III. $\lambda < 0 .$

For the sake of convenience we put

$$\lambda = - \omega^2$$

Then the general solution for (1.2) is

$$G(x) = C_1 \cos(\omega x) + C_2 \sin(\omega x)$$

We need to choose $C_1,\;\;C_2$ and $\omega$ in order to satisfy the boundary conditions

$$G(0) = 0,\;\; G(L) = 0$$

The first condition $G(0)=0$ implies that

$$C_1 = 0$$

Hence,

$$G(x) = C_2 \sin(\omega x)$$

Without loss of generality we can put $C_2=1.$ It is possible because we are looking for solutions having the form

$$u(t,x)= F(t) G(x).$$

That means the constant $C_2$ will be taken care of when we will deal with $F(t).$ Thus,

$$G(x) = \sin (\omega x)$$

and we need to choose $\omega$ in order to satisfy the condition

$$\sin (\omega L) = 0$$

which follows from

$$G(L)=0$$

Hence,

$$\omega L = \pi n$$

and

$$\omega = \frac{ \pi n }{ L }.$$

We found infinitely many solutions for G(x),

$$G_n (x) = \sin(\frac{ \pi n x }{ L } ),\;\;n=1,\;2,\;\dots$$