I have a struggle with eigen problem $\hat{L} u(x) = \lambda u(x)$ for a differential operator $\hat{L} = \frac{d^2}{d x^2} + 1$ with boundary conditions $u'(0) = 0$, $u'(l)=0$.

So far I solved an ODE and got the general solution

$$u(x) = C_1 \exp{\left(\sqrt{(1-\lambda)}x\right)} + C_2 \exp{\left(-\sqrt{(1-\lambda)}x\right)}$$

Then I applied boundary conditions and got two equations

$$\sqrt{1-\lambda}(C_1-C_2) = 0$$ $$\sqrt{1-\lambda}\left[C_1\exp{\left(\sqrt{(1-\lambda)}l\right)} - C_2\exp{\left(-\sqrt{(1-\lambda)}l\right)}\right] = 0$$

Solving the system for $C_1$ and $C_2$ I end up with $\lambda = 1$ and $\lambda_n = 1 - \frac{\pi^2(n-1/2)^2}{l^2}$, where $n \in \mathbb{Z}$. Is it correct? I'm not fully confident, especially about this $\lambda = 1$ eigenvalue

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    $\begingroup$ One small correction: if $u$ only depends on $x$ then the derivative is total. $\endgroup$ – Bcpicao Jun 8 '20 at 10:18
  • $\begingroup$ @bcpicao Sure, edited. $\endgroup$ – user464980 Jun 8 '20 at 10:22

Your final result is correct, but it is not clear to me how you got there.

You should look at the three cases $\lambda < 1$, $\lambda = 1$ and $\lambda > 1$ separately.

For $\mathbb{\lambda = 1}$ we have $\frac{d^2}{dx^2}u(x) = 0$, and it is easy to show that, with the boundary conditions you provided, any constant function is a solution. You can verify this by writing the general solution as

$$u(x) = C_1x + C_2.$$

The boundary conditions lead to $C_1=0$ and $C_2$ arbitrary. Therefore $\lambda = 1 $ is an eigenvalue of $\hat{L}$.

Now the zeros of the characteristic equation $r^2 + 1-\lambda = 0$ can be written as $r_{1,2} = \pm \sqrt{\lambda - 1}$ for $\lambda > 1$, and $r_{1,2} = \pm i\sqrt{1-\lambda}$ for $\lambda < 1$. Therefore nontrivial solutions (if they exist) are exponential for $\lambda > 1$ and oscillatory for $\lambda < 1$.

For $\mathbb{\lambda > 1}$ the general solution can be written as

$$u(x;\lambda) = C_1\exp\left( \sqrt{\lambda -1}x \right) + C_2\exp\left(- \sqrt{\lambda -1}x \right),$$

and using the boundary conditions we arrive at $C_1 = C_2$ and $C_2 = C_2\exp\left(2\sqrt{\lambda - 1}l \right)$, or $C_2=C_1=0$. Therefore there are no eigenvalues $\lambda > 1$.

For $\mathbb{\lambda < 1}$ the general solution can be written as

$$\begin{align} u(x;\lambda) &= A\exp\left( i\sqrt{1-\lambda}x \right) + B\exp\left(-i\sqrt{1-\lambda}x \right),\\ &=C_1\sin\left( \sqrt{1-\lambda}x \right) + C_2\cos\left(\sqrt{1-\lambda}x \right),\end{align}$$

where $C_2 = A+B$ and $C_1 = i(A-B)$. Using the boundary conditions we arrive at $C_1 = 0$ and $C_2\sin\left(\sqrt{1-\lambda}x\right) = 0$. Hence, nontrivial solutions exist for $\lambda = \lambda_n = 1-\frac{(n-1/2)^2\pi^2}{l^2}$ with $n=1,2,3,\dots$ and $\lambda_n$ are eigenvalues of $\hat{L}$.


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