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I would like to look at a 3d plot including an condition about the two variables of the plot

For this simple case, I only want to see the plot with the condition that x>0. Is that possible?


I am trying to simplify the eigenvalues of a 2x2 matrix [[a,b],[c,d]] subject to the condition a,b,c, and d are integers such that a+b = c+d. Why do the following commands not achieve this?



Eigenvalues(A) assuming a::integer,b::integer,c::integer,d::integer,a+b=d+c

How might I achieve what I need?


Since I was working in Matlab with Galerkin method which implies periodic boundary conditions I was wondering how to implement this in maple.

I tried this:


pde2 := diff(u(x, t), t)+3*(diff(u(x, t)^2, x))+diff(u(x, t),x$3) = 0

IBC := {u(0, t) = u(2, t), u(x, 0) = sech(50*(x-1/2))^2+2*sech(30*(x-1))^2, (D[1](u))(0, t) = (D[1](u))(2, t), (D[2](u))(0, t) = (D[2](u))(2, t)}

pds := pdsolve(pde2, IBC, numeric, time = t, range = 0 .. 2)

But it's telling me: 

Error, (in pdsolve/numeric/process_IBCs) initial/boundary conditions can only contain derivatives which are normal to the boundary, got (D[2](u))(0, t)

So what's wrong?

Dear all:

hello everybody;

I need your help to solve the system f(x,y)=0, and g(x,y)=0, such that there some parameter in the system, also all the parameter are positive and also our unkowns  x and y are also positive.

I try to write this code. I feel that under some condition we can have four solution or three or two. I need your help. Many thinks.

i am solving 3 ODE with boundary condition.. with boundary condition

Maple Worksheet - Error

Failed to load the worksheet /maplenet/convert/ .



then i got this error

Error, (in dsolve/numeric/bvp) initial Newton iteration is not converging

i dont know where i need to change.. could you help me..


i am solving 3 ODE question with boundary condition. when i running the programm i got this error.. any one could help me please.. :)


restart; with(plots); k := .1; E := 1.0; Pr := 7.0; Ec := 1.0; p := 2.0; blt := 11.5

Eq1 := diff(f(eta), eta, eta, eta)+f(eta)*(diff(f(eta), eta, eta))+Gr*theta(eta)-k*(diff(f(eta), eta))+2*E*g(eta) = 0;

diff(diff(diff(f(eta), eta), eta), eta)+f(eta)*(diff(diff(f(eta), eta), eta))+Gr*theta(eta)-.1*(diff(f(eta), eta))+2.0*g(eta) = 0


Eq2 := diff(g(eta), eta, eta)+f(eta)*(diff(g(eta), eta))-k*g(eta)-2*E*(diff(f(eta), eta)) = 0;

diff(diff(g(eta), eta), eta)+f(eta)*(diff(g(eta), eta))-.1*g(eta)-2.0*(diff(f(eta), eta)) = 0


Eq3 := diff(theta(eta), eta, eta)+Pr*(diff(theta(eta), eta))*f(eta)+Pr*Ec*((diff(f(eta), eta, eta))^2+(diff(g(eta), eta))^2) = 0;

diff(diff(theta(eta), eta), eta)+7.0*(diff(theta(eta), eta))*f(eta)+7.00*(diff(diff(f(eta), eta), eta))^2+7.00*(diff(g(eta), eta))^2 = 0


bcs1 := f(0) = p, (D(f))(0) = 1, g(0) = 0, theta(0) = 1, theta(blt) = 0, (D(f))(blt) = 0, g(blt) = 0;

f(0) = 2.0, (D(f))(0) = 1, g(0) = 0, theta(0) = 1, theta(11.5) = 0, (D(f))(11.5) = 0, g(11.5) = 0


L := [10, 11, 12];

[10, 11, 12]


for k to 3 do R := dsolve(eval({Eq1, Eq2, Eq3, bcs1}, Gr = L[k]), [f(eta), g(eta), theta(eta)], numeric, output = listprocedure); Y || k := rhs(R[3]) end do

Error, (in dsolve/numeric/bvp) initial Newton iteration is not converging





plot([Y || (1 .. 3)], 0 .. 10, labels = [eta, (D(f))(eta)]);

Warning, unable to evaluate the functions to numeric values in the region; see the plotting command's help page to ensure the calling sequence is correct





Download tyera(a).mw

With your help I have a solution to a system of three equations:

(parameters are calculated on the basis of the data (for different values) - one example below)
A1=0.00002072968491, A2=0, A3=0.001946449287, A4=0.01946449287

B1=, B2=0, B3=0.0004773383613, B4=0.00004773383613

C1=, C2=0, C3=, C4=0.00009087604510


eqa1: = A1 * (diff (Tg (x), x, x)) + A2 * (diff (Tg (x), x)) + (A3 + A4) * tan (x) + A3 * Tg (x) + A4 * Tw (x) = 0;

eqa2: = B1 * (diff (Tw (x), x, x)) + B2 * (diff (Tw (x), x)) + (B3 + B4) * Tw (x) + B3 * Tg (x) + B4 * tan (x) = 0;

eqa3: = C1 * (diff (Tz (x), x, x)) + (C3 + C4) * Tg (x) + C3 * tan (x) + C4 * Tw (x) = 0;


indets ({eqa1, eqa2, eqa3}) minus {x};

res: = Dsolve (eval ({eqa1, eqa2, eqa3}) union {boundary conditions ??}, numeric);


for k from 0 to 20 evalf (res (k), 4); from;

c1:= 0.524:


m: = 0;

for m from 0 to 20 and

T (m): = c1 * rhs (op (6, res (m))) + c2 * rhs (op (2, res (m))) + (1-c1-c2) * rhs (op (4, res (m))); print (m, T (m)); end to:


How and what type boundary conditions (I was thinking about the simplest or third type) to be able to determine the values on the y-axis on the graph. For example, the values started at -10, and ended at 10 (at a point (x, -10), (x, 10) in the coordinate system for a predetermined x, for example, from 0 to 20 which start at the point (0, -10 ) and stop at the point (20,10)). My main purpose is to collect these three solutions  to one equation T (x) = az * Tz (x) + and * Tw (x) + ag * Tg (x), and the ends of the graph, they should be in the above-mentioned points (0, -10 ) - start and (20,10) - stop.


Now thank you very much for the advice.


I'm trying to solve this system of ODEs by Laplace transform. 

> de1 := d^2*y(t)/dt^2 = y(t)+3*x(t)

> de2 := d^2*x(t)/dt^2 = 4*y(t)-4*exp(t)

with initial conditions 

> ICs := y(0) = 2, (D(y))(0) = 3, x(0) = 1, (D(x))(0) = 2



> deqns := de1, de2


> var := y(t), x(t)


I need to solve it for both y(t) and x(t), I have tried this by:

> dsolve({ICs, deqns}, var, method = laplace)


> dsolve({ICs, deqns}, y(t), method = laplace)

> dsolve({ICs, deqns}, x(t), method = laplace)


However I get this error message:

Error, (in dsolve/process_input) invalid initial condition


Any help is appreciated

I'm taking my first steps with maple and pdsolve, trying to run the example in the maplesoft support page:

which reads

> restart; with(PDEtools);
> U := diff_table(u(x, t));

and I get a solution that is different from the web page, and when i run

Im using maple 13. Any tips about what's wrong?



Hello everyone,

I'm trying to do some fitting using NonlinearFit, for the coefficients I know in advance, that they have to fulfill a condition (a+b>c+1). I couldn't find a way to make Maple take this condition into consideration while fitting my data. I tried to use Parameterrange to make the difference a+b-c+1 positive, this works for linear conditions like mine but leads to computational difficulties and errors like "no improved point could be found".

Thanks in advance,




please help me :



restart; eq := diff(T(x, y), x) = a*(diff(T(x, y), `$`(y, 2)))/u(y); u := proc (y) options operator, arrow; (-1)*1.218493440*10^11*y^2+4.244913600*10^6*y+0.33e-1 end proc; eq; ICs := (D[1](T))(x, 0) = 1000, (D[1](T))(x, 0.25e-4) = 2000, T(0, y) = 0; T_sol := pdsolve({ICs, eq}, T(x, y)); T_sol

diff(T(x, y), x) = a*(diff(diff(T(x, y), y), y))/u(y)


proc (y) options operator, arrow; (-1)*1.218493440*100000000000*(y^2)+4.244913600*1000000*y+0.33e-1 end proc


diff(T(x, y), x) = a*(diff(diff(T(x, y), y), y))/(-0.1218493440e12*y^2+4244913.600*y+0.33e-1)


(D[1](T))(x, 0) = 1000, (D[1](T))(x, 0.25e-4) = 2000, T(0, y) = 0


Error, (in PDEtools:-Library:-NormalizeBoundaryConditions) unable to isolate the functions {T(0, y), (D[1](T))(x, 0), (D[1](T))(x, 0.25e-4)} in the given boundary conditions {T(0, y) = 0, (D[1](T))(x, 0) = 1000, (D[1](T))(x, 0.25e-4) = 2000}






BC1 = diff(T(x, 0), y)=1000

BC2 = diff(T(x, 0.000025), y)=2000

IC = T(0,y)=0

where :



How can I solve this problem on Maple?
Can anyone help me please ... I wrote another post before but I can not solve the problem.

lambda is an experimental parameter. I have this initial condition n(x,0)=0.4, c(x,0)=0.

Thanks to everyone

Dear Maple Users,

I'm beginner in Maple.

I have this system of Pde:

with lambda experimental parameter and n,c,v dependent variables. I write this on Maple but I read on internet that the solution "float(undefined)" is an error.

I will insert this initial condition: c(x,0)=0,n(x,0)=0.4

Thanks everybody

Level: Idiot (Me)

I have a matrix of 3 columns and lots of rows M

  • First column is latitude in degrees
  • Second column is longitude in degrees
  • Third column is data

So I set lambda:=M(..,1) and phi_g:=M(..,2) giving me two column vectors.

I want to convert lambda and phi_g to polar coordinates theta and phi

theta:=90-lambda produces "Error, (in rtable/Sum) invalid arguments"


I also want to convert phi_g to phi where phi=phi_g when phi_g is 0...180 and phi=phi_g +360 when phi_g <0

How do I create a conditional function like this?

Is it possible to solve piecewise differential equations directly instead of separating the pieces and solving them separately.

like for example if i have a two dimensional function f(t,x) whose dynamics is as follows:

dynamics:= piecewise((t,x) in D1, pde1, pde2); where D1 is some region in (t,x)-plane

now is it possible to solve this system with one pde call numerically?

pde(dynamics, boundary conditions, numeric); doesnot work

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