# Eigenproblem for linear differential operator

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

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