Variational Analysis and Euler Equation of the Optimum Propeller Problem

Abstract

The problem of the optimum propeller with straight blades was first solved by Goldstein; in this paper, a variational formulation is proposed in order to extend the solution to non-planar blades. First, we find a class of functions (the circulation along the blade axis) for which the thrust and the aerodynamic drag moment are well defined. In this class, the objective functional is proved to be strictly convex and then the global minimum exists and is unique. Then we determine the Euler equation in the case of a general blade and show that the numerical results are consistent with the Goldstein’s solution. Finally, some numerical results with the Ritz method are presented for optimum propeller blades.

Appendices

Appendix 1: Singular Part of the Kernel

To obtain the singular part of the kernel is important for two reasons. First, the complete expression of the kernel is too complex to allow an analysis of existence, we study just the singular part (see Appendix 2). Furthermore, we have to demonstrate that the singularity is a second order pole in order to use the Monegato’s quadrature rule (see Sect. 5). The expression of the kernel, as defined in Sect. 5.1 by Eq. (30), is

$$\displaystyle{ \mathbf{g}_{i}(t,\xi,\eta ):= \frac{\mathrm{d}\mathbf{r}_{i}^{v}(t)} {\mathrm{d}t} \times \frac{\mathbf{r}(\eta ) -\mathbf{r}_{i}^{v}(t,\xi )} {\left \Vert \mathbf{r}(\eta ) -\mathbf{r}_{i}^{v}(t,\xi )\right \Vert ^{3}}\;, }$$

where i is the origin blade of the helicoidal wake. For i = 0 the blade is the same where we calculate the induced velocity. Therefore we have a singularity in (t, ξ) = (0, η), because

$$\displaystyle{ \lim _{\left (t,\xi \right )\rightarrow \left (0,\eta \right )}\mathbf{r}_{0}^{v}(t,\xi ) = \mathbf{r}(\eta )\;. }$$

We take into account just the kernel g ₀(t, ξ, η) (from now on the subscript₀ is dropped for simplicity), its components are

$$\displaystyle{ \left \{\begin{array}{rl} g_{x}(t,\xi,\eta ) = \frac{\left (z(\eta )-z^{v}(t,\xi )\right )y,_{ t}^{v}(t,\xi )-\left (y(\eta )-y^{v}(t,\xi )\right )z,_{ t}^{v}(t,\xi )} {\left \vert \left (x(\eta )-x^{v}(t,\xi )\right )^{2}+\left (y(\eta )-y^{v}(t,\xi )\right )^{2}+\left (z(\eta )-z^{v}(t,\xi )\right )^{2}\right \vert ^{3/2}} \\ g_{y}(t,\xi,\eta ) = \frac{\left (x(\eta )-x^{v}(t,\xi )\right )z,_{ t}^{v}(t,\xi )-\left (z(\eta )-z^{v}(t,\xi )\right )x,_{ t}^{v}(t,\xi )} {\left \vert \left (x(\eta )-x^{v}(t,\xi )\right )^{2}+\left (y(\eta )-y^{v}(t,\xi )\right )^{2}+\left (z(\eta )-z^{v}(t,\xi )\right )^{2}\right \vert ^{3/2}} \\ g_{z}(t,\xi,\eta ) = \frac{\left (y(\eta )-y^{v}(t,\xi )\right )x,_{ t}^{v}(t,\xi )-\left (x(\eta )-x^{v}(t,\xi )\right )y,_{ t}^{v}(t,\xi )} {\left \vert \left (x(\eta )-x^{v}(t,\xi )\right )^{2}+\left (y(\eta )-y^{v}(t,\xi )\right )^{2}+\left (z(\eta )-z^{v}(t,\xi )\right )^{2}\right \vert ^{3/2}}\\ \end{array} \right.\;. }$$

We analyse just the z component, for the other is nearly the same. We need the following Taylor expansions around the singularity.

$$\displaystyle{ \left \{\begin{array}{rl} &x^{v}(t,\xi ) = x(\eta ) + t\left.x,_{t}^{v}\right \vert _{(0,\eta )} + (\xi -\eta )\left.x,_{\xi }^{v}\right \vert _{(0,\eta )} + \mathcal{O}(t^{2} + (\xi -\eta )^{2}) \\ &y^{v}(t,\xi ) = y(\eta ) + t\left.y,_{t}^{v}\right \vert _{(0,\eta )} + (\xi -\eta )\left.y,_{\xi }^{v}\right \vert _{(0,\eta )} + \mathcal{O}(t^{2} + (\xi -\eta )^{2}) \end{array} \right. }$$

$$\displaystyle{ \left \{\begin{array}{rl} &x,_{t}^{v}(t,\xi ) = \left.x,_{t}^{v}\right \vert _{(0,\eta )} + t\left.x,_{tt}^{v}\right \vert _{(0,\eta )} + (\xi -\eta )\left.x,_{\xi t}^{v}\right \vert _{(0,\eta )} + \mathcal{O}(t^{2} + (\xi -\eta )^{2}) \\ &y,_{t}^{v}(t,\xi ) = \left.y,_{t}^{v}\right \vert _{(0,\eta )} + t\left.y,_{tt}^{v}\right \vert _{(0,\eta )} + (\xi -\eta )\left.y,_{\xi t}^{v}\right \vert _{(0,\eta )} + \mathcal{O}(t^{2} + (\xi -\eta )^{2})\end{array} \right. }$$

We move to polar coordinates around the singularity⁷

$$\displaystyle{ \left \{\begin{array}{rl} & t = r\cos \theta \\ &\xi -\eta = r\sin \theta \end{array} \right. }$$

and the Taylor expansions become

$$\displaystyle{ \left \{\begin{array}{rl} &x^{v}(r,\theta ) = x(\eta ) + \left.x,_{t}^{v}\right \vert _{(0,\eta )}r\cos \theta + \left.x,_{\xi }^{v}\right \vert _{(0,\eta )}r\sin \theta + \mathcal{O}(r^{2}) \\ &y^{v}(r,\theta ) = y(\eta ) + \left.y,_{t}^{v}\right \vert _{(0,\eta )}r\cos \theta + \left.y,_{\xi }^{v}\right \vert _{(0,\eta )}r\sin \theta + \mathcal{O}(r^{2}) \end{array} \right.\;, }$$

$$\displaystyle{ \left \{\begin{array}{rl} &x,_{t}^{v}(r,\theta ) = \left.x,_{t}^{v}\right \vert _{(0,\eta )} + \left.x,_{tt}^{v}\right \vert _{(0,\eta )}r\cos \theta + \left.x,_{\xi t}^{v}\right \vert _{(0,\eta )}r\cos \theta + \mathcal{O}(r^{2}) \\ &y,_{t}^{v}(r,\theta ) = \left.y,_{t}^{v}\right \vert _{(0,\eta )} + \left.y,_{tt}^{v}\right \vert _{(0,\eta )}r\cos \theta + \left.y,_{\xi t}^{v}\right \vert _{(0,\eta )}r\sin \theta + \mathcal{O}(r^{2})\end{array} \right.\;. }$$

In order to clarify the calculation we define the functions at the numerator and at the denominator of g _z :

$$\displaystyle{ g_{z}(r,\theta,\eta ) = \frac{\mathfrak{N}(t,\xi,\eta )} {\mathfrak{D}(t,\xi,\eta )}\;, }$$

where

$$\displaystyle\begin{array}{rcl} & & \mathfrak{N}(r,\theta,\eta ) = \left (y(\eta ) - y^{v}(t,\xi )\right )x,_{ t}^{v}(t,\xi ) -\left (x(\eta ) - x^{v}(t,\xi )\right )y,_{ t}^{v}(t,\xi ), {}\\ & & \mathfrak{D}(r,\theta,\eta ) = \left \vert \left (x(\eta ) - x^{v}(t,\xi )\right )^{2} + \left (y(\eta ) - y^{v}(t,\xi )\right )^{2} + \left (z(\eta ) - z^{v}(t,\xi )\right )^{2}\right \vert ^{3/2}\;. {}\\ \end{array}$$

For the terms of $\mathfrak{N}$ we have

$$\displaystyle\begin{array}{rcl} & & \left (x(\eta ) - x^{v}(t,\xi )\right )y,_{ t}^{v}(t,\xi ) = {}\\ & & = \left (-\left.x,_{t}^{v}\right \vert _{ (0,\eta )}r\cos \theta -\left.x,_{\xi }^{v}\right \vert _{ (0,\eta )}r\sin \theta + \mathcal{O}(r^{2})\right ) {}\\ & & \left (\left.y,_{t}^{v}\right \vert _{ (0,\eta )} + \left.y,_{tt}^{v}\right \vert _{ (0,\eta )}r\cos \theta + \left.y,_{\xi t}^{v}\right \vert _{ (0,\eta )}r\sin \theta + \mathcal{O}(r^{2})\right ) = {}\\ & & = -\left (\left.x,_{t}^{v}\right \vert _{ (0,\eta )}r\cos \theta + \left.x,_{\xi }^{v}\right \vert _{ (0,\eta )}r\sin \theta \right )\left.y,_{t}^{v}\right \vert _{ (0,\eta )} + \mathcal{O}(r^{2}) {}\\ \end{array}$$

and in the same way:

$$\displaystyle\begin{array}{rcl} & & \left (y(\eta ) - y^{v}(t,\xi )\right )x,_{ t}^{v}(t,\xi ) = {}\\ & & = -\left (\left.y,_{t}^{v}\right \vert _{ (0,\eta )}r\cos \theta + \left.y,_{\xi }^{v}\right \vert _{ (0,\eta )}r\sin \theta \right )\left.x,_{t}^{v}\right \vert _{ (0,\eta )} + \mathcal{O}(r^{2})\;. {}\\ \end{array}$$

Hence the expression of $\mathfrak{N}$:

$$\displaystyle\begin{array}{rcl} \mathfrak{N}(r,\theta,\eta )& =& \left (\left.x,_{t}^{v}\right \vert _{ (0,\eta )}r\cos \theta + \left.x,_{\xi }^{v}\right \vert _{ (0,\eta )}r\sin \theta \right )\left.y,_{t}^{v}\right \vert _{ (0,\eta )} {}\\ & -& \left (\left.y,_{t}^{v}\right \vert _{ (0,\eta )}r\cos \theta + \left.y,_{\xi }^{v}\right \vert _{ (0,\eta )}r\sin \theta \right )\left.x,_{t}^{v}\right \vert _{ (0,\eta )} + \mathcal{O}(r^{2}) = {}\\ & =& \left (\left.x,_{t}^{v}\right \vert _{ (0,\eta )}\left.y,_{t}^{v}\right \vert _{ (0,\eta )} -\left.y,_{t}^{v}\right \vert _{ (0,\eta )}\left.x,_{t}^{v}\right \vert _{ (0,\eta )}\right )r\sin \theta + \mathcal{O}(r^{2}) = {}\\ & =& Nr\sin \theta + \mathcal{O}(r^{2})\;. {}\\ \end{array}$$

where

$$\displaystyle{ N = \left.x,_{t}^{v}\right \vert _{ (0,\eta )}\left.y,_{t}^{v}\right \vert _{ (0,\eta )} -\left.y,_{t}^{v}\right \vert _{ (0,\eta )}\left.x,_{t}^{v}\right \vert _{ (0,\eta )}\;. }$$

Instead for the terms of $\mathfrak{D}$:

$$\displaystyle\begin{array}{rcl} & & \left (x(\eta ) - x^{v}(t,\xi )\right )^{2} = {}\\ & & = \left (-\left.x,_{t}^{v}\right \vert _{ (0,\eta )}r\cos \theta -\left.x,_{\xi }^{v}\right \vert _{ (0,\eta )}r\sin \theta + \mathcal{O}(r^{2})\right )^{2} = {}\\ & & = \left (\left.x,_{t}^{v}\right \vert _{ (0,\eta )}r\cos \theta + \left.x,_{\xi }^{v}\right \vert _{ (0,\eta )}r\sin \theta \right )^{2} + \mathcal{O}(r^{3}) {}\\ \end{array}$$

and nearly the same for the other components. The expression of $\mathfrak{D}$ becomes

$$\displaystyle\begin{array}{rcl} \mathfrak{D}(r,\theta,\eta )& =& \left \vert \left (\left.x,_{t}^{v}\right \vert _{ (0,\eta )}r\cos \theta + \left.x,_{\xi }^{v}\right \vert _{ (0,\eta )}r\sin \theta \right )^{2}\right. {}\\ & +& \left (\left.y,_{t}^{v}\right \vert _{ (0,\eta )}r\cos \theta + \left.y,_{\xi }^{v}\right \vert _{ (0,\eta )}r\sin \theta \right )^{2} {}\\ & +& \left.\left (\left.z,_{t}^{v}\right \vert _{ (0,\eta )}r\cos \theta + \left.z,_{\xi }^{v}\right \vert _{ (0,\eta )}r\sin \theta \right )^{2} + \mathcal{O}(r^{3})\right \vert ^{3/2} = {}\\ & =& r^{2}\left \vert D(\theta ) + \mathcal{O}(r)\right \vert ^{3/2} = r^{2}D^{\frac{3} {2} }(\theta )\left (1 + \mathcal{O}(r)\right )^{3/2}\;, {}\\ \end{array}$$

where

$$\displaystyle{ \begin{array}{rcl} D(\theta )& =&\left (\left.x,_{t}^{v}\right \vert _{(0,\eta )}\cos \theta + \left.x,_{\xi }^{v}\right \vert _{(0,\eta )}\sin \theta \right )^{2} \\ & + &\left (\left.y,_{t}^{v}\right \vert _{(0,\eta )}\cos \theta + \left.y,_{\xi }^{v}\right \vert _{(0,\eta )}\sin \theta \right )^{2} \\ & + &\left (\left.z,_{t}^{v}\right \vert _{(0,\eta )}\cos \theta + \left.z,_{\xi }^{v}\right \vert _{(0,\eta )}\sin \theta \right )^{2}\;. \end{array} }$$

(33)

Therefore g _z becomes

$$\displaystyle\begin{array}{rcl} g_{z}(r,\theta,\eta )& =& \frac{Nr\cos \theta + \mathcal{O}(r^{2})} {r^{2}D^{\frac{3} {2} }(\theta )\left (1 + \mathcal{O}(r)\right )^{3/2}} {}\\ & =& \frac{\left (N\sin \theta + \mathcal{O}(r)\right )\left (1 + \mathcal{O}(r)\right )^{-3/2}} {r^{2}D^{\frac{3} {2} }(\theta )} {}\\ & =& \frac{\left (N\sin \theta + \mathcal{O}(r)\right )\left (1 -\frac{3} {2}\mathcal{O}(r) + \mathcal{O}(r^{2})\right )} {r^{2}D^{\frac{3} {2} }(\theta )} {}\\ & =& \frac{N\sin \theta + \mathcal{O}(r)} {r^{2}D(\theta )} {}\\ & =& \frac{N\sin \theta } {r^{2}D^{\frac{3} {2} }(\theta )} + \mathcal{O}\left (\frac{1} {r}\right )\;. {}\\ \end{array}$$

Finally we obtain

$$\displaystyle{ \left \{\begin{array}{rl} &g_{x}(r,\theta,\eta ) = \frac{1} {r^{2}} \, \frac{\left (y,_{\xi }^{v}z,_{ t}^{v}-z,_{\xi }^{v}y,_{ t}^{v}\right )\sin \theta } {\left [D(\theta )\right ]^{3/2}} + \mathcal{O}\left (\frac{1} {r}\right ) \\ &g_{y}(r,\theta,\eta ) = \frac{1} {r^{2}} \, \frac{\left (z,_{\xi }^{v}x,_{ t}^{v}-x,_{\xi }^{v}z,_{ t}^{v}\right )\sin \theta } {\left [D(\theta )\right ]^{3/2}} + \mathcal{O}\left (\frac{1} {r}\right ) \\ &g_{z}(r,\theta,\eta ) = \frac{1} {r^{2}} \, \frac{\left (x,_{\xi }^{v}y,_{ t}^{v}-y,_{\xi }^{v}x,_{ t}^{v}\right )\sin \theta } {\left [D(\theta )\right ]^{3/2}} + \mathcal{O}\left (\frac{1} {r}\right ) \end{array} \right.\;, }$$

(34)

where all the partial derivatives are calculated in (0, η), and D(θ) is defined by (33). These expressions proved that the singularity is a second order pole. The partial derivative in (34), according to Eq. (9) are:

$$\displaystyle{ \left \{\begin{array}{l} x,_{t}^{v} = 1 \\ y,_{t}^{v} = -\mu _{0}z(\eta ) \\ z,_{t}^{v} =\mu _{0}y(\eta )\end{array} \right.\qquad \left \{\begin{array}{l} x,_{\xi }^{v} = x_{,\eta }(\eta ) \\ y,_{\xi }^{v} = y_{,\eta }(\eta ) \\ z,_{\xi }^{v} = z_{,\eta }(\eta )\end{array} \right.\;, }$$

thus Eq. (34) becomes

$$\displaystyle{ \left \{\begin{array}{rl} &g_{x}(t,\xi,\eta ) =\mu _{0} \frac{yy,_{\eta }+zz,_{\eta }} {\left [\overline{D}(t,\xi,\eta )\right ]^{3/2}} \left (\xi -\eta \right ) + \mathcal{O}\left ( \frac{1} {\sqrt{t^{2 } +(\xi -\eta )^{2}}} \right ) \\ &g_{y}(t,\xi,\eta ) = \frac{z,_{\eta }-\mu _{0}yx,_{\eta }} {\left [\overline{D}(t,\xi,\eta )\right ]^{3/2}} \left (\xi -\eta \right ) + \mathcal{O}\left ( \frac{1} {\sqrt{t^{2 } +(\xi -\eta )^{2}}} \right ) \\ &g_{z}(t,\xi,\eta ) = - \frac{y,_{\eta }+\mu _{0}zx,_{\eta }} {\left [\overline{D}(t,\xi,\eta )\right ]^{3/2}} \left (\xi -\eta \right ) + \mathcal{O}\left ( \frac{1} {\sqrt{t^{2 } +(\xi -\eta )^{2}}} \right ) \end{array} \right.\;, }$$

(35)

where

$$\displaystyle{ \begin{array}{rcl} \overline{D}(t,\xi,\eta )& =&\left [1 +\mu _{ 0}^{2}(y^{2} + z^{2})\right ]t^{2} + 2\left [x,_{\eta } +\mu _{0}(yz,_{\eta } - zy,_{\eta })\right ](\xi -\eta )t+ \\ & + &\left (x,_{\eta }^{2} + y,_{\eta }^{2} + z,_{\eta }^{2}\right )(\xi -\eta )^{2}\;. \end{array} }$$

(36)

Appendix 2: Existence of Functionals

In order to prove existence and continuity of the functionals we have to prove that the integrals, defining C _T and C _M , exist. Just the term containing induced velocity is singular; this is obtained as a summation over the blades and only the first term of this sum is singular, the one indicated by the apex⁰. Therefore we analyse the following integrals:

$$\displaystyle\begin{array}{rcl} C_{T_{ind}}^{0}(\varGamma )& =& \frac{1} {2\pi }\int _{0}^{1}\varGamma (\eta )\int _{ 0}^{1}\frac{\mathrm{d}\varGamma } {\mathrm{d}\xi }\int _{0}^{\infty }\mathbf{g}_{ 0}(t,\xi,\eta )\mathrm{d}t\mathrm{d}\xi \times \mathrm{ d}\mathbf{r}(\eta ) \cdot \mathbf{i}{}\end{array}$$

(37)

$$\displaystyle\begin{array}{rcl} C_{M_{ind}}^{0}(\varGamma )& =& -\frac{1} {2\pi }\int _{0}^{1}\varGamma (\eta )\mathbf{r}(\eta ) \times \left (\int _{ 0}^{1}\frac{\mathrm{d}\varGamma } {\mathrm{d}\xi }\int _{0}^{\infty }\mathbf{g}_{ 0}(t,\xi,\eta )\mathrm{d}t\mathrm{d}\xi \times \mathrm{ d}\mathbf{r}(\eta )\right ) \cdot \mathbf{i}\;,{}\end{array}$$

(38)

where g ₀(t, ξ, η) is the singular kernel⁸:

$$\displaystyle{ \mathbf{g}_{0}(t,\xi,\eta ):= \frac{\mathrm{d}\mathbf{r}_{0}^{v}(t)} {\mathrm{d}t} \times \frac{\mathbf{r}(\eta ) -\mathbf{r}_{0}^{v}(t,\xi )} {\left \Vert \mathbf{r}(\eta ) -\mathbf{r}_{0}^{v}(t,\xi )\right \Vert ^{3}}\;. }$$

Performing the operations between vectors we obtain the same kind of functional for both, momentum and thrust:

$$\displaystyle{ C(\varGamma ):=\int _{ 0}^{1}\varGamma (\eta )\int _{ 0}^{1}\varGamma ^{{\prime}}(\xi )\int _{ 0}^{\infty }\sum _{ j=1}^{3}f_{ j}(\eta )\left (\mathbf{g}_{0}(t,\xi,\eta )\right )_{j}\mathrm{d}t\mathrm{d}\xi \mathrm{d}\eta \;, }$$

where f _j (η) is a regular function depending on the blade line. We employ Eq. (35) and then we neglect the regular part of C(Γ) in order to obtain the following singular integral

$$\displaystyle{ \overline{C}(\varGamma ):=\int _{ 0}^{1}\varGamma (\eta )\int _{ 0}^{1}\varGamma ^{{\prime}}(\xi )\int _{ 0}^{\infty } \frac{f(\eta )(\xi -\eta )} {\left [a(\eta )t^{2} + 2b(\xi,\eta )t + c(\xi,\eta )\right ]^{\frac{3} {2} }} \mathrm{d}t\mathrm{d}\xi \mathrm{d}\eta \;, }$$

(39)

where

$$\displaystyle{ \begin{array}{rcl} a(\eta )& =&1 +\mu _{ 0}^{2}(y^{2} + z^{2}) \\ b(\xi,\eta )& =&\left [x,_{\eta } +\mu _{0}(yz,_{\eta } - zy,_{\eta })\right ](\xi -\eta ) \\ c(\xi,\eta )& =&\left (x,_{\eta }^{2} + y,_{\eta }^{2} + z,_{\eta }^{2}\right )(\xi -\eta )^{2} \end{array} }$$

(40)

and f(η) is a regular function depending on the blade line. This expression does not respect the definition of integral according to Riemann.⁹ We have to adopt Cauchy principal value. That means associating with the integral the value of a correspondent limit, if this limit exists finite then the integral converges according to Cauchy. Before proving the existence of this integral, we recall some useful definitions.

Definition 1.

A function $f: [a,b] \rightarrow \mathbb{R}$ is absolutely continuous in [a, b], and we write f ∈ AC[a, b] iff, for any ɛ > 0 it exists δ > 0 such that for any finite collections of disjoint intervals ]α _i , β _i [,  i = 1, …, k, included in [a, b] and with

$$\displaystyle{ \sum _{i=1}^{k}(\beta _{ i} -\alpha _{i}) <\delta \;,\quad \mbox{ it results }\quad \sum _{i=1}^{k}\left \vert f(\beta _{ i}) - f(\alpha _{i})\right \vert <\varepsilon \;. }$$

Definition 2.

Let $(Y,\mathcal{F},\mu )$ be a measure space and 1 ≤ p ≤ ∞. We put

$$\displaystyle\begin{array}{rcl} & & L^{p}(Y ) = \left \{f: Y \rightarrow \overline{\mathbb{R}}:\: f\mbox{ is measurable and }\int _{ Y }\left \vert f\right \vert ^{p}dy <\infty \right \}\;, {}\\ & & \Vert f\Vert _{L^{p}(Y )} = \left [\int _{Y }\left \vert f\right \vert ^{p}\mathrm{d}y\right ]^{\frac{1} {p} }\;. {}\\ \end{array}$$

Hölder Inequality. If q is conjugate exponent of p (i.e. 1∕p + 1∕q = 1, by stipulation, the conjugate exponent of 1 is ∞ ), if f ∈ L ^p(Y ) and g ∈ L ^q(Y ), then

$$\displaystyle{ fg \in L^{1}(Y )\quad \text{and}\quad \Vert fg\Vert _{ L^{1}(Y )} \leq \Vert f\Vert _{L^{p}(Y )}\Vert g\Vert _{L^{q}(Y )}\;. }$$

Proposition 1.

Let Γ ∈ AC[0;1] be such that

$$\displaystyle{ \varGamma (0) =\varGamma (1) = 0,\qquad \varGamma,\;\varGamma ^{{\prime}}\in L^{1+\varepsilon }(0;1)\:\text{with}\:\varepsilon> 0\;. }$$

Then $\overline{C}(\varGamma )$ , defined in (39) , is convergent as a Cauchy improper integral.

Proof.

Let us set

$$\displaystyle\begin{array}{rcl} S_{1}(h)&:=& \left \{(\xi,\eta ) \in \mathbb{R}^{2}:\, 0 \leq \xi \leq 1 - h,\,\xi +h \leq \eta \leq 1\right \} {}\\ S_{2}(h)&:=& \left \{(\xi,\eta ) \in \mathbb{R}^{2}:\, h \leq \xi \leq 1,\,0 \leq \eta \leq \xi -h\right \}\;, {}\\ \end{array}$$

with h ∈ [0; 1] (Fig. 19), and

$$\displaystyle\begin{array}{rcl} G_{S_{i}(h)}&:=& \iint _{S_{i}(h)}\varGamma ^{{\prime}}(\xi )\varGamma (\eta )\int _{ 0}^{\infty } \frac{f(\eta )(\xi -\eta )} {\left [a(\eta )t^{2} + 2b(\xi,\eta )t + c(\xi,\eta )\right ]^{\frac{3} {2} }} \mathrm{d}t\mathrm{d}\xi \mathrm{d}\eta {}\\ \end{array}$$

for i = 1, 2. The standard integration rules can be adopted for these integrals because they are regular.

Fig. 19

Regular integration sets

We solve the inner integral, in t:

$$\displaystyle{ T(\xi,\eta ):=\int _{ 0}^{\infty } \frac{\mathrm{d}t} {\left (at^{2} + 2bt + c\right )^{\frac{3} {2} }} =\int _{ 0}^{\infty } \frac{\mathrm{d}t} {\left [\left (\sqrt{a}t + b/\sqrt{a}\right )^{2} + c - b^{2}/a\right ]^{3/2}}\;, }$$

with the change of variable

$$\displaystyle\begin{array}{rcl} & & s = \sqrt{a}t + b/\sqrt{a}\quad \Rightarrow \quad \mathrm{d}t =\mathrm{ d}s/\sqrt{a} {}\\ & & t \rightarrow \infty \quad,\quad s \rightarrow \infty {}\\ & & t = 0\,,\,s = b/\sqrt{a}\;, {}\\ \end{array}$$

we have¹⁰

$$\displaystyle\begin{array}{rcl} T(\xi,\eta )& =& \frac{1} {\sqrt{a}}\int _{ \frac{b} {\sqrt{a}} }^{\infty } \frac{\mathrm{d}s} {\left [s^{2} + c - b^{2}/a\right ]^{3/2}} = \frac{1} {\sqrt{a}}\left [ \frac{s} {(c - b^{2}/a)\sqrt{c - b^{2 } /a + s^{2}}}\right ]_{ \frac{b} {\sqrt{a}} }^{\infty } = {}\\ & =& \frac{1} {\sqrt{c}\left (\sqrt{ac} + b\right )} = \frac{g(\eta )} {(\xi -\eta )^{2}}\;, {}\\ \end{array}$$

where, in the last passage, we used Eq. (40) and g(η) is a regular function depending on the blade line. We obtain

$$\displaystyle\begin{array}{rcl} G_{S_{i}(h)}& =& \iint _{S_{i}(h)}\frac{\varGamma ^{{\prime}}(\xi )\varGamma (\eta )\overline{f}(\eta )} {\eta -\xi } \mathrm{d}\xi \mathrm{d}\eta \;,\,\,\;i = 1,2\;, {}\\ \end{array}$$

where $\overline{f}(\eta ) = -f(\eta )g(\eta )$, that is a regular function depending on blade line.¹¹ Let us integrate by parts both $G_{S_{1}(h)}$ and $G_{S_{2}(h)}$. For the first integral we have

$$\displaystyle\begin{array}{rcl} & & G_{S_{1}(h)} =\int _{ 0}^{1-h}\int _{ \xi +h}^{1}\varGamma ^{{\prime}}(\xi )\varGamma (\eta )\overline{f}(\eta ) \frac{1} {\eta -\xi }\mathrm{d}\eta \mathrm{d}\xi {}\\ & & \qquad \quad =\int _{ 0}^{1-h}\left \{\left [\varGamma ^{{\prime}}(\xi )\varGamma (\eta )\overline{f}(\eta )\ln (\eta -\xi )\right ]_{\xi +h}^{1}\right. {}\\ & & \qquad \quad \left.-\int _{\xi +h}^{1}\varGamma ^{{\prime}}(\xi )\frac{\mathrm{d}} {\mathrm{d}\eta }\left [\varGamma (\eta )\overline{f}(\eta )\right ]\ln (\eta -\xi )\mathrm{d}\eta \right \}\mathrm{d}\xi \;, {}\\ \end{array}$$

because $\frac{\mathrm{d}} {\mathrm{d}\eta }\left [\ln \left (\eta -\xi \right )\right ] = \frac{1} {\eta -\xi }$. Then, by using the boundary condition Γ(1) = 0,

$$\displaystyle\begin{array}{rcl} & & G_{S_{1}(h)} =\int _{ 0}^{1-h}\left \{ -\varGamma ^{{\prime}}(\xi )\varGamma (\xi +h)\overline{f}(\xi +h)\ln h\right. {}\\ & & \qquad \quad \left.-\int _{\xi +h}^{1}\varGamma ^{{\prime}}(\xi )\frac{\mathrm{d}} {\mathrm{d}\eta }\left [\varGamma (\eta )\overline{f}(\eta )\right ]\ln (\eta -\xi )\mathrm{d}\eta \right \}\mathrm{d}\xi \;. {}\\ \end{array}$$

We proceed in the same way for the second integral:

$$\displaystyle\begin{array}{rcl} & & G_{S_{2}(h)} =\int _{ h}^{1}\int _{ 0}^{\xi -h}\varGamma ^{{\prime}}(\xi )\varGamma (\eta )\overline{f}(\eta ) \frac{1} {\eta -\xi }\mathrm{d}\eta \mathrm{d}\xi {}\\ & & \qquad \quad =\int _{ h}^{1}\left \{\left [\varGamma ^{{\prime}}(\xi )\varGamma (\eta )\overline{f}(\eta )\ln (\xi -\eta )\right ]_{ 0}^{\xi -h}\right. {}\\ & & \qquad \quad \left.-\int _{0}^{\xi -h}\varGamma ^{{\prime}}(\xi )\frac{\mathrm{d}} {\mathrm{d}\eta }\left [\varGamma (\eta )\overline{f}(\eta )\right ]\ln (\eta -\xi )\mathrm{d}\eta \right \}\mathrm{d}\xi \;, {}\\ \end{array}$$

because $\frac{\mathrm{d}} {\mathrm{d}\eta }\left [\ln \left (\xi -\eta \right )\right ] = -\frac{1} {\xi -\eta } = \frac{1} {\eta -\xi }$. Then, by using the other boundary condition Γ(0) = 0,

$$\displaystyle\begin{array}{rcl} & & G_{S_{2}(h)} =\int _{ h}^{1}\left \{\varGamma ^{{\prime}}(\xi )\varGamma (\xi -h)\overline{f}(\xi -h)\ln h\right. {}\\ & & \qquad \quad \left.-\int _{0}^{\xi -h}\varGamma ^{{\prime}}(\xi )\frac{\mathrm{d}} {\mathrm{d}\eta }\left [\varGamma (\eta )\overline{f}(\eta )\right ]\ln (\xi -\eta )\mathrm{d}\eta \right \}\mathrm{d}\xi \;. {}\\ \end{array}$$

It results that

$$\displaystyle\begin{array}{rcl} & & \overline{C}(\varGamma ) =\lim _{h\rightarrow 0}\left [G_{S_{1}(h)} + G_{S_{2}(h)}\right ] {}\\ & & \qquad \ \ =\lim _{h\rightarrow 0}\left [-\int _{0}^{1-h}\int _{ \xi +h}^{1}\varGamma ^{{\prime}}(\xi )\frac{\mathrm{d}} {\mathrm{d}\eta }\left [\varGamma (\eta )\overline{f}(\eta )\right ]\ln (\eta -\xi )\mathrm{d}\eta \mathrm{d}\xi \right. {}\\ & & \qquad \ \ +\ln h\left (\int _{h}^{1}\varGamma ^{{\prime}}(\xi )\varGamma (\xi -h)\overline{f}(\xi -h)\mathrm{d}\xi -\int _{ 0}^{1-h}\varGamma ^{{\prime}}(\xi )\varGamma (\xi +h)\overline{f}(\xi +h)\mathrm{d}\xi \right ) {}\\ & & \qquad \ \ \left.-\int _{h}^{1}\int _{ 0}^{\xi -h}\varGamma ^{{\prime}}(\xi )\frac{\mathrm{d}} {\mathrm{d}\eta }\left [\varGamma (\eta )\overline{f}(\eta )\right ]\ln (\xi -\eta )\mathrm{d}\eta \mathrm{d}\xi \right ]\;. {}\\ \end{array}$$

where

$$\displaystyle\begin{array}{rcl} & & \lim _{h\rightarrow 0}\:\ln h\left (\int _{h}^{1}\varGamma ^{{\prime}}(\xi )\varGamma (\xi -h)\overline{f}(\xi -h)\mathrm{d}\xi \right. {}\\ & & \phantom{vvvvvvvvvvvvvvvvvvvvvvvvvvvv}\left.-\int _{0}^{1-h}\varGamma ^{{\prime}}(\xi )\varGamma (\xi +h)\overline{f}(\xi +h)\mathrm{d}\xi \right ) = 0\;, {}\\ \end{array}$$

thus

$$\displaystyle\begin{array}{rcl} & & \overline{C}(\varGamma ) =\lim _{h\rightarrow 0}\left [ -\int _{0}^{1-h}\int _{ \xi +h}^{1}\varGamma ^{{\prime}}(\xi )\frac{\mathrm{d}} {\mathrm{d}\eta }\left [\varGamma (\eta )\overline{f}(\eta )\right ]\ln (\eta -\xi )\mathrm{d}\eta \mathrm{d}\xi \right. {}\\ & & \qquad \ \ \left.-\int _{h}^{1}\int _{ 0}^{\xi -h}\varGamma ^{{\prime}}(\xi )\frac{\mathrm{d}} {\mathrm{d}\eta }\left [\varGamma (\eta )\overline{f}(\eta )\right ]\ln (\xi -\eta )\mathrm{d}\eta \mathrm{d}\xi \right ] = {}\\ & & \qquad \ \ = -\int _{0}^{1}\int _{ 0}^{1}\varGamma ^{{\prime}}(\xi )\frac{\mathrm{d}} {\mathrm{d}\eta }\left [\varGamma (\eta )\overline{f}(\eta )\right ]\ln \left \vert \xi -\eta \right \vert \mathrm{d}\eta \mathrm{d}\xi \;. {}\\ \end{array}$$

The absolute value of sum is less than or equal to the sum of absolute values:

$$\displaystyle\begin{array}{rcl} & & \left \vert \int _{0}^{1}\int _{ 0}^{1}\varGamma ^{{\prime}}(\xi )\frac{\mathrm{d}} {\mathrm{d}\eta }\left [\varGamma (\eta )\overline{f}(\eta )\right ]\ln \left \vert \xi -\eta \right \vert \mathrm{d}\eta \mathrm{d}\xi \right \vert \leq \\ & &\phantom{vvvvvvvvvvvvvvvvvv}\int _{0}^{1}\left \vert \varGamma ^{{\prime}}(\xi )\right \vert \int _{ 0}^{1}\left \vert \frac{\mathrm{d}} {\mathrm{d}\eta }\left [\varGamma (\eta )\overline{f}(\eta )\right ]\right \vert \left \vert \ln \left \vert \xi -\eta \right \vert \right \vert \mathrm{d}\eta \mathrm{d}\xi.{}\end{array}$$

(41)

From Hölder inequality we obtain

$$\displaystyle\begin{array}{rcl} & & \int _{0}^{1}\left \vert \frac{\mathrm{d}} {\mathrm{d}\eta }\left [\varGamma (\eta )\overline{f}(\eta )\right ]\right \vert \left \vert \ln \left \vert \xi -\eta \right \vert \right \vert \mathrm{d}\eta \mathrm{d}\xi \leq {}\\ & &\phantom{vvvvvvvvvvvvvvvvvvvvvvvv}\left \Vert \frac{\mathrm{d}} {\mathrm{d}\eta }\left [\varGamma (\eta )\overline{f}(\eta )\right ]\right \Vert _{L^{1+\varepsilon }(0,1)}\left \Vert \ln \left \vert \xi -\eta \right \vert \right \Vert _{ L^{\frac{1+\varepsilon } {\varepsilon } }(0,1)}. {}\\ \end{array}$$

Observe that¹²

$$\displaystyle{ \left \Vert \ln \left \vert \xi -\eta \right \vert \right \Vert _{ L^{\frac{1+\varepsilon } {\varepsilon } }(0,1)} <c\quad \text{with }c \in \mathbb{R}\; }$$

and

$$\displaystyle\begin{array}{rcl} \left \Vert \frac{\mathrm{d}} {\mathrm{d}\eta }\left [\varGamma (\eta )\overline{f}(\eta )\right ]\right \Vert _{L^{1+\varepsilon }(0,1)}& \leq & \left \Vert \varGamma ^{{\prime}}(\eta )\overline{f}(\eta )\right \Vert _{ L^{1+\varepsilon }(0,1)} + \left \Vert \varGamma (\eta )\overline{f}^{{\prime}}(\eta )\right \Vert _{ L^{1+\varepsilon }(0,1)} \leq {}\\ &\leq & P\left \Vert \varGamma ^{{\prime}}(\eta )\right \Vert _{ L^{1+\varepsilon }(0,1)} + Q\left \Vert \varGamma (\eta )\right \Vert _{L^{1+\varepsilon }(0,1)}\;, {}\\ \end{array}$$

because, given the regularity of $\overline{f}(\eta )$, we have that

$$\displaystyle{ \exists P,Q \in \mathbb{R}\,: \quad \left \vert \overline{f}(\eta )\right \vert \leq P,\;\left \vert \overline{f}^{{\prime}}(\eta )\right \vert \leq Q\;\forall \eta \in [0;1]\;. }$$

From (41) we have, finally:

$$\displaystyle\begin{array}{rcl} & & \left \vert \overline{C}(\varGamma )\right \vert \leq cP\left \Vert \varGamma (\eta )\right \Vert _{L^{1}(0,1)}\left \Vert \varGamma ^{{\prime}}(\eta )\right \Vert _{ L^{1+\varepsilon }(0,1)} + \\ & & \phantom{vvvvvvvvvvvvvvvvvvvvvvvvvv} + cQ\left \Vert \varGamma (\eta )\right \Vert _{L^{1}(0,1)}\left \Vert \varGamma (\eta )\right \Vert _{L^{1+\varepsilon }(0,1)} <\infty {}\end{array}$$

(42)

as required. ⊓ ⊔

Appendix 3: Convexity of Momentum Functional

According to (15), (19) and (30) we define $C_{M_{1}}(\varGamma )$ and $C_{M_{2}}(\varGamma )$ such that

$$\displaystyle\begin{array}{rcl} C_{M_{1}}(\varGamma )&:=& 2N\int _{\gamma _{b}^{0}}\varGamma (\mathbf{r})\mathbf{r} \times \left [\left (\mathbf{i} +\mu _{0}r\mathbf{e}_{\theta }\right ) \times \mathrm{ d}\mathbf{r}\right ] \cdot \mathbf{i}\;, {}\\ C_{M_{2}}(\varGamma )&:=& -\frac{N} {2\pi } \sum _{i=0}^{N-1}\int _{ \gamma _{b}^{0}}\varGamma (\mathbf{r})\mathbf{r} \times \left (\int _{\gamma _{b}^{i}}\frac{\mathrm{d}\varGamma (\xi )} {\mathrm{d}\xi } \int _{\gamma _{v}^{i}}\mathbf{g}_{i}(t,\xi,\eta )\mathrm{d}t\mathrm{d}\xi \times \mathrm{ d}\mathbf{r}\right ) \cdot \mathbf{i}\;, {}\\ \end{array}$$

that means

$$\displaystyle{ C_{M}(\varGamma ) = C_{M_{1}}(\varGamma ) + C_{M_{2}}(\varGamma )\;. }$$

(43)

Before proving the convexity of the functional C _M , we recall the following:

Definition 3.

Let K be a vector space. A function $f: K \rightarrow \mathbb{R}$ is called convex, if and only if

$$\displaystyle{ (1-\alpha )f(x) +\alpha f(x) \geq f((1-\alpha )x +\alpha y),\quad \forall x,y \in K,\quad \forall \alpha \in [0,1]\;. }$$

(44)

We say that function f is strictly convex, if and only if the inequality (44) holds strictly.

Theorem 1.

Let K be a vector space and $f: K \rightarrow \mathbb{R}$ be a function whatever. Then f is strictly convex on K, if and only if $\forall x,y \in K$ the quotient ratio

$$\displaystyle{ t \rightarrow R_{y}(t) = \frac{f(x + ty) - f(x)} {t},\quad t \in \mathbb{R}_{+}\setminus \left \{0\right \} }$$

is an increasing function.

Proposition 2.

Let be:

$$\displaystyle{ I:= \left \{\varGamma \in AC[0;1],\:\varGamma,\;\varGamma ^{{\prime}}\in L^{1+\varepsilon }(0;1)\;\text{with}\;\varepsilon> 0,\:\varGamma (0) =\varGamma (1) = 0\right \} }$$

and C _M (Γ) defined by (43), so we have

• (a) the functional C _M is not strictly convex in I;

• (b) the functional C _M is strictly convex in $I^{+}:= \left \{\varGamma \in I\::\: C_{M_{2}}(\varGamma )> 0\right \}$ .

Proof.

We use a more compact expression for momentum functional.

$$\displaystyle{ \left \{\begin{array}{rcl} C_{M_{1}} & =&\int _{\gamma _{b}^{0}}\varGamma (\eta )H_{1}(\eta )\mathrm{d}\eta \\ C_{M_{2}} & =&\sum _{i=0}^{N-1}\int _{\gamma _{b}^{0}}\int _{\gamma _{b}^{i}}\varGamma (\eta )\varGamma ^{{\prime}}(\xi )H_{2}^{i}(\xi,\eta )\mathrm{d}\xi \mathrm{d}\eta \end{array} \right. }$$

Given Γ, g ∈ I and t > 0 we have

$$\displaystyle\begin{array}{rcl} R_{h}(t)& =& \frac{C_{M}\left (\varGamma +tg\right ) - C_{M}\left (\varGamma \right )} {t} {}\\ & =& \frac{C_{M_{1}}\left (\varGamma +tg\right ) - C_{M_{1}}\left (\varGamma \right )} {t} + \frac{C_{M_{2}}\left (\varGamma +tg\right ) - C_{M_{2}}\left (\varGamma \right )} {t} {}\\ & =& \frac{1} {t}\left [\int \left (\varGamma \left (\eta \right ) + tg\left (\eta \right )\right )H_{1}(\eta )\mathrm{d}\eta -\int \varGamma \left (\eta \right )H_{1}(\eta )\mathrm{d}\eta \right. {}\\ & & +\sum _{i=0}^{N-1}\iint _{ i}\left (\varGamma (\eta ) + tg(\eta )\right )\left (\varGamma ^{{\prime}}(\xi ) + tg^{{\prime}}(\xi )\right )H_{ 2}^{i}(\xi,\eta )\mathrm{d}\xi \mathrm{d}\eta {}\\ & & -\left.\sum _{i=0}^{N-1}\iint _{ i}\varGamma (\eta )\varGamma ^{{\prime}}(\xi )H_{ 2}^{i}(\xi,\eta )\mathrm{d}\xi \mathrm{d}\eta \right ] = {}\\ & =& \int g(\eta )H_{1}(\eta )\mathrm{d}\eta +\sum _{ i=0}^{N-1}\iint _{ i}\left (\varGamma (\eta )g^{{\prime}}(\xi ) + g(\eta )\varGamma ^{{\prime}}(\xi )\right )H_{ 2}^{i}(\xi,\eta )\mathrm{d}\xi \mathrm{d}\eta {}\\ & & +t\sum _{i=0}^{N-1}\iint _{ i}g(\eta )g^{{\prime}}\left (\xi \right )H_{ 2}^{i}(\xi,\eta )\mathrm{d}\xi \mathrm{d}\eta \;. {}\\ \end{array}$$

Hence the first derivative:

$$\displaystyle{ \frac{\mathrm{d}} {\mathrm{d}t}\left [R_{h}(t)\right ] =\int \int g\left (\eta \right )g^{{\prime}}\left (\xi \right )H_{ 2}(\xi,\eta )\mathrm{d}\xi \mathrm{d}\eta = C_{M_{2}}(g)\;. }$$

Note that the first derivative of quotient ratio, for g ∈ I is not, in general, positive. Whereas for g ∈ I ⁺ we do have $C_{M_{2}}(g)> 0$, then the function R _h (t) is strictly increasing. ⊓ ⊔

Appendix 4: Velocity Induced by the Bounded Vorticity

For wing the induction of the bounded vortex filament is in direction of the asymptotic velocity. That is the reason why this contribution is neglected in [3]. For airscrew the velocity seen by the blade is the composition of asymptotic and rotational velocity, thus, the velocity induced by the bounded filament does not have, in general, the same direction of the velocity seen by the blade. For straight blade propeller the velocity induced by the bounded vorticity, u _B ind (η), is null for reason of symmetry, but in general this contribution is not zero. According to the Biot-Savart law we have

$$\displaystyle{ \mathbf{u}_{B\,ind}(\eta ) = \frac{1} {4\pi }\sum _{i=0}^{N-1}\int _{ \gamma _{b}^{i}}\varGamma (\xi )\frac{\mathrm{d}\mathbf{r}_{i}} {\mathrm{d}\xi } \times \frac{\mathbf{r}(\eta ) -\mathbf{r}_{i}(\xi )} {\left \Vert \mathbf{r}(\eta ) -\mathbf{r}_{i}(\xi )\right \Vert ^{3}}\mathrm{d}\xi \;, }$$

(45)

where r(η) is the position vector on the induced blade and r _i (ξ) on the inducing blade. Note that for i = 0 the integrand is singular. We find that

$$\displaystyle{ \frac{\mathrm{d}\mathbf{r}_{0}} {\mathrm{d}\xi } \times \frac{\mathbf{r}(\eta ) -\mathbf{r}_{0}(\xi )} {\left \Vert \mathbf{r}(\eta ) -\mathbf{r}_{0}(\xi )\right \Vert ^{3}}\mathrm{d}\xi = \frac{f(\eta )} {(\xi -\eta )} + \frac{\mathcal{O}(\xi -\eta )} {(\xi -\eta )} \;, }$$

where f(η) is a regular function, thus there are not problem of existence in our class of functions $\mathcal{X}$, defined by (20). The following figures show the results obtained for a swept blade when the contribution of bounded vorticity is not neglected (Figs. 20, 21, 22, and 23). The comparison with the correspondent figures in Sect. 7, where bounded vorticity is neglected, is interesting because it shows the importance of the velocity induced by the bounded vorticity.

Fig. 20

Relative efficiency of blades swept in rotation plane for N = 1, Ω = 2 and C _T target  = 0. 1

Fig. 21

Circulation distribution for blades swept in rotation plane for N = 1, Ω = 2 and C _T target  = 0. 1

Fig. 22

Relative efficiency of blades swept in axial direction for N = 1, Ω = 2 and C _T target  = 0. 1

Fig. 23

Circulation distribution for blades swept in axial direction for N = 1, Ω = 2 and C _T target  = 0. 1