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Nonlinear operators

  1. Jul 18, 2006 #1
    Can someone give me an example of a nonlinear operator? My textbooks always proves that some operator is a linear operator, but I don't think I really know what a nonlinear operator looks like.

    One of my books defines an operator like [tex]\hat{B} \psi = \psi^2[/tex]. I see that this is a nonlinear operator because:

    [tex]\hat{B} (\psi_1 + \psi_2) = (\psi_1 + \psi_2)^2[/tex]

    ...and this is different from [tex]\psi_1^2 + \psi_2^2[/tex] which you would get by letting the operator B act on each function. But how can you define an operator like this? What would the mathematical form of such an operator be?
  2. jcsd
  3. Jul 18, 2006 #2


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    An operator that is not linear?? In other words, an operator, A, is nonlinear if and only if there exist some vectors, u and v, and some numbers, a and b, such that A(au+ bv) is NOT equal to aA(u)+ bB(v). That's a perfectly good "mathematical definition". I not sure what you mean by a "mathematical form". What is the "mathematical form" of a linear operator?

    Notice that if A(v)= v2[/sub] (assuming a one dimensional vector space for which squaring makes sense) then A(au+ bv)= (au+ bv)2= a2u2+ 2abuv+ b2v2 which is NOT the same as aA(u)+ bA(v)= au2+ bv2.

    A two-dimensional example might be A((x,y))= (x2,y2). Then A(a(x,y)+ b(p,q))= A((ax+bp,ay+ bq))= (ax+bp)2, (ay+ bq)2= (a2x2+2abxp+ b2p2,a2y2+ 2abyq+ b2q2) which is not the same as aA(x,y)+ bA(p,q)= a(x2,y2)+ b(p2,q2)= (ax2+ bp2,ay2+ bq2).

    Linear problems have the nice property that you can "take them apart", solve the simpler parts, and put those back together to get a solution to the original problem. With "non-linear" problems you can't do that. Essentially, "Linear Algebra" is the study of linear problems and so you very seldom have anything to do with non-linear operators. That is a much harder study!
  4. Jul 18, 2006 #3

    matt grime

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    they look like absolutely anything that is not linear. They are just arbitrary functions between spaces. f(x)=ax for some a are the only linear operators from R to R, for example, any other function, such as sin, x^2, log(x) and all the functions you know and love are non-linear operators.

    You just did define it.
  5. Jul 18, 2006 #4
    Thanks alot for the quick reply! What I meant was, what is the mathematical form of the operator [tex]\hat{B}[/tex] that I introduced? For example, the form of the momentum operator [tex]\hat{p}[/tex] in quamtum mechanics is [tex] - i \hbar \frac{d}{dx}[/tex], and the form of the laplace operator is:

    [tex]\frac{d^2}{dx^2} + \frac{d^2}{dy^2} + \frac{d^2}{dz^2}[/tex]

    But what is the form of the operator B that I introduced (I believe you called it A)? What would I write instead of the operator?
  6. Jul 18, 2006 #5


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    An operator is just a function. In this case, it sends one function to another function. For example, d/dx sends the function x^2 to the function 2x. But any well defined function works as an operator, even if we have not assigned a nice symbol like d/dx to it, and the operator you defined in your first post is perfectly valid.
  7. Jul 18, 2006 #6
    Thanks to all of you, I understand it now! :-)
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