# ParametricNDSolve without initial condition?

For example,

ParametricNDSolve[{y'[t] == a y[t]}, y, t, {a}]


returns a ParametricFunction which is actually inexplicable:

How to understand it?

Edited: It should be noticed that it differs from

ParametricNDSolve[{y'[t] == a y[t], y[0] == 1}, y, {t, 0, 10}, {a}]
y[1] /. sol


since

ParametricNDSolve[{y'[t] == a y[t]}, y, t, {a}]
y[1] /. sol


returns warning messages:

1. ParametricNDSolve::ndlim: Range specification t is not of the form {x, xend} or {x, xmin, xmax}.

2. ParametricNDSolve::ndnco: The number of constraints (0) (initial conditions) is not equal to the total differential order of the system plus the number of discrete variables (1).

I wonder why ParametricNDSolve runs on though there isn't any initial conditions or boundary conditions.

• To start understanding it: what happens if you try evaluating it? May 18 '15 at 2:56
• @J. M. Thank you for your attention! I don't know how to use it correctly. May 18 '15 at 2:56
• So, try something like ytest = y[1] /. sol, and then maybe try ytest[0]. May 18 '15 at 3:04
• But help shows how to use the result? The examples all show how to use it. Is the question about how to use the result or something else? May 18 '15 at 3:08
• @WateSoyan I don't think there is anything wrong; the ParametricFunction object is constructed without actually solving the equation. Once we try to evaluate it for a numeric value of the parameter a, we see that there is an insufficient number of initial conditions. May 18 '15 at 13:00

ParametricNDSolve is a numerical solver. You need to give it enough numerical values for the parameters and domain that the solver can come up with a unique InterpolatingFunction representing the numerical solution of your equation.

First, your differential equation of course has a family of $y(t)$ functions as solutions. How can ParametricNDSolve choose one among those? You need to give the numerical solver boundary conditions. In this case I chose y[0] == 1 arbitrarily.

Similarly, you need to specify over which interval of the independent variable you would like a solution.

Putting those together:

sol = ParametricNDSolve[{y'[t] == a y[t], y[0] == 1}, y, {t, 0, 10}, {a}]


Let us now choose an explicit value for the parameter $a$, e.g. $0.3$:

yinstance = y[0.3] /. sol


This returns:

Now we can plot this InterpolatingFunction, with $t$ ranging over the $(0,10)$ interval as specified in the call to the solver:

Plot[yinstance[t], {t, 0, 10}]


• Aw, I was hoping the OP would try the experiment himself of trying to evaluate a ParametricFunction[] sans initial conditions… :D (+1 nevertheless) May 18 '15 at 3:28
• Sorry.I have posted my comparison between that with initial conditions or boundary conditions and that without them.But I wonder why ParametricNDSolve runs on though there isn't any initial conditions or boundary conditions... May 18 '15 at 3:30
• @J. M. oopsie! and here I come, like the proverbial bull in the china shop, and spoil all the fun! :-) I'll have to be honest though: those error messages are really not the most illuminating, and they have driven me up the wall a few times as well... May 18 '15 at 3:31
• @WateSoyan I think that the catch is that no evaluation is actually attempted until you "force" it, e.g. by asking for a value etc. Having said that, I wholeheartedly agree with you there: it could fail A LOT more gracefully than that. I guess this is one more example of what I have seen the pro's refer to as Mathematica's "late to fail" philosophy. May 18 '15 at 3:33
• @WateSoyan I guess that in a way it is a kind attitude from the software's standpoint. It won't fail and throw up on you until it's really sure that you are without hope ;-) May 18 '15 at 3:35