Question: matrix in singular

Dear all

How to avoid in plot "matrix in singular" in my pde system

> a1 := -2;
-2
> b1 := 3;
3
> c1 := -1;
-1
> L := 15;
15
> b := .25;
0.25
> th := 1.5;
1.5
> pde := diff(u(x, t), x, x, x, x, t)+(6*a1-2*(3*a1*u(x, t)^2+2*b1*u(x, t)+c1))*(diff(u(x, t), x, x, x, x))-(2*(b^2-1))*(diff(u(x, t), x, x, t)+(3*a1*u(x, t)^2+2*b1*u(x, t)+c1+6*a1*u(x, t)+2*b1)*(diff(u(x, t), x, x)))+(b^2+1)^2 = 4*b*(b^2+1)^2*Heaviside(u(x, t)-th);
/ d / d / d / d / d \\\\\
|--- |--- |--- |--- |--- u(x, t)|||||
\ dx \ dx \ dx \ dx \ dt /////

/ 2 \ / d / d / d / d \\\\
+ \-10 + 12 u(x, t) - 12 u(x, t)/ |--- |--- |--- |--- u(x, t)||||
\ dx \ dx \ dx \ dx ////

/ d / d / d \\\
+ 1.8750 |--- |--- |--- u(x, t)|||
\ dx \ dx \ dt ///

/ 2 \ / d / d \\
+ 1.8750 \-6 u(x, t) - 6 u(x, t) + 5/ |--- |--- u(x, t)|| + 1.12890625 =
\ dx \ dx //

1.128906250 Heaviside(u(x, t) - 1.5)
> pde1 := diff(u(x, t), x, x, x, x, t)-v(x, t)+(6*a1-2*(3*a1*u(x, t)^2+2*b1*u(x, t)+c1))*(diff(u(x, t), x, x, x, x))-(2*(b^2-1))*(diff(u(x, t), x, x, t)+(3*a1*u(x, t)^2+2*b1*u(x, t)+c1+6*a1*u(x, t)+2*b1)*(diff(u(x, t), x, x)))+(b^2+1)^2 = 4*b*(b^2+1)^2*Heaviside(v(x, t)-th);
/ d / d / d / d / d \\\\\
|--- |--- |--- |--- |--- u(x, t)||||| - v(x, t)
\ dx \ dx \ dx \ dx \ dt /////

/ 2 \ / d / d / d / d \\\\
+ \-10 + 12 u(x, t) - 12 u(x, t)/ |--- |--- |--- |--- u(x, t)||||
\ dx \ dx \ dx \ dx ////

/ d / d / d \\\
+ 1.8750 |--- |--- |--- u(x, t)|||
\ dx \ dx \ dt ///

/ 2 \ / d / d \\
+ 1.8750 \-6 u(x, t) - 6 u(x, t) + 5/ |--- |--- u(x, t)|| + 1.12890625 =
\ dx \ dx //

1.128906250 Heaviside(v(x, t) - 1.5)
> pde2 := diff(v(x, t), t) = th*u(x, t)-b*v(x, t);
d
--- v(x, t) = 1.5 u(x, t) - 0.25 v(x, t)
dt
> IC1 := {u(-1, t) = -.7, u(1, t) = -.7, u(x, 0) = (1*1)*cos(L*x/(12.5*Pi))*exp(-(L*x/(12.5*Pi))^2), v(-1, t) = .1, v(1, t) = .1, v(x, 0) = x^2};
/
|
< u(-1, t) = -0.7, u(1, t) = -0.7,
|
\

/ 2\
/1.200000000 x\ | 1.440000000 x |
u(x, 0) = cos|-------------| exp|- --------------|, v(-1, t) = 0.1,
\ Pi / | 2 |
\ Pi /

\
2|
v(1, t) = 0.1, v(x, 0) = x >
|
/
> sys := {pde1, pde2};
// d / d / d / d / d \\\\\
{ |--- |--- |--- |--- |--- u(x, t)||||| - v(x, t)
\\ dx \ dx \ dx \ dx \ dt /////

/ 2 \ / d / d / d / d \\\\
+ \-10 + 12 u(x, t) - 12 u(x, t)/ |--- |--- |--- |--- u(x, t)||||
\ dx \ dx \ dx \ dx ////

/ d / d / d \\\
+ 1.8750 |--- |--- |--- u(x, t)|||
\ dx \ dx \ dt ///

/ 2 \ / d / d \\
+ 1.8750 \-6 u(x, t) - 6 u(x, t) + 5/ |--- |--- u(x, t)|| + 1.12890625 =
\ dx \ dx //

1.128906250 Heaviside(v(x, t) - 1.5),

d \
--- v(x, t) = 1.5 u(x, t) - 0.25 v(x, t) }
dt /
> pds1 := pdsolve(sys, IC1, numeric, time = t, range = -1 .. 1, spacestep = 1/4);
module () local INFO; export plot, plot3d, animate, value, settings; option ; end module

> p1 := pds1:-plot(t = 0.1e-1, x = -1 .. 1, numpoints = 50, color = black);

p2 := pds1:-plot(t = .2, linestyle = 2, numpoints = 50, color = black);

p3 := pds1:-plot(t = 10, numpoints = 50, color = black, linestyle = 3, labels = [u, x]);

plots[display]({p1, p2, p3});
Error, (in pdsolve/numeric/plot) unable to compute solution for t>0.:
matrix is singular
Algis
Please Wait...