How to proceed with the differential equation $x^{\left(x\right)}=0$, where $x\in\mathbb{N}$? I was thinking of some differential equation to do and I came up with this equation, parentheses mean how many times the function is derived. The first thing I did was the Laplace transform on both sides. (Here I will use the capital letter L to express the Laplace transform)$$L(x^{\left(x\right)})=L(0)$$Knowing from the Laplace Transform Table, $L(0)=\frac{0}{s}=0$, and that $L(x^{\left(y\right)})=s^yX-\sum _{n=1}^ys^{y-n}x^{\left(n-1\right)}\left(0\right)$, replacing x and y with x and putting it in the equation, we have $$s^xX-\sum _{n=1}^xs^{x-n}x^{\left(n-1\right)}\left(0\right)=0$$Isolating the X we have$$X=\frac{\sum _{n=1}^xs^{x-n}x^{\left(n-1\right)}\left(0\right)}{s^x}$$Cutting the top and bottom $s^x$, we have$$X=\sum _{n=1}^x\frac{x^{\left(n-1\right)}\left(0\right)}{s^n}$$Now doing the Inverse Laplace Transform on both sides we have$$L^{-1}\left(X\right)=L^{-1}\left(\sum _{n=1}^x\frac{x^{\left(n-1\right)}\left(0\right)}{s^n}\right)$$The Laplace Inverse of Laplace Transform is the original function, so $L^{-1}\left(X\right)=x$, and $L^{-1}\left(\sum _{n=1}^x\frac{x^{\left(n-1\right)}\left(0\right)}{s^n}\right)=\sum _{n=1}^xx^{\left(n-1\right)}\left(0\right)L^{-1}\left(\frac{1}{s^n}\right)$, knowing that x is greater than or equal to 0 and n goes from 1 to x, so n is a number that fits the rule $L^{-1}\left(\frac{\left(n-1\right)!}{s^n}\right)=t^{n-1}$, knowing that $L^{-1}\left(\frac{1}{s^n}\right)=\frac{L^{-1}\left(\frac{\left(n-1\right)!}{s^n}\right)}{\left(n-1\right)!}$, finally we have to $L^{-1}\left(\frac{1}{s^n}\right)=\frac{t^{n-1}}{\left(n-1\right)!}$, replacing these transforms in the equation we have$$x=\sum _{n=1}^x\frac{t^{n-1}x^{\left(n-1\right)}\left(0\right)}{\left(n-1\right)!}$$Unfortunately I don't know how to proceed from here, so can someone please help me?
 A: If $x$ is a natural valued function, but can take a real valued input, then it should be clear that $x$ is locally constant anywhere it is locally continuous. To see why, consider what a continuous $x$ would look like if it wasn't constant, if it went from $1$ to $2$, it would have to pass through $1.5$ to get there, which isn't allowed because $1.5$ isn't a natural number.
Since the derivative is only defined where the function is continuous, and our function is constant anywhere it is continuous, we only need to consider the derivative of constant valued functions. If $x = 0$ then $x^{(0)} = x = 0$. If $x > 0$ then $x^{(x)} = 0$ because $x$ is a constant being differentiated at least once.
Since every natural number constant satisfies this equation, and every continuous piece of a natural valued function is constant, then every natural valued function satisfies this equation wherever it is continuous. If $x$ is continuous everywhere, it is constant everywhere, and thus is a solution everywhere.
