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Following some comments from another question I wanted to ask: does an explicit method always require some sort of analytical function/solution?

Let's take Euler for example. You have a function $f$ which takes $y$ and $t$ and is equal to $y'$. But there is no "generic" or one $f$ as far as I know, you need an analytical solution from your problem. If I am computing velocity, perhaps it is $f(y, t) = y/t = \frac{dy(t)}{dt}$.

Suppose I do not have an analytic solution to a derivative of a particular function (or for a higher-order derivative). Can I not use an explicit method? Or am I missing something?

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    $\begingroup$ Not sure I fully understand the question, but for an ODE $y_t = F(t,y)$ you need the RHS function F. The function F does not have to be analytic, it can be a "black box" that takes numerical values of t and y and returns numerical values of F. $\endgroup$
    – Maxim Umansky
    Commented Feb 25, 2023 at 2:06

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You are confused about ODEs. You think that in order to solve an ODE, you need to know what $f=y'$ is. But it's the other way around: When you try to model something in the real work, you ask "how does a quantity $y(t)$ change with time?" Modeling how this quantity changes with time then gives you the $f$ in $$ y'(t) = f(t,y(t)), $$ and then you can in a next step solve the equation to obtain a function $y(t)$ that represents what $y(t)$ will be in the future.

In other words, to know $f$ you don't need to know $y$. But in order to know $y$, you need to know what $f$ is.

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  • $\begingroup$ That is what I mean, $y(t)$ can be a number, but we must have an analytic $f$ which describes how $y$ changes. So if I only know, for example $\psi(x) = \frac{-\hbar}{2m}\frac{d^2\psi}{dx^2} + V(x)\psi$ and I dont know $\frac{d^2\psi}{dx^2}$ I cant use an explicit method, right? $\endgroup$
    – cgbsu
    Commented Feb 26, 2023 at 2:08
  • $\begingroup$ Oh wait, so if I have the TISE, like above, and I solve $\frac{d^2\psi}{dx^2} = \psi - V(x)\psi$ as long as I have $\psi$ I know the relation! Is that it? If so I feel kinda dumb haha $\endgroup$
    – cgbsu
    Commented Feb 26, 2023 at 2:48
  • $\begingroup$ With regard too my last comment I tried it out, maybe it works for another solution, like velocity $v' = (dv/dt)$ however Schrodinger equation itself needs to be normalized so that particular solution I typed doesn't really work. Also I forgot E in it. But is my basic understanding correct here do you think? $\endgroup$
    – cgbsu
    Commented Feb 26, 2023 at 6:05
  • $\begingroup$ No, you still miss the idea. Think about population growth, as an example. Let's say that in every year, 1/80th of all people dies and 1/40th of all people has a child. Then if the number of people at time $t$ is $y(t)$, then the rate of change $y'(t)$ is $-y(t)/80 + y(t)/40$. In other words, you get the ODE $y'(t) = (-1/80+1/40)y(t)$ and the right hand side of your ODE will simply be $f(t,y)=(-1/80+1/40)y(t)$. $\endgroup$
    – Wolfgang Bangerth
    Commented Feb 26, 2023 at 10:36
  • $\begingroup$ Okay, yes, my bad, I wrote $v' = (dv/dt)$ as a function of $dv/dt$ instead of $v$. But ignoring normalization, is what I said about the Schrodinger equation in the previous comments roughly what I you said? I wrote $\frac{d^s\psi}{dx^2}$ as a function of $\psi$ and $x$ (though it should be a time dependent solution), which I could then take and plug into $\psi_{n+1} = \psi_n + step \cdot (\psi_n \cdot (E - V))$ where the ladder part is $f$ after $\psi_n +$. In your example you seem to do something similar, you don't take an analytic y then derive it to find $y'(t)$ you write f in terms of y $\endgroup$
    – cgbsu
    Commented Feb 26, 2023 at 19:17

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