# Linear Integral Equations (IEL2)

#### Recollect

An operator $L[f(x)]$ is called linear if $L[af(x)+bg(x)]=aL[f(x)]+bL[g(x)],a,b\in\mathbb R$.

#### Integral Operator

We can write an integral equations in terms of a linear operator, as in following examples.

Let $L[g(s)]=g(s)-\int_a^b {K(s,t)g(t)dt}\to(1)$ . Then we can write $g(s)=f(s)+\int_a^b {K(s,t)g(t)dt}$ as $L[g(s)]=f(s)$.

Similarly, $g(s)=\int_a^b {K(s,t)[g(t)]^2dt} \to(2)$ can be rewritten in terms of a operator as well.

#### Linear Integral Operator

An integral operator is linear if it satisfies above condition; that is, $L[af(x)+bg(x)]=aL[f(x)]+bL[g(x)],a,b\in\mathbb R$.

Note equation (1) above. Clearly,

$\displaystyle L[ af(s)+bg(s) ] \newline= af(s)+bg(s) -\int_a^b {K(s,t) [af(t)+bg(t)] dt} \newline = af(s) -\int_a^b {K(s,t) af(t)dt} + bg(s) -\int_a^b {K(s,t) bg(t) dt} \newline = af(s) - a\int_a^b {K(s,t) f(t)dt} + bg(s) -b\int_a^b {K(s,t) g(t) dt} \newline = a[f(s) -\int_a^b {K(s,t) f(t)dt}] + b[g(s) -\int_a^b {K(s,t) g(t) dt}] \newline = a L[f(s)] + b L[g(s)]$

Hence, that integral operator is linear. But note that operator in equation (2) is not linear, and you can prove it yourself.

In general, we will be studying equations of the form

$h(s)g(s)=f(s)+\lambda \int_a^b K(s,t)g(t)dt$

where

• h(s), f(s) and K(s,t) are known real/complex functions,
• g(s) is the unknown function
• K(s,t) is called the kernel
• $\lambda$ is a non-zero real or complex parameter called “eigenvalue”.
• Limits, a is constant and b may be fixed or a variable.

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