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

${\displaystyle \Delta \mathbf {L} ={\begin{bmatrix}1&-1&0\\0&1&-1\end{bmatrix}}\mathbf {x} -\mathbf {L} =B^{\top }\mathbf {x} -\mathbf {L} }$ 

where ${\displaystyle B^{\top }}$  is the matrix transpose of the incidence matrix

${\displaystyle B={\begin{bmatrix}1&0\\-1&1\\0&-1\end{bmatrix}},}$ 

relating each degree of freedom to the direction each spring pulls on it. The forces on the springs are

${\displaystyle F_{\text{springs}}=-W\Delta \mathbf {L} =-W(B^{\top }\mathbf {x} -\mathbf {L} )=-WB^{\top }\mathbf {x} +W\mathbf {L} }$ 

where W is a diagonal matrix giving the stiffness of every spring. Then the force on the nodes is given by left multiplying by ${\displaystyle B}$ , which we set to zero to find equilibrium:

${\displaystyle F_{\text{nodes}}=-BWB^{\top }\mathbf {x} +BW\mathbf {L} =0}$ 

which gives the linear equation:

${\displaystyle BWB^{\top }\mathbf {x} =BW\mathbf {L} }$ .

Now, the matrix ${\displaystyle BWB^{\top }}$  is singular, because all solutions are equivalent up to rigid-body translation. Let us prescribe a Dirichlet boundary condition, e.g., ${\displaystyle x_{1}=2}$ .

As an example, let W be the identity matrix then

${\displaystyle BWB^{\top }={\begin{bmatrix}1&-1&0\\-1&2&-1\\0&-1&1\end{bmatrix}}}$ 

is the Laplacian matrix. Plugging in ${\displaystyle x_{1}=2}$  we have

${\displaystyle BWB^{\top }\mathbf {x} ={\begin{bmatrix}1&-1&0\\-1&2&-1\\0&-1&1\end{bmatrix}}{\begin{bmatrix}2\\x_{2}\\x_{3}\end{bmatrix}}=BW\mathbf {L} ={\begin{bmatrix}-1\\-1\\2\end{bmatrix}}}$ .

Incorporating the 2 to the left-hand side gives

${\displaystyle {\begin{bmatrix}2\\-2\\0\end{bmatrix}}+{\begin{bmatrix}-1&0\\2&-1\\-1&1\end{bmatrix}}{\begin{bmatrix}x_{2}\\x_{3}\end{bmatrix}}={\begin{bmatrix}-1\\-1\\2\end{bmatrix}}}$ .

and removing rows of the system that we already know, and simplifying, leaves us with

${\displaystyle {\begin{bmatrix}-2\\0\end{bmatrix}}+{\begin{bmatrix}2&-1\\-1&1\end{bmatrix}}{\begin{bmatrix}x_{2}\\x_{3}\end{bmatrix}}={\begin{bmatrix}-1\\2\end{bmatrix}}}$ .

${\displaystyle {\begin{bmatrix}2&-1\\-1&1\end{bmatrix}}{\begin{bmatrix}x_{2}\\x_{3}\end{bmatrix}}={\begin{bmatrix}1\\2\end{bmatrix}}}$ .

so we can then solve

${\displaystyle {\begin{bmatrix}x_{2}\\x_{3}\end{bmatrix}}={\begin{bmatrix}2&-1\\-1&1\end{bmatrix}}^{-1}{\begin{bmatrix}1\\2\end{bmatrix}}={\begin{bmatrix}3\\5\end{bmatrix}}}$ .

That is, ${\displaystyle x_{1}=2}$ , as prescribed, and ${\displaystyle x_{2}=3}$ , leaving the first spring slack, and ${\displaystyle x_{3}=5}$ , leaving the second spring slack.

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• The Physics of Springs