$\pi:$ $Lx=\sum_{j=0}^{n}(p_{n-j}x^{(j)})^{(j)}$,$\,\,$ $x^{(j)}(a)=x^{(j)}(b)=0,\, j=0,1,...,n-1.$
where $p_{n-j}\in C^{n-j}[a,b]$ are real and $p_0(t)\neq0$ on $[a,b]$.
I want to show that the problem $\pi$ is self-adjoint. My approach as in the following:
$<Lx,y>=<\sum_{j=0}^{n}(p_{n-j}x^{(j)})^{(j)},y>=\sum_{j=0}^{n}\sum_{k=0}^{j}\dbinom{j}{k} <p_{n-j}^{(k)}x^{(2j-k)},y>$
Since $x^{(j)}(a)=x^{(j)}(b)=0,\, j=0,1,...,n-1$, I obtain the following formula by applying partial integration recursively,
$<Lx,y>=\sum_{j=0}^{n}\sum_{k=0}^{j}\dbinom{j}{k} <x,((-1)^kp_{n-j}^{(k)}y)^{(2j-k)}>$. Therefore I have
$L^*y=\sum_{j=0}^{n}\sum_{k=0}^{j}\dbinom{j}{k} ((-1)^kp_{n-j}^{(k)}y)^{(2j-k)}$.
Now at this point i need to show that $L=L^*$. But i couldn't can you give me any suggestion to show this result and can you correct my calculation above if it is incorrect.
Thanks in advance!...
You shouldn't open parantheses for Leibnitz rule, use the boundary conditions with integration by parts instead. Take $y$ that satisfies the same boundary data: $$y^{(j)}(a)=y^{(j)}(b)=0,\, j=0,1,\dots,n-1,$$then $$\int_a^b (px^{(j)})^{(j)}y\ dt =(px^{(j)})^{(j-1)}y\big|_{t=a}^{t=b} - \int_a^b (px^{(j)})^{(j-1)}y^{(1)}\ dt $$ $$=\dots= (-1)^j\int_a^b px^{(j)} y^{(j)}\ dt $$ $$=\dots =\int_a^b (py^{(j)})^{(j)}x\ dt $$