# self linear operator

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 $w(0)=w(1)=0$.
@ Didier, I understand what you wrote but the thing is, the boundary conditions does not play any role in the integration by parts since its in terms of $u, v$ whiles the derivative is in $x$. Is there is problem in the question? – wright Nov 16 '11 at 12:07