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Bi Linear Functionals and Symmetry

  1. Jul 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that ## \displaystyle B_1(u,v)=\int_a^b (p(x) u \cdot v + q(x) \frac{du}{dx} \cdot v)dx## is a bilinear functional and is NOT symmetric
    1. The problem statement, all variables and given/known data

    Bilinear relation ##B(\alpha u_1+\beta u_2,v)=\alpha B(u_1,v) +\beta B(u_2,v)## (1)
    ##B(u, \alpha v_1+ \beta v_2)=\alpha B(u,v_1) +\beta B(u,v_2)## (2)


    Symmetry ##B(u,v)=B(v,u)##

    where u and v are vectors and ##u\cdot v## is the dot product

    2. Relevant equations

    ##\displaystyle B(\alpha u_1+\beta u_2,v)=\int_a^b \left( p(x)(\alpha u_1+\beta u_2) \cdot (v)+q(x) \frac{d}{dx} (\alpha u_1+\beta u_2) \cdot v \right) dx##

    ## \displaystyle=\int_a^b \left ((p(x)\alpha u_1 \cdot v+p(x) \beta u_2 \cdot v +q(x) \alpha \frac{d u_1}{dx} \cdot v+q(x) \beta \frac{ d u_2}{dx}\cdot v \right) dx## which after re-arranging gives

    ##\alpha B(u_1,v) +\beta B(u_2,v)##

    Similarly for the RHS of (2). THis shows it is bilinear.

    However, I am not sure I understand the concept of symmetry in this case that it is not symmetric?
     
  2. jcsd
  3. Jul 12, 2012 #2
    You just need to show that B(u,v) ≠ B(v,u), aka showing that it is not true that B(u,v)=B(v,u)
    There's nothing more than plugging in involved here, what is it exactly you're having problems with?
     
  4. Jul 12, 2012 #3
    Is ##B(v,u)## to be written in the form ##B(\alpha v_1+\beta v_2,u)##?

    I dont get the idea of interchanging the vectors since ##u \cdot v= v \cdot u##....?
     
  5. Jul 12, 2012 #4
    You've already shown that it's bilinear so you now only need to show that it's not symmetric;
    Since you're only trying to show that B isn't symmetric you don't need to bother about plugging linear combinations in, you only need to compare B(u,v) with B(v,u).

    If B is symmetric, by definition, B(u,v) = B(v,u). You want to show that this is not the case.

    For example;
    Take [itex]A(x,y) = x^2 + y^2 [/itex] then
    [itex]A(x,y) = x^2 + y^2 = y^2 + x^2 = A(y,x)[/itex]
    So A is symmetric

    Take [itex]C(x,y) = x - y[/itex] then
    [itex]C(x,y) = x - y = -(y-x) = -C(y,x)[/itex]
    So C is NOT symmetric (if something has this property [itex]C(x,y) = - C(y,x)[/itex] it's called 'skew-symmetric')

    Take [itex]D(x,y) = 2x + y[/itex] then
    [itex]D(x,y) = 2x + y ≠ 2y +x = D(y,x)[/itex]

    Or you could even say, if F is symmetric then [itex]F(x,y) - F(y,x) = 0[/itex]
    then you'd get
    symmetric;
    [itex]A(x,y) - A(y,x) = x^2 + y^2 - y^2 - x^2 = 0[/itex]
    skew-symmetric (a subset of 'not symmetric');
    [itex]C(x,y) - C(y,x)= x - y - y + x) = 2x - 2y = 2C(x,y)[/itex]
    not symmetric
    [itex]D(x,y) - D(y,x)=2x + y - 2y -x = x - y ≠ 0[/itex]

    I know these are pretty simple examples but can you see the idea I'm trying to convey here?

    Actually, I just missed that last part so I guess the above was a little pointless -.-

    if you look at B, there's another dot product which WILL change if you interchange u and v
     
    Last edited: Jul 12, 2012
  6. Jul 12, 2012 #5
    This conveys the idea very nicely thanks. I have one quick question before I look at it properly tomorrow and apply it to the original question.

    Why do you write ##D(y,x)=2y+x##?
    I would have thought that ##D(y,x)=y+2x##?
    Isn't one just interchanging the variables?
     
  7. Jul 12, 2012 #6

    Mark44

    Staff: Mentor

    This is just basic function evaluation. If D(x, y) = 2x + y, then D(y, x) = 2y + x. The '2' coefficient goes with the first argument of the function, which in D(y, x) is y.
     
  8. Jul 12, 2012 #7
    nono, if I have [itex]D(x,y) = 2x + y[/itex] I'm pretty much saying 'take two times the first argument and add one times the second argument', using this [itex]D(y,x)[/itex] says 'take two times y and add one times x' which is 2y + x
     
  9. Jul 12, 2012 #8
    Thanks guys, that I didn't know..or perhaps it slowly crept out of my memory! I will look at the rest when I can.
     
  10. Jul 15, 2012 #9
    ## \displaystyle B_1(u,v)=\int_a^b (p(x) u \cdot v + q(x) \frac{du}{dx} \cdot v)dx \ne \int_a^b(p(x) v \cdot u+q(x) \frac{dv}{dx} \cdot u)dx=B_2(v,u)##..ok?
     
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