Please check ODE operator solutions

[ognik]

1) Given $\mathcal{L}u=0$ and $ g\mathcal{L}u$ is self-adjoint, show that for the adjoint operator $ \bar {\mathcal{L}}, \bar{\mathcal{L}}(gu)=0$

Is it enough to say that if self-adjoint, then $ \mathcal{L}= \bar {\mathcal{L}} $. I assume g represents a function of x (so no inner products with g) that transforms a non self-adjoint ODE into a self-adjoint ODE, therefore $ \bar{\mathcal{L}}(gu)=g\bar{\mathcal{L}}(u)=g{\mathcal{L}}(u)= 0 $?

2)For a 2nd order diff. operator $ \mathcal{L} $ that is self adjoint, show that $ <y_2 | \mathcal{L} y_1 > - <y_1 | \mathcal{L} y_2 > = \int_{a}^{b}[y_2 \mathcal{L}y_1 - y_1 \mathcal{L} y_2] \,dx$

My question is that the book has been using, for example, $ <y_2 | \mathcal{L} y_1 >= \int_{a}^{b}y_2^* \mathcal{L}y_1 dx $ so why no * (complex conjugates) in this problem? Is it a typo or am I about to learn something new?

Thanks

 

[jvicens]

.

1) Yes, it is enough to say that if $g\mathcal{L}u$ is self-adjoint, then $\mathcal{L}=\bar{\mathcal{L}}$. This is because for a self-adjoint operator, the adjoint operator is the same as the original operator. Therefore, $g\bar{\mathcal{L}}(u)=g\mathcal{L}(u)=0$.

2) It is not a typo; it is a different notation for the same thing. In the notation used in the book, $y_2^*$ represents the complex conjugate of $y_2$. In the notation used in the problem, $<y_2|\mathcal{L}y_1>$ and $<y_1|\mathcal{L}y_2>$ are already complex conjugates of each other, so there is no need for an additional $*$ symbol. Both notations are commonly used in mathematics, so it is just a matter of personal preference.