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Does this assumption cause problems in (many) cases?

  1. Sep 13, 2015 #1
    I was just looking over the method of variation of parameters again and remembered one part that made me always wonder if this method can be improved in any way. Here's a link for reference: http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx

    Now everything stated seems to make sense, but I am wondering if this assumption is known to make this method ineffective at times or not:

    ## u_{1}^{'} y_{1} + u_{2} ^{'} y_{2} = 0 ##

    If I'm not mistaken, this is simply done to simplify the equations later on. That is perfectly fine, but I'm just wondering, doesn't this restriction impose unnecessary constraints on the solutions? By doing this, aren't we only restricting ourselves to a few classes of solutions instead of all of them? In some cases, isn't it possible that:

    ## u_{1}^{'} y_{1} + u_{2} ^{'} y_{2} = C ## where ## C \neq 0 ##?

    I guess I just never quite understood the validity of this assumption and am wondering if this causes any major consequences when using this method. Any explanations for why or why not would be greatly appreciated!
     
  2. jcsd
  3. Sep 14, 2015 #2
    Keep in mind that we only have to find one particular solution once we know the general solution to the homogeneous equation. We don't need to find all possible functions u1 u2, we just need to find 1 pair of functions.

    Now consider the case where we make no assumptions on u1,u2. When you do this you'll eventually end up with 2 coupled second order inhomogeneous differential equations for u1 and u2. Our original goal was to solve 1 inhomogeneous differential equation. So in the absence of a simplifying assumptions we've actually made the problem harder!

    To avoid second order equations we have to make the general assumption*

    [itex] u_1' y_1 + u_2' y_2 =h(t) [/itex] where h(t) should be chosen to make the math as easy as possible. With this assumption you will get 2 coupled first order equations for u1 and u2. Thus we've reduced the problem to one we can solve.

    Now using an arbitrary h(t) you can turn the crank and eventually express both u1 and u2 as integrals that depend on y1,y2, y1',y2', g(t), q(t), h(t) and h'(t). The resulting integrals are sums over 4 terms, three of which depend on h(t) or h'(t). Setting h(t)=0 will give eliminate 3 of the terms giving us one integral to perform for each u. Setting h(t) to a constant will eliminate 1 of terms and give us 3 integrals for each u.

    In general integrating once is easier than integrating three or four times. However, there might be a few situations where a clever choice of h(t) will simplify the resulting integrals.

    *One might also imagine other assumptions that would result in 2 second order equations for u1 and u2 that are "easy" to solve. However, it's not clear to me how one would go about doing this is general.
     
  4. Sep 14, 2015 #3
    Thank you. That clarifies things a lot. I guess I was just wondering if using the assumption [itex] u_1' y_1 + u_2' y_2 = 0 [/itex] (or ## C ##) can restrict solutions, though. It definitely may help in simplifying the problem for most cases. I have not been able to come up with any in particular, but I can't help but think imposing these restrictions would render solving a DE impossible simply because there are no pairs of functions that satisfy the constraint. In general, this seems like a valid approach, but I guess I just don't recognize this method as being fully capable of solving EVERY inhomogeneous linear ordinary differential equations,
     
  5. Sep 14, 2015 #4
    If the ode is sufficently well behaved the uniqueness and existence apply then we know that any method that produces one particular solution is sufficent. The variation of parameters will always return a solution provided that the integrals for u1 and u2 are well defined.

    You could imagine situations where the forcing function g(t) is so ugly that the integrals for u1 and u2 aren't defined. Here the assumption u1'y1+u2'y2=0 is not the problem. Also, in such cases existence and uniqueness are probably not guarenteed.
     
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