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I am going through FEM lectures by University of Michigan, and in one of the lectures, the professor writes an ODE:

$$\frac{d^2u}{dx^2}+f(x)= 0, \ \ \ x \in (0,L)$$

Why are these differential equations only satisfied in open intervals, not in closed domains, when boundary conditions are specified? For example, one of the set of boundary conditions could be

$$u(0) = 0 \ \ and \ \ \frac{du}{dx}(x=L) = 0$$

Aakash Gupta
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1 Answers1

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The main reason for this is that the notion of differentiation breaks down in sets that are not open. Recall that the derivative of a function $f$ is given by $$f'(x) = \lim_{y\to x}\frac{f(y)-f(x)}{y-x}.$$ This definition makes no distinction about which direction $y$ approaches $x$ from, and for the limit to exists, it needs to exist for $y\to x^+$ and $y\to x^-$. Obviously on the boundaries of a closed interval, only one of these limits can be computed, so the derivative may not be well defined.

The other reason for this in differential equations is that (loosely speaking) when we integrate a derivative, we get information about the boundary. E.g. $\int_a^b f'(x)dx = f(b)-f(a)$. Also, the endpoints constitute a set of measure zero, which essentially means that $\int_{(0,1)}f(x)dx = \int_{[0,1]}f(x)dx$.

When analyzing differential equations, it is necessary to characterize the behavior at the boundary separately from the differential equation, since if we integrate the differential equation, the actual value of the differential equation at the end points get lost and we are left only with information about the solution at the boundary.

whpowell96
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  • I agree that the definition of the derivative will be affected by having a closed interval, but in the video it mentions that the system will be over constrained, that's why both boundary conditions and differential equation can't be satisfied at the boundary. Look at this, the relevant stuff starts at 1:35 - https://youtu.be/JfdcpgVTrco – Aakash Gupta Dec 13 '18 at 05:43
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    That makes sense. In finite differences this is easier to see. The boundary conditions will create an equation for each of the boundary points and the differential equation creates equations at each interior point. Enforcing both systems at the boundary points will obviously be overdetermined.

    Basically you can't have two different equations for the unknown function at a given point or it will be overdetermined.

     – whpowell96 Dec 13 '18 at 05:49
  • But isn't that the point, that we solve differential equation and get "constants" from the boundary conditions. So, there should be no problem of over determined system as far as I understand – Aakash Gupta Dec 13 '18 at 06:44
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    Exactly. If we required that the differential equations and and the BCs be satisfied at the boundary points, we have 4 equations for 2 unknowns and the system is overdetermined – whpowell96 Dec 13 '18 at 06:46
  • I still can't wrap my head around this thing, I will think for a while and update, thanks for the input – Aakash Gupta Dec 13 '18 at 06:48
  • @whpowell96 Although my reading mostly gave me the same perspective as your answer. I came across, this answer https://math.stackexchange.com/a/504567 were a part from Rudins book was quoted. Which seems to conflict with this view point. – Hjan Apr 23 '20 at 11:29
  • The answer under it in https://math.stackexchange.com/a/410435/471233 seems even more convincing to also formally consider also closed domains for boundary value problems. To specify specify u(x=0) if x is only in (0,1) seems inconsistent to me. – Hjan Apr 23 '20 at 11:43
  • Regardless of the validity of differentiation at $x=0$, boundary information is necessary to ensure uniqueness, and in some cases, existence of BVPs. In the case of OP's problem, you can construct explicit solutions to the problem that satisfy any boundary conditions you want. – whpowell96 Apr 23 '20 at 16:42
  • I suppose that one could view this entire debate in terms of integration, in particular, Stokes' theorem. There, one considers the boundary, which is the same for an open set and its closure. Maybe it is possible to define BVPs in this sense but if at the end of the day all you care about is an integral, then defining things on open sets (which is usually easier) is probably best practice since Stokes' theorem doesn't care about closed vs. open sets if they have the same boundary. – whpowell96 Apr 23 '20 at 16:44
  • Also solutions to differential equations typically have some sort of regularity meaning they can be continuously extended to the boundary easily – whpowell96 Apr 23 '20 at 16:44