668 
conform to the border conditions. For «= f(t, r= g(t) we can 
proceed analogously. 
Leaving the particular form of y (¢—t) undetermined, we shall 
now further treat an example that might be tested by experiment, 
which might, therefore, give us an opportunity to find the form of 
yw. In this case it is easy to write down the condition on which 
VoLTERRA’s scheme comes to damping. 
For the example that we now choose, the equation (1) holds 
again, but now let a force zero act from {= —o tot=0; wand v 
are then zero; a force A further constant suddenly begins to act at 
t=0. For the motion after the moment zero now the following 
equation holds: 
a 
Aa ae fet ae KK 
0 
2», 
de 
Practically the term ee be omitted (that is to say, we shall 
a 
presently examine ‘exactly what is tacitly assumed here). There then 
remains 
t 
a x (t) ~{ w(t) yp (t—t) dt = K, 
0 
being a linear integral equation of the 2"¢ order. The solution of 
this becomes: 
Ks 1 1 +e 
wv (t) =—(' + ‚fo dr +affr (t) w (t—t,) dr, r+.) 
at a at 
0 0 0 
When we now assume that « is great with regard to the hyste- 
resis, we may neglect terms with higher powers of u in the deno- 
minator. We get in first appr ee 
K 1 
2) = =(1 +— | y(t) ir) EEE 
if 
dz 
Is it really allowed to omit the term —— from the equation (A) 
dt 2 
above? When by differentiating the just found solution «(¢) twice, 
. 
= ne 
we determine ae for this purpose, we see at once that the right- 
K 
hand member contains the factor —. Hence when this may be ne- 
a 
glected compared with the hysteretical term: 
