Why can the deformation of a spring at a given point of the spring be considered directly proportional to the relative distance of the point? Hello i have been studying differential equations and in one example my professor tries to deduce the partial differential equation that describes the longitudinal displacement on a elastic, homogeneous, bar of constant transversal sections. In the text he gives a stiffness coefficient $K$ that apparently works as a type of normalized hook's constant in relation to the length of the bar.
Here begins my actual question: in the deduction of the partial differential equation he uses that if you take a elastic bar in its relaxed position as in the following figure:
 
And compress it with a force $F$ generating a displacement $S$ as in the figure below: 

But then he argues that if we take the middle point of the bar in its relaxed position, that is the point of distance $L/2$, in the compressed state this point "must have" a displacement of "$S/2$" as in the figure bellow:

So my question is why could he make the statement that the deformation is directly proportional to the ratio between the distance of a point and the length of the bar ?, implying that $K=Lk$, where $k$ is the hook's constant, is it always true or only in this specific case ? how can one demonstrate this fact ?

 A: If a rod of material is subjected to a tensile force at its ends, the internal tensile force at each location along the rod is, by equilibrium, equal to the force at the ends.  So the rod has no way of knowing that the internal force in any way differs from the end force.  It then stands to reason that the stretching response of each section of  material along the rod will be the same (i.e., uniform).  So all differential segments of the rod will stretch equally.  This automatically guarantees geometrically that the displacement of the center of the rod will be half that at the far end.
A: The formal way to treat a problem like this is to consider a small slice of the bar at a location $x$. The deformation amount as a function of location is $u(x)$ with the end deformation $\delta=u(\ell)$.
Take a slice of width ${\rm d}x$ and balance the forces, assuming the tension changes by ${\rm d}T$ across the slice. The amount of strain $\epsilon$ of the slice is given by Hook's law $$\left. \epsilon =\frac{\sigma}{E} = \frac{T}{E A} = \frac{u(x+{\rm d}x)-u(x)}{{\rm d}x} = \frac{\partial}{\partial x} u(x) \right\}\;\; T = E A \,\frac{\partial u}{\partial x} $$
Where $A$ is the sectional area and $E$ the Young's Modulus (or Modulus of Elasticity).

Now for the balance of forces:
$$ \left. (T + {\rm d}T) -T = 0 \right\} \;\; {\rm d}T = 0 $$
If ${\rm d}T=0$ then $$\left. T=E A \,\frac{\partial u}{\partial x}=\mbox{(const.)} \right\} \;\;u(x) \mbox{ is linear with }x$$
Given the end conditions in the problem, the solution is
$$ \boxed{ u(x) = \delta \,\frac{x}{\ell} }$$
Or, as you stated, the local deformation is a linear function of position. But for other end conditions, the solution might be different. What to take from this is if there is no external forces applied throughout the bar, the tension is constant, and the deformation is linear with position.
A: In this case you can consider Young’s modulus, $E$ which is the ratio of the tensile stress divided by the tensile strain, being constant throughout the material.  
$$E = \dfrac{\left(\frac{\text{tensile force}}{\text{cross sectional area}} \right )}{\left ( \frac{\rm extension}{\rm length}\right  )}$$. 
If tensile force, cross sectional area and Young’s modulus are constant then the extension is proportional to the length.
