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I have a differential equation (Or integral equation) of the form:

$$ f(x) = a e^{-x} + b \int_0^x f(cz+dx) e^{-z} dz$$

$a,b,c,d$ are constants.

I am considering whether the above equation has a closed form solution. If not, why it is the case? If so, I think guessing a functional form and using method of undetermined coefficients. But I am not sure how to guess.

Thanks so much!

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  • $\begingroup$ Any additional information concerning the function $f(c z+d x)$? Can this function be written in the form $q(z)\left( P(x)-P(z)\right)$ ? $\endgroup$
    – Upax
    Jul 31 '15 at 16:56
  • $\begingroup$ That's what we are solving for, right? $\endgroup$
    – Andy Xu
    Jul 31 '15 at 16:57
  • $\begingroup$ If it is possible to write the function f as q(z)\left( P(x)-P(z)\right) the integral equation can be written as a differential equation and its solution can be found using the Green function method. $\endgroup$
    – Upax
    Jul 31 '15 at 17:00
  • $\begingroup$ @Upax ?? Very very few functions $f$ are such that $f(cz+dx)=q(z)(p(x)-p(z))$ for every $(x,z)$... Actually, I know only one such function. $\endgroup$
    – Did
    Sep 4 '15 at 8:37
  • $\begingroup$ I would very much like to know what kind of application led to this equation? Seems interesting $\endgroup$
    – Yuriy S
    Apr 18 '18 at 9:52
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Not sure about the general methods for this kind of equation, but in this case we can guess the solution in the form:

$$f(x)= \sum_{n=0}^\infty p_n e^{-q_n x}$$

The convergence of this series depends on the values of $a,b,c,d$ and $x$. By using the ratio test, sufficient (but not necessary) condition would be:

$$\lim_{n \to \infty} \frac{\log |p_{n+1}|-\log |p_{n}|}{q_{n+1}-q_n}<x$$

Now how to get the explicit values of $p_n,q_n$? We can do it by directly substituting the series into the equation:

$$\sum_{n=0}^\infty p_n e^{-q_n x}=ae^{-x}+b \sum_{n=0}^\infty p_n e^{-dq_n x} \int_0^x e^{-(c q_n+1)z} \text{d}z $$

Integrating, we obtain:

$$\sum_{n=0}^\infty p_n e^{-q_n x}=ae^{-x}+\sum_{n=0}^\infty \frac{bp_n}{cq_n+1} \left(e^{-dq_n x}-e^{-\left((c+d\right)q_n+1 ) x} \right) \tag{1}$$

Now we can use the right hand side to find the parameters on the left hand side. Note: this choice is not unique, and thus the solution (if it exists) may not be unique either.

$$p_0=a \qquad q_0=1$$

$$p_1=\frac{ba}{c+1} \qquad q_1=d$$

$$p_2=-p_1 \qquad q_2=c+d+1$$


$$p_{2n+1}=\frac{bp_n}{c q_n+1} \qquad q_{2n+1}=d q_n$$

$$p_{2n+2}=-p_{2n+1} \qquad q_{2n+2}=(c+d)q_n+1$$

$$n \geq 1$$


Looking at the obtained recursion formulas, it seems to me that for a reasonable choice of constants the series will surely converge. Good choice would be for example $c,d,x>0$ and $|b|<1$.


As an example to show that this method works, here's computation for some 'random' parameters:

$$a=\pi, \qquad b=\frac12, \qquad c=3, \qquad d=\sqrt{7}$$

enter image description here

The red line is $f(x)$ computed for $40$ terms of the series, and blue points represent the right hand side of (1) using the same series.

Here's also the plot for $f(x)-a e^{-x}$, which is more interesting:

enter image description here

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