$$

where $\delta(\underline{r} - \underline{r}_p)$ is the Dirac delta function defined in $\mathrm{C}_c^\infty(V)$ and, $f \colon V \subseteq \mathbb{R}^3 \to \mathbb{R}$ and $f \in \mathrm{C}_c^\infty(V)$.

$\psi(\lVert \underline{r} - \underline{r}_p \rVert) = - \frac{1}{4 \pi} \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert}$ is called Fundamental Solution of the Laplace operator

Gauss's Divergence Theorem

Let $V \subset \mathbb{R}^3$ be a bounded domain, and his boundary $\partial V$. Let $\partial V$ be a smooth hypersurface and $\underline{n}$ the outward unit normal vector to $\partial V$. Supose $\underline{F} \colon V \subseteq \mathbb{R}^3 \to \mathbb{R}^3$ and $F \in \mathrm{C}^1(V) \cap \mathrm{C}^0(\partial V)$. It is true that

$$\iiint_V \nabla \cdot \underline{F} \mathrm{d}V = \iint_{\partial V} \underline{F} \cdot \underline{n} \mathrm{d}S$$

Green's 2nd Identity

Let $V \subset \mathbb{R}^3$ be a bounded domain, and his boundary $\partial V$. Let $\partial V$ be a smooth hypersurface and $\underline{n}$ the outward unit normal vector to $\partial V$. Supose $\underline{F} = \psi \nabla \phi - \phi \nabla \psi$, where $\psi \, , \phi \in \mathrm{C}^2(V) \cap \mathrm{C}^1(\partial V)$. It is true that

$$\begin{align*} &\iiint_V \nabla \cdot \underline{F} \mathrm{d}V = \iint_{S} \underline{F} \cdot \underline{n} \mathrm{d}S \implies \iiint_V \nabla \cdot \left( \psi \nabla \phi - \phi \nabla \psi \right) \mathrm{d}V = \iint_{\partial V} \left( \psi \nabla \phi - \phi \nabla \psi \right) \cdot \underline{n} \mathrm{d}S \implies \\\ &\iiint_V \left( \nabla \psi \cdot \nabla \phi + \psi \nabla^2 \phi - \nabla \phi \cdot \nabla \psi - \phi \nabla^2 \psi \right) \mathrm{d}V = \iint_{\partial V} \left[ \psi (\underline{n} \cdot \nabla) \phi - \phi (\underline{n} \cdot \nabla) \psi \right] \mathrm{d}S \implies \end{align*}$$ $$\begin{aligned} \iiint_V \left(\psi \nabla^2 \phi - \phi \nabla^2 \psi \right) \mathrm{d}V &= \iint_{\partial V} \left[ \psi (\underline{n} \cdot \nabla) \phi - \phi (\underline{n} \cdot \nabla) \psi \right] \mathrm{d}S \\\ &= \iint_{\partial V} \left( \psi \frac{\partial \phi}{\partial n} - \phi \frac{\partial \psi}{\partial n} \right)\mathrm{d}S \end{aligned}$$

Integral Equation of velocity potential $\phi$

• Let $V \subset \mathbb{R}^3$ be a bounded domain, and his boundary $\partial V$.
• Let $\partial V = S_\infty \cup S \cup S_w$ be a smooth hypersurface and $\underline{n} (= - \underline{e}_n )$ the outward unit normal vector to $\partial V$.
• Let $\phi \in \mathrm{C}^2(V) \cap \mathrm{C}^1(\partial V)$ and $\psi(\lVert \underline{r} - \underline{r}_p \rVert) = - \frac{1}{4 \pi} \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert}$
• Let $V_\epsilon = V - B[\underline{r}_p, \epsilon]$. Then $\partial V_\epsilon = \partial V \cup \partial B[\underline{r}_p, \epsilon] = S_\infty \cup S \cup S_w \cup S_\epsilon$

Using Green's 2nd Identity we have

$$\begin{align*} &\iiint_{V_\epsilon} \left(\psi \nabla^2 \phi - \phi \nabla^2 \psi \right) \mathrm{d}V = \iint_{\partial V_\epsilon} \left[ \psi (\underline{n} \cdot \nabla) \phi - \phi (\underline{n} \cdot \nabla) \psi \right] \mathrm{d}S \xRightarrow{\nabla^2 \psi(\underline{r}, \underline{r}_p) = 0 \, , \underline{r} \neq \underline{r}_p} \iiint_{V_\epsilon} \psi \nabla^2 \phi \mathrm{d}V = \iint_{\partial V_\epsilon} \left[ \psi (\underline{n} \cdot \nabla) \phi - \phi (\underline{n} \cdot \nabla) \psi \right] \mathrm{d}S \xRightarrow[ {\partial V_\epsilon = \partial V \cup \partial B[\underline{r}_p, \epsilon]} ]{ {V_\epsilon = V - B[\underline{r}_p, \epsilon]} } \\\ & \iiint_{V} \psi \nabla^2 \phi \mathrm{d}V - \iiint_{B[\underline{r}_p, \epsilon]} \psi \nabla^2 \phi \mathrm{d}V = \iint_{\partial V} \left[ \psi (\underline{n} \cdot \nabla) \phi - \phi (\underline{n} \cdot \nabla) \psi \right] \mathrm{d}S + \iint_{\partial B[\underline{r}_p, \epsilon]} \left[ \psi (\underline{n} \cdot \nabla) \phi - \phi (\underline{n} \cdot \nabla) \psi \right] \mathrm{d}S \xRightarrow{\epsilon \to 0} \end{align*}$$ $$\iiint_{V} \psi \nabla^2 \phi \mathrm{d}V - \lim_{\epsilon \to 0} \iiint_{B[\underline{r}_p, \epsilon]} \psi \nabla^2 \phi \mathrm{d}V = \iint_{\partial V} \left[ \psi (\underline{n} \cdot \nabla) \phi - \phi (\underline{n} \cdot \nabla) \psi \right] \mathrm{d}S + \lim_{\epsilon \to 0} \iint_{\partial B[\underline{r}_p, \epsilon]} \left[ \psi (\underline{n} \cdot \nabla) \phi - \phi (\underline{n} \cdot \nabla) \psi \right] \mathrm{d}S$$

• $\lim\limits_{\epsilon \to 0} \iiint_{B[\underline{r}_p, \epsilon]} \psi \nabla^2 \phi \mathrm{d}V = 0$
• $\lim\limits_{\epsilon \to 0} \iint_{\partial B[\underline{r}_p, \epsilon]} \psi (\underline{n} \cdot \nabla) \phi \mathrm{d}S = 0$
• $\lim\limits_{\epsilon \to 0} \iint_{\partial B[\underline{r}_p, \epsilon]} \phi (\underline{n} \cdot \nabla) \psi \mathrm{d}S = - \phi(\underline{r}_p)$

Important notes

• Since $\phi \in C^2(B[\underline{r}_p, \epsilon])$ and $B[\underline{r}_p, \epsilon]$ is a compact subset of $\mathbb{R}^3$, then $\nabla^2 \phi$ is bounded in $B[\underline{r}_p, \epsilon]$ $\left( \exists \, M \in \mathbb{R}^3: \lvert \nabla^2 \phi(\underline{r}) \rvert \leq M \quad \forall \underline{r} \in B[\underline{r}_p, \epsilon] \right)$
• $\psi(\lVert \underline{r} - \underline{r}_p \rVert) = - \frac{1}{4 \pi} \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert} = const \, , \qquad \forall \underline{r} \in \partial B[\underline{r}_p, \epsilon]$
• $\underline{n} = \frac{\underline{r}_p - \underline{r}}{\lVert \underline{r}_p - \underline{r} \rVert} = - \frac{\underline{r} - \underline{r}_p}{\lVert \underline{r} - \underline{r}_p \rVert} = - \underline{e}_n \, , \qquad \forall \underline{r} \in \partial B[\underline{r}_p, \epsilon] $

Proof

$$\begin{flalign*} \bullet \quad \lim_{\epsilon \to 0} \iiint_{B[\underline{r}_p, \epsilon]} \psi \nabla^2 \phi \mathrm{d}V &= \nabla^2 \phi \lim_{\epsilon \to 0} \iiint_{B[\underline{r}_p, \epsilon]} \psi \mathrm{d}V = \nabla^2 \phi \lim_{\epsilon \to 0} \iiint_{B[\underline{r}_p, \epsilon]} - \frac{1}{4 \pi} \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert} \mathrm{d}V = - \frac{1}{4 \pi} \nabla^2 \phi \lim_{\epsilon \to 0} \iiint_{B[\underline{r}_p, \epsilon]} \frac{1}{\rho} \mathrm{d}V && \\\ &= - \frac{1}{4 \pi} \nabla^2 \phi \lim_{\epsilon \to 0} \int_0^\epsilon \iint_{\partial B[ 0, 1]} \frac{1}{\rho} \rho^2 \mathrm{d}S \, \mathrm{d}\rho = - \frac{1}{4 \pi} \nabla^2 \phi \lim_{\epsilon \to 0} \int_0^\epsilon \rho \iint_{\partial B[0, 1]} \mathrm{d}S \, \mathrm{d}\rho = - \frac{1}{4 \pi} \nabla^2 \phi \lim_{\epsilon \to 0} \int_0^\epsilon 4\pi \rho \, \mathrm{d}\rho && \\\ &= - \nabla^2 \phi \lim_{\epsilon \to 0} \int_0^\epsilon \rho \, \mathrm{d}\rho = - \nabla^2 \phi \lim_{\epsilon \to 0} \int_0^\epsilon \frac{1}{2} \frac{\mathrm{d} \rho^2}{\mathrm{d} \rho} \, \mathrm{d}\rho = - \frac{1}{2} \nabla^2 \phi \lim_{\epsilon \to 0} \int_0^\epsilon \, \mathrm{d}\rho^2 = - \frac{1}{2} \nabla^2 \phi \lim_{\epsilon \to 0} \epsilon^2 = 0 \end{flalign*}$$ $$\begin{flalign*} \bullet \quad \lim_{\epsilon \to 0} \iint_{\partial B[\underline{r}_p, \epsilon]} \psi (\underline{n} \cdot \nabla) \phi \, \mathrm{d}S &= \lim_{\epsilon \to 0} \iint_{\partial B[\underline{r}_p, \epsilon]} - \frac{1}{4 \pi} \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert} (\underline{n} \cdot \nabla) \phi \, \mathrm{d}S = \lim_{\epsilon \to 0} - \frac{1}{4 \pi} \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert} \iint_{\partial B[\underline{r}_p, \epsilon]} (\underline{n} \cdot \nabla) \phi \, \mathrm{d}S && \\\ &= \lim_{\epsilon \to 0} - \frac{1}{4 \pi} \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert} \iiint_{B[\underline{r}_p, \epsilon]} \nabla \cdot \nabla \phi \, \mathrm{d}V = \lim_{\epsilon \to 0} - \frac{1}{4 \pi} \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert} \iiint_{B[\underline{r}_p, \epsilon]} \nabla^2 \phi \, \mathrm{d}V = \lim_{\substack{\epsilon \to 0 \\ \underline{r} \to \underline{r}_p}} - \frac{1}{4 \pi} \frac{\nabla^2 \phi(\underline{r})}{\lVert \underline{r} - \underline{r}_p \rVert} \iiint_{B[\underline{r}_p, \epsilon]} \mathrm{d}V && \\\ &= \lim_{\substack{\epsilon \to 0 \\ \underline{r} \to \underline{r}_p}} - \frac{1}{4 \pi} \frac{\nabla^2 \phi(\underline{r})}{\lVert \underline{r} - \underline{r}_p \rVert} \frac{4}{3} \pi \lVert \underline{r} - \underline{r}_p \rVert^3 = - \frac{1}{3} \nabla^2 \phi(\underline{r}_p) \lim_{\substack{\epsilon \to 0 \\ \underline{r} \to \underline{r}_p}} \lVert \underline{r} - \underline{r}_p \rVert^2 = 0 \end{flalign*}$$ $$\begin{flalign*} \bullet \quad \lim_{\epsilon \to 0} \iint_{\partial B[\underline{r}_p, \epsilon]} \phi (\underline{n} \cdot \nabla) \psi \mathrm{d}S &= \lim_{\epsilon \to 0} \iint_{\partial B[\underline{r}_p, \epsilon]} \phi \left(- \frac{\underline{r} - \underline{r}_p}{\lVert \underline{r} - \underline{r}_p \rVert} \cdot \nabla \right) \left(- \frac{1}{4 \pi} \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert} \right) \mathrm{d}S = \frac{1}{4 \pi} \lim_{\substack{\epsilon \to 0 \\ \underline{r} \to \underline{r}_p}} \iint_{\partial B[\underline{r}_p, \epsilon]} \phi \left( - \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert^2} \right) \mathrm{d}S && \\\ &= - \frac{1}{4 \pi} \lim_{\substack{\epsilon \to 0 \\ \underline{r} \to \underline{r}_p}} \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert^2} \phi(\underline{r}) \iint_{\partial B[\underline{r}_p, \epsilon]} \mathrm{d}S = - \frac{1}{4 \pi} \lim_{\substack{\epsilon \to 0 \\ \underline{r} \to \underline{r}_p}} \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert}^2 \phi(\underline{r}) 4 \pi \lVert \underline{r} - \underline{r}_p \rVert^2 = - \phi(\underline{r}_p) \end{flalign*}$$

By substituting the values of the limits into the integral equation, we have

$$\iiint_{V} \psi \nabla^2 \phi \mathrm{d}V = \iint_{\partial V} \left[ \psi (\underline{n} \cdot \nabla) \phi - \phi (\underline{n} \cdot \nabla) \psi \right] \mathrm{d}S + \phi(\underline{r}_p)$$

Internal Potential $\phi_i$

• Let $V_i \subset \mathbb{R}^3$ be a bounded domain, and his boundary $\partial V_i$.
• Let $\partial V_i = S \cup S_w$ be a smooth hypersurface and $\underline{n_i} (= \underline{e}_n )$ the outward unit normal vector to $\partial V_i$.
• Let $\phi_i \in \mathrm{C}^2(V_i) \cap \mathrm{C}^1(\partial V_i)$ and $\psi(\lVert \underline{r} - \underline{r}_p \rVert) = - \frac{1}{4 \pi} \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert}$
• $\underline{r}_p \notin V_i \quad ( \underline{r}_p \in V ) \implies \underline{r} \neq \underline{r}_p$

Using Green's 2nd Identity we have

$$\iiint_{V_i} \left(\psi \nabla^2 \phi_i - \phi_i \nabla^2 \psi \right) \mathrm{d}V = \iint_{\partial V_i} \left[ \psi (\underline{n_i} \cdot \nabla) \phi_i - \phi_i (\underline{n_i} \cdot \nabla) \psi \right] \mathrm{d}S \xRightarrow{\nabla^2 \psi(\underline{r}, \underline{r}_p) = 0 \, , \underline{r} \neq \underline{r}_p}$$ $$\iiint_{V_i} \psi \nabla^2 \phi_i \mathrm{d}V = \iint_{\partial V_i} \left[ \psi (\underline{n_i} \cdot \nabla) \phi_i - \phi_i (\underline{n_i} \cdot \nabla) \psi \right] \mathrm{d}S$$

• If $phi$ is the velocity potential of an conservative and incompressible velocity vector field $V$ then it it satisfies Laplace's equation $\left( \nabla^2 \phi = 0 \right)$
• If $\phi_i$ satisfies also Laplace's equation $\left( \nabla^2 \phi_i = 0 \right)$

Then combining these two integral equations we have

$$\begin{align*} &\iiint_{V} \psi \nabla^2 \phi \mathrm{d}V + \iiint_{V_i} \psi \nabla^2 \phi_i \mathrm{d}V = \iint_{\partial V} \left[ \psi (\underline{n} \cdot \nabla) \phi - \phi (\underline{n} \cdot \nabla) \psi \right] \mathrm{d}S + \phi(\underline{r}_p) + \iint_{\partial V_i} \left[ \psi (\underline{n_i} \cdot \nabla) \phi_i - \phi_i (\underline{n_i} \cdot \nabla) \psi \right] \mathrm{d}S \xRightarrow{\nabla^2 \phi = \nabla^2 \phi_i = 0} \\\ &\phi(\underline{r}_p) + \iint_{\partial V} \left[ \psi (\underline{n} \cdot \nabla) \phi - \phi (\underline{n} \cdot \nabla) \psi \right] \mathrm{d}S + \iint_{\partial V_i} \left[ \psi (\underline{n_i} \cdot \nabla) \phi_i - \phi_i (\underline{n_i} \cdot \nabla) \psi \right] \mathrm{d}S = 0 \xRightarrow{\underline{e}_n = - \underline{n} = \underline{n_i}} \\\ &\phi(\underline{r}_p) = \iint_{\partial V} \left[ \psi (\underline{e}_n \cdot \nabla) \phi - \phi (\underline{e}_n \cdot \nabla) \psi \right] \mathrm{d}S - \iint_{\partial V_i} \left[ \psi (\underline{e}_n \cdot \nabla) \phi_i - \phi_i (\underline{e}_n \cdot \nabla) \psi \right] \mathrm{d}S \xRightarrow[\partial V_i = S \cup S_w]{\partial V = S_\infty \cup S \cup S_w} \\\ &\phi(\underline{r}_p) = \iint_{S \cup S_w} \left[ \psi (\underline{e}_n \cdot \nabla) (\phi - \phi_i) - (\phi - \phi_i) (\underline{e}_n \cdot \nabla) \psi \right] \mathrm{d}S + \iint_{S_\infty} \left[ \psi (\underline{e}_n \cdot \nabla) \phi - \phi (\underline{e}_n \cdot \nabla) \psi \right] \mathrm{d}S \implies \end{align*}$$ $$\begin{align*} &\phi(\underline{r}_p) = \iint_{S \cup S_w} \left[ \psi (\underline{e}_n \cdot \nabla) (\phi - \phi_i) - (\phi - \phi_i) (\underline{e}_n \cdot \nabla) \psi \right] \mathrm{d}S + \phi_\infty(\underline{r}_p) \\\ &\text{where} \quad \psi(\lVert \underline{r} - \underline{r}_p \rVert) = - \frac{1}{4 \pi} \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert} \quad \text{and} \quad \phi_\infty(\underline{r}_p) = \iint_{S_\infty} \left[ \psi (\underline{e}_n \cdot \nabla) \phi - \phi (\underline{e}_n \cdot \nabla) \psi \right] \mathrm{d}S \end{align*}$$

Wake Assumption

By considering that the volume of space enclosed by the wake surface $S_w$ is very small, we can assume that $(\underline{e}_n \cdot \nabla) (\phi - \phi_i) = 0 \quad \forall \underline{r} \in S_w$.

$$\begin{align*} \phi(\underline{r}_p) &= \iint_S \left[ \psi (\underline{e}_n \cdot \nabla) (\phi - \phi_i) - (\phi - \phi_i) (\underline{e}_n \cdot \nabla) \psi \right] \mathrm{d}S + \iint_{S_w} - (\phi - \phi_i) (\underline{e}_n \cdot \nabla) \psi \mathrm{d}S + \phi_\infty(\underline{r}_p) \implies \\\ \phi(\underline{r}_p) &= \iint_S \left[ \frac{1}{4 \pi} (\phi - \phi_i) (\underline{e}_n \cdot \nabla) \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert} - \frac{1}{4 \pi} \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert} (\underline{e}_n \cdot \nabla) (\phi - \phi_i) \right] \mathrm{d}S + \iint_{S_w} \frac{1}{4 \pi} (\phi - \phi_i) (\underline{e}_n \cdot \nabla) \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert} \mathrm{d}S + \phi_\infty(\underline{r}_p) \end{align*}$$

Velocity Potential $\phi(\underline{r}_p)$

$$\begin{align*} \phi(\underline{r}_p) &= \iint_S \left[ \frac{\mu}{4 \pi} (\underline{e}_n \cdot \nabla) \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert} - \frac{\sigma}{4 \pi} \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert} \right] \mathrm{d}S + \iint_{S_w} \frac{\mu}{4 \pi} (\underline{e}_n \cdot \nabla) \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert} \mathrm{d}S + \phi_\infty(\underline{r}_p) \\\ \phi(\underline{r}_p) &= \iint_S - \frac{\sigma}{4 \pi} \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert} \mathrm{d}S + \iint_{S \cup S_w} \frac{\mu}{4 \pi} (\underline{e}_n \cdot \nabla) \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert} \mathrm{d}S + \phi_\infty(\underline{r}_p) \end{align*}$$

where:

• $\mu = \phi - \phi_i$
• $\sigma = (\underline{e}_n \cdot \nabla)(\phi - \phi_i)$
• $\phi_\infty(\underline{r}_p) = \iint_{S_\infty} \left[ \phi (\underline{e}_n \cdot \nabla) - \frac{1}{4 \pi} \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert} (\underline{e}_n \cdot \nabla) \phi \right] \mathrm{d}S$

if $\mu = \phi - \phi_i = \phi - \phi_\infty$ and $\sigma = (\underline{e}_n \cdot \nabla)(\phi - \phi_i) = (\underline{e}_n \cdot \nabla)(\phi - \phi_\infty )$, then for $P \in S^-$ :

$$\iint_S - \frac{\sigma}{4 \pi} \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert} \mathrm{d}S + \iint_{S \cup S_w} \frac{\mu}{4 \pi} (\underline{e}_n \cdot \nabla) \frac{1}{\lVert \underline{r} - \underline{r}_p \rVert} \mathrm{d}S = 0, \qquad \forall (x_p, y_p, z_p) \in S: (\underline{r} - \underline{r}_p) \cdot \underline{e}_n \to 0$$

note: $\sigma = (\underline{e}_n \cdot \nabla)(\phi - \phi_i) = (\underline{e}_n \cdot \nabla)(\phi - \phi_\infty ) = (\underline{e}_n \cdot \nabla)\phi - (\underline{e}_n \cdot \nabla)\phi_\infty = \underline{V} - \underline{e}_n \cdot \underline{V}_\infty = - \underline{e}_n \cdot \underline{V}_\infty$

Numerical Model (Panel Methods)

$$\sum_{j=0}^{N_s - 1} B_{ij} \sigma_j + \sum_{j=0}^{N_s + N_w - 1}C_{ij} \mu_j = 0 , \qquad 0 \le i < N_s$$

where:

• $B_{ij} = - \frac{1}{4 \pi} \iint_{S_j} \frac{1}{\lVert \underline{r} - \underline{r}_{cp_i} \rVert} \mathrm{d}{S_j} = - \frac{1}{4\pi} \iint_{S_j} \frac{1}{\sqrt{(l_{j} - l_{j_{cp_i}})^2 + (m_{j} - m_{j_{cp_i}})^2 + (n_{j} - n_{j_{cp_i}})^2}} \mathrm{d}S_j = - \frac{1}{4\pi} \iint_{S_j} \frac{1}{\sqrt{(l_{j} - l_{j_{cp_i}})^2 + (m_{j} - m_{j_{cp_i}})^2 + n_{j_{cp_i}}^2}} \mathrm{d}S_j$

• $C_{ij} = \frac{1}{4\pi} \iint_{S_j} (\underline{e}_n \cdot \nabla) \frac{1}{\lVert \underline{r} - \underline{r}_{cp_i} \rVert} \mathrm{d}{S_j} = - \frac{1}{4\pi} \iint_{S_j} \frac{n_{j} - n_{j_{cp_i}}}{\left( \sqrt{(l_{j} - l_{j_{cp_i}})^2 + (m_{j} - m_{j_{cp_i}})^2 + (n_{j} - n_{j_{cp_i}})^2} \right)^3} \mathrm{d}S_j = \frac{1}{4\pi} \iint_{S_j} \frac{n_{j_{cp_i}}}{\left( \sqrt{(l_{j} - l_{j_{cp_i}})^2 + (m_{j} - m_{j_{cp_i}})^2 + n_{j_{cp_i}}^2} \right)^3} \mathrm{d}S_j$

• $\sigma_j = - \underline{e}_{n_j} \cdot \underline{V}_\infty$

Steady Panel Method

from Kutta Condition: $\mu_w = const = \mu_U - \mu_L$

$$A_{ij} \mu_j = - B_{ij} \sigma_j , \qquad A_{ij} = \begin{cases} C_{ij} + \sum\limits_{k=0}^{NWP_j} C_{if_j(k)} & \text{if the $j$-th panel is located on the upper side of the surface and adjoins the trailing edge}\\\ C_{ij} & \text{if the $j$-th panel do not adjoins the trailing edge}\\\ C_{ij} - \sum\limits_{k=0}^{NWP_j} C_{if_j(k)} & \text{if the $j$-th panel is located on the lower side of the surface and adjoins the trailing edge} \end{cases}$$ $$0 \le i < N_s \qquad 0 \le j < N_s \qquad 0 \le k < NWP_j$$

where $NWP_j$ is the number of panels that the wake row shedding from the $j$-th panel consists of, and f_j(k) returns the id of the $k$-th panel of the wake row shedding from $j$-th panel

Features

1. Calculation of Non-Lifting Potential Flow about 3D arbitrarily-shaped rigid bodies

   1. Steady simulations
   2. Unsteady simulations

2. Calculation of Lifting Pseudo-Potential Flow around 3D arbitrarily-shaped rigid bodies

   1. Steady state simulations with flat rigid wake model
   2. Steady state iterative simulations with flexible wake model
   3. Unsteady simulations with a shedding wake model

Simulation Results

Potential Flow around a Sphere

Potential Flow around a Low Aspect Ratio Rectangular Wing

Steady Simulation with rigid wake

Steady Simulation with iterative wake

Unsteady Simulation with wake roll up

Potential Flow around a BWB-UAV