How can we show that for $\lambda <0$ we get the trivial solution $X(x)=0$? Find the solution of the problem $$u_t(x, t)-u_{xx}(x, t)=0, 0<x<1, t>0 \tag {*} \\ u(0, t)=0, t>0 \\ u_x(1,t)+u_t(1,t)=0, t>0$$ 
I have done the following: 
We are looking for solutions of the form $$u(x, t)=X(x) \cdot T(t)$$ 
$$u(0, t)=X(0) \cdot T(t)=0 \Rightarrow X(0)=0 \\ X'(1) \cdot T(t)+X(1) \cdot T'(t)=0 \Rightarrow \frac{X'(1)}{X(1)}=-\frac{T'(t)}{T(t)}$$ 
$$(*) \Rightarrow X(x) \cdot T'(t)-X''(x) \cdot T(t)=0 \\ \Rightarrow \frac{X(x) \cdot T'(t)}{X(x) \cdot T(t)}-\frac{X''(x) \cdot T(t)}{X(x) \cdot T(t)}=0 \\ \Rightarrow \frac{T'(t)}{T(t)}=\frac{X''(x)}{X(x)}=-\lambda$$ 
So, we get the following two problems: 
$$\left.\begin{matrix}
X''(x)+\lambda X(x)=0, 0<x<1\\ 
X(0)=0 \\
\frac{X'(1)}{X(1)}=\lambda \Rightarrow X'(1)-\lambda X(1)=0
\end{matrix}\right\}(1)
$$ 
$$\left.\begin{matrix}
T'(t)+\lambda T(t)=0, t>0
\end{matrix}\right\}(2)$$ 
For the problem $(1)$ we do the following: 
The characteristic polynomial is $d^2+\lambda=0$. 


*

*$\lambda<0$: 
General solution: $X(x)=c_1 e^{\sqrt{-\lambda }x}+c_2e^{-\sqrt{-\lambda}x}$ 
$$X(0)=0 \Rightarrow c_1+c_2=0 \Rightarrow c_1=-c_2$$ 
$$X(1)=c_1e^{\sqrt{-\lambda}}+c_2e^{-\sqrt{-\lambda}}=c_1(e^{\sqrt{-\lambda}}-e^{-\sqrt{-\lambda}})$$ 
$$X'(x)=\sqrt{-\lambda}c_1e^{\sqrt{-\lambda }x}-\sqrt{-\lambda}c_2 e^{-\sqrt{-\lambda}x} \\ X'(1)=\sqrt{-\lambda}c_1e^{\sqrt{-\lambda }}-\sqrt{-\lambda}c_2 e^{-\sqrt{-\lambda}}=\sqrt{-\lambda}c_1(e^{-\lambda}+e^{-\sqrt{-\lambda}}$$ 
$$X'(1)-\lambda X(1)=0 \Rightarrow \sqrt{-\lambda}c_1(e^{\sqrt{-\lambda}}+e^{-\sqrt{-\lambda}})-\lambda c_1(e^{\sqrt{-\lambda}}-e^{-\sqrt{-\lambda}})=0 \Rightarrow c_1 [ e^{\sqrt{-\lambda}}(\sqrt{-\lambda}-\lambda)+e^{-\sqrt{-\lambda}}(\sqrt{-\lambda}+\lambda)]=0$$ 
How could we continue to show that for $\lambda <0$ we get the trivial solution $X(x)=0$ ?? 
$$$$ 
EDIT: 


*

*$\lambda <0$ : 
$X(x)=c_1 \sinh (\sqrt{-\lambda} x)+c_2 \cosh (\sqrt{-\lambda}x)$ 
Using the initial values we get that $X(x)=0$, trivial solution. 

*$\lambda=0$ : 
$X(x)=c_1 x+c_2$ 
Using the initial values we get that $X(x)=0$, trivial solution. 

*$\lambda >0$ : 
$X(x)=c_1 cos (\sqrt{\lambda}x)+c_2 \sin (\sqrt{\lambda}x)$ 
$X(0)=0 \Rightarrow c_1=0 \Rightarrow X(x)=c_2=\sin (\sqrt{\lambda}x)$ 
$X'(1)-\lambda X(1)=0 \Rightarrow \tan (\sqrt{\lambda})=\frac{1}{\sqrt{\lambda}}$ 
That means that the eigenvalue problem $(1)$ has only positive eigenvalues $0<\lambda_1 < \lambda_2 < \dots < \lambda_k < \dots $ that are the positive roots of the equation $\tan \sqrt{x}=\frac{1}{\sqrt{x}}$. 
Is this correct?? 
Why can we say that the number of the eigenvalues is countable ?? 
How can we show that $$\lim_{k \rightarrow +\infty} \frac{\sqrt{\lambda_k}}{k \pi}=1$$ ??
 A: It may be easier in this case to take the general solution as
$$
X(x)=c_1\sinh \mu x+c_2\cosh \mu x,\quad \mu:=\sqrt{-\lambda}.
$$
Then the first boundary condition implies that $c_2=0$ and the second is that either
$$
c_1=0
$$ 
or
$$
\tanh \mu =-\frac{1}{\mu}.
$$
But the last equation has no solutions, hence $c_1=0$.
A: Using an integrating factor, we can show that the solution to
$$
X''+\lambda X=0\tag{1}
$$
for $\lambda\gt0$, is
$$
X(x)=a\cos(\sqrt\lambda\,x)+b\sin(\sqrt\lambda\,x)\tag{2}
$$
for $\lambda=0$, is
$$
X(x)=a+bx\tag{3}
$$
and for $\lambda\lt0$, is
$$
X(x)=a\cosh(\sqrt{-\lambda}\,x)+b\sinh(\sqrt{-\lambda}\,x)\tag{4}
$$
Furthermore, we also have
$$
X(0)=0\tag{5}
$$
and
$$
X'(1)-\lambda X(1)=0\tag{6}
$$

If $\lambda\gt0$, $(2)$ and $(5)$ say that $a=0$. Then $(2)$ and $(6)$ say that
$$
b(\sqrt\lambda\cos(\sqrt\lambda)-\lambda\sin(\sqrt\lambda))=0\tag{7}
$$
if $b=0$, then $X(x)=0$, otherwise, we need
$$
\tan(\sqrt\lambda)=\frac1{\sqrt\lambda}\tag{8}
$$
which has an infinite set of solutions, the smallest being $\lambda=0.7401738843949670422$.

If $\lambda=0$, $(3)$ and $(5)$ say that $a=0$. Then $(3)$ and $(6)$ say that $b=0$. Therefore, $X(x)=0$.

If $\lambda\lt0$, $(4)$ and $(5)$ say that $a=0$. Then $(4)$ and $(6)$ say that
$$
b(\sqrt{-\lambda}\cosh(\sqrt{-\lambda})-\lambda\sinh(\sqrt{-\lambda}))=0\tag{9}
$$
if $b=0$, then $X(x)=0$, otherwise, we need
$$
\tanh(\sqrt{-\lambda})=-\frac1{\sqrt{-\lambda}}\tag{10}
$$
which has no solutions since $\tanh(x)\gt0$ when $x\gt0$ and $\tanh(x)\lt0$ for $x\lt0$.

Thus, as you have added since I started writing this up, there is a sequence of solutions with $\lambda\gt0$, where $\lambda$ satisfies $(8)$.
As $\lambda$ gets larger, $\frac1{\sqrt\lambda}\to0$. The places were $\tan(x)$ is positive and near $0$ are slightly greater than multiples of $\pi$. Thus, $\sqrt\lambda$ must be slightly greater than $k\pi$ for some non-negative $k\in\mathbb{Z}$. Certainly, we can label each such $\lambda$ as $\lambda_k$. In that case, we have
$$
\lim_{k\to\infty}\frac{\sqrt{\lambda_k}}{k\pi}=1\tag{11}
$$
