I am getting the following expression when I partially differentiate an expression:

PDE11 := diff(theta(z, p), z, z, p)+2*lambda(p)*theta(z, p)*(diff(lambda(p), p))+lambda(p)^2*(diff(theta(z, p), p))+lambda(p)^2*(sin(theta(z, p))-theta(z, p))+2*p*lambda(p)*(sin(theta(z, p))-theta(z, p))*(diff(lambda(p), p))+p*lambda(p)^2*(cos(theta(z, p))*(diff(theta(z, p), p))-(diff(theta(z, p), p)))

I differentiate the above equation to get each term in the form of :table([f=......])

(table([f = 1+sum(Lambda[n](0)/factorial(n), n = 1 .. infinity)]))(p)^2

It is difficult to understand the expression. Maple does not show any error. Can you please tell me what the error is?

hi every one

please see attached file blew and help me for dsolve set equations.i dont know aboue errors with in

thanks

equation.mw

Take for instance the signal [1,0,-3,2,1,0,1,2] (considered in Wavelets for Kids, Vidakivic & Mueler, AMS, 1991).

I want to anlalyse it down to the level zero with Haar wavelets. It seems that MAPLE's commnad only applies once.

Hi !

I am trying to solve a pde with initial/boundary conditions, in the numeric mode. It works very well when I provide 3 numerical conditions : 1 initial condition (uniform distribution at first), and two boundary conditions.

Now I want to switch the initial condition to a function of r. It's a polynome I obtained after interpolation of another result. I tested it and the function interp20(r) works. But the pdsolve doesn't seem to evaluate the function, when it comes to start the procedure and pdsolve doesn't return a module as it should, although it doesn't send any message error.

You can see the code following, with the error messages I get.

Is it strictly impossible to use a function as an initial condition ?
Is it just a problem I could solve by converting the function or its result to something else ?
Is float the right type of input ? For example, if I write a:=283.15, is it a float ?
Have you already had similar problems ? How did you solve them ? Where could I find working code examples on this ?
Could I solve this problem with non-uniform initial condition with another Maple function ?

Then you very much for your help !

hi.i have problem for dsolve equation.please help me again

thanks alotequation.mw

Hi everybody.

In the following attached file, I have created 3 matrices with 13*13 and 13*6 dimensions and their components are numeric. I closed Maple and ran it again. Surprisingly, when I try to browse the matrices again, I encounter with this problem " empty RTable structure" while I imported the components, manually.

Thanks in advance for your answers.

Ho.mw

I think that I found a bug in Maple! Please run the following command:

I need the Generators of above Ideal. What is your idea?!

I have a problem about the integration of a function. The maple returns mathematical form of the given command

the function is

fn :=(-4.079067798*10^(-16)+3.422708023*10^(-16)*I)*(3.363377947*10^(-11)+5.977507284*10^(-12038)*I+(3.363377947*10^(-11)+4.678081798*10^(-22)*I)*erf(1.664331698*10^15*qq-0.2503507367e-1-4.649313602*I)) *exp(-2.77*(qq/(tau*tau))^2)*exp(-I*w0*qq)*exp(-(ss-qq)/T_pop)

qq is the integration variable qq=-inf..ss

w0=10^15

tau=10^(-15)

T_pop=30e-15

Hi everyone,

I am trying to solve the equation of heat tranfer, time dependent, with particular Initial and boundary conditions but I am stuck by technical problems both in getting an analytical solution and a numerical one.

The equation

the equation.

I defined a and b numerically. domain is : and I defined surf_power numerically.

The initial condition is : , T0 defined numerically

The boundary condition is : , because it has a shperical symetry.

To me, it looks like a well posed problem. Does it look fine ?

Problem in analytical solution :

It doesn't accept the boundary condition so I only input the initial condition and it actually gives me back an expression that can be evaluated but it never does : I can't reduce it more than an expression of fourier which I can't eval. The solution :
The solution calculated in (0,0). I was hoping T0...

Are you familiar with these problems ? What would be the perfect syntax you would use to solve this ?

The numerical solution problems :

Sometimes it tells me that my boundary condition is equivalent  to 0 = 0, and I don't see why. Some other times it tells me I only gave 1 boundary/initial condition even if I wrote both. Here is what I wrote for example :

(because it kept asking me to add these two options : 'time' and 'range')

Are you familiar with these problems ? What would be the perfect syntax you would use to solve this ? I must at least have syntax problems because even if I keep reading the Help, it's been a long time since I used Maple.

Thank very much for any indication you could give me !

Simon

Hello Mapleprime members

I am having an issue with a Maple script I have written that reads in a separate Maple script at the start of the code which deals with regular Jacobian transformations I apply to the problem and got tired writing it each time so externalised it and read it in each time.

On Maple 2015 on my Mac there is nothing wrong with the script and it runs without issue.

When I run it on Maple 17 on my linux machine It says the read in file does not exist and the only difference between both machines is the home directory path. (or perhaps version of Maple, but I have used thew following commands in older Maple versions)

The Maple script has this at the top:

cwd:=currentdir():

home:=getenv("HOME");
Expspath:=cat(home,"Pathway to the file to be read in");
currentdir(Expspath);

I am simply getting Maple to find the home directory and apply it to the start of the file path which i thought would universally work between machines, and allow the running of Maple scripts between operating systems.

When I run the program on the linux Maple 17, printing Expspath from above gives  exactly where the file is located as it finds the home directory correctly and uses the correct filepath, it just says it does not exist. is there something obvious I am missing?

The Maple file to be read in is on Dropbox so just after the home directory is /Dropbox. If that makes any difference. In both instances the file path is where the file is located (exact match) but the mac will read in the file and linux says it does not exist even though it is the correct filepath.

Any help would be appreciated

- Yeti

 my answer are 143.2030067,143.2030339,143.2030610

but, it is wrong. can anyone tell me where is the wrong part?

please!!

I am attempting to solve a system of second order ODEs. I place conditions on the solutions and use the solve command to figure the correct constants for the general solutions of the ODEs; however, the conditions do not appear to hold after I substitute the constants back into the general solutions. Any help would be greatly appreciated. Here's the code and an explanation:

First some constants

> A := 1; B := 9/10;
> j := 1-1/B;

 This is our homogeneous odes. I will give the general solutions of the inhomogeneous system momentarily 

> eqnv1 := diff(v1(x), `$`(x, 2)) = (1-1/(j+1))*v1(x)+v2(x)/(j+1);
> eqnv2 := diff(v2(x), `$`(x, 2)) = -v1(x)/(A*(j+1))+(B/A+1/(A*(j+1)))*v2(x);

Next we get the general solution of this sytem of odes.

> soln := dsolve([eqnv1, eqnv2])

Next we have our solutions of the inhomogeneous problem1. Basically solution v1neg, v2neg on [0,xi] and v1pos, v2pos on [xi,1]. We will assume v1,v2 are C^1 across xi; however, the location of xi is not known at this time so they must remain split.

> v1neg := op([1, 2], soln)-1;
> v2neg := op([2, 2], soln)-1/B;
> v1pos := op([1, 2], soln)+1;
> v2pos := op([2, 2], soln)+1/B;

There's probably a better way to do this, but I relabeled the constants:

> v1negc := subs([_C1 = a[1], _C2 = a[2], _C3 = a[3], _C4 = a[4]], v1neg);
> v2negc := subs([_C1 = a[1], _C2 = a[2], _C3 = a[3], _C4 = a[4]], v2neg);
>
> v1posc := subs([_C1 = a[5], _C2 = a[6], _C3 = a[7], _C4 = a[8]], v1pos);
> v2posc := subs([_C1 = a[5], _C2 = a[6], _C3 = a[7], _C4 = a[8]], v2pos);

Next we have eight conditions the solutions must satisfy. Namely v1, v2 are C^1 across xi and v1',v2' are 0 at {0,1}.

> syscon1 := subs(x = xi, v1negc) = subs(x = xi, v1posc);
> syscon2 := subs(x = xi, v2negc) = subs(x = xi, v2posc);
> syscon3 := subs(x = xi, diff(v1negc, x)) = subs(x = xi, diff(v1posc, x));
> syscon4 := subs(x = xi, diff(v2negc, x)) = subs(x = xi, diff(v2posc, x));
> syscon5 := subs(x = 0, diff(v1negc, x)) = 0;
> syscon6 := subs(x = 0, diff(v2negc, x)) = 0;
> syscon7 := subs(x = 1, diff(v1posc, x)) = 0;
> syscon8 := subs(x = 1, diff(v2posc, x)) = 0;

We solve to get the constants for the solutions.

> constants := simplify(evalf(solve({syscon1, syscon2, syscon3, syscon4, syscon5, syscon6, syscon7, syscon8}, {a[1], a[2], a[3], a[4], a[5], a[6], a[7], a[8]})));
>

We substitute the values for the constants.

> a[1] := op([1, 2], constants); a[2] := op([2, 2], constants); a[3] := op([3, 2], constants); a[4] := op([4, 2], constants); a[5] := op([5, 2], constants); a[6] := op([6, 2], constants); a[7] := op([7, 2], constants); a[8] := op([8, 2], constants);

Lastly we try to verify that the conditions from earlier hold:

> evalf(subs(xi = .2, subs(x = xi, v1negc-v1posc)));
-1.7597825261536669519
> evalf(subs(xi = .2, subs(x = xi, v2negc-v2posc)));
-1.8936659961101033997
> evalf(subs([x = 0, xi = .2], diff(v1negc, x)));
-0.38633519704430619686

They should hold for any xi, but they don't appear to. All of these should be 0. For a large xi, the numbers get very large so I was thinking perhaps roundoff error, but even when I do an exact solution and then evalf just at the end, I still have large error so I'm not sure what the problem is. Sorry for the long question. Thanks so much for the help.

Hi

Anyone could help me in solving the following system of equations to get constants C1, C2, C3 and C4. MALPE give me this "soution may have been lost".  The MAPLE sheet is also attached.

Download solution_lost.mw

Hi all,

I have this system

> system1D := H = alpha*gamma[2, 2]*d[2, 1]-beta*d[1, 2]*gamma[1, 2]^2-gamma*d[1, 2]*gamma[2, 1]^2+alpha*gamma[2, 2]^2*d[2, 2]-beta*d[2, 2]*gamma[2, 2]^2-gamma*d[2, 2]*gamma[2, 2]^2, E = alpha*gamma[2, 1]*d[1, 1]-beta*d[1, 2]*gamma[1, 1]-gamma*d[1, 1]*gamma[2, 1]+alpha*gamma[2, 1]^2*d[1, 2]-beta*d[2, 2]*gamma[2, 1]-gamma*d[2, 1]*gamma[2, 2], B = alpha*gamma[1, 1]*d[2, 1]-beta*d[1, 1]*gamma[1, 1]^2-gamma*d[1, 1]*gamma[1, 1]^2+alpha*gamma[1, 1]^2*d[2, 2]-beta*d[2, 1]*gamma[2, 1]^2-gamma*d[2, 1]*gamma[1, 2]^2, D = alpha*gamma[1, 2]*d[2, 1]-beta*d[1, 1]*gamma[1, 2]^2-gamma*d[1, 2]*gamma[1, 1]^2+alpha*gamma[1, 2]^2*d[2, 2]-beta*d[2, 1]*gamma[2, 2]^2-gamma*d[2, 2]*gamma[1, 2]^2, A = alpha*gamma[1, 1]*d[1, 1]-beta*d[1, 1]*gamma[1, 1]-gamma*d[1, 1]*gamma[1, 1]+alpha*gamma[1, 1]^2*d[1, 2]-beta*d[2, 1]*gamma[2, 1]-gamma*d[2, 1]*gamma[1, 2], C = alpha*gamma[1, 2]*d[1, 1]-beta*d[1, 1]*gamma[1, 2]-gamma*d[1, 2]*gamma[1, 1]+alpha*gamma[1, 2]^2*d[1, 2]-beta*d[2, 1]*gamma[2, 2]-gamma*d[2, 2]*gamma[1, 2], F = alpha*gamma[2, 1]*d[2, 1]-beta*d[1, 2]*gamma[1, 1]^2-gamma*d[1, 1]*gamma[2, 1]^2+alpha*gamma[2, 1]^2*d[2, 2]-beta*d[2, 2]*gamma[2, 1]^2-gamma*d[2, 1]*gamma[2, 2]^2, G = alpha*gamma[2, 2]*d[1, 1]-beta*d[1, 2]*gamma[1, 2]-gamma*d[1, 2]*gamma[2, 1]+alpha*gamma[2, 2]^2*d[1, 2]-beta*d[2, 2]*gamma[2, 2]-gamma*d[2, 2]*gamma[2, 2], H = alpha*delta[2, 2]*d[2, 1]-beta*d[1, 2]*delta[1, 2]^2-gamma*d[1, 2]*delta[2, 1]^2+alpha*delta[2, 2]^2*d[2, 2]-beta*d[2, 2]*delta[2, 2]^2-gamma*d[2, 2]*delta[2, 2]^2, E = alpha*delta[2, 1]*d[1, 1]-beta*d[1, 2]*delta[1, 1]-gamma*d[1, 1]*delta[2, 1]+alpha*delta[2, 1]^2*d[1, 2]-beta*d[2, 2]*delta[2, 1]-gamma*d[2, 1]*delta[2, 2], B = alpha*delta[1, 1]*d[2, 1]-beta*d[1, 1]*delta[1, 1]^2-gamma*d[1, 1]*delta[1, 1]^2+alpha*delta[1, 1]^2*d[2, 2]-beta*d[2, 1]*delta[2, 1]^2-gamma*d[2, 1]*delta[1, 2]^2, D = alpha*delta[1, 2]*d[2, 1]-beta*d[1, 1]*delta[1, 2]^2-gamma*d[1, 2]*delta[1, 1]^2+alpha*delta[1, 2]^2*d[2, 2]-beta*d[2, 1]*delta[2, 2]^2-gamma*d[2, 2]*delta[1, 2]^2, A = alpha*delta[1, 1]*d[1, 1]-beta*d[1, 1]*delta[1, 1]-gamma*d[1, 1]*delta[1, 1]+alpha*delta[1, 1]^2*d[1, 2]-beta*d[2, 1]*delta[2, 1]-gamma*d[2, 1]*delta[1, 2], C = alpha*delta[1, 2]*d[1, 1]-beta*d[1, 1]*delta[1, 2]-gamma*d[1, 2]*delta[1, 1]+alpha*delta[1, 2]^2*d[1, 2]-beta*d[2, 1]*delta[2, 2]-gamma*d[2, 2]*delta[1, 2], F = alpha*delta[2, 1]*d[2, 1]-beta*d[1, 2]*delta[1, 1]^2-gamma*d[1, 1]*delta[2, 1]^2+alpha*delta[2, 1]^2*d[2, 2]-beta*d[2, 2]*delta[2, 1]^2-gamma*d[2, 1]*delta[2, 2]^2, G = alpha*delta[2, 2]*d[1, 1]-beta*d[1, 2]*delta[1, 2]-gamma*d[1, 2]*delta[2, 1]+alpha*delta[2, 2]^2*d[1, 2]-beta*d[2, 2]*delta[2, 2]-gamma*d[2, 2]*delta[2, 2];


> subs({A = 0, B = 0, C = 0, D = 0, E = 0, F = 0, G = 0, H = 0, delta[1, 1] = 1, delta[1, 2] = 0, delta[2, 1] = 0, delta[2, 2] = 0, gamma[1, 1] = 1, gamma[1, 2] = 0, gamma[2, 1] = 0, gamma[2, 2] = 0, delta[1, 1]^2 = 0, delta[1, 2]^2 = 0, delta[2, 1]^2 = 1, delta[2, 2]^2 = 0, gamma[1, 1]^2 = 0, gamma[1, 2]^2 = 1, gamma[2, 1]^2 = 0, gamma[2, 2]^2 = 0}, {system1D});

The problem is: there is any simple way to use command "subs" when some expression such that delta[1,1]=1, gamma[1,1]=1, gamma[1,2]^2=1 have value and others are zero.

Can someone please advice and help me on this?

thanks

witribm

Hello all,

Yesterday, I upgraded to Mac OS X El Capitan.

Now, when working with Maple 2015, I feel the gui is very slow and sometimes irresponsive, when trying to scroll through my worksheets as well as through help pages (e.g., "help plot3d"). When I do the same within Maple 18, it works without any problems.

Does anybody else have the same issue?

Cheers,

Franky