666 jvbasha

javid basha jv

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These are questions asked by 666 jvbasha

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

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));
 

Dear maple users,

Greetings.

When converting the maple figure into EPS format (for latex) which shows white patches.

How to avoid such patches.

Dear maple users,

Greetings.


JVB.mw
 

restart; with(plots)

fcns := {f(x), g(x), t(x)}

m := .2; pa := 3.14*(1/3); aa := .1; bb := .3; ta := .2; kt := .4; h2 := 1+m*ax+bb*sin((2*3.14)*(ax-ta)); h1 := -1-m*ax-aa*sin((2*3.14)*(ax-ta)+pa); a2 := kt+aa*sin((2*3.14)*(ax-ta))+bb*sin((2*3.14)*(ax-ta)+pa)

1+.2*ax+.3*sin(6.28*ax-1.256)

 

-1-.2*ax-.1*sin(6.28*ax-.209333333)

 

.4+.1*sin(6.28*ax-1.256)+.3*sin(6.28*ax-.209333333)

(1)

eq1 := 1.5*(diff(f(x), `$`(x, 4)))-.40*(diff((diff(f(x), `$`(x, 2)))^3, `$`(x, 2)))-.20*(diff(f(x), `$`(x, 2)))+.20*(diff(t(x), `$`(x, 1)))+.20*(diff(g(x), `$`(x, 1))) = 0

1.5*(diff(diff(diff(diff(f(x), x), x), x), x))-2.40*(diff(diff(f(x), x), x))*(diff(diff(diff(f(x), x), x), x))^2-1.20*(diff(diff(f(x), x), x))^2*(diff(diff(diff(diff(f(x), x), x), x), x))-.20*(diff(diff(f(x), x), x))+.20*(diff(t(x), x))+.20*(diff(g(x), x)) = 0

(2)

eq2 := 1.75*(diff(t(x), `$`(x, 2)))+(diff(t(x), `$`(x, 1)))*(diff(g(x), `$`(x, 1)))+2*(diff(t(x), `$`(x, 1)))^2+.7*((diff(f(x), `$`(x, 2)))^2-(diff(f(x), `$`(x, 2)))^4)+.1*(diff(f(x), `$`(x, 1)))^2+.1*t(x) = 0

1.75*(diff(diff(t(x), x), x))+(diff(t(x), x))*(diff(g(x), x))+2*(diff(t(x), x))^2+.7*(diff(diff(f(x), x), x))^2-.7*(diff(diff(f(x), x), x))^4+.1*(diff(f(x), x))^2+.1*t(x) = 0

(3)

eq3 := diff(g(x), `$`(x, 2))+2*(diff(t(x), `$`(x, 2)))-.5*g(x) = 0

diff(diff(g(x), x), x)+2*(diff(diff(t(x), x), x))-.5*g(x) = 0

(4)

bc := f(h2) = (1/2)*a2, (D(f))(h2) = 0, f(h1) = -(1/2)*a2, (D(f))(h1) = 0, t(h2) = 1, t(h1) = 0, g(h2) = 1, g(h1) = 0

f(1+.2*ax+.3*sin(6.28*ax-1.256)) = .2000000000+0.5000000000e-1*sin(6.28*ax-1.256)+.1500000000*sin(6.28*ax-.209333333), (D(f))(1+.2*ax+.3*sin(6.28*ax-1.256)) = 0, f(-1-.2*ax-.1*sin(6.28*ax-.209333333)) = -.2000000000-0.5000000000e-1*sin(6.28*ax-1.256)-.1500000000*sin(6.28*ax-.209333333), (D(f))(-1-.2*ax-.1*sin(6.28*ax-.209333333)) = 0, t(1+.2*ax+.3*sin(6.28*ax-1.256)) = 1, t(-1-.2*ax-.1*sin(6.28*ax-.209333333)) = 0, g(1+.2*ax+.3*sin(6.28*ax-1.256)) = 1, g(-1-.2*ax-.1*sin(6.28*ax-.209333333)) = 0

(5)

L := [.1]; AP := NULL; NN := nops(L); for k to NN do R := dsolve(eval({bc, eq1, eq2, eq3}, ax = L[k]), fcns, type = numeric, method = bvp[midrich], maxmesh = 1000, abserr = 10^(-4), AP); AP := approxsoln = R; X1 || k := rhs(R(0)[4]) end do; ba3 := [(X1 || (1 .. NN))(0)]

[.1]

 

1

 

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

 

R(0)

(6)

``


 

Download JVB.mw

 

How to obtain a solution for various values of "ax"

waiting for your reply.

 

 

Dear maple users @acer @Carl Love @Kitonum @Preben Alsholm @dharr @tomleslie

Greeting.

I have solved some PDEs in analytically and numerically.

But the numerical and analytical results are not matching.

I hope there is some problem with an analytical solution, especially in the first order and second-order boundary conditions.

Both the codes have enclosed here, waiting for a reply.

AN.mw

NUM.mw

Note: The PDEs are performed in the Maple 18 version. 

When compiling an analytical solution, in clarify expression click the remember table assignment.

 

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