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gardner_Fex.txt
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=====F-Expansion Method======
Equations are written in MATHEMATICA language.
----------------------------------------------------------------------------------------------------
Input equation is: D[u[x,t],{x,3}]+2*a*u[x,t]*D[u[x,t],{x,1}]-3*b*u[x,t]^2*D[u[x,t],{x,1}]+D[u[x,t],{t,1}] = 0;
The Diff. Equ. is integrable;
The assigned Value[s] to the integration Constant[s]->
ic1: 0
The Diff. Equ. becomes: a*U[xi]^2*k1-b*U[xi]^3*k1+U[xi]*k0+k1^3*D[U[xi],{xi,2}] = 0;
u[x,t] = U[xi],
where xi = k1*x+t*k0;
The value of N is: 1;
U = a1*F+a0+b1*F^(-1);
The first-order nonlinear ODE: D[F[xi],xi] = Sqrt[A2*F^2+A3*F^3+A4*F^4];
****************************************************************************************************
The system of algebraic equations are:
Diff[F,xi,1]^0*F^0: -2*b*b1^3*k1 = 0;
Diff[F,xi,1]^0*F^1: -6*b*b1^2*k1*a0+2*a*b1^2*k1 = 0;
Diff[F,xi,1]^0*F^2: 2*b1*k0+4*a*b1*k1*a0-6*b*b1*k1*a0^2-6*b*b1^2*a1*k1+2*b1*k1^3*A2 = 0;
Diff[F,xi,1]^0*F^3: -2*b*k1*a0^3-12*b*b1*a1*k1*a0+2*a0*k0+b1*A3*k1^3+2*a*k1*a0^2+4*a*b1*a1*k1 = 0;
Diff[F,xi,1]^0*F^4: 2*a1*k0-6*b*b1*a1^2*k1-6*b*a1*k1*a0^2+4*a*a1*k1*a0+2*a1*k1^3*A2 = 0;
Diff[F,xi,1]^0*F^5: -6*b*a1^2*k1*a0+2*a*a1^2*k1+3*A3*a1*k1^3 = 0;
Diff[F,xi,1]^0*F^6: 4*A4*a1*k1^3-2*b*a1^3*k1 = 0;
In the following results Const is an arbitrary constant.
****************************************************************************************************
solving above system of equations for variables {k0,k1,a0,a1,b1,A2,A3,A4}->
{b1=0,k1=-a0^(-1)*(a-b*a0)^(-1)*k0,a1=0}
Solution[s] of input Diff. Equ. is (are)=>
solution #1 u[x,t] = a0;
{b1=0,a0=0,a1=0}
Solution[s] of input Diff. Equ. is (are)=>
solution #2 u[x,t] = 0;
{A4=1/2*b*a1^2*a0^2*(a-b*a0)^2*k0^(-2),b1=0,A3=-2/3*(a-3*b*a0)*a1*a0^2*(a-b*a0)^2*k0^(-2),k1=-a0^(-1)*(a-b*a0)^(-1)*k0,A2=-(a-2*b*a0)*a0^3*(a-b*a0)^2*k0^(-2)}
U = a1*F+a0,
where F is the solution of
D[F[xi],xi] = Sqrt[1/2*b*a1^2*a0^2*(a-b*a0)^2*k0^(-2)*F^4-(a-2*b*a0)*a0^3*(a-b*a0)^2*k0^(-2)*F^2-2/3*(a-3*b*a0)*a1*a0^2*(a-b*a0)^2*k0^(-2)*F^3];
Solution[s] of input Diff. Equ. is (are)=>
solution #3 u[x,t] = a0-12*(9*(-1+Tanh[1/2*Sqrt[-5*b^2*a*a0^5*k0^(-2)+4*b*a^2*a0^4*k0^(-2)-a^3*a0^3*k0^(-2)+2*b^3*a0^6*k0^(-2)]*(t*k0-a0^(-1)*(a-b*a0)^(-1)*x*k0)])^2*(b^3*a1^2*a0^4*k0^(-2)+b*a^2*a1^2*a0^2*k0^(-2)-2*b^2*a*a1^2*a0^3*k0^(-2))*(5*b^2*a*a0^5*k0^(-2)-4*b*a^2*a0^4*k0^(-2)+a^3*a0^3*k0^(-2)-2*b^3*a0^6*k0^(-2))+8*(a^3*a1*a0^2*k0^(-2)+7*b^2*a*a1*a0^4*k0^(-2)-5*b*a^2*a1*a0^3*k0^(-2)-3*b^3*a1*a0^5*k0^(-2))^2)^(-1)*Sech[1/2*Sqrt[-5*b^2*a*a0^5*k0^(-2)+4*b*a^2*a0^4*k0^(-2)-a^3*a0^3*k0^(-2)+2*b^3*a0^6*k0^(-2)]*(t*k0-a0^(-1)*(a-b*a0)^(-1)*x*k0)]^2*a1*(a^3*a1*a0^2*k0^(-2)+7*b^2*a*a1*a0^4*k0^(-2)-5*b*a^2*a1*a0^3*k0^(-2)-3*b^3*a1*a0^5*k0^(-2))*(5*b^2*a*a0^5*k0^(-2)-4*b*a^2*a0^4*k0^(-2)+a^3*a0^3*k0^(-2)-2*b^3*a0^6*k0^(-2));
solution #4 u[x,t] = a0-6*Sech[Sqrt[-5*b^2*a*a0^5*k0^(-2)+4*b*a^2*a0^4*k0^(-2)-a^3*a0^3*k0^(-2)+2*b^3*a0^6*k0^(-2)]*(t*k0-a0^(-1)*(a-b*a0)^(-1)*x*k0)]*a1*(3*Sqrt[2*(b^3*a1^2*a0^4*k0^(-2)+b*a^2*a1^2*a0^2*k0^(-2)-2*b^2*a*a1^2*a0^3*k0^(-2))*(5*b^2*a*a0^5*k0^(-2)-4*b*a^2*a0^4*k0^(-2)+a^3*a0^3*k0^(-2)-2*b^3*a0^6*k0^(-2))+4/9*(a^3*a1*a0^2*k0^(-2)+7*b^2*a*a1*a0^4*k0^(-2)-5*b*a^2*a1*a0^3*k0^(-2)-3*b^3*a1*a0^5*k0^(-2))^2]+2*Sech[Sqrt[-5*b^2*a*a0^5*k0^(-2)+4*b*a^2*a0^4*k0^(-2)-a^3*a0^3*k0^(-2)+2*b^3*a0^6*k0^(-2)]*(t*k0-a0^(-1)*(a-b*a0)^(-1)*x*k0)]*(a^3*a1*a0^2*k0^(-2)+7*b^2*a*a1*a0^4*k0^(-2)-5*b*a^2*a1*a0^3*k0^(-2)-3*b^3*a1*a0^5*k0^(-2)))^(-1)*(5*b^2*a*a0^5*k0^(-2)-4*b*a^2*a0^4*k0^(-2)+a^3*a0^3*k0^(-2)-2*b^3*a0^6*k0^(-2));
{A2=-k1^(-3)*k0,A3=-2/3*a*a1*k1^(-2),A4=1/2*b*a1^2*k1^(-2),b1=0,a0=0}
U = a1*F,
where F is the solution of
D[F[xi],xi] = Sqrt[-2/3*a*a1*k1^(-2)*F^3+1/2*b*a1^2*k1^(-2)*F^4-k1^(-3)*k0*F^2];
Solution[s] of input Diff. Equ. is (are)=>
solution #5 u[x,t] = -12*(8*a^2*a1^2*k1^(-4)+9*b*(-1+Tanh[1/2*(k1*x+t*k0)*Sqrt[-k1^(-3)*k0]])^2*a1^2*k1^(-5)*k0)^(-1)*a*Sech[1/2*(k1*x+t*k0)*Sqrt[-k1^(-3)*k0]]^2*a1^2*k1^(-5)*k0;
solution #6 u[x,t] = -6*Sech[(k1*x+t*k0)*Sqrt[-k1^(-3)*k0]]*a1*k1^(-3)*(2*a*Sech[(k1*x+t*k0)*Sqrt[-k1^(-3)*k0]]*a1*k1^(-2)+3*Sqrt[4/9*a^2*a1^2*k1^(-4)+2*b*a1^2*k1^(-5)*k0])^(-1)*k0;
{k0=0,k1=0}
U = a1*F+a0+b1*F^(-1),
where F is the solution of
D[F[xi],xi] = Sqrt[A2*F^2+A3*F^3+A4*F^4];
Solution[s] of input Diff. Equ. is (are)=>
solution #7 u[x,t] = a0-b1*(A3^2-A4*A2)*A3^(-1)*A2^(-1)-(A3^2-A4*A2)^(-1)*A3*a1*A2;
solution #8 u[x,t] = 1/2*b1*(Sqrt[A3^2-4*A4*A2]-A3)*A2^(-1)+2*(Sqrt[A3^2-4*A4*A2]-A3)^(-1)*a1*A2+a0;
****************************************************************************************************
****************************************************************************************************
Time: 1.066 seconds