# Is there a way to solve explicitly the following functional equation?

I want to find an unknown function (actually CDF) $F(p)$ which solves

$1 - \lambda F(\frac{q_L}{q_H}p) - (1-\lambda)F(p-[q_H-q_L]) - \frac{K}{p-c_H} = 0$,

where $0<\lambda<1$, $q_H > q_L > 0$, $q_H > c_H > 0$, $K>0$, and $p \in (c_H, q_H]$.

Unfortunately, I don't really have an idea how to proceed, apart from randomly guessing functional forms (I'm note even sure about which tags to choose for this problem). So any suggestions would be greatly appreciated. Thanks!

You wish to solve $$1 - \lambda F(\frac{q_L}{q_H}p) - (1-\lambda)F(p-[q_H-q_L]) - \frac{K}{p-c_H} = 0$$ Let $S$ be a scaling operator $S F(p)=F(s p)$ with $s=\frac{q_L}{q_H}$. Let $T$ be a translation operator with $T f(p)=F(p-t)$ with $t=q_H-q_L$. Then your equation becomes $$\lambda S F + (1-\lambda)T F = -\frac{K}{p-c_H}+1$$ $$\left( \lambda S + (1-\lambda)T \right) F = -\frac{K}{p-c_H}+1$$ $$(1-\lambda)\left( \frac{\lambda}{1-\lambda} S + T \right) F = -\frac{K}{p-c_H}+1$$ Let us assume that $\epsilon=\frac{\lambda}{1-\lambda}$ is small. $$(1-\lambda)\left( T + \epsilon S\right) F = -\frac{K}{p-c_H}+1$$ $$(1-\lambda) F = -\left( T + \epsilon S\right)^{-1}\frac{K}{p-c_H}+\left( T + \epsilon S\right)^{-1}1$$ For constant functions, such as $1$, both $T$ and $S$ reduce to the identity operator $$(1-\lambda) F = -\left(\left( T + \epsilon S\right)^{-1}\frac{K}{p-c_H}\right)+\frac{1}{1 + \epsilon}$$ What remains to be done is calculating $$\left( T + \epsilon S\right)^{-1}\frac{K}{p-c_H}$$ We can hopefully use a series expansion $$\left( \sum_{i=0}^{\infty} (-\epsilon)^i(T^{-1} S)^i T^{-1}\right)\frac{K}{p-c_H}$$ Now

• $T^{-1}f(x)=f(x+t)$
• $T^{-1} S T^{-1}f(x)=f(s(x+t)+t)$
• $T^{-1} S T^{-1} S T^{-1}f(x)=f(s(s(x+t)+t)+t)$
• $(T^{-1} S)^i T^{-1}f(x)=f(s^i x+t \sum_{j=0}^i s^j)=f\left(s^i x+t \frac{s^{i+1}-1}{s-1}\right)$

thus $$\left( \sum_{i=0}^{\infty} (-\epsilon)^i(T^{-1} S)^i T^{-1}\right)\frac{K}{p-c_H} = K \sum_{i=0}^{\infty} \frac{(-\epsilon)^i}{s^i p+t \frac{s^{i+1}-1}{s-1}-c_H}$$ $$\left( \sum_{i=0}^{\infty} (-\epsilon)^i(T^{-1} S)^i T^{-1}\right)\frac{K}{p-c_H} = K \sum_{i=0}^{\infty} \frac{(s-1)(-\epsilon)^i}{(s-1)(s^i p-c_H)+t (s^{i+1}-1)}$$ Now the question becomes, can this sum be evaluated in a closed form? Wolfram alpha is of little help, though it happily calculates this related sum.

Summary: under various convergence conditions, $F$ can be written as follows $$F = 1 -\frac{K}{1-\lambda} \sum_{i=0}^{\infty} \frac{\left(-\frac{\lambda}{1-\lambda}\right)^i}{s^i p+t \frac{s^{i+1}-1}{s-1}-c_H}$$ We can verify that this is correct in the $\lambda=0$ case $$F(p) = 1 -K \frac{1}{ p+t-c_H}$$ indeed obeys $$1 - F(p-t) - \frac{K}{p-c_H} = 0$$

Three more points:

• The ratio test will tell you that the sum we derived here will only converge for $\epsilon<1$ i.e. $\lambda<1/2$. To get the other half of the solution (the $\lambda$ near 1 case), write instead $$\lambda \left( S + \frac{1-\lambda}{\lambda}T \right) F = -\frac{K}{p-c_H}+1$$ and expand in the new small parameter $\frac{1-\lambda}{\lambda}$.
• This function has an infinite number of exponentially-spaced singularities on the real axis, the largest of which is at $p=c_H-t$. They do not occur within your region of interest $p\in ]c_H,q_H]$, which is good, but they make it implausible that a simple closed-form expression exists.

• This function is a valid CDF between its highest zero and $\infty$, i.e. it is nondecreasing and approaches 1. However its highest zero does not equal $c_H$, and may be $>c_H$ (this happens when $K$ is large), in which case it is not a valid CDF within the region $p\in ]c_H,q_H]$.
• Thanks for the amazing answer! I will go through it in detail and hope that I can continue from there. – Martin Apr 4 '16 at 11:58
• I am curious now: what is this a CDF for? – Wouter Apr 4 '16 at 17:46
• The CDF is the "solution" (Nash equilibrium) of an economic model where we are interested in firms' optimal pricing strategy when there are two types of consumers in some specific market. The "standard" case, which is easy to solve, would be when $\lambda = 1$. – Martin Apr 5 '16 at 8:26