Integral Equations

Abstract

Integral equations of Volterra type arise quite naturally in physical applications modelled by initial-value problems. Consider the linear Volterra equation of the second kind. (Fredholm equations of the second kind which are associated with boundary value problems for a fmite interval [a,b], are similar except that the upper limit is b.)

EquationSource% MathType!MTEF!2!1!+- % feaagCart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiabgU % caRiaadkeaaaa!3864! ]></EquationSource><EquationSource Format="TEX"><![CDATA[$$\varphi \left( x \right)=f\left( x \right)+\lambda \int_{a}^{x}{K\left( x,y \right)}\varphi \left( y \right)dy$$

with a ≤ x, y ≤ b, and let λ = 1. Using decomposition EquationSource% MathType!MTEF!2!1!+- % feaagCart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiabgU % caRiaadkeaaaa!3864!]></EquationSource><EquationSource Format="TEX"><![CDATA[$$\varphi =\sum\nolimits_{n=0}^{\infty }{{{\varphi }_{n}}}\left( x \right)$$, we identify φ ₀ = f(x) assuming f(x) ≠ 0,

EquationSource% MathType!MTEF!2!1!+- % feaagCart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiabgU % caRiaadkeaaaa!3864! ]></EquationSource><EquationSource Format="TEX"><![CDATA[$$\phi \left( x \right)={{\phi }_{0}}+\int_{a}^{x}{K}\left( x,y \right)\sum\limits_{n=0}^{\infty }{{{\phi }_{n}}}\left( y \right)dy$$

then and write EquationSource% MathType!MTEF!2!1!+- % feaagCart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiabgU % caRiaadkeaaaa!3864!]></EquationSource><EquationSource Format="TEX"><![CDATA[$$\varphi =\sum\nolimits_{n=0}^{\infty }{{{\varphi }_{n}}}$$ as the solution, or with an m-term approximant EquationSource% MathType!MTEF!2!1!+- % feaagCart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiabgU % caRiaadkeaaaa!3864!]></EquationSource><EquationSource Format="TEX"><![CDATA[$$\varphi =\sum\nolimits_{m=0}^{m-1}{{{\varphi }_{n}}}$$. In some cases, exact solutions are determinable. Consider an example:

EquationSource% MathType!MTEF!2!1!+- % feaagCart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiabgU % caRiaadkeaaaa!3864! ]></EquationSource><EquationSource Format="TEX"><![CDATA[$$K\left( x,y \right)=|y-x|$$

EquationSource% MathType!MTEF!2!1!+- % feaagCart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiabgU % caRiaadkeaaaa!3864! ]></EquationSource><EquationSource Format="TEX"><![CDATA[$$f\left( x \right)=x$$

Then

EquationSource% MathType!MTEF!2!1!+- % feaagCart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiabgU % caRiaadkeaaaa!3864! ]></EquationSource><EquationSource Format="TEX"><![CDATA[$${{\varphi }_{1}}=\int_{0}^{x}{|y-x|}{{\varphi }_{0}}\left( y \right)dy=\int_{0}^{x}{|y-x|ydy}=-{{x}^{3}}/3!$$

Thus the two-term approximant is

EquationSource% MathType!MTEF!2!1!+- % feaagCart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiabgU % caRiaadkeaaaa!3864! ]></EquationSource><EquationSource Format="TEX"><![CDATA[$$\varphi \simeq {{\varphi }_{1}}+{{\varphi }_{2}}=x-\left( {{x}^{3}}/3! \right)$$

or φ = sin x as is easily verified either by calculating more terms or by substitution.