If the nominal lengths, L, of the springs are known to be 1 and 2 units respectively, then the system can be solved as follows:
Consider the simple case of three nodes connected by two springs. Then the stretching of the two springs is given as a function of the positions of the nodes by
where is the matrix transpose of the incidence matrix
relating each degree of freedom to the direction each spring pulls on it.
The forces on the springs are
where W is a diagonal matrix giving the stiffness of every spring. Then the force on the nodes is given by left multiplying by , which we set to zero to find equilibrium:
which gives the linear equation:
- .
Now, the matrix is singular, because all solutions are equivalent up to rigid-body translation. Let us prescribe a Dirichlet boundary condition, e.g., .
As an example, let W be the identity matrix then
is the Laplacian matrix. Plugging in we have
- .
Incorporating the 2 to the left-hand side gives
- .
and removing rows of the system that we already know, and simplifying, leaves us with
- .
- .
so we can then solve
- .
That is, , as prescribed, and , leaving the first spring slack, and , leaving the second spring slack.