Integral equation type 1

In mathematics , an integral equation in which the function you are looking for occurs only under the integral sign is called an integral equation of the first type . For example , are, and given, then is ${\displaystyle A,B\subseteq \mathbb {R} }$ ${\displaystyle k\colon A\times B\times \mathbb {R} \to \mathbb {R} }$ ${\displaystyle f\colon A\to \mathbb {R} }$

${\displaystyle \int _{B}k\left(s,t,x(t)\right)\,dt=f(s)}$

an integral equation of the 1st kind for the unknown function . ${\displaystyle x\colon B\to \mathbb {R} }$

Classification

If the limits of the integral are fixed, the integral equation is called Fredholmsch ; if the free variable occurs in the integral limits, the integral equation is called Volterrasch .

A simple example of a Volterra integral equation of the first kind is the equation

${\displaystyle \int _{0}^{s}x(t)\,dt=f(s)}$,

the solution of the apparent first derivative is: . ${\displaystyle x(t)=f'(t)\!}$

Solvability

Integral equations of the 1st type are usually so-called incorrectly posed problems , i.e. problems that cannot be solved in a canonical way. Is namely

${\displaystyle K\colon x\mapsto \int _{B}k\left(s,t,x(t)\right)\,dt}$

a compact operator between Banach spaces X and Y and if K has infinite-dimensional image space, then the image of is of the first category in . This means that it cannot be continuously inverted or at least open. Regularization procedures are therefore required to solve integral equations of the first type. ${\displaystyle K}$ ${\displaystyle Y}$ ${\displaystyle K}$

example

The above-mentioned formation of the first derivative is also an incorrectly posed problem: If one considers, for example, the normalized vector space

${\displaystyle \left(C^{\infty }\left([0,1]\right),\|.\|_{\sup }\right)}$

the infinitely often continuously differentiable functions of the interval with respect to the supremum norm, then is the operator ${\displaystyle [0,1]}$

${\displaystyle D\colon {\begin{cases}C^{\infty }\left([0,1]\right)\to C^{\infty }\left([0,1]\right)\\f\mapsto f'\end{cases}}}$,

which of the function is the solution of the integral equation ${\displaystyle f}$

${\displaystyle \int _{0}^{s}x(t)dt=f(s)}$

assigns, a discontinuous linear operator :

${\displaystyle f_{n}:={\frac {1}{n}}\sin \left(n^{2}x\right)}$

is a null sequence in the sense of the supreme norm, there

${\displaystyle \|f_{n}\|_{\sup }={\frac {1}{n}},}$

but for

${\displaystyle Df_{n}=f'_{n}=n\cos \left(n^{2}x\right)}$

applies:

${\displaystyle \|Df_{n}\|_{\sup }=\|f'_{n}\|_{\sup }=n}$.

The sequence of functions thus converges against the function , but the sequence of images diverges. ${\displaystyle f_{n}(t)\!}$ ${\displaystyle f(t)=0\!}$ ${\displaystyle Df_{n}\!}$

Numerical differentiation

This property is also reflected when one tries to numerically differentiate approximately given functions. For example, if you numerically calculate the derivative of at the point by forming the difference quotients for different step sizes , you typically get the following result: ${\displaystyle e^{x}}$ ${\displaystyle x=0{,}5}$ ${\displaystyle {\frac {e^{0,5+h}-e^{0,5}}{h}}}$ ${\displaystyle h}$

${\displaystyle h\!}$	${\displaystyle {\frac {e^{0,5+h}-e^{0,5}}{h}}}$
${\displaystyle 1}$	2.83297
${\displaystyle 0{,}1}$	1.73398
${\displaystyle 0{,}01}$	1.65699
${\displaystyle 0{,}001}$	1.64955
${\displaystyle 0{,}0001}$	1.6488
${\displaystyle 10^{-5}}$	1.64873
${\displaystyle 10^{-6}}$	1.64872
${\displaystyle 10^{-7}}$	1.64872
${\displaystyle 10^{-8}}$	1.64872
${\displaystyle 10^{-9}}$	1.64872
${\displaystyle 10^{-10}}$	1.64872
${\displaystyle 10^{-11}}$	1.64873
${\displaystyle 10^{-12}}$	1.64868
${\displaystyle 10^{-13}}$	1.64979
${\displaystyle 10^{-14}}$	1.64313
${\displaystyle 10^{-15}}$	1.55431
${\displaystyle 10^{-16}}$	2.22045

The exact value of the derivative is . The error first decreases for smaller and smaller increments until you get practically the correct value, but surprisingly the error increases again for even smaller steps . This is explained by the fact that for small ones the discretization error, i.e. the difference between the difference quotient and the derivative, becomes smaller and smaller, but the error increases that arises from the fact that one does not have exactly available, but only a numerical one Approximation of this function. Since differentiation is a discontinuous linear operator, this second error can be arbitrarily large. ${\displaystyle e^{0{,}5}\approx 1{,}64872}$ ${\displaystyle h}$ ${\displaystyle h}$ ${\displaystyle h}$ ${\displaystyle e^{x}}$

This behavior is generally typical for incorrectly posed problems.

Further examples

Other examples of integral equations of the first type are the inverse Laplace transform and the inverse Radon transform named after Johann Radon , which plays an important role in computed tomography . Both are incorrectly posed problems.

The inverse Fourier transformation is also an integral equation of the first type, but in contrast to the other examples, it is a correctly posed problem.

swell

• Dirk Werner : Functional Analysis . 6th, corrected edition, Springer-Verlag, Berlin 2007, ISBN 978-3-540-72533-6 .