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Question: Error, invalid input: diff received tau/omega, which is not valid for its 2nd argument

Question:Error, invalid input: diff received tau/omega, which is not valid for its 2nd argument

torabi 80 Maple 2017
syntax diff
January 05 2018
0

hello

How I can remove these errors?

Thanks

erorr.mw
 

NULL

 

 
> 

restart; eq31g := diff(u(t), `$`(t, 2))+u(t)+mu[s]*(diff(u(t), t))^3 = (1-mu[s])*(diff(u(t), t))*(diff(u(t), `$`(t, 2)))

diff(diff(u(t), t), t)+u(t)+mu[s]*(diff(u(t), t))^3 = (1-mu[s])*(diff(u(t), t))*(diff(diff(u(t), t), t))

(1.1)
> 

 

 

 
> 

eq33a:=subs(t=tau/omega,value(subs(u(t)=u(omega*t),eq31g)));

diff(diff(u(tau), tau/omega), tau/omega)+u(tau)+mu[s]*(diff(u(tau), tau/omega))^3 = (1-mu[s])*(diff(u(tau), tau/omega))*(diff(diff(u(tau), tau/omega), tau/omega))

 (2.1)
> 

omgRule:=omega=1+add(epsilon^i*omega[i],i=1..2);

omega = epsilon^2*omega[2]+epsilon*omega[1]+1

(2.2)
> 

uExpRule:=u(tau)=add(epsilon^i*u[i](tau),i=1..3);

u(tau) = epsilon*u[1](tau)+epsilon^2*u[2](tau)+epsilon^3*u[3](tau)

(2.3)
> 

temp:=subs(uExpRule,omgRule,convert(eq33a,diff));

Error, invalid input: diff received tau/omega, which is not valid for its 2nd argument

 
> 

eq33b:=convert(series(lhs(temp),epsilon,4),polynom)=0;

Error, invalid input: lhs received temp, which is not valid for its 1st argument, expr

 
> 

eqEps:=seq(coeff(lhs(eq33b),epsilon,i)=0,i=1..3);

Error, invalid input: lhs received eq33b, which is not valid for its 1st argument, expr

 

The general solution of the first-order equation, eqEps[1], can be expressed as

> 

sol1:=dsolve({eqEps[1],u[1](0)=a,D(u[1])(0)=0},u[1](tau));

Error, (in dsolve) not a system with respect to the unknowns [u[1](tau)]

 
> 

eq33c:=subs(u[1]=u[2],lhs(eqEps[1])=lhs(eqEps[1]))-subs(sol1,0=lhs(eqEps[2]));

Error, invalid input: lhs received eqEps[1], which is not valid for its 1st argument, expr

 

Expanding the right-hand side of eq33c in a Fourier series using trigonometric identities yields

> 

eq33c_RHS:=combine(rhs(eq33c));

Error, invalid input: rhs received eq33c, which is not valid for its 1st argument, expr

 

Eliminating the terms,  and , demands that . Then, the particular solution of eq33c can be expressed as

> 

sol2:=combine(subs(_C1=0,_C2=0,dsolve(subs(omega[1]=0,eq33c),u[2](tau))));

Error, (in dsolve) expecting an ODE or a set or list of ODEs. Received eq33c

 

Substituting sol1 and sol2 into the third-order equation, eqEps[3], and using the fact that , we obtain

> 

eq33d:=subs(u[1]=u[3],lhs(eqEps[1])=lhs(eqEps[1]))-subs(sol1,sol2,omega[1]=0,0=lhs(eqEps[3]));

Error, invalid input: lhs received eqEps[1], which is not valid for its 1st argument, expr

 

Expanding the right-hand side of eq33d in a Fourier series using trigonometric identities, we have

> 

eq33d_RHS:=combine(rhs(eq33d));

Error, invalid input: rhs received eq33d, which is not valid for its 1st argument, expr

 

Eliminating the terms that lead to secular terms from eq33d_RHS demands that

> 

omg2Rule:=omega[2]=solve(coeff(eq33d_RHS,cos(beta+tau)),omega[2]);

omega[2] = omega[2]

(2.4)

As discussed above, for a second-order uniform expansion, we do not need to solve for . Combining the first- and second-order solutions, we obtain, to the second approximation, that

> 

combine(subs(sol1,sol2,uExpRule));

Error, invalid input: subs received sol1, which is not valid for its 1st argument

 

where

> 

tau=subs(omega[1]=0,omg2Rule,subs(omgRule,omega*t));

tau = (epsilon^2*omega[2]+1)*t

(2.5)

``


 

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