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Differential Equation with Function of Shifted Variable

  1. Aug 5, 2013 #1
    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!

  2. jcsd
  3. Aug 5, 2013 #2

    Simon Bridge

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    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##: $$\frac{dy}{dt}\frac{dt}{dx} = y(x)\frac{dt}{dx}$$
    niggle: for clarity, the prime goes with the function and not the operand thus: ##y^\prime(t)## and not ##y(t)^\prime##
  4. Aug 5, 2013 #3


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

    Continue like that.
  5. Aug 5, 2013 #4
    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:

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

    Now, try and use the Euler method to solve that IVP for say t=0 to 5. Just a suggestion.
  6. Aug 5, 2013 #5


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    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.
  7. Aug 5, 2013 #6

    Simon Bridge

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    Yeah - I stuffed up :(
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