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Hadamard regularization

From Wikipedia, the free encyclopedia
Mathematical method extending convergence
Renormalization and regularization

In mathematics,Hadamard regularization (also calledHadamard finite part orHadamard's partie finie) is a method ofregularizing divergent integrals by dropping some divergent terms and keeping the finite part, introduced byJacques Hadamard (1923, book III, chapter I,1932).Marcel Riesz (1938,1949) showed that this can be interpreted as taking themeromorphic continuation of a convergent integral.

Description

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If theCauchy principal value integralC∫abf(t)t−xdt(for a<x<b){\displaystyle {\mathcal {C}}\int _{a}^{b}{\frac {f(t)}{t-x}}\,dt\quad ({\text{for }}a<x<b)}exists, then it may be differentiated with respect tox to obtain the Hadamard finite part integral as follows:ddx(C∫abf(t)t−xdt)=H∫abf(t)(t−x)2dt(for a<x<b).{\displaystyle {\frac {d}{dx}}\left({\mathcal {C}}\int _{a}^{b}{\frac {f(t)}{t-x}}\,dt\right)={\mathcal {H}}\int _{a}^{b}{\frac {f(t)}{(t-x)^{2}}}\,dt\quad ({\text{for }}a<x<b).}

Note that the symbolsC{\displaystyle {\mathcal {C}}} andH{\displaystyle {\mathcal {H}}} are used here to denote Cauchy principal value and Hadamard finite-part integrals respectively.

The Hadamard finite part integral above (fora <x <b) may also be given by the following equivalent definitions:H∫abf(t)(t−x)2dt=limε→0+{∫ax−εf(t)(t−x)2dt+∫x+εbf(t)(t−x)2dt−f(x+ε)+f(x−ε)ε},{\displaystyle {\mathcal {H}}\int _{a}^{b}{\frac {f(t)}{(t-x)^{2}}}\,dt=\lim _{\varepsilon \to 0^{+}}\left\{\int _{a}^{x-\varepsilon }{\frac {f(t)}{(t-x)^{2}}}\,dt+\int _{x+\varepsilon }^{b}{\frac {f(t)}{(t-x)^{2}}}\,dt-{\frac {f(x+\varepsilon )+f(x-\varepsilon )}{\varepsilon }}\right\},}H∫abf(t)(t−x)2dt=limε→0+{∫ab(t−x)2f(t)((t−x)2+ε2)2dt−πf(x)2ε−f(x)2(1b−x−1a−x)}.{\displaystyle {\mathcal {H}}\int _{a}^{b}{\frac {f(t)}{(t-x)^{2}}}\,dt=\lim _{\varepsilon \to 0^{+}}\left\{\int _{a}^{b}{\frac {(t-x)^{2}f(t)}{((t-x)^{2}+\varepsilon ^{2})^{2}}}\,dt-{\frac {\pi f(x)}{2\varepsilon }}-{\frac {f(x)}{2}}\left({\frac {1}{b-x}}-{\frac {1}{a-x}}\right)\right\}.}

The definitions above may be derived by assuming that the functionf (t) is differentiable infinitely many times att =x fora <x <b, that is, by assuming thatf (t) can be represented by its Taylor series aboutt =x. For details, see Ang (2013). (Note that the term−⁠f (x)/2⁠(⁠1/b −x⁠ −⁠1/a −x⁠) in the second equivalent definition above is missing in Ang (2013) but this is corrected in the errata sheet of the book.)

Integral equations containing Hadamard finite part integrals (withf (t) unknown) are termed hypersingular integral equations. Hypersingular integral equations arise in the formulation of many problems in mechanics, such as in fracture analysis.

Example

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Consider the divergent integral∫−111t2dt=(lima→0−∫−1a1t2dt)+(limb→0+∫b11t2dt)=lima→0−(−1a−1)+limb→0+(−1+1b)=+∞{\displaystyle \int _{-1}^{1}{\frac {1}{t^{2}}}\,dt=\left(\lim _{a\to 0^{-}}\int _{-1}^{a}{\frac {1}{t^{2}}}\,dt\right)+\left(\lim _{b\to 0^{+}}\int _{b}^{1}{\frac {1}{t^{2}}}\,dt\right)=\lim _{a\to 0^{-}}\left(-{\frac {1}{a}}-1\right)+\lim _{b\to 0^{+}}\left(-1+{\frac {1}{b}}\right)=+\infty }ItsCauchy principal value also diverges sinceC∫−111t2dt=limε→0+(∫−1−ε1t2dt+∫ε11t2dt)=limε→0+(1ε−1−1+1ε)=+∞{\displaystyle {\mathcal {C}}\int _{-1}^{1}{\frac {1}{t^{2}}}\,dt=\lim _{\varepsilon \to 0^{+}}\left(\int _{-1}^{-\varepsilon }{\frac {1}{t^{2}}}\,dt+\int _{\varepsilon }^{1}{\frac {1}{t^{2}}}\,dt\right)=\lim _{\varepsilon \to 0^{+}}\left({\frac {1}{\varepsilon }}-1-1+{\frac {1}{\varepsilon }}\right)=+\infty }To assign a finite value to this divergent integral, we may considerH∫−111t2dt=H∫−111(t−x)2dt|x=0=ddx(C∫−111t−xdt)|x=0{\displaystyle {\mathcal {H}}\int _{-1}^{1}{\frac {1}{t^{2}}}\,dt={\mathcal {H}}\int _{-1}^{1}{\frac {1}{(t-x)^{2}}}\,dt{\Bigg |}_{x=0}={\frac {d}{dx}}\left({\mathcal {C}}\int _{-1}^{1}{\frac {1}{t-x}}\,dt\right){\Bigg |}_{x=0}}The inner Cauchy principal value is given byC∫−111t−xdt=limε→0+(∫−1−ε1t−xdt+∫ε11t−xdt)=limε→0+(ln⁡|ε+x1+x|+ln⁡|1−xε−x|)=ln⁡|1−x1+x|{\displaystyle {\mathcal {C}}\int _{-1}^{1}{\frac {1}{t-x}}\,dt=\lim _{\varepsilon \to 0^{+}}\left(\int _{-1}^{-\varepsilon }{\frac {1}{t-x}}\,dt+\int _{\varepsilon }^{1}{\frac {1}{t-x}}\,dt\right)=\lim _{\varepsilon \to 0^{+}}\left(\ln \left|{\frac {\varepsilon +x}{1+x}}\right|+\ln \left|{\frac {1-x}{\varepsilon -x}}\right|\right)=\ln \left|{\frac {1-x}{1+x}}\right|}Therefore,H∫−111t2dt=ddx(ln⁡|1−x1+x|)|x=0=2x2−1|x=0=−2{\displaystyle {\mathcal {H}}\int _{-1}^{1}{\frac {1}{t^{2}}}\,dt={\frac {d}{dx}}\left(\ln \left|{\frac {1-x}{1+x}}\right|\right){\Bigg |}_{x=0}={\frac {2}{x^{2}-1}}{\Bigg |}_{x=0}=-2}Note that this value does not represent the area under the curvey(t) = 1/t2, which is clearly always positive. However, it can be seen where this comes from. Recall the Cauchy principal value of this integral, when evaluated at the endpoints, took the formlimε→0+(1ε−1−1+1ε)=+∞{\displaystyle \lim _{\varepsilon \to 0^{+}}\left({\frac {1}{\varepsilon }}-1-1+{\frac {1}{\varepsilon }}\right)=+\infty }

If one removes the infinite components, the pair of1ε{\displaystyle {\frac {1}{\varepsilon }}} terms, that which remains, thefinite part, islimε→0+(−1−1)=−2{\displaystyle \lim _{\varepsilon \to 0^{+}}\left(-1-1\right)=-2}

which equals the value derived above.

References

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