- #1
bugatti79
- 794
- 1
Homework Statement
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
Homework Statement
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
Homework 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?