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Intuitive understanding of the reference solution (numerical analysis)

  1. Apr 5, 2014 #1

    jjr

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    Hello

    I am having some trouble understanding what the difference is between the reference solution and the "true" solution of a set of certain differential equation. The book I'm working with is Numerical Analysis by Walter Gautschi. Equation 5.16 from this book is a set of first order differential equations with a set of initial conditions: [tex] \frac{d\textbf{y}}{dx} = \textbf{f} (x, \textbf{y} ), a \leq x \leq b; \textbf{y} (a) = \textbf{y}_0. [/tex]

    3 pages later he introduces this reference solution that is confusing me a bit. I quote:
    "[...] we consider the solution [itex] \textbf{u} (t) [/itex] of the differential equation (5.16) passing through the point [itex] (x, \textbf{y} ) [/itex], that is, the local intial value problem [tex] \frac{d\textbf{u}}{dt} = \textbf{f}(t, \textbf{u} ), x \leq t \leq x + h; \textbf{u} (x) = \textbf{y}." [/tex]

    This is for discrete-variable methods of approximation, so h here is the step length in x between the points one is trying to approximate.

    So as I stated in the beginning, I'm having some trouble understanding the difference between these solutions. The function [itex] \textbf{f} [/itex] is the same in both cases, but in the former case it is valid for the whole interval [itex] [a,b] [/itex], whereas in the latter it is only valid locally between [itex] x [/itex] and [itex] x+h [/itex]. Does this mean that one can find a solution of the differential equation 5.16 that is only valid on a small interval, but is not equal to a true solution on the whole interval? Is [itex] \textbf{f} = \textbf{u} [/itex] for the whole interval [itex] [x,x+h] [/itex] or just at the very start of the interval? In the book they are talking about a local description of one-step method, in which one is trying to find an approximation to [itex] \textbf{u} [/itex], which suggests that [itex] \textbf{u} [/itex] is indeed an acceptable solution (but only on a small interval?). So does one use different [itex] \textbf{u} [/itex]'s for every new interval [itex] [x,x+h] [/itex] in one's attempt to reach an approximation to the global solution?

    I hope I made the source of my confusion clear, and I will be happy to elaborate if needed.

    J
     
  2. jcsd
  3. Apr 6, 2014 #2

    AlephZero

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    The difference between the solutions y() and u() is they don't have the same boundary conditions. In general, u() will not satisfy u(a) = y_0.

    If the approximate solution arrives at a point (x,y), which is not exactly on the true solution, the only data you have available to compute the next step is based on the point where you are, not the point where you would have been knowing the true solution. His "reference solution" is an exact solution of the differential equation, but passing through the "wrong" point.

    In general, you can only do an error analysis based on the reference solution, for the (obvious) reason that you don't know what the true solution is. So, you can demonstrate that the "local" error in one step of the numerical solution is bounded in some way, but extending that result to estimate the "global" error involves making some more assumptions about whether two solutions that are "close" to each other at one point stay close together at other points, or whether they diverge from each other.

    I don't have the book so I don't know the context here, but hopefully that will be helpful.
     
  4. Apr 7, 2014 #3

    jjr

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    Yes, that was quite helpful. Thank you!
     
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