# Solution of an integral equation

Consider a simple Wiener-Hopf integral equation of the first kind with unknown function $\phi(x)$ for $x\geq 0$:

$$f(x)=\int_0^\infty \phi(y)\min\{x,y\}\,\mathrm{d}y$$

where $f(x)=x-a$ and $a \geq 0$.

Although it arose from a well stated physical problem, I suggest there are no solutions of any of these equations in $\phi(x)$. Or are they?

Note: Would there be a solution if we choose $f(x)= \begin{cases} 0 &;x<a\\ x-a &;x\geq a \end{cases}$ instead?

• Function should be $$f(y)=\int_0^\infty \phi(y)\min\{x,y\}\,\mathrm{d}x$$. Jul 13, 2016 at 23:43
• - It was a mistake in differentials, I will correct that Jul 20, 2016 at 0:17

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I'll assume the equation is $\ds{\,\mathrm{f}\pars{x} = \int_{0}^{\infty}\phi\pars{y}\min\braces{x,y}\,\dd y}$.

\begin{align} \,\mathrm{f}\pars{x} & = \int_{0}^{x}\phi\pars{y}y\,\dd y + x\int_{x}^{\infty}\phi\pars{y}\,\dd y \\[3mm] \imp\ \,\mathrm{f}'\pars{x} & = \phi\pars{x}x + \int_{x}^{\infty}\phi\pars{y}\,\dd y - x\phi\pars{x} \\[3mm] \imp\ \,\mathrm{f}''\pars{x} & = -\phi\pars{x}\quad\imp\quad\color{#f00}{\phi\pars{x}} = \color{#f00}{-\,\mathrm{f}''\pars{x} = -\delta\pars{x - a}} \end{align}

$$\mbox{because}\ \mathrm{f}\pars{x} = \Theta\pars{x - a}\pars{x - a}\ \imp\ \mathrm{f}'\pars{x} = \Theta\pars{x - a}\ \imp\ \mathrm{f}''\pars{x} = \delta\pars{x - a}$$

Checking ?: $$\int_{0}^{\infty}\bracks{-\delta\pars{y - a}}\min\braces{x,y}\,\dd y = -\min\pars{x,a} = \left\lbrace\begin{array}{rcl} \ds{-x} & \mbox{if} & \ds{x < a} \\ \ds{-a} & \mbox{if} & \ds{x > a} \end{array}\right.$$

Because the solution is expresed in terms of the second derivative $\,\mathrm{f}''\pars{x}$, it's still a solution of $\ds{\,\mathrm{f}\pars{x} + cx + d}$ for some arbitrary constants $\ds{c}$ and $\ds{d}$. It means the 'above checking' becomes $$\left\lbrace\begin{array}{rcl} \ds{-x + cx + d} & \mbox{if} & \ds{x < a} \\ \ds{-a + cx + d} & \mbox{if} & \ds{x > a} \end{array}\right.$$ The original $\ds{\,\mathrm{f}\pars{x}}$ is recovered with $\color{#f00}{\ds{c = 1}}$ and $\color{#f00}{\ds{d = 0}}$.

In general, it means there should be additional conditions besides the integral equation.