WHEN TUCK'S DOUBLE INTEGRAL WHICH SHOULD (?) VANISH DOES

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.

References

Faculty of Mathematical Studies
University of Southampton
Southampton SO17 1BJ, ENGLAND