# A strange, but very simple, differential equation

1. Jan 20, 2006

### Aleph-0

Hi,
What do you think about the possible (nontrivial) solutions of this equation ?
(1) f '(x) = f(x-1) (f is a smooth function R -> R)
More generally, what do we know about these ?
(2) f '(x) = f(x+a) (a fixed in R)
(3) f '(x) = f(g(x)) (g smooth diffeomorphism R -> R )

2. Jan 20, 2006

### 7447

As i understand from you i think that f(x) is some kind of function,so that
f(x+a) is obtained by replace every x in f(x) by (x+a)
another expansion about that fuction: f(x+a) is f(x) itself delayed by (a) so it has the same shape of f(x)
note: f(x+a) shift f(x) to the left if a>0
f(x+a) shift f(x) to the right if a<0

if you want more clarify, do post here.

3. Jan 21, 2006

### Aleph-0

I think there is an misunderstanding. I know what f(x+a) means...
I know that if a>0 then x+a > x ... (Was it a joke? )
Maybe was I not clear : I'd like to know the solutions of this differential equation :
df/dx(x) = f (x-1) (for example) .
And if there are no other solution but the function x -> 0, I'd like to have a proof of this.
any idea anyone ?

4. Jan 21, 2006

### Hurkyl

Staff Emeritus
These are called delayed and advanced differential equations.... well, (1) and (2) are called that anyways. They're not anywhere near as nice as ordinary differential equations.

5. Jan 21, 2006

### mathwonk

thats all i recall, they were called "delay equations" and supposed to be hard. i would try the tack of drawing the dirwction fields and see how hard they are qualitatively. maybe it is only explicitly that theya re hard, or maybe even existence is ahrd to rpve, but you could makje some simple assumoptions and easily drawe the direction field and thus get the impression that a solution exists.

or you could try power series methods, just substituting in. seems a little tedious though, but a good opportunity to test the usefulness of various non explicit methods.

6. Jan 21, 2006

### Hurkyl

Staff Emeritus
I think for f'(x) = f(x-1), it's very easy to build solutions. You just treat it as a recurrence relation -- you start with just about any smooth function on [0, 1), and use the equation to extend the solution.

I -think- that the only thing you need to worry about is to get (all of) the derivatives at 0 and 1 to line up, and you're golden.

7. Jan 21, 2006

### Aleph-0

I didn't even know that these equations had a name. With that I think I'll find some information, thanks.
I tried to use a power serie expansion ot see what kind of coefficients one would have, but the relations are not quite simple, and I was unable to prove any existence.
As for drawing, I didnt convince myself that I was really drawing a solution. It looked like an exponential (which seems right) but I am not really sure.
For example, if I start with a smooth function on [0,1), I can write the function on [1,2) But there is a problem with f'(1). And then, there is problem with f '(2), f'(3), etc... Therefore, It seems that even if I have some kind of an approximation of a solution, I am not really drawing a solution. A trivial example :
on [0,1) : x->1
on [1,2) : x->x
on [2,3) : x->(1/2) x^2
on [3,4) : x->(1/6) x^3
etc.
I would be happy with a real, rigorous, proof of the existence, I'll try to find that on the www.

8. Jan 22, 2006

### saltydog

Delay-differential equations (DDEs) often involve rather than an initial "value", an initial "function" for values of t<$t_0$. For example, consider this DDE IVP:

$$f^{'}=f(x-1);\quad f_0(x)=x^2+2x+2\;\text{for}\;x\in [-1,0]$$

then for the interval:

$$x\in[0,1]$$

we have:

$$f^{'}=f_0(x-1)$$

the RHS being a known quantity and thus:

$$f_1(x)=\int_{-1}^{x-1}f_0(v)dv+f_0(0);\quad x\in [0,1]$$

Now, what then is the general expression for:

$$f_n(x)=?\quad x\in [n-1,n]$$

Edit: Oh yea, kindly post a plot for the solution of the above IVP for the interval [0,3].

Edit: One more: Verify that the solution thus obtained satisfies the differential equation. That is show:

$$f^{'}(2.7)=f(1.7)$$

Last edited: Jan 22, 2006
9. Jan 23, 2006

### Aleph-0

Thanks saltydog, but I am not sure that
f would be smooth on R. (or on [0, +infinity) ) .
And my question is : is there a real function which is not everywhere zero,
smooth on R such that :
f '(x) = f(x-1) on R
I've been searching for an answer on the web but the only things I've found are problems and solutions of that kind :
Given a function on a initial interval (for example [0,1) ) f_0
one search the solutions of f '(x) = f(x-1) on [1, +inifinity)
with f = f_0 on [0,1). And there is nothing about smoothness or extensions to R.
Wouldn't be strange that the question I am asking is an open problem ?
I think it's not, but I dont know yet...

10. Jan 23, 2006

### saltydog

I hold that the solution (in recursive form):

$$f_n(x)=\int_{n-2}^{x-1} f_{n-1}(u)du+f_{n-1}(n-1)$$

is smooth.

Surely it's smooth within each interval so the only points of concern are at the end points of each interval. Take for example the first two intervals:

$$f_1(x)=2+x+\frac{x^3}{3}$$

$$f_2(x)=\frac{23}{12}+\frac{2}{3}x+x^2-\frac{x^3}{3}+\frac{x^4}{12}$$

The functions and it's derivatives within each interval are continuous and:

$$f_1(1)=10/3=f_2(1)$$

$$f_1^{'}(1)=2=f_2^{'}(1)$$

Edit: corrected equations

Last edited: Jan 23, 2006
11. Jan 23, 2006

### Aleph-0

For example, you have :
$$f_1^{'}(0)=f_0(-1)=1$$
$$f_0^{'}(0)=2$$
so this is not smooth (not even C1) in 0.
And therefore you have no smoothness in 1, in 2 etc..
Indeed, you have :
$$f_2^{''}(1)=f_1^{'}(0)=f_0(-1)=1$$
$$f_1^{''}(1)=f_0^{'}(0)=2$$
But If you find a smooth function $$f_0$$ on [-1,0) such that :
for all n>0 $$f_0^{(n-1)}(-1)=f_0^{(n)}(0)$$
I think this would be right (do you know such a function ? It should exist).But even then I wouldn't know how to build a smooth solution on R.

Last edited: Jan 23, 2006
12. Jan 23, 2006

### saltydog

Ok Aleph. Thanks for clearing that up for me. Just for fun, I plotted the third derivatives of the first three functions (f1,f2,f3). The plot is below. It's discontinuous nature clearly indicates the derivatives are not continuous thus not smooth.

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