# Find $f(x)$ where $f(x)(A-\frac{B}{x+B/A})+Cf(x+\frac{B}{A})=0$.

$$A, B, C > 0$$, $$x$$ is complex and $$Re(x)>0$$. My guess is that $$f(x)=0$$ but I don't know how to prove it.

• do you have any hypothesis about $f$, continuous, differentiable, analytic...? – Marsan Feb 8 at 20:26
• Unfortunately, no. $f(x)$ is actually the Laplace transform of a function I am trying to find. I originally had an integral equation which included convolutions so I took the Laplace transform of the whole thing in the hopes of solving it. The resulting equation had some terms in the RHS as well. I was hoping to use an approach similar to that used for differential equations by first solving the resulting equation with RHS equal to 0, which is what I posted here. – Zaeem Hussain Feb 11 at 16:22

For $$f(x)\left(A-\dfrac{B}{x+C}\right)+Df(x+C)=0$$ ,

$$Df(x+C)=\dfrac{B-A(x+C)}{x+C}f(x)$$

$$f(Cx+C)=\dfrac{B-A(Cx+C)}{D(Cx+C)}f(Cx)$$

$$f(C(x+1))=\dfrac{B-AC(x+1)}{CD(x+1)}f(Cx)$$

$$f(C(x+1))=\dfrac{-AC\left(x-\dfrac{B}{AC}+1\right)}{CD(x+1)}f(Cx)$$

$$f(C(x+1))=-\dfrac{A}{D}\dfrac{x-\dfrac{B}{AC}+1}{x+1}f(Cx)$$

With reference to http://eqworld.ipmnet.ru/en/solutions/fe/fe1105.pdf,

The general solution is $$f(Cx)=\Theta_1(x)\dfrac{A^x\Gamma\left(x-\dfrac{B}{AC}+1\right)}{D^x\Gamma(x+1)}$$, where $$\Theta_1(x)$$ is an arbitrary unit antiperiodic function

$$f(x)=\Theta(x)\dfrac{A^\frac{x}{C}\Gamma\left(\dfrac{x}{C}-\dfrac{B}{AC}+1\right)}{D^\frac{x}{C}\Gamma\left(\dfrac{x}{C}+1\right)}$$, where $$\Theta(x)$$ is an arbitrary antiperiodic function with period $$C$$

Similarly, for $$f(x)\left(A-\dfrac{B}{x+\dfrac{B}{A}}\right)+Cf\left(x+\dfrac{B}{A}\right)=0$$ ,

The general solution is $$f(x)=\Theta(x)\dfrac{A^\frac{Ax}{B}\Gamma\left(\dfrac{Ax}{B}\right)}{C^\frac{Ax}{B}\Gamma\left(\dfrac{Ax}{B}+1\right)}=\Theta(x)\dfrac{A^{\frac{Ax}{B}-1}B}{C^\frac{Ax}{B}x}$$, where $$\Theta(x)$$ is an arbitrary antiperiodic function with period $$\dfrac{B}{A}$$

• Thanks, but I think you cannot apply the result in eqworld.ipmnet.ru/en/solutions/fe/fe1105.pdf directly for $f(Cx)$. If we change the variables to $y=Cx$, we get an equation of the form $f(y+C)=R(y)f(y)$, whereas the result in the link only applies if $f(y+1)=R(y)f(y)$. Do you know of a more general result that would apply to $f(y+C)=R(y)f(y)$? – Zaeem Hussain Apr 17 at 15:56