Square Integrable Functions, Norms, Trial Method in Integral Equations

Square Integrable function or quadratically integrable function $\mathfrak{L}_2$ function

A function $y(x)$ is said to be square integrable or $\mathfrak{L}_2$ function on the interval $(a,b)$ if $$\displaystyle {\int_a^b} {|y(x)|}^2 dx <\infty$$ or $$\displaystyle {\int_a^b} y(x) \bar{y}(x) dx <\infty$$.

For further reading, I suggest this Wikipedia page.

$y(x)$ is then also called ‘regular function’.

The kernel $K(x,t)$ , a function of two variables is an $\mathfrak{L_2}$ – function if atleast one of the following is true:

  • $\int_{x=a}^b \int_{t=a}^b |K(x,t)|^2 dx dt <\infty$
  • $\int_{t=a}^b |K(x,t)|^2 dx <\infty$
  • $\int_{x=a}^b |K(x,t)|^2 dt <\infty$

Inner Product of two $\mathfrak{L}_2$ functions

The inner product or scalar product $(\phi, \psi)$ of two complex $\mathfrak{L}_2$ functions $\phi$ and $\psi$ of a real variable $x$ ; $a \le x \le b$ is defined as

$(\phi, \psi) = \int_a^b \phi(x) \bar{\psi}(x) dx$  .

Where $\bar{\psi}(x)$ is the complex conjugate of  $\psi(x)$.

When $(\phi, \psi) =0$, or $\int_a^b \phi(x) \bar{\psi}(x) dx =0$ then $\phi$ and $\psi $ are called orthogonal to each other.

Norm of a function

The norm of a complex- function $y(x)$ of a single real variable $x$ is given by

$|| y(x) || = \sqrt{\int_a^b y(x) \bar{y(x)} dx}=\sqrt{\int_a^b |y(x)|^2 dx}$

Where $\bar{y(x)}$ represents the complex conjugate of $y(x)$.

The norm of operations between any two functions $\phi$ and $\psi$  follows Schwarz and Minkowski’s triangle inequalities, provided

$|| \phi \cdot \psi || \le ||\phi|| \cdot ||\psi||$ —- Schwarz’s Inequality

$|| \phi +\psi || \le ||\phi|| + ||\psi||$ ——-Triangle Inequality/Minkowski Inequality

Solution of Integral Equations by Trial Method

A solution of an equation is the value of the unknown function which satisfies the complete equation. The same definition is followed by the solution of an integral equation too. First of all we will handle problems in which a value of the unknown function is given and we are asked to verify whether it’s a solution of the integral equation or not. The following example will make it clear:

  • Show that $y(x)= {(1+x^2)}^{-3/2}$ is a solution of $$y(x) = \dfrac{1}{1+x^2} – \int_0^x \dfrac{t}{1+x^2} y(t) dt$$.

This is a Volterra’s equation of second kind with lower limit $a=0$ and upper limit being the variable $x$.

Solution: Given $$y(x) = \dfrac{1}{1+x^2} – \int_0^x \dfrac{t}{1+x^2} y(t) dt \ldots (1)$$

where $y(x)= {(1+x^2)}^{-3/2} \ldots (2)$

and therefore, $y(t)= {(1+t^2)}^{-3/2} \ldots (3)$ (replacing x by t).

The Right Hand Side of (1)

$=\dfrac{1}{1+x^2} – \int_0^x \dfrac{t}{1+x^2} y(t) dt$

$=\dfrac{1}{1+x^2} – \int_0^x \dfrac{t}{1+x^2} {(1+t^2)}^{-3/2} dt$ [putting the value of $y(t)$ from (3)]

$=\dfrac{1}{1+x^2} -\dfrac{1}{1+x^2} \int_0^x \dfrac{t}{{(1+t^2)}^{3/2}} dt$

since $\dfrac{1}{1+x^2}$ is independent quantity as the integration is done with respect to $t$ i.e., dt only, therefore $\dfrac{1}{1+x^2}$ can be excluded outside the integration sign.

$=\dfrac{1}{1+x^2} +\dfrac{1}{1+x^2} \left({\dfrac{1}{\sqrt{1+x^2}} -1}\right)$

       Since $\int_0^x \dfrac{t}{{1+t^2}^{3/2}} dt $=$1-\dfrac{1}{\sqrt{1+x^2}}$



=The Left Hand Side of (2)

Hence, $y(x) ={(1+x^2)}^{-3/2}$ is a solution of (1). $\Box$

Trial method isn’t exactly the way an integral equation can be solved, it is however very important for learning and pedagogy point of views. In upcoming articles, we’ll learn various other techniques to solve an integral equation. But, for now, in next two parts of this series, we shall be reading how ordinary differential equations can be converted into integral equation and vice-versa.


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