How do I show that the strain tensor
$$\epsilon_{ij}=\partial_i u_j + \partial_j u_i$$
in the case of a one-dimensional spring is reduced to
$$\epsilon= \frac{x-L}{L},$$
where L is the initial length of the spring. I can see that the only component that survives in this case is $\epsilon_{xx}=\partial_x u_x$, but I don't see how $L$ appears at the denominator.
For a homogeneous deformation of the spring, $$u=\epsilon_{xx}x$$where $\epsilon_{xx}$ is a constant. So, at x=L, $$u(L)=\epsilon_{xx}L$$Therefore, $$\epsilon_{xx}=\frac{u(L)}{L}=\frac{[L+u(L)]-L}{L}$$
In the reference configuration, the spring is a straight line:
$$X(s)=s,\ s\in[0,L]$$
its length being $$\ell_0=\intop_0^L |\dot X(s)| ds=L.$$
When the spring has stretched to have length $x$, its expression will be
$$X'(s)=\lambda s$$
The length has now changed, and by requiring that it is indeed $x$ we can find $\lambda$:
$$\ell(x)=\intop_0^L \lambda ds = x\ \ \Longrightarrow \lambda = x/L$$
Therefore $$X'(s) = \frac{x}{L}\ s.$$
The displacement is defined as $u(s)=X'(s)-X(s)$, which in this case becomes:
$$u(s)=\left(\frac{x}{L}-1\right)\ s$$
The strain is $$\epsilon_{xx}\equiv\epsilon=\partial_s u=\frac{x-L}{L}$$