The recent interesting letters and article of Tuck [1], McLean [2]
and Love [3] have discussed the circumstances under which the double integral
*\int_a^b g(x) dx \sint_a^b \frac{f(t)}{t - x} dt \tag1*

is zero. (The bar denotes a Cauchy principal value, defined in the
normal way.) The main results of the discussion so far have been to
establish the following.

The purpose of this note is to point out that, in many cases arising from practical applied mathematical models, the vanishing or otherwise of integrals such as (1) may provide information that is crucial to the final resolution of the problem. Also, there are cases where integrals related to (1) may be exploited in a different fashion to gain valuable information.

A first example is provided by the model developed by Childress [4] for
inviscid incompressible flow past a backward-facing step. Space does
not permit anything but the most cursory explanations of the fluid
mechanics involved, but the crucial physical assumption (which may be
verified in many flows by direct observation) is that the region
immediately downstream of the step consists of recirculating flow.
Assuming that the recirculating eddy has constant vorticity and is of
high aspect ratio (i.e much longer than it is wide), thin aerofoil
theory may be used to recover the nonlinear singular integro
differential equation
*
\frac{1}{\pi}\sint_{-1}^1\frac{S'(t)}{x - t} dt = h + \frac{\omega^2}
{8}S^2(x) (x \in [0,1]).
\tag2
*

Here *S(x)* is the height of the eddy boundary from the downstream
wall, *\omega* is the eddy vorticity and *h* is related to the jump
in Bernoulli constant from the eddy to the free stream flow. The
boundary conditions are *S(-1) = 1*, *S(1) = S'(-1) = S'(1) = 0*. As it
stands, (2) contains too many unknowns, for neither *\omega* nor
*h* are specified. However, by multiplying by *S'(x)*, integrating
between *-1* and *1* and assuming that (1) is zero by the usual
argument, we find, upon using the boundary conditions, that
*h + \frac{\omega^2}{24} = 0.*

This result was established (with some effort) in [4] by performing a global force balance; the existence of such a relationship is essential if the model is to be closed.

A similar case where properties of (1) may be used to advantage
occurs in the model developed by Fitt et al. [5] for inviscid
irrotational slot injection into a cross flow. Again neglecting most of
the modelling details, we denote by *S(x)* the equation of the
streamline that divides the injected flow from the free stream. After
some analysis it transpires that all the important details of the flow
may be determined by consideration of the nonlinear singular integro
differential equation
*
\frac{1}{\pi}\sint_0^\infty \frac{S'(t)}{t - x} dt =
\cases
\frac{1}{2} & (0 < x < 1) \\
\frac{1}{2} - \frac{M^2}{2S^2(x)}& (1 < x < \infty).
cases
*

Physically motivated boundary conditions assert that *S(0) = S'(0) = 0*,
whilst *S \rightarrow M* (where *M* is the unknown total mass flow) as
*x \rightarrow \infty*. As in the previous case, although it is not
known a priori if the smoothness of *S* is sufficient to allow
(1) to be zero, we proceed anyway by multiplying both sides of
the equation by *S'(x)* and integrating between *0* and *\infty*. This
gives
*
S(1) = \frac{M}{2} = \frac{S(\infty)}{2}.\tag3
*

It is quite surprising that such a simple relationship exists between
*S(1)* and *S(\infty)* for such an involved governing equation.
In this case the nonlinearity of the governing equation means that,
although some of the asymptotic properties of *S(x)* may be established
without too much trouble, a numerical solution of the equation is
ultimately required. This will inevitably require a finite
truncation of the infinite range, and (3) turns out to be an
invaluable result for the purposes of inferring *S(\infty)*, and
therefore the mass flow *M*.

Experience shows that for practical purposes it is nearly always a good
idea to perform the integration (1), as information may be
gained even in cases where the integral does not exist. Consider, for
example, the Stewartson lifting-line equation (see [6])
*\frac{1}{\pi}\sint_0^\infty \frac{S'(t)}{t-x} dt =
S(x),\quad S(0) = 1,\quad S(\infty) = 0.*

Multiplying both sides of the equation by *S'(x)* and integrating from
*0* to *\infty* gives the contradictory result that *S(0) = S(\infty)*.
This strongly suggests that *S'(x)* is unlikely to be square-integrable
at *x = 0* (evidently *S(x) \sim x^{-1}* for large *x*). The obvious
guess that *S(x) \sim 1 + O(x^{1/2})* for small *x* proves to be
correct.

Further examples where the integral may be performed abound. In [7] the
equations for a cavitating aerofoil that traps an eddy were shown to be
*-\frac{1}{2\pi}\sint_\alpha^\beta \frac{S'(t)}{t - x} dt = -\lambda +
\frac{\Gamma^2}{8}S^2 + \frac{\nu}{2\pi}\log\left(\frac{x - \alpha}
{x}\right) - 2\pi v(x) (x \in [\alpha,\beta])*

*\sint_0^1 \frac{v(t)}{t-x} dt = -\frac{\mu}{2} -
\frac{S'}{4} (x \in [0,1]).*

Here the boundary conditions are *S(\alpha) = \nu\alpha*, *S'(\alpha) =
\nu*, *S(\beta) = S'(\beta) = 0* and *v(1) = 0*. Once again, a
relationship is required between the physical parameters *\lambda* and
*\gamma*. Multiplying the first equation by *S'* and integrating yields
*
\frac{\nu}{2\pi}\int_\alpha^\beta S'(x)\log\left(\frac{x - \alpha}{x}
\right) dx + \lambda S(\alpha) - \frac{\Gamma^2}{24}S^2(\alpha) -
2\pi\int_\alpha^\beta v(x)S'(x) dx.\tag4
*

Although some integrals of the unknown functions are still involved,
this relationship may easily be implemented numerically to relate
*\Gamma* and *\lambda*. It is also worth making the point that
frequently it is possible to perform integrations similar to (1)
which also give interesting results. For example, a further relationship
between *\Gamma* and *\lambda* may be derived by multiplying the first
equation above by *((x - \alpha)/(\beta - x))^{\pm 1/2}* and integrating
with respect to *x* from *\alpha* to *\beta*. Once again, the left-hand
side integral is identically zero. Although this process gives rise to a
somewhat unwieldy relationship, it may be used with (4) to
completely eliminate *\Gamma* and *\lambda* from the problem, allowing
the solution process to proceed.

Finally, it is noted that the use of such integrations is not restricted
to Cauchy type integrals (though for other kernels rigorous results may
be far harder to establish). For example, it was shown in [8] that the
equation determining the streamline *T(x)* that separates an
axisymmetric cavity flow around a thin body from a free stream is given
by
*
\align
& -\frac{1}{2}\int_0^\alpha \frac{T''(\xi) - T''(x)}{\mid \xi - x\mid}
d\xi + \frac{T''(x)}{2}\log\left[\frac{\epsilon^2T(x)}{4\pi x(\alpha-x)}
\right] + \frac{(T'(x))^2}{4T(x)} \\
&=
\cases
\pi & (0 \leq x \leq 1) \\
\pi - \beta^2(T^2(1) - T^2(x))/16\pi & (1 \leq x \leq \alpha)
cases
align
*

with *T(0) = \pi H_1^2/(\epsilon^2 L^2)*, *T(\alpha) = \pi H_2^2/(
\epsilon^2 L^2)* and *T'(0) = T'(\alpha) = 0*. Multiplying through by
*T'(x)* and integrating (assuming that this is permitted) allows much
of the right-hand to be integrated completely, whilst an integration by
parts shows that the left-hand side is identically zero. Thence
*\beta^2 = \frac{48\pi^2(T(\alpha) - T(0))}{3T^2(1)T(\alpha) -
T^3(\alpha) - 2T^3(1)}.*

This allows *\beta* to be eliminated from the problem. In this case,
another similar relationship may also be derived, since the integral
term in the equation vanishes identically when integrated between *0*
and *\alpha*. This yields
*
\align
& \int_0^\alpha \frac{T''(x)}{2}\left[\frac{1}{2}\log T(x) +
\log\left(\frac{\epsilon^2}{4\pi x(\alpha - x)}\right) \right] dx \\
&= \pi\alpha - \int_1^\alpha \frac{\beta^2}{16\pi}[T^2(1) - T^2(x)] dx,
align
*

which proves to be a useful check on any numerical results produced.

Faculty of Mathematical Studies

University of Southampton

Southampton SO17 1BJ, ENGLAND

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