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I have to consider the eigenvalue problem: $$ L[u] := \frac{d^2 u}{dx^2}= λu,x \in (0,1)\quad u(0)-\frac{du}{dx}(0)=0, u(1)=0.$$ I need to show that the eigenvalues are negative.

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Suppose we have $\lambda > 0$, then $u'' = \lambda u$ gives us $$u(t) = \alpha\exp(-\sqrt\lambda t) + \beta\exp(\sqrt\lambda t)$$ with $\alpha, \beta \in \mathbb R$. Now \begin{align*} u(0) - u'(0) &= \alpha + \beta -\sqrt \lambda \alpha + \sqrt \lambda \beta\\ &= \alpha(1 - \sqrt \lambda) + \beta(1 + \sqrt\lambda)\\ \text{and } u(1)&= \alpha\exp(-\sqrt \lambda) + \beta\exp(\sqrt\lambda) \end{align*} So we must have by the boundary conditions $$\beta = -\alpha \exp(2\sqrt\lambda), \quad \beta = -\alpha\cdot\frac{1-\sqrt \lambda}{1+\sqrt \lambda} $$ that is $\alpha = \beta = 0$ or $$ \exp(2\sqrt \lambda) = \frac{1-\sqrt\lambda}{1+ \sqrt\lambda} $$ But this is impossible for $\lambda > 0$ as $\lambda \mapsto \exp(2\sqrt\lambda)$ is strictly increasing and $\lambda \mapsto \frac{1-\sqrt\lambda}{1+\sqrt \lambda}$ is decreasing and they are both equal at $0$. So $\alpha = \beta = 0$ and no $\lambda > 0$ is an eigenvalue.

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