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I've got this nasty-looking integral equation involving taking two minimums:


where $\delta(\cdot)$ is the Dirac delta function and $a$, $b$, $c$, and $d$ are constants. I am trying to find $f(x)$.

I recognize that this is a convolution of sorts and that the minimum on the LHS is easy to deal with (at least in theory) by breaking this thing apart into two cases. However, I am lost trying to figure the RHS out. Is there a way to solve with a transform of some kind? Any help would be appreciated!

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Well, integrals of delta functions are always nice, so let's start there.

$$\begin{align} a+c\min(b,x) & =\int_{-\infty}^\infty f(x-t)[a+d\delta(t-(x-b))+c\min(b,t)]dt \\ & =\int_{-\infty}^\infty f(x-t)d\delta(t-(x-b))dt+\int_{-\infty}^\infty f(x-t)[a+c\min(b,t)]dt \\ & =df(b)+\int_{-\infty}^b f(x-t)(a+ct)\,dt+\int_b^\infty f(x-t)(a+cb)\,dt \end{align}$$

Now that's not as bad, but it's still an integral equation. Let's treat this for $x<b$ and take a derivative:

$$\begin{align} c & =\int_{-\infty}^b f'(x-t)(a+ct)\,dt+\int_b^\infty f'(x-t)(a+cb)\,dt \\ & =a[f(-\infty)-f(x-b)]+(a+cb)[f(x-b)-f(\infty)]+\int_{-\infty}^b f'(x-t)ct\,dt \\ & =af_{-\infty}+cbf(x-b)-(a+cb)f_\infty+f(x-t)ct\,\bigg|_{-\infty}^b-\int_{-\infty}^b f'(x-t)c\,dt \\ & =af_{-\infty}+cbf(x-b)-(a+cb)f_\infty+f(x-b)cb-f_\infty c\cdot\infty+cf(x-b)\Rightarrow f_\infty=0 \\ & =af_{-\infty}+c(2b+1)f(x-b) \end{align}$$

Note that because this is equals a constant, it follows that $f(x)=f_{-\infty}$ is constant for all $x<0$, so

$$c=af_{-\infty}+c(2b+1)f_{-\infty}\Rightarrow f_{-\infty}=\frac c{a+2bc+c}=f(x),\qquad x<0$$

If we do this again for $x>b$, the only change will be on the LHS in the different choice for the min. If we take a derivative, we will just get $0$ on the LHS, so

$$0 =af_{-\infty}+c(2b+1)f(x-b)\Rightarrow f(x)=\frac {-a}{(2b+1)(a+2bc+c)},\qquad x>0$$

But this contradicts $f_\infty=0$, so either there is no solution, or $f(x)$ violates some properties that make it amenable to integration by parts or the FTC used here. I'll let others figure that out.

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