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Superimposition principle in solving ODEs

  1. Aug 20, 2012 #1
    Hi all,

    first of all, I have to admit I have often used this richness of knowledge that permeates through the posts of this forum to find answers to questions that I have come across in my studies. Thanks for all! Now, I have a question to post, for the first time.

    I am trying to teach myself diff. equations using the MIT OpenCourseware materials and have bumped into a statement to which I have not found a rigorous answer.

    One of the ways to identify the general solution to an ODE of the form [itex]\dot{x}+p(t)x=q(t)[/itex] is by the sum of a particular solution and the solution to a homogeneous "version" of the differential equation where [itex]q(t)=0[/itex].

    While I can see how the solution "works" by comparing it to a solution derived from the evaluation by integrating factors and the resulting integral expression, I have not been able to prove the principle (and it's bothering me). I also understand that the linearity of ODEs allows us to treat the solution to [itex]\dot{x}+p(t)x=q(t)[/itex] as equivalent to the sum of the solutions of [itex]\dot{x}+p(t)x=q(t)[/itex] and [itex]\dot{x}+p(t)x=C*0[/itex], which is generally referred to a as the superimposition principle.

    I have a hunch that the two solutions are linearly independent and sufficient to span the solution space, but I cannot come up with some formal reasoning to justify it. In other words, why does such a sum always hold? Is it true for all ODEs? Can someone provide some depth to this matter?

    Many thanks for reading.
  2. jcsd
  3. Aug 20, 2012 #2


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    No, the set of all solutions to a non-homogeneous equation do not form a vector space. The set of all solutions to a homogeous d.e. form a vector space which is why it is best to go back to that first.

    A simple analogy is this- the set of points, (x, y), in a coordinate system form a (two dimensional) vector space with "coordinatewise" addition and scalar multiplication. The points on a straight line through the origin form a subspace. Points on a line not through the origin do NOT form a vector space. For example, [itex](x_1, x_1+1)[/itex] and [itex](x_2, x_2+1)[/itex] are in the set of all (x, y) such that y= x+ 1. But the sum [itex](x_1+ x_2, x_1+ x_2+ 2)[/itex] does not satisfy that equation (it satisfies y= x+ 2) nor does the scalar product [itex]a(x, x+1)= (ax, ax+ a)[/itex] (it satisfies y= x+ a).

    But we can look at the line through the origin parallel to that one: y= x. Choose any single point on the line y= x+ 1, (0, 1), say, or (1, 2). Any point on y= x+ 1 can be written as a point on y= x plus that single point: (x, x+ 1)= (x, x)+ (0, 1) or (x, x+ 1)= (x-1, x-1)+ (1, 2).

    Similarly, If "L" is an nth order homogeneous differential operation, then the set of all solutions to Ly= 0 forms and n-dimensional vector space. Now suppose that [itex]y_1(x)[/itex] and [itex]y_2(x)[/itex] both satisfy Ly= f(x) for some function f. Then if we let [itex]y(x)= y_1(x)- y_2(x)[/itex] we have [itex]L(y_1- y_2)= L(y_1)- L(y_2)= f(x)- f(x)= 0[/itex]. That is, the difference between any two solutions satisfies Ly= 0 and so can be written as a linear combination of n independent solutions to the Ly= 0. Let's call that [itex]y_a[/itex]. The [itex]y_1(x)- y_2(x)= y_a(x)[/itex] from which [itex]y_1(x)= y_a(x)+ y_2(x)[/itex]. Since [itex]y_1[/itex] can be any solution to the entire equation, that shows that any solution to the entire equation can be written as a solution to the homogeneous equation plus this [itex]y_2(x)[/itex], a solution to the homogeneous equation.
    Last edited by a moderator: Aug 20, 2012
  4. Aug 20, 2012 #3


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    Actually, it's the superposition principle: super + position. Superimposition principle sounds like the linear sum of a bunch of inconvenient burdens to form a super-inconvenient burden. ;)

    (The term superimposition is also an actual printing press term, referring to "the placement of an image or video on top of an already-existing image or video, usually to add to the overall image effect, but also sometimes to conceal something (such as when a different face is superimposed over the original face in a photograph).")
  5. Aug 20, 2012 #4
    @ HallsofIvy: Thanks for the clear explanation.

    @ Mute: <embarrassed> I should have read more carfully. I got carried away by "superimpose", which is used commonly in my mother tongue, Italian.
  6. Aug 20, 2012 #5


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    Your English is far better than my Italian (or just about any language).
  7. Aug 20, 2012 #6


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    No need to be embarrassed, it's an understandable mistake anyone could have made. I pointed it out because I didn't want you to be googling "superimposition principle" and wondering why you were only finding information about printing presses!

    Although actually, google.com assumes you are looking for "superposition principle" and gives results based on that phrase, too. However, the second hit is a page allegedly about physics which uses the term "superimposition principle", but a quick glance at the website shows that the author seems to have his own "fringe" theories of unified physics. So I guess that's another reason to point out the error - you might stumble upon pages by people promoting their own fringe theories.
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