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To solve a a DE rigorously, I'm not really sure how to set up the problem, in terms of definitions and premises.

So as a simple example, suppose we're given a function $f : \mathbb{R}^2 \rightarrow \mathbb{R},$ and we wish find the general solution to the first-order DE of the form $y'=f(x,y).$ My questions are as follows.

  1. Should we immediately define a function $S$ that takes a set $X$ and returns the set of all solutions on $X$, denoted $S_X$? In the sense of: $S_X = \{y : X \rightarrow \mathbb{R} \mid y'=f(x,y), \;y \mbox{ diff}\}.$

  2. Should we assume that we have a set $X$ (on which we're looking for solutions) that is fixed but arbitrary? If so, should we furthermore assume that $X$ is an interval? That its open? Does it need to be non-empty and/or have two or more elements?

  3. If the answer to any of the above questions is 'yes,' should we furthermore assume that we have a function $y : X \rightarrow \mathbb{R}$ that is fixed but arbitrary? If so, should we furthermore assume that $y$ is differentiable? That $y$ satisfies the DE? Note that, if the answer to this final question is 'yes', we can write this more simply as "Assume $y \in S_X$" so long as we've defined $S_X$.

  4. Is there anything else that needs to be done at the outset?

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  • $\begingroup$ If people are being careful, they'll tell you exactly what they want when they ask you to solve a differential equation. For example they might give you an open interval $I \subset \mathbb R$, a number $t_0 \in I$, a number $y_0 \in \mathbb R$, and a continuous function $f:I \times \mathbb R \to \mathbb R$, and ask you to find a differentiable function $y:I \to \mathbb R$ such that $y'(t) = f(t,y(t))$ for all $t \in I$ and $y(t_0) = y_0$. Or something else like that. If you read about existence and uniqueness theorems for ODEs, hypotheses like this are stated carefully. $\endgroup$ – littleO Aug 8 '13 at 6:06
  • $\begingroup$ @littleO, I don't think that's an efficient way of going about it. Then you're solving every equation anew every time. You're better off first finding the general solution, and then plucking out the particular solution you're looking for. That way, if you're later looking for another particular solution, you don't have to do all that work all over again. So, I still don't know how to proceed to find the general solution. $\endgroup$ – goblin Aug 8 '13 at 6:17
  • $\begingroup$ Yes, perhaps. Two comments that don't directly answer your question: 1) keep in mind that often there may not be any formula for a solution that you can write down. But given an initial condition, we can still solve the ODE numerically. $\endgroup$ – littleO Aug 8 '13 at 6:29
  • $\begingroup$ 2) The standard existence and uniqueness theorems for ODEs shed some light on the situation here. Assume your $f$ is continuous. Given $t_0,y_0 \in \mathbb R$, there exists an open interval containing $t_0$ on which a solution to the ODE satisfying $y(t_0) = y_0$ exists. Let $I$ be the largest such interval. On this interval $I$, the solution (satisfying $y(t_0) = y_0$) is unique. (Please correct me if I'm stating these theorems wrong.) $\endgroup$ – littleO Aug 8 '13 at 6:30
  • $\begingroup$ @littleO, it sounds about right, I don't know the theorems very well myself. So perhaps we should be assuming that $X$ is a maximum interval on which the solution exists? But I don't really think this is the right way to set the problem up, since it doesn't generalize to problems whose solutions have domains of dimension 2-or-higher. $\endgroup$ – goblin Aug 8 '13 at 6:34
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Is there anything else that needs to be done at the outset?

First thing to do is to ask yourself why you want to solve this equation. Where does the equation come from? Where will solving it lead you? The choice of how to interpret "solution" depends on the above. Here are some options to choose from:

  1. $y$ is a solution of $y'=f(x,y)$ if $y$ is differentiable at every point and $y'=f(x,y)$ holds at every point.

  2. $y$ is a solution of $y'=f(x,y)$ if $y$ is absolutely continuous, and its derivative $y'$ (which is defined almost everywhere) satisfies $y'=f(x,y)$ at almost every point (that is, outside of a set of measure zero).

  3. $y$ is a solution of $y'=f(x,y)$ is it is a distribution for which $f(x,y)$ makes sense and coincides with the distributional derivative of $y$.

For example, $y'=H(x)$ (where $H$ is the Heaviside function) has no solutions in the sense 1, but has solutions in the sense 2 or 3. If the equation is $y'=y^2+H(x)$, then 3 is not a good option because we can't square distributions in general. This leaves 2. But if we want to allow discontinuous solutions (which may be necessary when the $\delta$-function gets involved), then 2 is not an option but 3 could be. In Finding Weak Solutions to ODEs you will find other examples of solutions that are interpreted in sense 3.

In general, I think your questions stem from an excessively formal approach to ODE, from putting rigour ahead of practicality. Most of analysis is amorphous and does not naturally assume such crystalline form.

Should we immediately define a function $S$ that takes a set $X$ and returns the set of all solutions on $X$, denoted $S_X$?

No, unless you can think of a way how this can be useful. I can't.

Should we assume that we have a set $X$ (on which we're looking for solutions) that is fixed but arbitrary? If so, should we furthermore assume that X is an interval? That it's open? Does it need to be non-empty and/or have two or more elements?

Do you have any use for a solution whose domain is the empty set? Probably not.

It turns out to be convenient to look for solutions defined on some open interval containing a point of interest. One need not specify this interval in advance. The interval of existence of a solution is not easy to find (as this unanswered question demonstrates), and it may well be different for different solutions. For example, the solution of $y'=y^2$ with $y(0)>0$ has the interval of existence $(-\infty, 1/y(0))$.

Should we assume...

Assume whatever you want, if it helps you. The results you obtain will be based on those assumptions. So, the results will be conditional until and unless you can demonstrate that the assumptions indeed hold. Whether or not any particular assumption, such as differentiability, is helpful, depends on the context.

For example, you can assume that the equation has a solution that is locally represented by a power series. Based on this assumption, you may be able to obtain a power series formula for the solution. At this stage the result is conditional, because it uses an assumption the validity of which is not known. But if you prove that the power series converges and its sum indeed satisfies the equation, the assumption will be validated.

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