How Can the Spring Equation for a Chain of Springs be Solved?

[MathematicalPhysicist]

There is a chain of springs, such that q_k is the displacement of the kth mass in this chain.

We have the next equation of Newton second law, where T is the tension force:
$$\frac{d^2q_k}{dt^2}=T(q_{k+1}-q_k)-T(q_k-q_{k-1})$$
Let Q_k=q_k+1-q_k

I showed that if Q(t)_k=Q(k-ct)=Q(s)

then we get the next ODE:
(1)$$c^2 Q''(s)=T(s+1)-2T(s)+T(s-1)$$
I need to show that the solution is:
$$Q(s)=\int_{s-1}^{s+1} (1-|s-z|)T(z)dz$$
by direct integration i.e integrating (1) or by Fourier transform.

I tried both but I get stuck, by direct integration I get:

$$c^2Q(s)= \int^{s} \int^{x} [T(z+1)-2T(z)+T(z-1)] dz dx$$
don't know how to proceed from here, can anyone give a helping hand? (-:

 

[MathematicalPhysicist]

I also tried the Fourier transform I got to:
$$Q(s)=FT(FT^-1(T(s))*\frac{sin^2(w/2)}{(w/2)^2})$$

I tried mathematica to solve this, but I am not sure how to write it with an implicit function as FT^-1(T(s)).

 

[ross_tang]

I wonder about equation (1). Since:

$$\frac{d^2Q_k}{\text{dt}^2}=T\left(Q_{k+1}\right)-2T\left(Q_k\right)+T\left(Q_{k-1}\right)$$

So, we have:

$$c^2 Q''(s)=T(Q(s+1))-2T(Q(s))+T(Q(s-1))$$

instead of

$$c^2 Q''(s)=T(s+1)-2T(s)+T(s-1)$$

Please clarify.

 

[LCKurtz]

What are the equations if you have k = 2 springs? Using your equations you would have displacements q₀, q₁, q₂, and q₃ involving 4 displacements and only two equations.