The formulation of a “digital waveguide” simulation model for an idealised bowed string is very similar to the equivalent model for the clarinet, detailed in section 8.5.3. There are two main differences: first, the excitation of the clarinet occurs at the end of the tube while a string is bowed somewhere in the middle; and second, the nonlinear function associated with the simplest friction model is more severely non-smooth than the mouthpiece characteristic we assumed for the clarinet. The first issue makes the formulation slightly more complicated, but makes no fundamental change in the way the model operates.

The second issue makes a bigger difference: it means that stick-slip motion of a bowed string always involves non-smooth behaviour. There is no equivalent of the linearised argument leading to a threshold of blowing pressure, as we found in the clarinet case. This fits with experience of playing a violin: there is no possibility of quasi-sinusoidal motion when you press lightly or bow slowly: if the string vibrates it is always stick-slip motion of some kind with a rich spectrum of harmonics.

The starting point for modelling is similar to the clarinet case: we assume that in the immediate vicinity of the single point where we want the bow to act, the transverse vibration of the string can be approximated by that of an ideal string (as analysed in section 3.1.1). First, we think about an *infinite* string, extending away on both sides from the bowed point. If we apply a force $f(t)$ to such an infinite string, the velocity response of the string at the bowed point is $v(t)= f(t)/(2Z_0)$ where $Z_0=\sqrt{T/m}$ is the wave impedance of the string, of tension $T$ and mass per unit length $m$. In other words, our infinite string behaves like an ideal dashpot, with velocity proportional to applied force. This motion at the bowed point generates two outgoing waves of magnitude $f(t)/(2Z_0)$, sent off symmetrically in the two directions.

Of course, our real string has a finite length. The situation is then as sketched in Fig. 1. The force $f(t)$ generates outgoing waves as on the infinite system (shown in red in the figure), but now after a delay reflected waves (shown in blue) return to the bowed point from the two sides. Each separate reflected wave can be computed by convolution of the outgoing wave with a *reflection function* similar to the one used in the clarinet model. This time we need two reflection functions, which we can call $r_1(t)$ and $r_2(t)$, describing reflection behaviour from the bridge side and the finger side respectively. They will have different delays, which will depend on where the player chooses to place the bow on the string. This bowing position will play an important role in later discussions: it is usually described via a non-dimensional parameter $\beta$ which is the bow-bridge distance expressed as a proportion of the string length.

We can call the combined velocity resulting from these incoming waves

$$v_h(t)=\frac{1}{2Z_0} \left[r_1 * f(t) +r_2 * f(t) \right] \tag{1}$$

where * denotes the operation of convolution and the subscript ‘h’ suggests “history”, just as in the clarinet model of section 8.5.3. At a given time $t$ we can calculate $v_h(t)$ from our knowledge of the past history of the force $f(t)$. Now the actual velocity at the bowed point is the sum of the infinite-string response and the returning reflected waves:

$$v(t)=\frac{f(t)}{2Z_0} +v_h(t) . \tag{2}$$

This is the direct equivalent for the bowed-string problem of equation (9) from section 8.5.3. It is the equation of a straight line in the $v-f$ plane, with slope $2Z_0$ and horizontal position governed by the value of $v_h$.

The new values of $v(t)$ and $f(t)$ are given by the intersection of this straight line with the friction force — velocity curve. If the coefficient of friction is denoted $\mu(v)$ and the normal bow force is $F_b$, then this curve is given by $f=F_b \mu(v)$. For the particular plots and simulations shown in section 9.2, the curve-fitted coefficient of friction takes the form

$$\mu(v)=(\mu_{ss}-\mu_d) e^{-v/v_{cc}} + (\mu_s-\mu_d) e^{-v/v_c} + \mu_d \tag{3}$$

with $\mu_{ss}=1.2$, $\mu_s=0.8$, $\mu_d=0.35$, $v_{cc}=0.01$ m/s and $v_c=0.1$ m/s.