Dear maple users 

Greetings.

In this code, I am solving the PDEs via perturbation method.

There is some mistake in the boundary condition and pdsolve.

Kindly help me that to get the solution for this PDE via perturbation method.

Wating for your replay.

BC: 

Code: JVB.mw

eq1 := diff(h(t), [`$`(t, nu)])+(1/4)*h(t) = 8/3*(1/(sqrt(Pi)*t^(1/2))-(3/32)*t^2+(3/16)*exp(-t)-1);

with ics h^k (0)=0 for k=0..n

how are please I want to help In some graphs in maple 

the graph in the picture

thank you 

)

Download analy_numer.mw

Hello,
I am trying to compute the convolution integral numerically and compare the result with the analytic result. My numerics is giving terrible results:

clc;clear;
% Define i1(t) and i2(t) as symbolic variable
syms i1(t) i2(t)

% Given differential equation
ode1 = diff(i1) == 0.5*i1 + -3*i2 +5*exp(-2*t);
ode2 = diff(i2) == 2*i1 - 6*i2;
odes = [ode1; ode2];

% Define initial conditions
cond1 = i1(0) == 1;
cond2 = i2(0) == -1;
conds = [cond1; cond2];

% Solution of system of differential equation
[i1(t), i2(t)] = dsolve(odes,conds)

fplot(i1(t))
hold on
fplot(i2(t))
legend('i1(t)','i2(t)')

Output:

i1(t) =
exp((t*(73^(1/2) - 11))/4)*(73^(1/2)/8 + 13/8)*((57*73^(1/2))/292 - exp((3*t)/4 - (73^(1/2)*t)/4)*((15*73^(1/2))/292 + 5/4) + 3/4) - exp(-(t*(73^(1/2) + 11))/4)*(73^(1/2)/8 - 13/8)*(exp((3*t)/4 + (73^(1/2)*t)/4)*((15*73^(1/2))/292 - 5/4) + (3*73^(1/2)*(73^(1/2) - 19))/292)

i2(t) =
exp(-(t*(73^(1/2) + 11))/4)*(exp((3*t)/4 + (73^(1/2)*t)/4)*((15*73^(1/2))/292 - 5/4) + (3*73^(1/2)*(73^(1/2) - 19))/292) + exp((t*(73^(1/2) - 11))/4)*((57*73^(1/2))/292 - exp((3*t)/4 - (73^(1/2)*t)/4)*((15*73^(1/2))/292 + 5/4) + 3/4)

Simulations.mw

Hey everyone,

f_1 and f_2  are satisfying the set of non-linear integral equations I have attached to this message.
I know that I need to solve them numerically by iterations. Probably, the first guest of the function f_1 and f_2  is the driving term. a is just a parameter which can be fixed (I guess smaller than \pi/4). * is the convolution product and k is the momentum space parameter. I learnt that in order to solve them I should solve them in the Fourier space. I know also that I need to discretize these function in the “real ” space between {-L,+L} before applying the FFT or one of its relatives. Thank you for any suggestions or leads.

fitting_with_Statistics_package_V2(1)_-_Copy.mw

Dear maple users.

Greetings for the day.

I hope you are all fine and safe.

In the below mention code, I need to plot "ax" at 0..1 when N=1 and N=2.

But the code only working for the N=0 case.

How to tackle this situation and plot the function for various values of "ax" at ax=0..1.

waiting for your reply.

JBV.mw

Code:

restart;
PDEtools[declare](f(x), t(x), g(x), prime = (x));

N := 2; m := .2; pa := 3.14*(1/3); ax := ax; h2 := 1+.2*ax+.3*sin((2*3.14)*(ax-.2)); h1 := -1-.2*ax-.1*sin((2*3.14)*(ax-.2)+pa); a2 := 1.4+.1*sin((2*3.14)*(ax-.2))+.3*sin((2*3.14)*(ax-.2)+pa);

f(x):=sum(p^j*f[j](x),j=0..N);  t(x):=sum(p^j*t[j](x),j=0..N);

g(x):=sum(p^j*g[j](x),j=0..N);

Eq1 := (1-p)*(diff(f(x), `$`(x, 4)))+p*((1+.2)*(diff(f(x), `$`(x, 4)))-(.2*(1/3))*(diff((diff(f(x), `$`(x, 2)))^3, `$`(x, 2)))-2*(diff(f(x), `$`(x, 2)))+diff(t(x), `$`(x, 1))+diff(g(x), `$`(x, 1)));

Eq2 := (1-p)*(1+1.2)*(diff(t(x), `$`(x, 2)))+p*((1+1.2)*(diff(t(x), `$`(x, 2)))+.1*(diff(t(x), `$`(x, 1)))*(diff(g(x), `$`(x, 1)))+.2*(diff(t(x), `$`(x, 1)))^2+.5*(diff(f(x), `$`(x, 1)))^2);

Eq3 := (1-p)*(diff(g(x), `$`(x, 2)))+p*(diff(g(x), `$`(x, 2))+diff(t(x), `$`(x, 2)));

for j from 0 to N do
equ1[j] := coeff(Eq1, p, j) = 0; 
equ2[j] := coeff(Eq2, p, j) = 0;
equ3[j] := coeff(Eq3, p, j) = 0;
end do;

con[1][0] := f[0](h2) = (1/2)*a2, (D(f[0]))(h2) = 0, f[0](h1) = -(1/2)*a2, (D(f[0]))(h1) = 0; con[2][0] := t[0](h2) = 1, t[0](h1) = 0; con[3][0] := g[0](h2) = 1, g[0](h1) = 0;

for i to N do

con[1][i] := f[i](h2) = 0, (D(f[i]))(h2) = 0, f[i](h1) = 0, (D(f[i]))(h1) = 0; con[2][i] := t[i](h2) = 0, t[i](h1) = 0; con[3][i] := g[i](h2) = 0, g[i](h1) = 0 end do;

for i from 0 to N do
P:=dsolve({equ1[i],equ2[i],equ3[i],con[1][i],con[2][i],con[3][i]},{f[i](x),t[i](x),g[i](x)}):
f[i](x):=rhs(P[1]);
t[i](x):=rhs(P[2]);
g[i](x):=rhs(P[3]);
end do:

f(x):=evalf(simplify(sum(f[n](x),n=0..N)));
Am := (1+.2)*(diff(f(x), `$`(x, 3)));
with(plots);

display(plot(eval(Am, x = .6), ax = 0 .. 1, numpoints = 200, color = blue));
 

blessing.mw please i really need your help on this, i have been trying to fix this error for a while now. 

Dear all

I have  Lie commutations for vectors e1, e2, e3, e4, e5, e6 as follow:

[e1, e3] = e3, [e1, e4] = e4, [e1, e5] = e5, [e1, e6] = e6, [e2, e3] = -e5, [e2, e4] = e6, [e3, e5] = e6

for which the command 

Query("Jacobi")

returns the false result, which means, the vectors are not closed under Jacobi's identity. How can I find vector triplets for which Jacobi's identity does not hold?

Please find Maple file.Jacobi_identity.mw