J4James

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11 years, 313 days

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These are replies submitted by J4James

@Preben Alsholm Thanks dear.

Here I shall share the background.

ODE:=diff(u(y),y)=y*m*exp(Lambda*(y*m)^2);

bcs:=D(u)(0)=0,u(h)=-1;

dsolve({ODE,bcs});

uu:=rhs(%);

Eq1:=F-int(uu,y=0..h)=0; 

DDP:=[solve(FF1,m)];#Now here I get the need to solve this equation for m

m:=%[];

U:=(evalf(uu)); #

h:=1+phi*cos(2*Pi*x):

F:=A-1:

Lambda:=2*B*W^2:


A:=2:x:=1:B:=3:W:=2:

plot([subs(phi=0.0,U),subs(phi=0.2,U)],y=-1..1,axes=box,color=[red,green],adaptive= false);

I need m to be in a usable form.

 

@Carl Love Thanks. How to save the animation as .avi extension?

@Carl Love 

For the case Q = 2 view= [DEFAULT, DEFAULT, -0.6..0.3] in the animate worked just fine 

but for the case Q = 0, I lost the comparison again.

animate( plot3d, [TSol,y=-1..1,x=0..1], Q=0..2,contours=65,style=patchnogrid, grid=[80,80], orientation=[-360,-180], lightmodel=light4, shading=zhue, transparency=0.3,style=contour,view= [DEFAULT, DEFAULT, -0.6..0.3],frames=45);

plot3d(subs(Q=0,TSol),y=-1..1,x=0..1,contours=65,style=patchnogrid, grid=[80,80], orientation=[-360,-180], lightmodel=light4, shading=zhue,  transparency=0.3,style=contour);

What I am missing here?

I am want that on each frame the animate( plot3d) and plot3d must have the same ouput.

@Carl Love 

Worked like a charm!

Cheers!!!

@Carl Love Thanks for your input.

How we can get the same results from the two?

forexample?

Cheers!

@Carl Love Thanks.

Also need your input on the issue with match while having a parameter in the Eq1.

Eq1:= :=a*diff(psi(y),y$4)-diff(psi(y),y$2)=0; # a is parameter

 

@Carl Love 

Why eval? In this case, it seems that eval doesn't work

Cs:=solve({Eq4,Eq5,Eq6,Eq7},[_C1,_C2,_C3,_C4]):
C1 :=eval(_C1,Cs):

I have updated my previous comment.

@Preben Alsholm match worked in this case but if

Eq1:=a*diff(psi(y),y$4)-diff(psi(y),y$2)=0; # a is parameter

or 

Eq1:=diff(psi(y),y$4)-b*diff(psi(y),y$2)=0; # b is parameter

res1:=dsolve(Eq1);
bcs:=psi(h1)=F,D(psi)(h1)=-1,psi(h2)=-F,D(psi)(h2)=-1;
res2:=simplify(dsolve({Eq1,bcs},psi(y)));
match(rhs(res2)=rhs(res1),y,s);

and then match doesn't work.

 

How to single out/call _C1 only from s?

 

@Preben Alsholm Thanks, it worked! 

Is there a way to find/extract those 4 constants of integration in a compressed form? 

@mehdi jafari 

paste it here

@mehdi jafari 

Nowhere I mentioned about getting a numerical sol. 

The question here is how to use maple in a robust way to find those 4 constants of int for a complete exact sol? 

@mehdi jafari 

Sorry for the confusion.

I am not trying to have a numeric solution. I am looking for an exact solution.

dsolve({Eq1});

 gives me an exact solution of the form _C1+_C2*y+_C3*exp(y)+_C4*exp(-y) with 4 unkowns, which I can then find using the bcs but ...

The point is, why can't the dsolve handle itself in the case of finding an exact solution?

 Moreover, Is there a rubost way to find these unknowns?

@Carl Love 

f:=1; #is the same as eq1

let assume that eq1 has a solution of the form

f(x)=1-exp(-x);

which happens to satisfy the conditions f(0)=0 and D(f)(10) =1.

What I am trying here is to get a solution (numerical one) for eq1 which should satisfy both conditions.

 

 

@Preben Alsholm 

plot(eval(lambda1,{alpha=-1,S=-0.5}),K=0.0..2);

complex values were detected

How to impose this condition alpha>=-(1/2)*S^2/(K+2) in plot (not in 3d)?

 

@Markiyan Hirnyk Thanks, I will try your suggested approach.

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