# Homework Help: Bi Linear Functionals and Symmetry

1. Jul 12, 2012

### bugatti79

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. Jul 12, 2012

### genericusrnme

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?

3. Jul 12, 2012

### bugatti79

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 $u \cdot v= v \cdot u$....?

4. Jul 12, 2012

### genericusrnme

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 $A(x,y) = x^2 + y^2$ then
$A(x,y) = x^2 + y^2 = y^2 + x^2 = A(y,x)$
So A is symmetric

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

Take $D(x,y) = 2x + y$ then
$D(x,y) = 2x + y ≠ 2y +x = D(y,x)$

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

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
5. Jul 12, 2012

### bugatti79

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?

6. Jul 12, 2012

### 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.

7. Jul 12, 2012

### genericusrnme

nono, if I have $D(x,y) = 2x + y$ I'm pretty much saying 'take two times the first argument and add one times the second argument', using this $D(y,x)$ says 'take two times y and add one times x' which is 2y + x

8. Jul 12, 2012

### bugatti79

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.

9. Jul 15, 2012

### bugatti79

$\displaystyle B_1(u,v)=\int_a^b (p(x) u \cdot v + q(x) \frac{du}{dx} \cdot v)dx \ne \int_a^b(p(x) v \cdot u+q(x) \frac{dv}{dx} \cdot u)dx=B_2(v,u)$..ok?