Math 2.0

Integral equations Volterra equations Example

Abel's integral equation

Builds on Weakly singular kernelBeta function

Example

Let b>0b > 0 and let ff be a given function on [0,b][0, b]. Abel's integral equation is the equation ∫0xφ(t)x−t dt=f(x),0≤x≤b,\int_0^x \frac{\varphi(t)}{\sqrt{x - t}} \, dt = f(x), \quad 0 \le x \le b, for an unknown function φ\varphi on [0,b][0, b]. It is a Volterra equation of the first kind, of convolution type, with the weakly singular kernel k(x−t)=(x−t)−1/2k(x - t) = (x - t)^{-1/2}. More generally, Abel's equation with exponent α∈(0,1)\alpha \in (0, 1) has the kernel (x−t)−α(x - t)^{-\alpha}.

Check on an instance: take φ=1\varphi = 1. Then ∫0x(x−t)−1/2 dt=2x\int_0^x (x - t)^{-1/2} \, dt = 2 \sqrt{x}, so φ=1\varphi = 1 solves the equation with f(x)=2xf(x) = 2\sqrt{x}. Take φ(t)=t\varphi(t) = t. The substitution t=xut = xu gives ∫0xt(x−t)−1/2 dt=x3/2∫01u(1−u)−1/2 du=43x3/2\int_0^x t (x - t)^{-1/2} \, dt = x^{3/2} \int_0^1 u (1 - u)^{-1/2} \, du = \tfrac43 x^{3/2}, using the Beta function value B(2,12)=43B(2, \tfrac12) = \tfrac43.

Remarks

The kernel is not continuous on the diagonal, so the reduction to an equation of the second kind by differentiation does not apply. The solution is instead given by the Abel inversion formula, φ(x)=1πddx∫0xf(t)x−t dt\varphi(x) = \frac{1}{\pi} \frac{d}{dx} \int_0^x \frac{f(t)}{\sqrt{x - t}} \, dt. For f(x)=2xf(x) = 2\sqrt{x} this gives 1πddx(πx)=1\frac{1}{\pi} \frac{d}{dx} (\pi x) = 1, as expected. Abel met this equation in a mechanical problem that generalizes the tautochrone problem.

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Further reading

  1. Web https://en.wikipedia.org/wiki/Abel_transform
About this page
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Kindexample
WritingW0 stub
MSC45D05, 45E10
Instance ofVolterra equation of the first kind, Volterra equation of convolution type
UsesWeakly singular kernel, Beta function

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