# Solve differential equation $-f'(x)= a_1 f(a_2 x+a_3)$ with $f(0)=1$.

How to solve the following differential equation

\begin{align} -f'(x)= a_1 f(a_2 x+a_3), \end{align} where $$f(0)=1$$.

I looked around I think this falls under the category of discrete delayed differential equations.

• Your problem is ill posed. Defining just one point of $f$ will not be enough to determine the solution. – Artem Feb 23 at 22:39
• @Artem would value as $x \to -\infty$ be good enough? Or the value of $f'(0)$? – Lisa Feb 24 at 0:11

If it's okay, I will assume $$a_2>0$$. (But if you're keen to have $$a_2<0$$ as well, write a comment and I'll see if I have time to work out that case.) The term "(discrete) delay differential equation" would apply to the case that $$a_2=1$$ and $$a_3<0$$.

As stated in one of the comments under your question, to uniquely determine a solution it is not enough to specify the value of $$f$$ at just a single point. Instead, as is typically the case for delay differential equations, an "initial condition" consists of specifying $$f$$ on a suitable interval! If this "initial condition" satisfies a certain requirement then it gives rise to a solution and this solution is unique. All this will be described below.

If $$a_2 \neq 1$$, then one needs to consider your differential equation separately over the range $$x \in (\frac{a_3}{1-a_2},\infty)$$ and the range $$x \in (-\infty,\frac{a_3}{1-a_2})$$. For each of these two cases one needs to specify an "initial condition" consisting of the values of $$f$$ over a suitable interval.

Let us define $$T \colon \mathbb{R} \to \mathbb{R}$$ by $$T(x)=a_2x+a_3$$, and if $$a_2 \neq 1$$ then we write $$X_0$$ for the fixed point of $$T$$, namely $$X_0=\frac{a_3}{1-a_2}$$. We write $$T^n$$ for the $$n$$-th iterate of $$T$$ if $$n>0$$, or for the $$|n|$$-th iterate of the inverse of $$T$$ if $$n<0$$ (and we let $$T^0$$ be the identity function). Explicitly, for all $$n \in \mathbb{Z}$$, $$T^n(x) \ = \ \left\{ \begin{array}{c l} a_2^nx+\frac{a_3(1-a_2^n)}{1-a_2} & a_2 \neq 1 \\ x + na_3 & a_2=1 \end{array} \right.$$

(I) First suppose $$a_2 \in (0,1)$$. To construct a solution $$f(x)$$ on the range $$x \in (X_0,\infty)$$, you need to specify a suitable "initial condition". To specify a suitable "initial condition", you choose an $$x_0 \in (X_0,\infty)$$ and specify the function $$f$$ on $$[x_0,T^{-1}(x_0)]$$. In order for this initial condition $$\left. f \right|_{[x_0,T^{-1}(x_0)]}$$ to give rise to a solution on $$(X_0,\infty)$$, we need that $$\left. f \right|_{[x_0,T^{-1}(x_0)]}$$ is $$C^\infty$$ with the additional boundary requirement $$f^{(k+1)}(T^{-1}(x_0)) = -a_1f^{(k)}(x_0)$$ for all $$k \geq 0$$. In this case, we construct the unique solution on $$(X_0,\infty)$$ as follows: define $$f$$ on $$(X_0,x_0)$$ by $$f(x) = (-a_1)^{-n}f^{(n)}(T^{-n}(x))$$ for $$x \in [T^n(x_0),T^{n-1}(x_0))$$ with $$n \geq 1$$; and define $$f$$ on $$(T^{-1}(x_0),\infty)$$ by constructing $$f$$ recursively on the intervals $$(T^{-n}(x_0),T^{-(n+1)}(x_0)]$$ with $$n \geq 1$$, by $$f(x) \ = \ f(T^{-n}(x_0)) - a_1 \int_{T^{-n}(x_0)}^x f(T(y)) \, dy.$$ Thus we have constructed $$f$$ on $$(X_0,\infty)$$; it is easy to check that our constructed function $$f$$ solves the differential equation, and it is not hard to see that it is the only possible solution on $$(X_0,\infty)$$ given the initial condition. (And in a similar way, it is not hard to see that if the initial condition fails to satisfy the boundary requirement, then it cannot give rise to a solution on $$(X_0,\infty)$$.)

Likewise one can construct a solution on $$(-\infty,X_0)$$ given a suitable "initial condition". To define a suitable initial condition, you choose an $$x_1 \in (-\infty,X_0)$$ and specify $$f$$ on $$[T^{-1}(x_1),x_1]$$. For this initial condition $$\left. f \right|_{[T^{-1}(x_1),x_1]}$$ to give rise to a solution on $$(-\infty,X_0)$$, we need that $$\left. f \right|_{[T^{-1}(x_1),x_1]}$$ is $$C^\infty$$ with the additional boundary requirement $$f^{(k+1)}(T^{-1}(x_1)) = -a_1f^{(k)}(x_1)$$ for all $$k \geq 0$$. In this case, to construct the solution, define $$f$$ on $$(x_1,X_0)$$ by $$f(x) = (-a_1)^{-n}f^{(n)}(T^{-n}(x))$$ for $$x \in (T^{n-1}(x_1),T^n(x_1)]$$ with $$n \geq 1$$, and define $$f$$ on $$(-\infty,T^{-1}(x_1))$$ by constructing $$f$$ recursively on the intervals $$[T^{-(n+1)}(x_1),T^{-n}(x_1))$$ with $$n \geq 1$$, by $$f(x) \ = \ f(T^{-n}(x_1)) + a_1 \int_x^{T^{-n}(x_1)} f(T(y)) \, dy.$$ Thus we have constructed a solution on $$(-\infty,X_0)$$.

(II) Suppose $$a_2>1$$. To define a suitable "initial condition" for a solution on $$(X_0,\infty)$$, you choose an $$x_0 \in (X_0,\infty)$$ and specify $$f$$ on $$[x_0,T(x_0)]$$ with $$\left. f \right|_{[x_0,T(x_0)]}$$ being $$C^\infty$$ and satisfying the boundary requirement $$f^{(k+1)}(x_0) = -a_1f^{(k)}(T(x_0))$$ for all $$k \geq 0$$. With this, define $$f$$ on $$(T(x_0),\infty)$$ by $$f(x) = (-a_1)^{-n}f^{(n)}(T^{-n}(x))$$ for $$x \in (T^n(x_0),T^{n+1}(x_0)]$$ with $$n \geq 1$$, and define $$f$$ on $$(X_0,x_0)$$ by constructing $$f$$ recursively on the intervals $$[T^{-n}(x_0),T^{-(n-1)}(x_0))$$ with $$n \geq 1$$, by $$f(x) \ = \ f(T^{-(n-1)}(x_0)) + a_1 \int_x^{T^{-(n-1)}(x_0)} f(T(y)) \, dy.$$ To define a suitable "initial condition" for a solution on $$(-\infty,X_0)$$, you choose an $$x_1 \in (\infty,X_0)$$ and specify $$f$$ on $$[T(x_1),x_1]$$ with $$\left. f \right|_{[T(x_1),x_1]}$$ being $$C^\infty$$ and satisyfing the boundary requirement $$f^{(k+1)}(x_1) = -a_1f^{(k)}(T(x_1))$$ for all $$k \geq 0$$. With this, define $$f$$ on $$(-\infty,T(x_1))$$ by $$f(x) = (-a_1)^{-n}f^{(n)}(T^{-n}(x))$$ for $$x \in [T^{n+1}(x_1),T^n(x_1))$$ with $$n \geq 1$$, and define $$f$$ on $$(x_1,X_0)$$ by constructing $$f$$ recursively on the intervals $$(T^{-(n-1)}(x_1),T^{-n}(x_1)]$$ with $$n \geq 1$$ by $$f(x) \ = \ f(T^{-(n-1)}(x_1)) - a_1 \int_{T^{-(n-1)}(x_1)}^x f(T(y)) \, dy.$$ Finally, given a solution $$f$$ on $$(-\infty,X_0) \cup (X_0,\infty)$$, it is not hard to show [I think!] that the left-sided and right-sided limits of $$f$$ at $$X_0$$ exist; if they are equal to each other, then setting $$f(X_0)$$ to be equal to this limit gives a solution on the whole of $$\mathbb{R}$$ (and this solution is $$C^\infty$$ on the whole of $$\mathbb{R}$$).

(III) Suppose $$a_2=1$$ and $$a_3>0$$. Given a suitable initial condition, one can construct a solution on the whole of $$\mathbb{R}$$. To define a suitable initial condition, you choose an $$x_0 \in \mathbb{R}$$ and specify $$f$$ on the interval $$[x_0,T(x_0)]=[x_0,x_0+a_3]$$. The boundary requirement for this initial condition to have a solution, and the construction of the solution on $$\mathbb{R}$$, are exactly the same as given in Case (II) for constructing solutions on $$(X_0,\infty)$$ when $$a_2>1$$. To be precise, the initial condition $$\left. f \right|_{[x_0,x_0+a_3]}$$ has to be $$C^\infty$$ with the boundary requirement $$f^{(k+1)}(x_0) = -a_1f^{(k)}(x_0+a_3)$$ for all $$k \geq 0$$. With this, define $$f$$ on $$(x_0+a_3,\infty)$$ by $$f(x) = (-a_1)^{-n}f^{(n)}(x-na_3)$$ for $$x \in (x_0+na_3,x_0+(n+1)a_3]$$ with $$n \geq 1$$, and define $$f$$ on $$(-\infty,x_0)$$ by constructing $$f$$ recursively on the intervals $$[x_0-na_3,x_0-(n-1)a_3)$$ with $$n \geq 1$$, by $$f(x) \ = \ f(x_0-(n-1)a_3) + a_1 \int_x^{x_0-(n-1)a_3} f(y+a_3) \, dy.$$

(IV) Suppose $$a_2=1$$ and $$a_3<0$$. Given a suitable initial condition, one can construct a solution on the whole of $$\mathbb{R}$$. To define a suitable initial condition, you choose an $$x_0 \in \mathbb{R}$$ and specify $$f$$ on the interval $$[x_0,T^{-1}(x_0)]=[x_0,x_0+|a_3|]$$. The boundary requirement for this initial condition to have a solution, and the construction of the solution on $$\mathbb{R}$$, are exactly the same as given in Case (I) for constructing solutions on $$(X_0,\infty)$$ when $$a_2 \in (0,1)$$. To be precise, the initial condition $$\left. f \right|_{[x_0,x_0+|a_3|]}$$ has to be $$C^\infty$$ with the boundary requirement $$f^{(k+1)}(x_0+|a_3|) = -a_1f^{(k)}(x_0)$$ for all $$k \geq 0$$. With this, define $$f$$ on $$(-\infty,x_0)$$ by $$f(x) = (-a_1)^{-n}f^{(n)}(x+n|a_3|)$$ for $$x \in [x_0-n|a_3|,x_0-(n-1)|a_3|)$$ with $$n \geq 1$$, and define $$f$$ on $$(x_0+|a_3|,\infty)$$ by constructing $$f$$ recursively on the intervals $$(x_0+n|a_3|,x_0+(n+1)|a_3|]$$ with $$n \geq 1$$, by $$f(x) \ = \ f(x_0+n|a_3|) - a_1 \int_{x_0+n|a_3|}^x f(y-|a_3|) \, dy.$$

Remark. Case (IV) is a simple case of a "delay differential equation". For this case (as with delay differential equations in general) one is likely to be concerned only with "forward-time solutions". Assuming $$a_2=0$$ and $$a_3<0$$, a "forward-time solution" of your differential equation is a function $$f \colon [x_0,\infty) \to \mathbb{R}$$ (for some $$x_0 \in \mathbb{R}$$) such that your differential equation is satisfied on $$[x_0+|a_3|,\infty)$$ (with only the right-sided derivative considered at the left endpoint $$x_0+|a_3|$$). To define a forward-time solution, the "initial condition" that one needs to specify is the function $$f$$ on the interval $$[x_0,x_0+|a_3|]$$; this initial condition $$\left. f \right|_{[x_0,x_0+|a_3|]}$$ may be any continuous function!$$^\ast$$ From here, one defines $$f$$ on $$(x_0+|a_3|,\infty)$$ by the recursive procedure described above for Case (IV). An example in the case that $$x_0=a_3$$ (so $$x_0+|a_3|=0$$) and $$\left. f \right|_{[a_3,0]}$$ is constant at $$1$$ is given in the Wikipedia article https://en.wikipedia.org/wiki/Delay_differential_equation.

$$^\ast$$In fact, for "weak" forward-time solutions (meaning that differentiability on $$[x_0+|a_3|,\infty)$$ can be weakend to weak differentiability on $$[x_0+|a_3|,\infty)$$), $$\left. f \right|_{[x_0,x_0+|a_3|]}$$ can be any $$L^1$$ function, although one would additionally need to specify the value of $$\lim_{x \searrow x_0+|a_3|} f(x)$$.

• I realise my initial reply may not have been so clear; so I have made changes with a slightly more "pedagogical" style. – Julian Newman Mar 6 at 3:12
• For case (I) ($0<a_2<1$), what I wrote about solutions on the whole of $\mathbb{R}$ is wrong, so I have removed it. I can think about that further. The answer by Yuri Negometyanov constructs one possible solution for this case via Taylor series [the unique analytic solution when $f$ is specified at a single point - unless the specified value happens to be at a point $x$ where the Taylor series around $X_0$ converges to $0$], although I expect that there will also be other non-analytic solutions. – Julian Newman Mar 6 at 14:36
• On second thoughts, I have no idea whether analytic solutions do exist. – Julian Newman Mar 6 at 15:21
• Nothing specific solutions, common ideas only. – Yuri Negometyanov Mar 7 at 10:30
• To "solve" a differential equation does not mean to give some examples of solutions, but rather to obtain (for given parameter-values) the complete set of solutions, ideally in such a manner that if you give me sufficient "initial data" of any solution, I can tell you the entire solution. For "initial data" taking the form of specifying $f$ on an interval, my answer achieves this for $a_2=1$ when the DE is interpreted on $\mathbb{R}$, and for $a_2\in (0,\infty)\setminus\{1\}$ when the DE is interpreted on $(-\infty,X_0)$ or $(X_0,\infty)$. – Julian Newman Mar 7 at 17:24

Case $$1$$: $$a_2>0$$

Let $$f(x)=\int_0^\infty e^{-xt}K(t)~dt$$ ,

Then $$\int_0^\infty te^{-xt}K(t)~dt=a_1\int_0^\infty e^{-(a_2x+a_3)t}K(t)~dt$$

$$\int_0^\infty te^{-xt}K(t)~dt-a_1\int_0^\infty e^{-a_2xt}e^{-a_3t}K(t)~dt=0$$

$$a_2^2\int_0^\infty te^{-a_2xt}K(a_2t)~dt-a_1\int_0^\infty e^{-a_2xt}e^{-a_3t}K(t)~dt=0$$

$$\int_0^\infty e^{-a_2xt}(a_2^2tK(a_2t)-a_1e^{-a_3t}K(t))~dt=0$$

$$\therefore a_2^2tK(a_2t)-a_1e^{-a_3t}K(t)=0$$

Let $$\begin{cases}t_1=\log_{a_2}t\\K_1(t_1)=K(t)\end{cases}$$ ,

Then $$a_2^{t_1+2}K_1(t_1+1)=a_1e^{-a_3a_2^{t_1}}K_1(t_1)$$

Obviously $$f(x)$$ is in $$C^{\infty}(\textbf{R})$$. Also it hold by induction $$f^{(\nu)}(x)=(-a_1)^{\nu}a_2^{\nu(\nu-1)/2}f\left(g_n(x)\right)\textrm{, }\forall x\in\textbf{R}$$ I.e if $$g(x)=a_2x+a_3$$, then $$g_n(x)=(a_3+a_2(a_3+a_2(a_3+\ldots+a_2(a_3+a_2x\underbrace{)\ldots)))}_{n-parenthesis}.$$ Hence $$f^{(\nu)}(x)=(-1)^{\nu}a_1^{\nu}a_2^{\nu(\nu-1)/2}f\left(a_3(1+a_2^1+a_2^2+\ldots+a_2^{\nu-1})+a_2^{\nu} x\right)\textrm{, }\forall \nu=1,2,\ldots.$$ Hence when $$\nu=1,2,\ldots$$, we get $$f^{(\nu)}(-\frac{a_3}{a_2-1})=(-a_1)^{\nu}a_2^{\nu(\nu-1)/2}f\left(-\frac{a_3}{a_2-1}\right).$$ If we set $$c=\frac{a_3}{1-a_2}$$,then $$f(x)=\sum^{\infty}_{n=0}\frac{f^{(n)}(c)}{n!}(x-c)^{n}=f(c)\left(1+\sum^{\infty}_{n=1}\frac{a_2^{n(n-1)/2}(-a_1)^{n}}{n!}(x-c)^n\right),$$ for all $$x\in\textbf{R}$$, when $$|a_2|<1$$.

• This is the same as Yuri Negometyanov's answer before he edited it. As in my initial comment to Yuri Negometyanov: you've constructed an analytic function on $\mathbb{R}$ that satisfies the requirement of the differential equation at the single point $c$, but how do you know that it solves the differential equation everywhere else? – Julian Newman Mar 7 at 2:52
• No. The point $c$ is known constant $a_3/(1-a_2)$. Also the convergent of the function $f(x)$ as given by the Taylor series has ratio all $\textbf{R}$, when $|a_2|<1$, and diverges (has no solution) when $|a_2|>1$. The case $|a_2|=1$ gives also solution. It hapens to be, that in particular point $c$ the Taylor coefficients simplified much more and thats all. In any other point we have the same function. – Nikos Bagis Mar 7 at 8:09
• If $f(x)$ is analytic in $\textbf{R}$, then Taylor's theorem read as $f(x)=\sum_{n\geq0}f^{(n)}(x_0)/n! (x-x_0)^n$, for every $x_0$ in $\textbf{R}$ and $x$ in $\textbf{R}$. For example with $f(x)=e^x$ it does not matter what is $x_0$, we have $e^x=\sum_{n\geq 0}e^{x_0}/n!(x-x_0)^n$. – Nikos Bagis Mar 7 at 8:22
• This is my solution and my previous approach. – Yuri Negometyanov Mar 7 at 10:32
• Ok. I got it. I search in the Web for time delay equations. There have infinite number of solutions and mainly are solved with numerical methods. I was confused by the above thoughts, in which to me seems too natural. I have written in the past an article in arXiv about these kind of equations without knowing about the related theory maybe will be helpful: arxiv.org/abs/1009.2570 – Nikos Bagis Mar 7 at 20:37

Edition of 06.03.19

First, if $$\underline{a_1=0}$$ or $$\underline{a_2=0},$$ then $$f(x)=0.$$

If $$\underline{a_2=1},$$ then finding of the solution in the form of $$f(x)= \mathrm{const}\cdot e^{-kx}\tag1$$ gives $$ke^{-kx} = a_1e^{-kx-ka_3},$$ $$ke^{ka_3}=a_1,$$ $$k= \begin{cases} a_1,\quad\text{if}\quad a_3=0,\\[4pt] \dfrac{\log(a_1a_3)}{a_3 W(\log(a_1a_3))},\quad\text{otherwize}, \end{cases}\tag2$$ where $$W(t)$$ is the Lambert W-function.

Assume $$\underline{a_1\not=0},\quad \underline{a_2\not = 0},\quad \underline{a_2\not= 1}.$$

Taking in account that $$f'\left(\frac{a_3}{1-a_2} - x\right) = a_1f\left(\frac{a_2a_3}{1-a_2}-a_2x+a_3\right) = a_1f\left(a_2\left(\frac{a_3}{a_2(1-a_2)}-x\right)\right),$$ denote $$r=\frac{a_3}{a_2(1-a_2)},\quad y=r-x,\quad g(y) = f\left(\dfrac y{a_1}\right).\tag3$$ Then $$g'(y) = \dfrac1{a_1}f'\left(\dfrac y{a_1}\right) = f\left(\dfrac{a_2}{a_1}\,y\right),$$ $$g'(y) = g(a_2y),\quad g(r) = 1.\tag4$$

Boundary condition defines factor of $$g(y).$$ Behavior of $$g(y)$$ is linked with the parameter $$a_2.$$

Denote $$v=g(0),$$ and let us calculate derivatives of $$g(y):$$ \begin{align} &g(0) = v,\\[4pt] &g'(y) = g(a_2y),\quad g'(0) = v,\\[4pt] &g''(y) = a_2 g'(a_2 y) = a_2g(a_2^2y),\quad g''(0) = a_2v,\\[4pt] &g'''(y) = a_2^3 g'(a_2^2y) = a_2^3 g(a_2^3y),\quad g'''(0) = a_2^3v,\dots,\\[4pt] &g^{(n)}(y) =a_2^{\frac12n(n-1)}g\left(a_2^{\frac12n(n-1)}\right),\quad g^{(n)}(0) =a_2^{\frac12n(n-1)}v,\dots \end{align} Maclaurin series are $$g(y) = v \left(1+y+\frac12a_2y^2+\frac16a_2^3y^3+\dots+\dfrac1{n!}a_2^{\frac12n(n-1)}y^n+\dots\right).\tag5$$

If $$\underline{|a_2|>1}$$ then $$n! \le n^n = e^{n\log n},$$ and the series $$(5)$$ diverge.

If $$\underline{a_2 = 1}$$ then the series $$(5)$$ converge to $$ve^x,$$ and then $$g(y) = e^{x-r}.\tag6$$ Formula $$(6)$$ cannot be used in the solution, but it gives a limit case for $$g(y).$$

If $$\underline{a_2 = -1}$$ then the series $$(5)$$ converge to $$v(\sin x + \cos x),$$ and then $$g(y)=\dfrac{\cos\left(\dfrac\pi4-y\right)}{\cos\left(\dfrac\pi4-r\right)}.\tag7$$

If $$\underline{|a_2|<1}$$ then the series $$(5)$$ converge, wherein $$g(y) = \dfrac {\sum\limits_{n=0}^\infty \dfrac1{n!}a_2^{\frac12n(n-1)}y^n} {\sum\limits_{n=0}^\infty \dfrac1{n!}a_2^{\frac12n(n-1)}r^n}.\tag8$$

The Wolfram Alpha plot of the numerators of $$(8)$$ shown above, illustrates the behavior of the function $$g(y)$$ for $$a_2=\pm 0.9, \pm 0.6, \pm 0.3,\quad (y=-5,5).$$

The solution of the issue equation in the case $$|a_2| < 1$$ is $$f\left(x,\vec a\right) = g(a_1(r-x)).\tag9$$

• For a2 in (-1,1), how do you know that the function you constructed solves the differential equation? You've only constructed it to be the analytic function fulfilling the requirement at r itself. – Julian Newman Mar 6 at 15:26
• @JulianNewman Thanks for the comment. See detalized solution. – Yuri Negometyanov Mar 6 at 21:27
• There seems to be a fundamental error in your new answer: in the line $f'\left(\frac{a_3}{1-a_2}-x\right)=\ldots$, the rightmost term in the equality is missing "$+a_3$"; i.e. it reads "$\ldots=a_1f\left(a_2\left(\frac{a_3}{1-a_2}-x\right)\right)$" but it should read $\hspace{20mm}$"$\ldots=a_1f\left(a_2\left(\frac{a_3}{1-a_2}-x\right)+a_3\right)$". This error goes all the way through. [Minor points: There should also be a $-$ sign somewhere if we follow the notation of the question; and in the middle term of that line, the $-a_3$ should be $+a_3$.] – Julian Newman Mar 7 at 2:45
• @JulianNewman It was a typo without further influence ( "$-a_3$" instead "$+a_3$".Thanks for attention, fixed. – Yuri Negometyanov Mar 7 at 7:44
• You have corrected my "Minor point" which, as you say, has no further influence. But you have not corrected my main concern, namely that the +a3 is still missing from the rightmost term of that line. This error does have further influence. You have persuaded me that for |a2| less than 1, your Taylor series approach constructs a solution of the differential equation if a3=0, but I'm not yet convinced for non-zero a3. – Julian Newman Mar 7 at 12:52