Either the integrator setup routine has not been called prior to the first call of this routine, or a communication array has become corrupted.
Either the linear algebra setup routine has not been called prior to the first call of this routine, or a communication array has become corrupted.
Either the routine was entered on a continuation call without a prior call of this routine, or a communication array has become corrupted.
Either the value of
njcpvt is not the same as the value supplied to the setup routine or a communication array has become corrupted.
${\mathbf{njcpvt}}=\u27e8\mathit{\text{value}}\u27e9$,
njcpvt (setup)
$=\u27e8\mathit{\text{value}}\u27e9$.
Either the value of
nwkjac is not the same as the value supplied to the setup routine or a communication array has become corrupted.
${\mathbf{nwkjac}}=\u27e8\mathit{\text{value}}\u27e9$,
nwkjac (setup)
$=\u27e8\mathit{\text{value}}\u27e9$.
Either the value of
sdysav is not the same as the value supplied to the setup routine or a communication array has become corrupted.
${\mathbf{sdysav}}=\u27e8\mathit{\text{value}}\u27e9$,
sdysav (setup)
$=\u27e8\mathit{\text{value}}\u27e9$.
Failure during internal time interpolation.
tout and the current time are too close.
${\mathbf{itask}}=\u27e8\mathit{\text{value}}\u27e9$ and
${\mathbf{tout}}=\u27e8\mathit{\text{value}}\u27e9$ and the current time is
$\u27e8\mathit{\text{value}}\u27e9$.
${\mathbf{itask}}=3$ and
tout is more than an integration step behind the current time.
${\mathbf{tout}}=\u27e8\mathit{\text{value}}\u27e9$, current time minus step size:
$\u27e8\mathit{\text{value}}\u27e9$.
On entry, an illegal (negative) maximum number of steps was provided in a prior call to a setup routine. ${\mathbf{maxstp}}=\u27e8\mathit{\text{value}}\u27e9$.
On entry, an illegal (negative) maximum stepsize was provided in a prior call to a setup routine. ${\mathbf{hmax}}=\u27e8\mathit{\text{value}}\u27e9$.
On entry, an illegal (negative) minimum stepsize was provided in a prior call to a setup routine. ${\mathbf{hmin}}=\u27e8\mathit{\text{value}}\u27e9$.
On entry, ${\mathbf{atol}}=\u27e8\mathit{\text{value}}\u27e9$.
Constraint: ${\mathbf{atol}}\ge 0.0$.
On entry,
${\mathbf{ires}}=\u27e8\mathit{\text{value}}\u27e9$ and
$\frac{dy}{dt}=0.0$ for all elements.
Check the evaluation of the residual for this value of
ires.
On entry, ${\mathbf{irevcm}}=\u27e8\mathit{\text{value}}\u27e9$.
Constraint: $0\le {\mathbf{irevcm}}\le 11$.
On entry, ${\mathbf{itask}}=\u27e8\mathit{\text{value}}\u27e9$.
Constraint: $1\le {\mathbf{itask}}\le 5$.
On entry,
${\mathbf{itask}}=4$ or
$5$ and
tcrit is before the current time in the direction of integration.
${\mathbf{itask}}=\u27e8\mathit{\text{value}}\u27e9$,
${\mathbf{tcrit}}=\u27e8\mathit{\text{value}}\u27e9$ and the current time is
$\u27e8\mathit{\text{value}}\u27e9$.
On entry,
${\mathbf{itask}}=4$ or
$5$ and
tcrit is before
tout in the direction of integration.
${\mathbf{itask}}=\u27e8\mathit{\text{value}}\u27e9$,
${\mathbf{tcrit}}=\u27e8\mathit{\text{value}}\u27e9$ and
${\mathbf{tout}}=\u27e8\mathit{\text{value}}\u27e9$.
On entry, ${\mathbf{itol}}=\u27e8\mathit{\text{value}}\u27e9$.
Constraint: $1\le {\mathbf{itol}}\le 4$.
On entry, ${\mathbf{neq}}=\u27e8\mathit{\text{value}}\u27e9$.
Constraint: ${\mathbf{neq}}\ge 1$.
On entry, ${\mathbf{neq}}=\u27e8\mathit{\text{value}}\u27e9$ and ${\mathbf{ldysav}}=\u27e8\mathit{\text{value}}\u27e9$.
Constraint: ${\mathbf{neq}}\le {\mathbf{ldysav}}$.
On entry, ${\mathbf{rtol}}=\u27e8\mathit{\text{value}}\u27e9$.
Constraint: ${\mathbf{rtol}}\ge 0.0$.
On entry,
tout is less than
t with respect to the direction of integration given by the sign of
h0 in a prior call to a setup routine.
${\mathbf{tout}}=\u27e8\mathit{\text{value}}\u27e9$,
${\mathbf{t}}=\u27e8\mathit{\text{value}}\u27e9$ and
${\mathbf{h0}}=\u27e8\mathit{\text{value}}\u27e9$.
On entry,
tout is too close to
t to start integration.
${\mathbf{tout}}=\u27e8\mathit{\text{value}}\u27e9$ and
${\mathbf{t}}=\u27e8\mathit{\text{value}}\u27e9$.
On re-entry, ${\mathbf{imon}}=\u27e8\mathit{\text{value}}\u27e9$.
Constraint: $\mathrm{-2}\le {\mathbf{imon}}\le 4$.
On re-entry, ${\mathbf{inln}}=\u27e8\mathit{\text{value}}\u27e9$ for the case ${\mathbf{irevcm}}=9$ and ${\mathbf{imon}}=0$.
Constraint: ${\mathbf{inln}}=3$.
On re-entry,
ires has been set to an illegal value during initialization.
${\mathbf{ires}}=\u27e8\mathit{\text{value}}\u27e9$ at time
$\u27e8\mathit{\text{value}}\u27e9$.
On re-entry, the solution vector appears to have been overwritten.
Further integration will not be attempted.
The initial stepsize, $h=\u27e8\mathit{\text{value}}\u27e9$, is too small.
Weight number
$i=\u27e8\mathit{\text{value}}\u27e9$ used in the local error test is too small. Check the values of
rtol and
atol.
${\mathbf{atol}}\left(i\right)$ and
${\mathbf{y}}\left(i\right)$ may both be zero.
Weight
$i=\u27e8\mathit{\text{value}}\u27e9$.