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General Question About a System of DE's

  1. Oct 29, 2015 #1
    We were looking at some examples of systems of differential equations in class a couple days ago, and some of them were simple looking systems. One such was

    ##\frac{dr}{dt} = 4j##
    ##\frac{dj}{dt} = r##

    Why is it that we can't treat this as a scenario where both sides of each equation could just be integrated immediately or something like in physics where the first equation would become ##dr = 4jdt## and then integrate? (I have a question current posted in the calculus section about splitting the derivative if anyone cares to answer when this technique DOES work)
     
  2. jcsd
  3. Oct 29, 2015 #2

    Mark44

    Staff: Mentor

    If j and r were constants, you could do this, but the implication is that both j and r are unknown functions of t.
    It's exactly the same as trying to evaluate this integral: ##\int f(x) dx##, without knowing anything about f.

    Also, if this were your system...
    ##\frac{dr}{dt} = 4r##
    ##\frac{dj}{dt} = j##

    ... then you could separate each equation like so:
    ##dr = 4rdt \Rightarrow \frac{dr}r = 4 dt \Rightarrow \int \frac{dr}{r} = 4 \int dt \Rightarrow \ln|r| = 4t + C##
    Exponentiating both sides would yield ##|r| = Ae^{4t}##, where ##A = e^C##.
    The other equation could be solved in a like manner.
     
  4. Oct 30, 2015 #3
    Perfect, that explanation helped a lot. Thanks!
     
  5. Oct 30, 2015 #4
    Well couldn't you rewrite on vector form and then integrate to get a matrix exponential?
    Set ## v =(r,j) ## such that we get the equation
    ## \frac{d}{dt}v = (0,4;1,0)v ##,
    and then integrate.
     
  6. Oct 30, 2015 #5

    Mark44

    Staff: Mentor

    It's a lot more complicated than this. If you integrate as you suggest you get a solution that contains this factor:
    $$e^{\begin{bmatrix} 0 & 4 \\ 1 & 0 \end{bmatrix}~t}$$
    Expanding that exponential is difficult, as it involves an infinite number of terms, each requiring a higher power of that matrix.

    The usual technique involves finding a diagonal matrix D that is similar to the matrix above; that is, a matrix ##D = P^{-1}AP##, where the eigenvalues of matrix A appear on the diagonal in D, and the eigenvectors of A are the columns of P. For more information, see any differential equations textbook that has a section on solution of systems of diff. equations using matrix diagonalization.
     
  7. Oct 30, 2015 #6
    As far as I know solving simple matrix exponentials is taught very early on at universities. I would not call it more complicated but certainly less used.
     
  8. Oct 30, 2015 #7

    Mark44

    Staff: Mentor

    Yes, that's about right.
    It's quite complicated in comparison to matrix addition and multiplication, and even finding the inverse of a matrix. For diagonalization, you need to find the eigenvalues, find the eigenvectors, form a matrix whose columns are the eigenvectors, find the inverse of that matrix, and so on.
     
  9. Oct 31, 2015 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    And, not all matrices can be diagonalized.
     
  10. Oct 31, 2015 #9
    But you can still find the matrix exponential. At any rate. The question was why can't we just integrate both sides and my answer is: you can but you will have to solve something different instead.
     
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