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How do we define "linear" for single and multivariable?

  1. Mar 2, 2015 #1

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    In my multivariable calculus class, we briefly went over Taylor polynomial approximations for functions of two variables. My professor said that the second degree terms include any of the following:

    $$x^2, y^2, xy$$

    What surprised me was the fact that [itex]xy[/itex] was listed as a nonlinear term.

    In my Ordinary Differential Equations class, an example of a linear differential equation was

    $$y'=xy$$

    I was stuck wondering how one case could be linear and the other polynomial. I suppose the differences between the examples comes from the fact that the former example dealt with a function of two variables whereas the latter is only in terms of one.

    If anyone can provide an explanation as to why these two examples don't contradict each other, or on a broader scale, how the term "linear" is defined with more than one variable, I would greatly appreciate it.
     
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  3. Mar 2, 2015 #2

    Simon Bridge

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    The term "linear" does not mean anything by itself, it has to refer to something... so an equation may be linear in non linear terms.
    The term xy is 2nd order, but the equation z=xy is linear in x, and also linear in y.
    I don't think we would commonly see z=xy as a 2nd order equation, but if you evaluate z along a line like x=y you'll see the quadratic.

    I think your example is of a 1st order non linear DE: a separable equation.
    http://tutorial.math.lamar.edu/Classes/DE/Separable.aspx
     
    Last edited: Mar 2, 2015
  4. Mar 2, 2015 #3

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    Thanks, Simon!

    The fact that the terms in the DE are not linear clarifies some things.
     
  5. Mar 2, 2015 #4

    Simon Bridge

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    I didnt read your post properly and edited my reply to cope.
     
  6. Mar 2, 2015 #5

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    Thanks for the updated feedback, Simon.

    According to WolframAlpha, the DE is a first-order linear ordinary differential equation
    http://www.wolframalpha.com/input/?i=y'=xy

    Also, could you please define second order? Is the term in this given context synonymous with second degree?
     
  7. Mar 2, 2015 #6

    Simon Bridge

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    order and degree are often interchanged in this context, it does not pay to be rigid about semantics, it is the maths that matters.
     
  8. Mar 2, 2015 #7

    mathwonk

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    i think "linear" usually means linear in the unknown function, which here is y. so this equation can be written as y' - xy = 0, or (D-x)y=0. the fact that it is linear in y is equivalent to the fact that sums of solutions y1+y2 = y are again solutions and also scalar multiples cy1 = y of solutions, are again solutions.
     
  9. Mar 2, 2015 #8

    Simon Bridge

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    I wasnt too happy with Pauls approach, the notes never actually define the terms... looking closely, his nonlinear general form is a form of his linear one. Probably need to check the examples. Its generally a good resource.

    I'd normally take a DE to be linear if the DE operator is linear.
    When applying the linear definition (superposition) you have to work out what counts as addition etc.
    Easy to do with functions of one variable, with two or more, you can usually treat them as components of a vector. From this you see how the term may apply varioisly depending on which part of the equation you are interested in.

    The way to understand what you are being told is to look at the context. What exactly is being called linear in each case.
     
  10. Mar 2, 2015 #9

    Simon Bridge

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    take f(x, y)=xy

    if f is linear then f((a, b)+(c, d)) =f(a, b)+ f(c, d)
    i.e. (a+c)(b+d) = ab + cd

    In the context of a polynomial expansion in more than one variable, your confusion is in the name for the degree or order of the term. The order (as in "calculate to nth order") is usually the number of independent variables multiplied together that make up the variable part of the term. I.e if all the variables were x, what power would it be? This is separate from considerations of linearity.

    Another way, if xy were 1st order, then what order would xy2 be?
     
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