# Differential Equation with Function of Shifted Variable

1. Aug 5, 2013

### Gackhammer

Hey guys, I have a question for you... how would one go about solving an equation like this...

or this...

This came across my mind the other day. I was wondering how to solve that equation if the function is of t-1, instead of t. Obviously, if it was f(t), the solution would be Ce^t, but I was wondering if there is a way to find stuff for f(t-1).

Is there any literature that would help lead meh into da right direction? Obviously my title doesn't do much justice, so I wanted to know if stuff like this has already been dun before!

Thanks!

2. Aug 5, 2013

### Simon Bridge

It's easier to work in Leibnitz notation: $$\frac{d}{dt}y(t)=y(t-1)$$ I'd change the variable ...$\small x=t-1\Rightarrow t=x+1\Rightarrow dt/dx = 1$ - multiply both sides by $\small dt/dx$: $$\frac{dy}{dt}\frac{dt}{dx} = y(x)\frac{dt}{dx}$$
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niggle: for clarity, the prime goes with the function and not the operand thus: $y^\prime(t)$ and not $y(t)^\prime$

3. Aug 5, 2013

### HallsofIvy

Staff Emeritus
I am not sure what you are suggesting with this. If you are suggesting that the left side becomes
dy/dx so that, since dt/dx= 1, dy/dx(x)= y(x), you are mistaken. With the change of variable, this becomes dy/dx(x+1)= y(x). Either way, this is a "differential-difference" equation which can be very difficult to solve.

Typically, rather than an "initial value" at specific value of t, the initial value is over an interval, say 0 to 1, and you can solve the equation "interval by interval".

For example, if we are given that f(t)= t for t in [0, 1], then, for t between 1 and 2, the equation is f'(t)= t-1 so that f(t)= t^2/2- t+ C. Since f is continuous at t= 1, we have f(1)= 1= 1/2- 1+ C so that C= 2- 1/2= 3/2. f(t)= t^2- t+ 3/2 for t in [1, 2].

For t in [1, 2], $f'(t)= (t-1)^2- (t-1)+ 3/2= t^2- 3t+ 7/2$. Integrating, $f(t)= (1/3)t^3- (3/2)t^2+ (7/2)t+ C$. Since f is continuous at t= 2, $f(2)= 4- 2+ 3/2= 7/2= 8/3- 6+ 7+ C$ so that C= -11/3. $f(t)= (1/3)t^3- (3/2)t^2+ (7/2)t- 11/3$ for x in [1, 2]

Continue like that.

4. Aug 5, 2013

### jackmell

That's a delay-differential equation. You can look that up. You can solve this one numerically for a particular initial function value in the interval (-1,0) for example, y(t)=t+1. We'd then solve:

$$y'(t)=y(t-1),\quad y(t)=t+1, t\in [-1,0]$$

Now, try and use the Euler method to solve that IVP for say t=0 to 5. Just a suggestion.

5. Aug 5, 2013

### hilbert2

One solution of the equation $f'(x)=f(x-1)$ is $f(x)=e^{W(1)x}$, where $W(x)$ is the Lambert W function. Also that solution multiplied by any constant is a solution.

That kind of DE:s with a shifted variable don't appear much in physics, as they describe non-local dynamics.

6. Aug 5, 2013

### Simon Bridge

Yeah - I stuffed up :(