Differential Equation with Shifted Input

Question

So I'm fairly new to differential equations, and while tinkering around with graphs, I came up with one that is confusing me quite a bit. I have both a simpler form and a more complicated form. The simpler one is:

$$ f(x)=f'(x-c) $$

I see that when $c=0$, a solution is $f(x)=e^x,$ and that when $c=\frac{\pi}{2},$ solutions are $f(x) = \sin x$ and $f(x) = \cos x,$ but I don't see how to generalize it to any c. Conceptually, I view this equation as saying that when you take the derivative, the function is shifted over by $c.$

[Edit:

I couldn't get anywhere assuming the function was a sinusoidal, but I was able to get some results assuming an exponential:

If we assume $f(x) = Ae^{bx}$, where $A$ and $b$ are real constants, we get the equation

$$ Ae^{bx}=Abe^{b(x-c)} $$

Taking the natural log of both sides and simplifying yields the following:

$$ c=\frac{\ln b}{b} $$

However, I can't figure out how to solve this for $b.$ ]

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The more complicated form is:

$$ f(x)=f^{(n)}(x-nc) $$

This reduces to the simpler case when $n = 1,$ and when $c=\frac{\pi}{2},$ regardless of $n,$ solutions are again $f(x) = \sin x$ and $f(x) = \cos x$. However, again, I don't see how to generalize to both $n$ and $c,$ nor how to determine whether such a general solution exists. Conceptually, I view this equation as saying that each time you take the derivative of a function, it gets shifted $c$.

Answer 1

You may want to look into delay differential equations. If you let $t = x-c$ and let $k=-c$, then $x = t-k$, so your equation is equivalent to $f^\prime(t)=f(t-k)$. This is the quintessential delay differential equation. Here, the derivative at some time is affected by the value at another time. The term "delay" comes from the fact that usually, $k>0$, so the derivative is dependent on the function's past values. If $k<0$, we call it an "advanced" delay differential equation, since it needs to know in advance what it's future value will be to determine the derivative at the current time $-$ hence why these kinds of equations are much more rare in real life situations.

Research in delay differential equations is very active currently, as the applications to neural nets and the like are extremely pertinent today. There are some resources you can look at to get acquainted with the basics (for example, this short survey and the references listed at the end), but I would say to first take the time to master differential equations without delays before you really start to dig into this material. Most people who study delay differential equations do so after at least 2 or 3 semesters of ordinary differential equations, and probably at least one of them at the graduate level.

Of course, what you've done here is great for just poking around, and I encourage you to continue poking around to see what you can find on your own. Go ahead and do some research of your own (now that you have a jumping off point), but just know that any hurdles you run into in your journey will likely be overcome by a more fundamental understanding of differential equations as a whole.

Answer 2

Notice that $f'$ is a solution to the one dimension wave equation: $$\dfrac{\partial^2 u}{\partial x^2}(x,c)=\dfrac{\partial^2 u}{\partial c^2}(x,c)$$ So $$u(x,c)=(k_1e^{\lambda x}+k_2e^{-\lambda x})(k_3e^{\lambda c}+k_4e^{-\lambda c})=c_1e^{\lambda(x+c)}+c_2e^{\lambda(x-c)}+c_3e^{-\lambda(x+c)}+c_4e^{-\lambda(x-c)}$$ Where $k_i,c_i$ are constant, that may be complex, and $\lambda$ is a constant that can be either pure real or pure imaginary numbers(because this equation does not Describes a wave we can, and should, let $\lambda$ be a real number).

To be sure let's check our result with the $3$ known examples you gave:

If $c=0$ we get $u(x,c)=k_1e^{\lambda x}+k_2e^{-\lambda x}$, because any $\lambda$ solve the wave equation the only thing I need to take into account is the condition that the anti derivative of $u$ in respect to $x$ is $u$, hence $\lambda=1,k_2=0$ and we get $Ae^{x}$

Let's $c$ be $\pi/2$:$$c_1e^{\lambda(x+\pi/2)}+c_2e^{\lambda(x-\pi/2)}+c_3e^{-\lambda(x+\pi/2)}+c_4e^{-\lambda(x-\pi/2)}$$ Let $\lambda=i$ to get $$c_1\cos(x+\pi/2)+ic_1\sin(x+\pi/2)+c_2\cos(x-\pi/2)+ic_2\sin(x-\pi/2)+c_3\cos(x+\pi/2)-ic_3\sin(x+\pi/2)+c_4\cos(x-\pi/2)-ic_4\sin(x-\pi/2)=A_1\sin(x)+A_2\cos(x)$$ Now just check your initial conditions.

You can see that the solution of the wave equation gives you a lot of functions that most of them does not solve your equation, sadly I don't have a good way to find $\lambda$, I also don't know if this is all the solutions or just subset.

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Your assumption that the function is exponential is very good, but still missing a lot of solutions that the above explanation showed, I'll keep search for (1) is the above explanation gives you all of the solution (2) if not, what the other are.

I'll add information if I found something

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Let me explain why did I create new function: $u(x,c)$ and how I got to solution:

You asked how to solve for $f$ using $c$ given $f(x)=f'(x-c)$, I don't know how to solve this so I search for things I do know and I recognize that $f'(x-c)$ is similar to something I do know, if I regard $c$ to be another variable and create the new function $u(x,c)=f'(x-c)$ I have a wave function:\begin{align}\dfrac{\partial^2 f'(x-c)}{\partial x^2}=\dfrac{\partial \dfrac{d[x-c]}{dx}f''(x-c)}{\partial x}=\frac{d[x-c]}{dx}f^{(3)}(x-c)=f^{(3)}(x-c)\\ \dfrac{\partial^2 f'(x-c)}{\partial c^2}=\dfrac{\partial \dfrac{d[x-c]}{dc}f''(x-c)}{\partial c}=-\frac{d[x-c]}{dc}f^{(3)}(x-c)=f^{(3)}(x-c)\end{align}

To solve the PDE $\dfrac{\partial^2 u}{\partial x^2}(x,c)=\dfrac{\partial^2 u}{\partial c^2}(x,c)$ is not an easy task but assuming that $u(x,c)=F(x)G(c)$(I won't explain why we can assume this here, if you want read more about the wave equation) we can find that: $$\dfrac{\partial^2 u}{\partial x^2}(x,c)=F''(x)G(c)\\\dfrac{\partial^2 u}{\partial c^2}(x,c)=G''(c)F(x)\\\implies\frac{F''(x)}{F(x)}=\frac{G''(c)}{G(c)}$$ Because every side is depends on a different variable we conclude that those fractions are equal to a constant: $\lambda^2$, with this we get $2$ ODE equations:$$F''(x)-\lambda^2 F(x)=0\\G''(c)-\lambda^2 G(c)=0$$

After solving those and multiplying the solutions you'll get $$u(x,c)=(k_1e^{\lambda x}+k_2e^{-\lambda x})(k_3e^{\lambda c}+k_4e^{-\lambda c})=c_1e^{\lambda(x+c)}+c_2e^{\lambda(x-c)}+c_3e^{-\lambda(x+c)}+c_4e^{-\lambda(x-c)}$$

Now this part is new, I did not want to confused you but @AlexR is mostly right:

Because of the linearity of the solution we get:

$$u(x,c)=\sum_{\lambda=-\infty}^\infty c_{\lambda,1}e^{\lambda(x+c)}+c_{\lambda,2}e^{\lambda(x-c)}+k_{\lambda,1}e^{i\lambda(x+c)}+k_{\lambda,2}e^{i\lambda(x-c)}$$

Again, this does not answer your question because there are few questions that I did not gave the answer, because I don't know the answer to them yet

Answer 3

What you have is a delayed differential equation. You can look for solutions in the form of exponential functions as above, but with the added condition that $b$ can be complex.

The equation

$$ be^{-bc} = 1 $$

is transcendental. The solution exists in the form of the Lambert W function, as follows

$$ -bc e^{-bc} = -c $$ $$ -bc = W(-c) $$ $$ b = -\frac{1}{c}W(-c) $$

where $b \in \mathbb C$. Due to the multi-valued property of the complex exponential function, there exists infinitely many (complex) solutions in $b$ for any given $c \ne 0, \dfrac{1}{e}$. You can compute them numerically.

You already know that $c=0$ gives $b=1$. The other special case is on the branch point of the complex Lambert W function, $c = \dfrac{1}{e}$, which gives $b=e$

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If you are unfamiliar with complex exponentials, just know that they encompass both real exponentials and real trig functions, due to Euler's identity $$ e^{bx} = e^{(\alpha + i\beta)x} = e^{\alpha x} \big(\cos (\beta x) + i\sin (\beta x)\big) $$

where $\alpha$ and $\beta$ are the real and imaginary parts of $b$, respectively. Then the equation has two linearly independent solutions $$y(x) = c_1 e^{\alpha x}\cos (\beta x) + c_2 e^{\alpha x}\sin (\beta x)$$

for any non-real $b$