Find all possible $(\lambda,u)$ where $\lambda \in \mathbb R$ and $u\ne0$, to the boundary value problem:
$u''(x)+\lambda u(x)=0,x\in (0,1),$
$u(0)=u(1); u'(0)=u'(1).$
My Effort: for $\lambda>0,u(x)=A_n\cos\sqrt\lambda_nx+B_n\sin\sqrt\lambda_nx$, where$\lambda_n=4n\pi^2$(After Calculation), $n=\pm1,\pm2,....$ is it correct? Please verify.What will be the case if $\lambda<0$. I am confused. Please help.
The boundary conditions for this type of problem can often be formulated as a matrix equation yielding a fairly general and automatic way to find the eigenvalues. Here's how to do so for this problem.
First, as you notice, the general solution of the differential equation is $$u(x)=a\cos(\sqrt\lambda x)+b\sin(\sqrt\lambda x)$$ so that $$u'(x)=-a\sqrt{\lambda}\sin(\sqrt\lambda x)+b\sqrt{\lambda}\cos(\sqrt\lambda x).$$ From here, it's easy to write down the boundary conditions, namely $$u(0)=u(1) \: \text{ iff } \: a=a\cos(\sqrt{\lambda})+b\sin(\sqrt{\lambda})$$ and $$u'(0)=u'(1) \: \text{ iff } \: b\sqrt{\lambda}=-a\sqrt{\lambda}\sin(\sqrt{\lambda})+b\sqrt{\lambda}\cos(\sqrt{\lambda}).$$ It's easy to see that $\lambda=0$ yields the constant eigenfunction so, accounting for that, we divide off the $\sqrt{\lambda}$ from the second equation to obtain the pair $$ \begin{align} a&=a\cos(\sqrt{\lambda})+b\sin(\sqrt{\lambda}) \\ b&=-a\sin(\sqrt{\lambda})+b\cos(\sqrt{\lambda}). \end{align} $$ This can be written as the matrix equation $$ \left( \begin{array}{cc} \cos(\sqrt{\lambda}) & \sin(\sqrt{\lambda}) \\ -\sin(\sqrt{\lambda}) & \cos(\sqrt{\lambda}) \end{array} \right) \left( \begin{array}{c} a \\ b \end{array} \right) = \left( \begin{array}{c} a \\ b \end{array} \right) $$ or $$ \left( \begin{array}{cc} \cos(\sqrt{\lambda})-1 & \sin(\sqrt{\lambda}) \\ -\sin(\sqrt{\lambda}) & \cos(\sqrt{\lambda})-1 \end{array} \right) \left( \begin{array}{c} a \\ b \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \end{array} \right). $$
Now, the question is, for which values of $\lambda$ does the preceding matrix equation have a non-trivial solution? The answer, of course, is when the determinant of the matrix is zero. This yields a simple equation whose roots are exactly the eigenvalues you're looking for. Computing the determinant and then simplifying, we get $$\sin^2\left(\sqrt{\lambda }\right)+\cos^2\left(\sqrt{\lambda }\right)-2 \cos \left(\sqrt{\lambda }\right)+1 = 2-2 \cos \left(\sqrt{\lambda }\right).$$ This last expression equals zero precisely when $\sqrt{\lambda}=2n\pi$ for some $n\in\mathbb Z$. The possible values of $\lambda$ are then $4n^2\pi^2$.