# Solving a specific differential equation

I have been trapped in solving the following ODE for a long time. I wonder if it has unique analytical solution $$$$[b+c_B(\bar{\beta}^H-\bar{\beta}^L)]\frac{dF(x)}{dx}+c_BF(x)-c_BF(x+\bar{\beta}^H-\bar{\beta}^L-b/c_B)-c_B=0.$$$$

I could try to assume $$F(x)$$ is linear. But I was wondering if there is a way to prove or disprove the uniqueness of the solution of this ODE? Is there a systematic way to find all the solutions to this equation? Thanks for the comments. Now I know that this is a delay differential equation. In my problem, I have $$x\geq b/c_B+\bar{\beta}^L-\bar{\beta}^H$$. And in fact, $$F(x)$$ is a cumulative distribution function in my case.

• It’s not an ODE but a delay differential equation. – Thomas Rot May 5 '19 at 7:32
• I didn't know this. I am gonna search for it. Thanks! – FTXX May 5 '19 at 7:44
• Try Fourier/Laplace transform. – Alexandre Eremenko May 5 '19 at 11:40
• please use the answer box only for an answer to the question; this is a comment, not an answer. (also, writing in English is strongly recommended on this site, to ensure that everyone can participate in the discussion) – Carlo Beenakker May 5 '19 at 13:25
• @user140272 The whole point of this site is that questions, answers and comments are accessible to any person, such as yourself, who understands English. – Liviu Nicolaescu May 5 '19 at 13:36

To continue Liviu Nicolaescu's simplification: put $$F(x):=f(x/B)$$ so the equation writes $$f'(x)+f(x)-f(x+1)+1=0.$$ A particular solution of it is simply $$f(x):=x^2$$, so we are left with the homogeneous equation $$u'(x)+u(x)-u(x+1)=0.$$ If we put $$u(x)=v(x)e^{-x}$$ this becomes the (well-known) $$v'(x)=\lambda v(x+1)$$ with $$\lambda=1/e$$.
• A solution of the homogeneous equation is $u(x)=ax+b$, or $v(x)=(ax+b)e^x$, btw. – Pietro Majer May 5 '19 at 18:48
• Thank you so much! I just guess $u(x)=ax+b$ and I plug it into my DDE and solve for the value of $a$. Here, can I claim the unique solution to this DDE has to have the form $u(x)=ax+b+x^2$? I am a little bit lost in the part where you say "If we put $u(x)=v(x)e^{-x}$, this becomes...". This seems to suggest that $u(x)=v(x)e^{-x}$ is the only way to solve this homogenous equation. Is this related to characteristic functions? – FTXX May 6 '19 at 0:14
• Could you please provide some references on how to solve $v'(x)=\lambda v(x+1)$, to get $v(x)=(ax+b)e^x$? I can verify that it is correct, but how to solve this delay differential equation in a systematic way and is the uniqueness guaranteed? Thanks! – FTXX May 6 '19 at 1:34
• @FTXX: One general approach to analyse delay differential equations in a systematic way (and to obtain uniqueness) is to use the theory of $C_0$-semigroups; see for instance Sections II.3.29 and VI.6 in "Engel, Nagel: One-Parameter Semigroups for Linear Evolution Equations (2000)", or the entire monograph "Bátkai, Piazzera: Semigroups for Delay Equations (2005)". – Jochen Glueck May 6 '19 at 6:27
This is not an answer but a suggestion. There is a lot of "noise" in your equation. You can simplify it a bit. Set $$A:= b+c_B(\bar{\beta}^H-\bar{\beta}^L),\;\;c:=c_B.$$ Then your equation reads $$AF'(x)+cF(x)-cF(x+A/c)-c=0,$$ or $$BF'(x)+F(x)-F(x+B)+1=0,\;\;B=A/c.$$ This is a linear autonomous delay equation. As some commented, try Fourier transform. Alternatively, have a look at Chapter VII in the book Ordinary and Delay Differential Equations, by R.D. Driver.