Find exponential solutions of the delay differential equation f'(x) = f(x - 1).

Step-by-step solution

1. 1

   Notice what makes this equation unusual. The derivative at $x$ depends on the value one unit earlier, so this is a delay differential equation rather than an ordinary one. The exponential trial still works, but the resulting characteristic equation will not be polynomial.

2. 2

   Substitute $f(x)=e^{rx}$. Then $f'(x)=re^{rx}$, while the delayed term picks up a constant factor:

   $f(x-1)=e^{r(x-1)}=e^{-r}e^{rx}$

3. 3

   Cancel $e^{rx}$ to get the characteristic equation.

   $re^{rx}=e^{-r}e^{rx}\quad\Longrightarrow\quad r=e^{-r}\quad\Longrightarrow\quad re^{r}=1$

4. 4

   Solve $re^r=1$ with the Lambert W function. By definition $W$ inverts $w\mapsto we^{w}$, so

   $r=W(1)\approx 0.5671432904$

   This number is the omega constant. There is no elementary closed form, which is characteristic of delay equations.

5. 5

   Write the solutions and note the infinitely many branches. The real exponential solution is

   $f(x)=Ce^{W(1)x}\approx Ce^{0.567143x}$

   and since $re^r=1$ has infinitely many complex roots $r_k=W_k(1)$, the general form is a superposition $f(x)=\sum_k C_ke^{W_k(1)x}$, with complex conjugate pairs combined to stay real-valued.

6. 6

   Verify the constant numerically. With $r=0.5671432904$: $re^{r}=0.5671432904\times 1.763222834=1.000000000$ $\checkmark$, and equivalently $e^{-r}=0.567143=r$ $\checkmark$. Substituting back, $f'(x)=rCe^{rx}$ and $f(x-1)=Ce^{rx}e^{-r}=rCe^{rx}$, which agree exactly.