alaloush

solving nonlinear algebraic system...

Asked: alaloush 35 Product: Maple 2018

March 06 2020

1 0

I am trying to solve a set of equations

Why are the results not the same as the following results?

Is there any other way to get the correct answer?

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(1)

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>	$vars := {a__0, a__1, b__1, k}$

${a__0, a__1, b__1, k}$

(6)

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sys1 := {}; SolsT := {}; for i to 5 do sys1 := `union`(sys1, {T[i]}) end do; sys := sys1

${3*a__0*a__1^2*q = 0, b__1^3*q+2*b__1*k^2 = 0, 2*a__1*k^2*m^2+a__1^3*q = 0, a__0^3*q+6*a__0*a__1*b__1*q+a__0*p = 0, -a__1*b__1*k^2*m^2+3*a__1^2*b__1^2*q+3*a__0^2*a__1*b__1-a__1*b__1*k^2+a__1*b__1*p = 0}$

(7)

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for i to 5 do indets(T[i]) end do

${a__0, a__1, q}$

${a__1, k, m, q}$

${a__0, a__1, b__1, k, m, p, q}$

${a__0, a__1, b__1, p, q}$

${b__1, k, q}$

(8)

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$[{a__0 = 0, a__1 = a__1, b__1 = 0, k = (1/2)*(-2*q)^(1/2)*a__1/m}, {a__0 = 0, a__1 = a__1, b__1 = 0, k = -(1/2)*(-2*q)^(1/2)*a__1/m}, {a__0 = (-q*p)^(1/2)/q, a__1 = 0, b__1 = b__1, k = (1/2)*(-2*q)^(1/2)*b__1}, {a__0 = -(-q*p)^(1/2)/q, a__1 = 0, b__1 = b__1, k = (1/2)*(-2*q)^(1/2)*b__1}, {a__0 = (-q*p)^(1/2)/q, a__1 = 0, b__1 = b__1, k = -(1/2)*(-2*q)^(1/2)*b__1}, {a__0 = -(-q*p)^(1/2)/q, a__1 = 0, b__1 = b__1, k = -(1/2)*(-2*q)^(1/2)*b__1}, {a__0 = (-q*p)^(1/2)/q, a__1 = 0, b__1 = 0, k = k}, {a__0 = -(-q*p)^(1/2)/q, a__1 = 0, b__1 = 0, k = k}, {a__0 = 0, a__1 = 0, b__1 = b__1, k = (1/2)*(-2*q)^(1/2)*b__1}, {a__0 = 0, a__1 = 0, b__1 = b__1, k = -(1/2)*(-2*q)^(1/2)*b__1}, {a__0 = 0, a__1 = 0, b__1 = 0, k = k}, {a__0 = 0, a__1 = (-2*q*(m^2-6*m+1)*p)^(1/2)*m/(q*(m^2-6*m+1)), b__1 = -(-2*q*(m^2-6*m+1)*p)^(1/2)/(q*(m^2-6*m+1)), k = ((m^2-6*m+1)*p)^(1/2)/(m^2-6*m+1)}, {a__0 = 0, a__1 = -(-2*q*(m^2-6*m+1)*p)^(1/2)*m/(q*(m^2-6*m+1)), b__1 = (-2*q*(m^2-6*m+1)*p)^(1/2)/(q*(m^2-6*m+1)), k = ((m^2-6*m+1)*p)^(1/2)/(m^2-6*m+1)}, {a__0 = 0, a__1 = (-2*q*(m^2-6*m+1)*p)^(1/2)*m/(q*(m^2-6*m+1)), b__1 = -(-2*q*(m^2-6*m+1)*p)^(1/2)/(q*(m^2-6*m+1)), k = -((m^2-6*m+1)*p)^(1/2)/(m^2-6*m+1)}, {a__0 = 0, a__1 = -(-2*q*(m^2-6*m+1)*p)^(1/2)*m/(q*(m^2-6*m+1)), b__1 = (-2*q*(m^2-6*m+1)*p)^(1/2)/(q*(m^2-6*m+1)), k = -((m^2-6*m+1)*p)^(1/2)/(m^2-6*m+1)}, {a__0 = 0, a__1 = (-2*q*(m^2+6*m+1)*p)^(1/2)*m/(q*(m^2+6*m+1)), b__1 = (-2*q*(m^2+6*m+1)*p)^(1/2)/(q*(m^2+6*m+1)), k = ((m^2+6*m+1)*p)^(1/2)/(m^2+6*m+1)}, {a__0 = 0, a__1 = -(-2*q*(m^2+6*m+1)*p)^(1/2)*m/(q*(m^2+6*m+1)), b__1 = -(-2*q*(m^2+6*m+1)*p)^(1/2)/(q*(m^2+6*m+1)), k = ((m^2+6*m+1)*p)^(1/2)/(m^2+6*m+1)}, {a__0 = 0, a__1 = (-2*q*(m^2+6*m+1)*p)^(1/2)*m/(q*(m^2+6*m+1)), b__1 = (-2*q*(m^2+6*m+1)*p)^(1/2)/(q*(m^2+6*m+1)), k = -((m^2+6*m+1)*p)^(1/2)/(m^2+6*m+1)}, {a__0 = 0, a__1 = -(-2*q*(m^2+6*m+1)*p)^(1/2)*m/(q*(m^2+6*m+1)), b__1 = -(-2*q*(m^2+6*m+1)*p)^(1/2)/(q*(m^2+6*m+1)), k = -((m^2+6*m+1)*p)^(1/2)/(m^2+6*m+1)}]$

(9)

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for i to nops(Solll) do SOlls[i] := simplify(Solll[i], 'symbolic') end do

${a__0 = 0, a__1 = a__1, b__1 = 0, k = ((1/2)*I)*2^(1/2)*q^(1/2)*a__1/m}$

${a__0 = 0, a__1 = a__1, b__1 = 0, k = -((1/2)*I)*2^(1/2)*q^(1/2)*a__1/m}$

${a__0 = I*p^(1/2)/q^(1/2), a__1 = 0, b__1 = b__1, k = ((1/2)*I)*2^(1/2)*q^(1/2)*b__1}$

${a__0 = -I*p^(1/2)/q^(1/2), a__1 = 0, b__1 = b__1, k = ((1/2)*I)*2^(1/2)*q^(1/2)*b__1}$

${a__0 = I*p^(1/2)/q^(1/2), a__1 = 0, b__1 = b__1, k = -((1/2)*I)*2^(1/2)*q^(1/2)*b__1}$

${a__0 = -I*p^(1/2)/q^(1/2), a__1 = 0, b__1 = b__1, k = -((1/2)*I)*2^(1/2)*q^(1/2)*b__1}$

${a__0 = I*p^(1/2)/q^(1/2), a__1 = 0, b__1 = 0, k = k}$

${a__0 = -I*p^(1/2)/q^(1/2), a__1 = 0, b__1 = 0, k = k}$

${a__0 = 0, a__1 = 0, b__1 = b__1, k = ((1/2)*I)*2^(1/2)*q^(1/2)*b__1}$

${a__0 = 0, a__1 = 0, b__1 = b__1, k = -((1/2)*I)*2^(1/2)*q^(1/2)*b__1}$

${a__0 = 0, a__1 = 0, b__1 = 0, k = k}$

${a__0 = 0, a__1 = I*2^(1/2)*p^(1/2)*m/(q^(1/2)*(m^2-6*m+1)^(1/2)), b__1 = -I*2^(1/2)*p^(1/2)/(q^(1/2)*(m^2-6*m+1)^(1/2)), k = p^(1/2)/(m^2-6*m+1)^(1/2)}$

${a__0 = 0, a__1 = -I*2^(1/2)*p^(1/2)*m/(q^(1/2)*(m^2-6*m+1)^(1/2)), b__1 = I*2^(1/2)*p^(1/2)/(q^(1/2)*(m^2-6*m+1)^(1/2)), k = p^(1/2)/(m^2-6*m+1)^(1/2)}$

${a__0 = 0, a__1 = I*2^(1/2)*p^(1/2)*m/(q^(1/2)*(m^2-6*m+1)^(1/2)), b__1 = -I*2^(1/2)*p^(1/2)/(q^(1/2)*(m^2-6*m+1)^(1/2)), k = -p^(1/2)/(m^2-6*m+1)^(1/2)}$

${a__0 = 0, a__1 = -I*2^(1/2)*p^(1/2)*m/(q^(1/2)*(m^2-6*m+1)^(1/2)), b__1 = I*2^(1/2)*p^(1/2)/(q^(1/2)*(m^2-6*m+1)^(1/2)), k = -p^(1/2)/(m^2-6*m+1)^(1/2)}$

${a__0 = 0, a__1 = I*2^(1/2)*p^(1/2)*m/(q^(1/2)*(m^2+6*m+1)^(1/2)), b__1 = I*2^(1/2)*p^(1/2)/(q^(1/2)*(m^2+6*m+1)^(1/2)), k = p^(1/2)/(m^2+6*m+1)^(1/2)}$

${a__0 = 0, a__1 = -I*2^(1/2)*p^(1/2)*m/(q^(1/2)*(m^2+6*m+1)^(1/2)), b__1 = -I*2^(1/2)*p^(1/2)/(q^(1/2)*(m^2+6*m+1)^(1/2)), k = p^(1/2)/(m^2+6*m+1)^(1/2)}$

${a__0 = 0, a__1 = I*2^(1/2)*p^(1/2)*m/(q^(1/2)*(m^2+6*m+1)^(1/2)), b__1 = I*2^(1/2)*p^(1/2)/(q^(1/2)*(m^2+6*m+1)^(1/2)), k = -p^(1/2)/(m^2+6*m+1)^(1/2)}$

${a__0 = 0, a__1 = -I*2^(1/2)*p^(1/2)*m/(q^(1/2)*(m^2+6*m+1)^(1/2)), b__1 = -I*2^(1/2)*p^(1/2)/(q^(1/2)*(m^2+6*m+1)^(1/2)), k = -p^(1/2)/(m^2+6*m+1)^(1/2)}$

(10)

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(11)

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Find Hirota Bilinear form using Maple...

Asked: alaloush 35 Product: Maple 2018

February 25 2020

1 1

I have read the article attached in the following

https://drive.google.com/open?id=1BLzklSFx3nPovLC_2L1muAWY1kFnbwLV

I wrote the code mentioned in the article in order to find the Hirota formula This is the code that I wrote For Hirota Method But when running the program it does not work well

Could you help me to fix the mistake and run the program

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BD := proc (FF, DD) local f, g, x, m, opt; if nargs = 1 then return `*`(FF[]) end if; f, g := FF[]; x, m := DD[]; opt := args[3 .. -1]; if m = 0 then return procname(FF, opt) end if; procname([diff(f, x), g], [x, m-1], opt)-procname([f, diff(g, x)], [x, m-1], opt) end proc

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print_BD := proc (FF, DD) local f, g, x, m, i; f, g := FF[]; f := cat(f, '', g); g := product(D[args[i][1]]^args[i][2], i = 2 .. nargs); if g <> 1 then f := g*(`*`(f)) end if; f end proc

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printHBT := proc (uf, u, f, i, j, PL, alpha := 1) local DD, g, C, tmp, pl; pl := printHBF(PL); if 0 < j then print(u = 2*alpha*('diff')(ln(f), `$`(x, j))) else print(u = 2*alpha*ln(f)) end if; if 0 < i then print(('diff')(pl/f^2, `$`(x, i)) = 0) else print(pl/f^2 = 0) end if; NULL end proc

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$guessalpha := proc (Res, uf, u, f, i, j, PL, alpha) local tmp, res, pl, flag, k; flag := 1; tmp := [op(Res)]; tmp := map(numer, tmp); tmp := gcd(tmp[1], tmp[-1]); if type(tmp, `*`) then tmp := remove(has, tmp, f) end if; if tmp <> 0 and has(tmp, {alpha}) then tmp := solve(tmp/alpha^difforder(uf), {alpha}); if tmp <> NULL and has(tmp, {alpha}) then lprint(tmp); for k to nops([tmp]) while flag = 1 do res := collect(expand(subs(tmp[k], Res)), f, factor); if res = 0 then pl := subs(tmp[k], PL); printHBT(uf, u, f, i, j, pl, rhs(tmp[k])); flag := 0 end if end do end if end if; PL end proc$

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eq := diff(u(x, t), t, t)-(diff(u(x, t), x, x))-3*(diff(u(x, t), x, x))^2-(diff(u(x, t), x, x, x, x)) = 0

(5)

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diff(diff(u(x, t), t), t)-(diff(diff(u(x, t), x), x))-3*(diff(diff(u(x, t), x), x))^2-(diff(diff(diff(diff(u(x, t), x), x), x), x)) = 0

Error, (in getFnumer) expected 1 st argument be type `+`.

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Download Hirota_to_send.mw

How do I simplify KdV equation in Maple ...

Asked: alaloush 35 Product: Maple 2018

February 04 2020

0 2

How do I simplify KdV equation in Maple by using =fxt))xx)?)

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   I am by using =2*difffxtxx)
    My aim is to get the form
    diff((f*(diff(f, x, t))-(diff(f, x))*(diff(f, t))+f*(diff(f, x, x, x))-4*(diff(f, x, x, x))*(diff(f, x))+3*(diff(f, x, x))^2)/f^2, x) = 0

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Download KdV_simplify

How do I find Lax pair for this equatio...

Asked: alaloush 35 Product: Maple 2018

January 25 2020

0 0

Hello
In this example, we have the KdV equation
t] - 6 uux] + xxx] = 0
I would like to find the Lax pair for the KdV equation, which are
Lψ=λψ
ψ[t] = Mψ

Lt+ML-LM = 0 called a compatibility condition
So, I will start from this purpose
Then we will assume M in the form
will assume M in the form
M := a3*Dx^3+a^2+a1*Dx+a0
thenb using M and L in the for L[tL-LM = 0can find
Dx^5+( ) Dx^4+( ) Dx^3+( ) Dx^2+( ) Dx+( )=0
then wean find a_i =0,1,2,3
In the following maple code to do that
my question is
.How I canoue the soluo get a_i2,3 usinmaple code
any maple packge to find Lax pair for PDE -

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in this exampile we have KdV equation

I would likeind the Lax pair for the KdV equation, which are

Lψ=λψ

where

+ML-LM = 0 called apatibility condition

So, I will start this purpose

L:=-Dx^2+u;

then we will assume M the m

Ma3*Dx^3+a2*Dx^2+Dx+a0

then busing in the form +ML-LM = 0 can find

( ) Dx^5+( ) Dx^4+( ) Dx^3+( ) Dx^2+( ) Dx+( )=0

then we can find a_i ;i=,2,3

the fllowile code to

my queion is ;

1) How I can continue the solution to get a_i ;i=0,1,2,3 using maple code ?

2) isir any maple packge to find Lax pair for PDE ?

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alias(u = u(x, t)); declare(u(x, t)); alias(a3 = a3(x, t)); declare(a3(x, t)); alias(a2 = a2(x, t)); declare(a2(x, t)); alias(a1 = a1(x, t)); declare(a1(x, t)); alias(a0 = a0(x, t)); declare(a0(x, t))

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-a3*Dx^5-2*Dx^4*(diff(a3, x))-a2*Dx^4+Dx^3*u*a3-Dx^3*(diff(diff(a3, x), x))-2*Dx^3*(diff(a2, x))-Dx^3*a1+Dx^2*u*a2-Dx^2*(diff(diff(a2, x), x))-2*Dx^2*(diff(a1, x))-Dx^2*a0+Dx*u*a1-Dx*(diff(diff(a1, x), x))-2*Dx*(diff(a0, x))+u*a0-(diff(diff(a0, x), x))

(5)

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-a3*Dx^5-a2*Dx^4+Dx^3*u*a3-Dx^3*a1+3*Dx^2*a3*(diff(u, x))+Dx^2*u*a2-Dx^2*a0+3*Dx*a3*(diff(diff(u, x), x))+2*Dx*a2*(diff(u, x))+Dx*u*a1+a3*(diff(diff(diff(u, x), x), x))+a2*(diff(diff(u, x), x))+a1*(diff(u, x))+u*a0

(6)

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a3*(diff(diff(diff(u, x), x), x))+(3*Dx*a3+a2)*(diff(diff(u, x), x))+diff(diff(a0, x), x)+Dx*(diff(diff(a1, x), x))+Dx^2*(diff(diff(a2, x), x))+Dx^3*(diff(diff(a3, x), x))+(3*Dx^2*a3+2*Dx*a2+a1)*(diff(u, x))+2*Dx^4*(diff(a3, x))+2*Dx^3*(diff(a2, x))+2*Dx^2*(diff(a1, x))+2*Dx*(diff(a0, x))

(7)

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diff(u, t)-a3*(diff(diff(diff(u, x), x), x))-(3*Dx*a3+a2)*(diff(diff(u, x), x))-(diff(diff(a0, x), x))-Dx*(diff(diff(a1, x), x))-Dx^2*(diff(diff(a2, x), x))-Dx^3*(diff(diff(a3, x), x))-(3*Dx^2*a3+2*Dx*a2+a1)*(diff(u, x))-2*Dx^4*(diff(a3, x))-2*Dx^3*(diff(a2, x))-2*Dx^2*(diff(a1, x))-2*Dx*(diff(a0, x)) = 0

(8)

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-2*Dx^4*(diff(a3, x))+(-(diff(diff(a3, x), x))-2*(diff(a2, x)))*Dx^3+(-3*a3*(diff(u, x))-(diff(diff(a2, x), x))-2*(diff(a1, x)))*Dx^2+(-2*a2*(diff(u, x))-3*a3*(diff(diff(u, x), x))-(diff(diff(a1, x), x))-2*(diff(a0, x)))*Dx-a1*(diff(u, x))-a2*(diff(diff(u, x), x))-a3*(diff(diff(diff(u, x), x), x))-(diff(diff(a0, x), x))+diff(u, t) = 0