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I'm trying to solve this question on self-adjoint in complex numbers. I'm stuck and need help. Consider $ L =\frac{d^3}{dx^{3}}$ to be a linear operator which acts on a function from $[0,1]$ to $\Bbb R$. Given the boundary conditions

x(0) =1,\quad x'(0) - 2x''(1)=0, \quad x(1) -x'(1) = 0, where all the quantities are real. An inner product is defined as $ \langle u(x), v(x) \rangle =\int_{0}^{1} u(x)v(x) dx.$ I'm to determine whether $L$ is self-adjoint.

This is what I have done so far , please help me as to what to do next.

\langle u(x), Lv(x) \rangle = u(x)v''(x)|_{0}^{1} - u'(x)v'(x)|_{0}^{1} + u''(x)v(x)|_{0}^{1} - \langle Lu(x), v(x) \rangle. That is after integrating by parts three times. Now, I don't know what to do since after plugging the upper and lower limits, there is nothing left since . Any guidelines?

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The adjoint $L^*$ of $L$ is defined by $\langle L^*u,v\rangle=\langle u,Lv\rangle$ for every suitable $u$ and $v$. Hence $L^*=L$ if $\langle u,Lv\rangle=\langle Lu,v\rangle$ for every $u$ and $v$. Hence your task is to evaluate $\langle u,Lv\rangle$ as the sum of $\langle Lu,v\rangle$ and some boundary terms, and then to check that these boundary terms are identically zero (and to do that without errors of signs or otherwise, you could check this).

To see what is going on in another case, consider the operator $K$ defined by (Ku)(x)=u'(x). Then \langle u,Kv\rangle=\langle u,v'\rangle=\left.u(x)v(x)\right|_{x=0}^{x=1}-\langle u',v\rangle=u(1)v(1)-u(0)v(0)-\langle Ku,v\rangle hence $K^*=-K$ on the space of the functions $w$ defined on $[0,1]$ such that $w(0)=w(1)=0$.

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    The boundary conditions are crucial to make the boundary terms disappear. Please check your integrations by parts (at the moment they are wrong) and see the added paragraph in my post.2011-11-16