Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

Hi all,

I will use the following dummy example.

with function,f

f:=(xid,yid)->sum(x[i],i=1..xid)*sum(y[i],i=1..yid);

 

and a complicated term, myterm


myterm:=(f(3,4)+f(2,2))*f(1,1):
myterm:=expand(myterm);

 

'if' i have some previous knowledge, or know a bit of the term, i can find the structure by doing this


repar:=[f(1,1),f(2,2),f(3,4)];  # Or with more f(xid,yid) terms

tmp:=seq(repar[i]=ff[i],i=1..3);
simplify(myterm,{tmp});          # This is fine, gives me what i want

 

But, can we go further, and more 'obvious'

 

Given the fucntion f, same as before, and the same 'myterm'

can I have this
restart:
iwant:=(f(3,4)+f(2,2))*f(1,1);  # as a result, straightforward

so I dont have to go back to 'repar' and find that the terms exactly are.

 

Thanks,

 


restart; printf("Factory Simulation: %s\n\n", Ghana*Chocolates); printf("Chocolate Factory Simulation Project .\n\n"); printf("%5a  %10a  %10a  %10a  %10a  %10a   \n", W, O(t, t+1), B(t), T(t), F(t), R(t)); printf(" ---------------------------------------------------------------------------\n"); _local(T, R, F, B, O, Q); initialise := proc (n) local i, t, a; T := proc (t) options operator, arrow; 250 end proc; R := proc (t) options operator, arrow; 150 end proc; F := proc (t) options operator, arrow; 250 end proc; B := proc (t) options operator, arrow; 50 end proc; for i to n do t := i; if B(t) < F(t) then Ds(t, t+1) := B(t) else Ds(t, t+1) := F(t) end if; M(t, t+1) := 50; O(t, t+1) := 50; P(t, t+1) := 50; if t = 2 then O(t, t+1) := 100 elif t = 3 then O(t, t+1) := 0 end if; P(t, t+1) := T(t)-F(t)+Ds(t, t+1); if R(t) < P(t, t+1) then P(t, t+1) := R(t) elif R(t) < 0 then P(t, t+1) := 0 end if; R(t+1) := R(t)+M(t, t+1)-P(t, t+1); F(t+1) := F(t)+P(t, t+1)-Ds(t, t+1); B(t+1) := B(t)+O(t, t+1)-Ds(t, t+1); R(t) := R(t+1); F(t) := F(t+1); B(t) := B(t+1); for a to t do Q(t, t+1) := O(t, t+1)+O(t-t+a, t-t+a+1) end do; T(t+1) := (t+1)*Q(t, t+1)/t; printf("%5a  %10a  %10a  %10a  %10a  %10a   \n", i, evalf[3](O(t, t+1)), evalf[3](B(t)), evalf[3](T(t)), evalf[3](F(t)), evalf[3](R(t))) end do; printf(" ---------------------------------------------------------------------------\n\n\n"); Restart; printf("%5a  %10a  %10a  %10a     \n", W, M(t, t+1), P(t, t+1), Ds(t, t+1)); printf(" ---------------------------------------------------------------------------\n") end proc; initialise(10)

    1         50.         50.        250.        250.        150.   
    2        100.        100.        200.        200.        200.   
    3          0.          0.        300.        300.         50.   
    4         50.         50.          0.          0.        400.   
    5         50.        100.        125.        125.        325.   
    6         50.         50.        120.        120.        280.   
    7         50.         50.        117.        117.        283.   
    8         50.         50.        114.        114.        286.   
    9         50.         50.        112.        112.        288.   
   10         50.         50.        111.        111.        289.   
 ---------------------------------------------------------------------------


    W          50      875/18          50     
 ---------------------------------------------------------------------------

 

NULL

NULL


Download suppress_warnins.mw

Hi All,

 

I need to suppress the warning messages when i try to pring my Table. Any help will be much apreciated.

 

Regards,

Xcyborg

I’m trying to figure out how to find a basis for a subspace, V, of Rdefined by V = {(x, y, z)l(2x-3y+6z=0)}

 

I’m using the student linear algebra module for maple 17

 

I’ve tried defining the subspace and asking for the basis of V but I always get an error code.

 

I’ve tried consulting the maple website and looking through their help menu, but can’t find anything that answers how to find a basis... At least a basis from the subspace defined in my problem.

I know how to find a basis for the subspace by hand but not with maple.

Any help will be greatly appreciated. 

(1/2)*(-x-y+sqrt(-3*x^2-6*x*y-3*y^2))/(x+y)

 

the complex value is always a constant whatever x and y 

i suspected that this is a wrong function

Hello: I'm looking over the Help section but I can not find a Maple package that has a command to compute the symmetric sum or the cyclic sum. I just started working with inequalities. Please could anyone recommend a package that allows me to compute expressions related to Muirhead's Inequality (see part 2 of  the answer:  http://math.stackexchange.com/questions/488967/majorization-muirhead)

I want to introduce a matrix of order M by M as for any m, M, pl help as show in file

 

Help.mw

 

 

Hi All,

I have a stupid question.

I wanted to distinguish the fuction of map and Map and tried some test.

Sometimes, the Map is identical to the map:

But sometimes not:

 


Maple Help seems not to explain the difference very clearly.

map_and_Map.mw

I'd appreciate any help on this topic. Thank a lot.

i want to find initial condition for F(0),G(0), H(0) and thetap(0) which is are missing in this problem.. then i facing this error Error, (in dsolve/numeric/bvp) Newton iteration is not converging



Maple Worksheet - Error

Failed to load the worksheet /maplenet/convert/hydronew2.mw .

Download hydronew2.mw

Hi, with a list

l:=[1,1,1,2,3,3,4];

What's the best way to get the index(s) for the values equal to '1'?

Say for x=1, we want

[1,2,3]

for x=2, we want

[4]

ect.

 

I'm calculating the geodesics to a parametrized system in R3. When trying to solve the geodesic equations for a surface of revolution, I'm getting a strange error. The goal is to write the code for any parametrized surface, not just revolution (hence I didn't use the short cut for surfaces of revolution).

For the code, I find the first fundamental for, then the geodesics, then turn the christoffel symbols into a time-variant array. (Gamma for the position of (u,v) in the uv-plane, and C as the array so I can take derivatives with respect to time.)

The code for creating the Christoffel Symbols and the parametrization of the paraboloid:

restart; with(LinearAlgebra):
V := (u, v) -> <v*cos(u), v*sin(u), v^2+1>;

Christoff := proc (X)
local x1, x2, M, N, i, j, k, s, E, F, G, g, Q, Delta, Prelim, cyclicPrelim;
global Gamma, C; #GAMMA
x1 := (u, v) -> <diff(X[1], u), diff(X[2], u), diff(X[3], u)>;
x2 := (u, v)-> <(diff(X[1], v), diff(X[2], v), diff(X[3], v))>;
E :=  (u, v) -> DotProduct(x1(u, v), x1(u, v), conjugate = false);
F := (u, v) -> DotProduct(x1(u, v), x2(u, v), conjugate = false);
G := (u, v) -> DotProduct(x2(u, v), x2(u, v), conjugate = false);
simplify([E(u, v), F(u, v), G(u, v)]);
M := (u, v) -> <E(u, v), F(u, v); F(u, v), G(u, v)>;
simplify(M(u,v));
printlevel := 3;
Delta := simplify(Determinant(M(u, v)));
N := (1/Delta)*<G(u, v), -F(u, v); -F(u, v), E(u, v)>;
Q[1] := simplify(map(diff, M(u, v), u));
Q[2] := simplify(map(diff, M(u, v), v));
for i to 2 do for j to 2 do for k to 2 do
Prelim[i, j, k] := Q[k][i, j];
simplify(Prelim[i, j, k] ); end do end do end do;
#(OPTIONAL PRINTOUT) print(Prelim);
for i to 2 do for j to 2 do for k to 2 do
cyclicPrelim[i, j, k] := Prelim[i, j, k]+Prelim[j, k, i]-Prelim[k, i, j] ;
end do end do end do;
#(OPTIONAL PRINTOUT)  print(cyclicPrelim);
for i to 2 do for j to 2 do for k to 2 do
Gamma[i, j, k] := simplify((1/2)*add(N[i, s]*cyclicPrelim[j, s, k], s = 1 .. 2));
end do end do end do;
# for k from 1 to 2 do
#print(`GAMMA[i,j,k] =` Matrix([[ `%a` , `%a` ],[ `%a` , `%a` ]])` \n`, Gamma[1,1,k], [Gamma[1,2,k], Gamma[2,1,k], Gamma[2,2,k]);
# end do;
#printf('GAMMA[%a,%a,%a] = %a \n', i,j,k, Gamma[i,j,k]);

print([Gamma[1,1,1], Gamma[1,2,1], Gamma[2,1,1], Gamma[2,2,1], Gamma[1,1,2], Gamma[1,2,2], Gamma[2,1,2], Gamma[2,2,2]]);

for i from 1 to 2 do
for j from 1 to 2 do
for k from 1 to 2 do
C[i,j,k]:= apply(Gamma[i,j,k],t);
end do end do end do;
C:=Array(1..2,1..2,1..2,[ [ [ apply(Gamma[1,1,1],t), apply(Gamma[1,2,1],t) ], [ apply(Gamma[1,1,2],t), apply(Gamma[1,2,2],t) ] ], [ [ apply(Gamma[2,1,1],t), apply(Gamma[2,2,1],t) ],  [ apply(Gamma[2,1,2],t), apply(Gamma[2,2,2],t) ] ] ]);
print(C);

end proc;

 

The differential equation solver:

inits:=[u(0)=1, D(u(t))(0)=1,v(0) = 1, D(v(t))(0)=1];

sys1:= [D[1$2](u(t))+C[1,1,1]*(D(u(t)))^(2)+2*C[1,1,2]*(D(u(t)))*(D(v(t)))+C[1,2,2]*(D(v(t)))^2=0, D[1$2](v(t))+C[2,1,1]*(D(u(t)))^(2)+2*C[2,1,2]*(D(u(t))*D(v(t)))^(2)+C[2,2,2]*(D(v(t)))^2=0];

L:=dsolve({sys1} union {inits});

 

The error message that comes up is:

"Error, (in unknown) invalid input: op expects 1 or 2 arguments, but received 0"

 

 

Any help would be greatly appreciated.

Hi say I have the vector V1.

V1:=Vector([a,b,c,d,e,f,g]):

and function myfun.

 

how do i use it as the input to the function my fun, by taking away each element in turn?
myfun(V1[2..]);              # 1st element removed
myfun(V1[[1,3..]]);         # 2nd element removed
myfun(V1[[1,2,4..]]);      # 3rd element removed

and so go

is there a more efficient way?

 

Many thanks,

i am trying to solve 6 ODE with boundary condition


restart

with*plots

with*plots

(1)

Eq1 := (1-theta(eta)/theta[r])*(diff(f(eta), eta, eta, eta))+(diff(f(eta), eta, eta))*(diff(theta(eta), eta))/theta[r]+(1-theta(eta)/theta[r])^2*(f(eta)*(diff(f(eta), eta, eta))-(diff(f(eta), eta))^2-M*(diff(f(eta), eta))+B*H(eta)*(F(eta)-(diff(f(eta), eta)))) = 0

(1-theta(eta)/theta[r])*(diff(diff(diff(f(eta), eta), eta), eta))+(diff(diff(f(eta), eta), eta))*(diff(theta(eta), eta))/theta[r]+(1-theta(eta)/theta[r])^2*(f(eta)*(diff(diff(f(eta), eta), eta))-(diff(f(eta), eta))^2-M*(diff(f(eta), eta))+B*H(eta)*(F(eta)-(diff(f(eta), eta)))) = 0

(2)

Eq2 := G(eta)*(diff(F(eta), eta))+F(eta)^2+B*(F(eta)-(diff(f(eta), eta))) = 0

G(eta)*(diff(F(eta), eta))+F(eta)^2+B*(F(eta)-(diff(f(eta), eta))) = 0

(3)

Eq3 := G(eta)*(diff(G(eta), eta))+B*(f(eta)+G(eta)) = 0

G(eta)*(diff(G(eta), eta))+B*(f(eta)+G(eta)) = 0

(4)

Eq4 := G(eta)*(diff(H(eta), eta))+H(eta)*(diff(G(eta), eta))+F(eta)*H(eta) = 0

G(eta)*(diff(H(eta), eta))+H(eta)*(diff(G(eta), eta))+F(eta)*H(eta) = 0

(5)

Eq5 := (1+s*theta(eta))*(diff(theta(eta), eta, eta))+(diff(theta(eta), eta))^2*s+Pr*(f(eta)*(diff(theta(eta), eta))-(diff(f(eta), eta))*theta(eta))+(2/3)*B*H(eta)*(theta[p](eta)-theta(eta)) = 0

(1+s*theta(eta))*(diff(diff(theta(eta), eta), eta))+(diff(theta(eta), eta))^2*s+Pr*(f(eta)*(diff(theta(eta), eta))-(diff(f(eta), eta))*theta(eta))+(2/3)*B*H(eta)*(theta[p](eta)-theta(eta)) = 0

(6)

Eq6 := 2*F(eta)*theta[p](eta)+G(eta)*(diff(theta[p](eta), eta))+L0*B*(theta[p](eta)-theta(eta)) = 0

2*F(eta)*theta[p](eta)+G(eta)*(diff(theta[p](eta), eta))+L0*B*(theta[p](eta)-theta(eta)) = 0

(7)

bcs1 := f(0) = 0, (D(f))(0) = 1, (D(f))(10) = 0;

f(0) = 0, (D(f))(0) = 1, (D(f))(10) = 0

(8)

fixedparameter := [M = .5, B = .5, theta[r] = -10, L0 = 1, s = .1, Pr = 1];

[M = .5, B = .5, theta[r] = -10, L0 = 1, s = .1, Pr = 1]

(9)

Eq7 := eval(Eq1, fixedparameter);

(1+(1/10)*theta(eta))*(diff(diff(diff(f(eta), eta), eta), eta))-(1/10)*(diff(diff(f(eta), eta), eta))*(diff(theta(eta), eta))+(1+(1/10)*theta(eta))^2*(f(eta)*(diff(diff(f(eta), eta), eta))-(diff(f(eta), eta))^2-.5*(diff(f(eta), eta))+.5*H(eta)*(F(eta)-(diff(f(eta), eta)))) = 0

(10)

Eq8 := eval(Eq2, fixedparameter);

G(eta)*(diff(F(eta), eta))+F(eta)^2+.5*F(eta)-.5*(diff(f(eta), eta)) = 0

(11)

Eq9 := eval(Eq3, fixedparameter);

G(eta)*(diff(G(eta), eta))+.5*f(eta)+.5*G(eta) = 0

(12)

Eq10 := eval(Eq5, fixedparameter);

(1+.1*theta(eta))*(diff(diff(theta(eta), eta), eta))+.1*(diff(theta(eta), eta))^2+f(eta)*(diff(theta(eta), eta))-(diff(f(eta), eta))*theta(eta)+.3333333333*H(eta)*(theta[p](eta)-theta(eta)) = 0

(13)

Eq11 := eval(Eq6, fixedparameter);

2*F(eta)*theta[p](eta)+G(eta)*(diff(theta[p](eta), eta))+.5*theta[p](eta)-.5*theta(eta) = 0

(14)

bcs2 := F(10) = 0;

F(10) = 0

(15)

bcs3 := G(10) = -f(10);

G(10) = -f(10)

(16)

bcs4 := H(10) = n;

H(10) = n

(17)

bcs5 := theta(10) = 0;

theta(10) = 0

(18)

bcs6 := theta[p](10) = 0;

theta[p](10) = 0

(19)

L := [.2];

[.2]

(20)

for k to 1 do R := dsolve(eval({Eq10, Eq11, Eq4, Eq7, Eq8, Eq9, bcs1, bcs2, bcs3, bcs4, bcs5, bcs6}, n = L[k]), [f(eta), F(eta), G(eta), H(eta), theta(eta), theta[p](eta)], numeric, output = listprocedure); Y || k := rhs(R[5]); YP || k := rhs(R[6]); YJ || k := rhs(R[7]); YS || k := rhs(R[2]) end do

``


Download hydro.mw

restart

with*plots

with*plots

(1)

Eq1 := (1-theta(eta)/theta[r])*(diff(f(eta), eta, eta, eta))+(diff(f(eta), eta, eta))*(diff(theta(eta), eta))/theta[r]+(1-theta(eta)/theta[r])^2*(f(eta)*(diff(f(eta), eta, eta))-(diff(f(eta), eta))^2-M*(diff(f(eta), eta))+B*H(eta)*(F(eta)-(diff(f(eta), eta)))) = 0

(1-theta(eta)/theta[r])*(diff(diff(diff(f(eta), eta), eta), eta))+(diff(diff(f(eta), eta), eta))*(diff(theta(eta), eta))/theta[r]+(1-theta(eta)/theta[r])^2*(f(eta)*(diff(diff(f(eta), eta), eta))-(diff(f(eta), eta))^2-M*(diff(f(eta), eta))+B*H(eta)*(F(eta)-(diff(f(eta), eta)))) = 0

(2)

Eq2 := G(eta)*(diff(F(eta), eta))+F(eta)^2+B*(F(eta)-(diff(f(eta), eta))) = 0

G(eta)*(diff(F(eta), eta))+F(eta)^2+B*(F(eta)-(diff(f(eta), eta))) = 0

(3)

Eq3 := G(eta)*(diff(G(eta), eta))+B*(f(eta)+G(eta)) = 0

G(eta)*(diff(G(eta), eta))+B*(f(eta)+G(eta)) = 0

(4)

Eq4 := G(eta)*(diff(H(eta), eta))+H(eta)*(diff(G(eta), eta))+F(eta)*H(eta) = 0

G(eta)*(diff(H(eta), eta))+H(eta)*(diff(G(eta), eta))+F(eta)*H(eta) = 0

(5)

Eq5 := (1+s*theta(eta))*(diff(theta(eta), eta, eta))+(diff(theta(eta), eta))^2*s+Pr*(f(eta)*(diff(theta(eta), eta))-(diff(f(eta), eta))*theta(eta))+(2/3)*B*H(eta)*(theta[p](eta)-theta(eta)) = 0

(1+s*theta(eta))*(diff(diff(theta(eta), eta), eta))+(diff(theta(eta), eta))^2*s+Pr*(f(eta)*(diff(theta(eta), eta))-(diff(f(eta), eta))*theta(eta))+(2/3)*B*H(eta)*(theta[p](eta)-theta(eta)) = 0

(6)

Eq6 := 2*F(eta)*theta[p](eta)+G(eta)*(diff(theta[p](eta), eta))+L0*B*(theta[p](eta)-theta(eta)) = 0

2*F(eta)*theta[p](eta)+G(eta)*(diff(theta[p](eta), eta))+L0*B*(theta[p](eta)-theta(eta)) = 0

(7)

bcs1 := f(0) = 0, (D(f))(0) = 1, (D(f))(10) = 0;

f(0) = 0, (D(f))(0) = 1, (D(f))(10) = 0

(8)

fixedparameter := [M = .5, B = .5, theta[r] = -10, L0 = 1, s = .1, Pr = 1];

[M = .5, B = .5, theta[r] = -10, L0 = 1, s = .1, Pr = 1]

(9)

Eq7 := eval(Eq1, fixedparameter);

(1+(1/10)*theta(eta))*(diff(diff(diff(f(eta), eta), eta), eta))-(1/10)*(diff(diff(f(eta), eta), eta))*(diff(theta(eta), eta))+(1+(1/10)*theta(eta))^2*(f(eta)*(diff(diff(f(eta), eta), eta))-(diff(f(eta), eta))^2-.5*(diff(f(eta), eta))+.5*H(eta)*(F(eta)-(diff(f(eta), eta)))) = 0

(10)

Eq8 := eval(Eq2, fixedparameter);

G(eta)*(diff(F(eta), eta))+F(eta)^2+.5*F(eta)-.5*(diff(f(eta), eta)) = 0

(11)

Eq9 := eval(Eq3, fixedparameter);

G(eta)*(diff(G(eta), eta))+.5*f(eta)+.5*G(eta) = 0

(12)

Eq10 := eval(Eq5, fixedparameter);

(1+.1*theta(eta))*(diff(diff(theta(eta), eta), eta))+.1*(diff(theta(eta), eta))^2+f(eta)*(diff(theta(eta), eta))-(diff(f(eta), eta))*theta(eta)+.3333333333*H(eta)*(theta[p](eta)-theta(eta)) = 0

(13)

Eq11 := eval(Eq6, fixedparameter);

2*F(eta)*theta[p](eta)+G(eta)*(diff(theta[p](eta), eta))+.5*theta[p](eta)-.5*theta(eta) = 0

(14)

bcs2 := F(10) = 0;

F(10) = 0

(15)

bcs3 := G(10) = -f(10);

G(10) = -f(10)

(16)

bcs4 := H(10) = n;

H(10) = n

(17)

bcs5 := theta(10) = 0;

theta(10) = 0

(18)

bcs6 := theta[p](10) = 0;

theta[p](10) = 0

(19)

L := [.2];

[.2]

(20)

for k to 1 do R := dsolve(eval({Eq10, Eq11, Eq4, Eq7, Eq8, Eq9, bcs1, bcs2, bcs3, bcs4, bcs5, bcs6}, n = L[k]), [f(eta), F(eta), G(eta), H(eta), theta(eta), theta[p](eta)], numeric, output = listprocedure); Y || k := rhs(R[5]); YP || k := rhs(R[6]); YJ || k := rhs(R[7]); YS || k := rhs(R[2]) end do

``


then i get this error

Error, (in dsolve/numeric/bvp/convertsys) unable to convert to an explicit first-order system

i dont know where i need to change after view it one by one..

Download hydro.mw

I'll start with a quick positive.  One of the great advantages of upgraded software is the wealth of new features that we all get to play around with.  .. and then I will counter that with a great disadvantage, and that is, we all just about get familiar and comfortable with all the new features then BAM! a new version is released.  Of course we're then mesmorized once again by all the new bells and whistles and maybe even a couple of great celebrations occur with nice small updates throughout the year.  The other downside is that even though a large number of bugs may have been fixed a number of new ones are broght in with those new features. 

A side effect of a fast release is there are fewer and fewer applications associated with a release, and that is apparent in the application center.  Although mobius apps and the maple cloud may have also had some impact on that as well.

Now this is pale in comparison to book writers who scramble to keep their books current with new software.  I will quote a section from the introduction in the book Essential Maple 7 which highlights the problems the author had way back then .. I can't imagine how they feel now but here's the passage ...

"Indeed, one reason that there was so much time between the first and second
editions of this book is precisely that Maple has been evolving so rapidly in the
last few years, too rapidly for me to revise this book (much less complete my
others) while coping with my other duties."

That just hits the nail on the head, if you think Maple was evolving fast back then, the furious rate that upgrades are released now I would think authors have an almost impossible task to keep up. 

There are many that would agree with the author, that Maple is advancing so rapidly that we barely have time to gather our thoughts.  Maybe a solution is that we should slow down and create a much more polished piece of software, but again the caveat to that is our competition might just jump out in front.  However the norm today is that each new year represents a new release of software and we all celebrate when that happens.  If life seemed rushed back when Maple 7 was released I can't imagine what it'll be like 10 years from now when Maple 28 rolls around. 

Dear Mapleprimes,

I have been struggling with a problem in the last couple of days. I wish to export a Maple plot to LaTeX while ensuring font consistency. While searching for solutions online, I found the psfrag package in LaTeX. So far, however, I have been unsuccesful in making this work. As as test, I attempted to export plot(x^2) to LaTeX. I used the following code to convert to .eps which worked fine:

plotsetup(ps, plotoutput = `plot1.eps`, plotoptions = `portrait, noborder,height=5in,width=5in`);plot(x^2);

Then in LaTeX, I have:

\documentclass{article}

\usepackage{graphicx}

\usepackage{psfrag}

\begin{document}

\begin{figure}[!h]
\centering
\psfrag{x}{$ \alpha $}
\includegraphics[scale=0.5]{plot1.eps}
\end{figure}
\end{document}

However, no replacements are made. After intense Google searching I found the following post http://www.mapleprimes.com/posts/43255-Trouble-Replacing-Maple-Axes-Labels which to sum up argues that this was only possible with earlier versions of Maple.

Does anyone know if the problem has been resolved?

Does anyone know any other ways to ensure font consistency for plots imported from Maple to LaTeX?

Thank you very much in advance!

C

Is it possible to solve (numerically or symbolically) the system of PDEs
sys:={diff(Y(x, t), x$2) = exp(-2*x*b)*(A(x, t)-Y(x, t)), diff(A(x, t), t) = exp(-2*x*b)*(Y(x, t)-A(x, t)) }
under the conditions
ibc:={A(x, 0) = 0, Y(0, t) = 0.1, D[1](Y)(0, t) = 0},
 where the parameter b takes the values 0,0.05,0.1, in Maple? The ranges are t=0..7, x=0..20.

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