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Homework Help: Initial Value problem

  1. Aug 2, 2009 #1
    What is the general rule for determining how many initial conditions one needs to uniquely determine a solution of a set of ODEs or PDEs?

    is it simply the number of derivatives? How about for coupled differential equations?


  2. jcsd
  3. Aug 2, 2009 #2
    From what I understand you are correct, an equation with n derivatives requires n initial conditions. Not too sure if the same applies for coupled DEs though.
  4. Aug 2, 2009 #3
    Generally speaking, and I mean as general as you can get, the number of initial conditions needed is equal to the order of the DE. Idk anything about PDE's but that rule holds true for all linear, homogeneous ODEs. Non-linear DEs often are unsolvable. It holds true for systems of ODEs, also.
  5. Aug 2, 2009 #4
    HallsofIvy posted this in another forum:

    The set of all solutions to any second order, linear, homogeneous, equation forms a vector space of dimension two. That means, if you can find two independent solutions (that form a basis for the space) you can write any solution as a linear combination of the two solutions. That's all [tex]y(x)= c_1y_1(x)+ c_2y_2(x)[/tex] means.

    You can take that beyond second order and say for all linear, homogeneous equations.
  6. Aug 2, 2009 #5
    I guess my question is really in regards to a set of coupled differential equations, could you have a set of say N coupled nonlinear ODEs which have a unique solution without specifying any initial conditions?


  7. Aug 3, 2009 #6


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    No, solving a single nth order equation will involve n unknown constants. Solving, say, m nth order equations will involve mn unknown constants. More generally, if you have a system of m differential equations, each of order [itex]n_1, n_2, \cdot\cdot\cdot, n_m[/itex], the solution will involve [itex]n_1+ n_2+ \cdot\cdot\cdot+ n_m[/itex] unknown constants that have to be determined by additional conditions.
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