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Linear Functionals

  1. Jul 11, 2012 #1
    1. The problem statement, all variables and given/known data

    Why does this not qualify as a linear functional based on the relation ##l(\alpha u+\beta v)=\alpha l(u)+\beta l(v)##?

    ##\displaystyle I(u)=\int_a^b u \frac{du}{dx} dx##

    2. Relevant equations
    where ##\alpha## and ##\beta## are real numbers and ##u## , ##v## are dependant variables.
    3. The attempt at a solution
    If we let ##\displaystyle I(v)=\int_a^b v \frac{dv}{dx} dx##

    then ##l(\alpha u+\beta v)=##

    ##\displaystyle \int_a^b ( \alpha u \frac{du}{dx} dx + \beta v \frac{dv}{dx} dx)=\displaystyle \int_a^b \alpha u \frac{du}{dx} dx +\int_a^b \beta v \frac{dv}{dx} dx=\displaystyle \alpha\int_a^b u \frac{du}{dx} dx +\beta \int_a^b v \frac{dv}{dx} dx=\alpha l(u)+\beta l(v)##...? THanks
     
  2. jcsd
  3. Jul 11, 2012 #2
    [tex] I(\alpha u) = \int_a^b \alpha u \frac{d (\alpha u)}{d x} dx [/tex]
     
  4. Jul 11, 2012 #3

    HallsofIvy

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    Very cursorily, it is not "linear" because it is a product of two terms involving u.

    Your formula for [itex]l(\alpha u+ \beta v)[/itex] is incorrect. You are doing as if [itex](a+ b)^2[/itex] were equal to [itex]a^2+ b^2[/itex] and that is not true.

    You need
    [tex]l(\alpha u+ \beta v)= \int_a^b (\alpha u+ \beta v)\frac{d(\alpha u+ \beta v)}{dx}dx[/tex]

    [tex]= \int_a^b (\alpha u+ \beta v)(\alpha\frac{du}{dx}+ \beta\frac{dv}{dx})dx[/tex]
    [tex]= \alpha^2 \int_a^b u\frac{du}{dx}dx+ \alpha\beta\int_a^b u\frac{dv}{dx}dx+ \alpha\beta\int_a^b v\frac{du}{dx}dx+ \beta^2\int_a^bv\frac{dv}{dx}dx[/tex]
     
  5. Jul 11, 2012 #4
    This is what I thought so too and that this non linearity has nothing to do with the realtion in my first thread...however

    ......in the book (which I am self studying finite element theory) it states

    " a functional ##l(u)## is said to be linear in u iff it satisfies the relation..."

    ##l(\alpha u+\beta v)= \alpha l(u)+\beta l(v)##.....? How is this wrong?
     
  6. Jul 11, 2012 #5

    HallsofIvy

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    Well, as I showed in my first response, it is NOT [itex]\alpha l(u)+ \beta l(v)[/itex],
    it is [itex]\alpha^2l(u)+ \beta^2l(v)[/itex] plus two additional terms!
     
  7. Jul 12, 2012 #6
    Ok, what about this one. Using the same relation ##l(αu+βv)=αl(u)+βl(v)## for a functional ##l(u)=\displaystyle \int_a^b f(x) u dx +c##? The book states this is not a linear functional...? Why? Heres my attempt..

    let ##l(v)=\displaystyle \int_a^b g(x) v dx +d## then the LHS of the relation can be written as

    ##\displaystyle \int_a^b \alpha f(x) u dx +\alpha c + \displaystyle \int_a^b \beta g(x) v dx +\beta d=\alpha (\int_a^b f(x) u dx +c) + \beta (\displaystyle \int_a^b g(x) v dx +d)=αl(u)+βl(v)##...Why is this not a linear functional?
     
  8. Jul 12, 2012 #7
    Nonononononono!

    First of all [itex]l(v)=\displaystyle\int\limits_a^b f(x)vdx+c[/itex], NOT [itex]l(v)=\displaystyle\int\limits_a^b g(x)vdx+d[/itex]. Who give you rights to say [itex]g(x)[/itex] or [itex]d[/itex]? NOBODY!
    [itex]l(\alpha u + \beta v) = \displaystyle\int\limits_a^b f(x)(\alpha u + \beta v)dx + c[/itex], not [itex]+\alpha c + \beta c[/itex]
     
  9. Jul 12, 2012 #8
    ok...the the RHS would end up like
    ## \displaystyle \alpha \int\limits_a^b f(x)u dx + \beta \int\limits_a^b f(x)vdx + c=\alpha l(u)+\beta l(v)+c##

    What about the c though, thats not in the relation?
     
  10. Jul 12, 2012 #9
    This isn't good, because

    [itex]\alpha l(u) + \beta l(v)+c = \alpha \left(\displaystyle\int\limits_a^b f(x)udx+c\right)+\beta \left(\displaystyle\int\limits_a^b f(x)vdx+c\right)+c = \alpha\displaystyle\int\limits_a^b f(x)udx + \alpha c + \beta \displaystyle\int\limits_a^b f(x)vdx + \beta c + c \neq \displaystyle \alpha \int\limits_a^b f(x)u dx + \beta \int\limits_a^b f(x)vdx + c[/itex]
     
  11. Jul 12, 2012 #10
    So it cannot be a linear functional then...right? Looks good.THanks!!
     
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