How to obtain and work with steady-state part of DE solution?
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I'm trying to obtain resonance curves of the system. System can be described as
F,m,k:=2,1,4:
lambda:= beta/(2*m):
omega:=sqrt(k/m):
de:=diff(x(t),t$2)+2*lambda*diff(x(t),t)+omega^2*x(t)=F*cos(gamma1*t):
cond:=x(0)=0, D(x)(0)=0:
sol := dsolve({cond, de});
Solving gives sum of terms, some of which "die out" with time (since these terms have exp(-...*t)
) and some of which form steady-state solution (solution for t -> ∞
). This solution will be in form xstst=f(gamma1)*sin(...)
. In order to obtain resonance curves, I need to plot f(gamma1)
(for chosen constant betas
, say, 2,1,0.5,0.25,
etc.).
I've solved this "by hand" and found f := F/(sqrt((-gamma1^2+omega^2)^2+4*lambda^2*gamma1^2))
. Plotting this for any chosen beta gives the result needed, for example, for beta:=0.5
the plot is
I wonder if I can obtain these curves using maple functions only (without solving anything "by hand" at all).
[edited]
maple
add a comment |
I'm trying to obtain resonance curves of the system. System can be described as
F,m,k:=2,1,4:
lambda:= beta/(2*m):
omega:=sqrt(k/m):
de:=diff(x(t),t$2)+2*lambda*diff(x(t),t)+omega^2*x(t)=F*cos(gamma1*t):
cond:=x(0)=0, D(x)(0)=0:
sol := dsolve({cond, de});
Solving gives sum of terms, some of which "die out" with time (since these terms have exp(-...*t)
) and some of which form steady-state solution (solution for t -> ∞
). This solution will be in form xstst=f(gamma1)*sin(...)
. In order to obtain resonance curves, I need to plot f(gamma1)
(for chosen constant betas
, say, 2,1,0.5,0.25,
etc.).
I've solved this "by hand" and found f := F/(sqrt((-gamma1^2+omega^2)^2+4*lambda^2*gamma1^2))
. Plotting this for any chosen beta gives the result needed, for example, for beta:=0.5
the plot is
I wonder if I can obtain these curves using maple functions only (without solving anything "by hand" at all).
[edited]
maple
add a comment |
I'm trying to obtain resonance curves of the system. System can be described as
F,m,k:=2,1,4:
lambda:= beta/(2*m):
omega:=sqrt(k/m):
de:=diff(x(t),t$2)+2*lambda*diff(x(t),t)+omega^2*x(t)=F*cos(gamma1*t):
cond:=x(0)=0, D(x)(0)=0:
sol := dsolve({cond, de});
Solving gives sum of terms, some of which "die out" with time (since these terms have exp(-...*t)
) and some of which form steady-state solution (solution for t -> ∞
). This solution will be in form xstst=f(gamma1)*sin(...)
. In order to obtain resonance curves, I need to plot f(gamma1)
(for chosen constant betas
, say, 2,1,0.5,0.25,
etc.).
I've solved this "by hand" and found f := F/(sqrt((-gamma1^2+omega^2)^2+4*lambda^2*gamma1^2))
. Plotting this for any chosen beta gives the result needed, for example, for beta:=0.5
the plot is
I wonder if I can obtain these curves using maple functions only (without solving anything "by hand" at all).
[edited]
maple
I'm trying to obtain resonance curves of the system. System can be described as
F,m,k:=2,1,4:
lambda:= beta/(2*m):
omega:=sqrt(k/m):
de:=diff(x(t),t$2)+2*lambda*diff(x(t),t)+omega^2*x(t)=F*cos(gamma1*t):
cond:=x(0)=0, D(x)(0)=0:
sol := dsolve({cond, de});
Solving gives sum of terms, some of which "die out" with time (since these terms have exp(-...*t)
) and some of which form steady-state solution (solution for t -> ∞
). This solution will be in form xstst=f(gamma1)*sin(...)
. In order to obtain resonance curves, I need to plot f(gamma1)
(for chosen constant betas
, say, 2,1,0.5,0.25,
etc.).
I've solved this "by hand" and found f := F/(sqrt((-gamma1^2+omega^2)^2+4*lambda^2*gamma1^2))
. Plotting this for any chosen beta gives the result needed, for example, for beta:=0.5
the plot is
I wonder if I can obtain these curves using maple functions only (without solving anything "by hand" at all).
[edited]
maple
maple
edited Jan 17 at 13:39
Kelly Shepphard
asked Jan 3 at 21:38
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Kelly ShepphardKelly Shepphard
11616
11616
add a comment |
add a comment |
2 Answers
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It does not make sense to expect Maple to represent the result in terms of either sin(theta)
or cos(theta)
, using some formulas for those terms that appear nowhere in the problem specification and are entirely introduced by you.
The following is obtained using a radical (square root) in each of cond1
and cond2
.
restart;
de := diff(x(t),t,t)+2*lambda*diff(x(t),t)+omega^2*x(t)=F*cos(gamma1*t):
cond := x(0)=0, D(x)(0)=0:
sol := dsolve({cond, de}):
E,T := selectremove(hastype,rhs(sol),specfunc(anything,exp)):
lprint(T);
F*(2*sin(gamma1*t)*gamma1*lambda-cos(gamma1*t)*gamma1^2
+cos(gamma1*t)*omega^2)/(gamma1^4+4*gamma1^2*lambda^2-
2*gamma1^2*omega^2+omega^4)
cond1 := cos(theta) = (-gamma1^2+omega^2)
/((-gamma1^2+omega^2)^2+4*lambda^2*gamma1^2)^(1/2):
cond2 := sin(theta) = (-2*lambda*gamma1)
/((-gamma1^2+omega^2)^2+4*lambda^2*gamma1^2)^(1/2):
frontend(algsubs, [numer(rhs(cond1))=lhs(cond1)*denom(rhs(cond1)),
numer(T)],
[{`+`,`*`,`=`},{}]):
frontend(algsubs, [numer(rhs(cond2))=lhs(cond2)*denom(rhs(cond2)),
%],
[{`+`,`*`,`=`},{}]):
ans := collect(combine(%, trig),cos)/denom(T):
lprint(ans);
F*cos(gamma1*t+theta)/(gamma1^4+4*gamma1^2*lambda^2-
2*gamma1^2*omega^2+omega^4)^(1/2)
subsans := eval(eval(ans,[lambda=beta/(2*m),omega=sqrt(k/m)]),
[F=2,m=1,k=4]):
lprint(subsans);
2*cos(gamma1*t+theta)
/(beta^2*gamma1^2+gamma1^4-8*gamma1^2+16)^(1/2)
add a comment |
I'm not sure I fully understand your question, but when I run your code I get some terms with exponentials, some with sines and some with cosines. You can grab the coefficient of the sine terms with
coeff( collect( rhs( sol ) , sin( gamma1 * t ) ) , sin( gamma1 * t ) , 1 )
Thank you for the answer, but I need to find coeff only for the last summand (which doesn't contain exp(-t*...)) and term will be like f(gamma1)*sin(gamma1*t+theta), where theta is also a function of lambda,omega and gamma1. I've tried to collect this, but it seems that it won't work (collecting gives the same 3rd term as before). At least, now I know that it's impossible. +1 for the help, thank you. This idea was very useful for understanding.
– Kelly Shepphard
Jan 17 at 6:43
@KellyShepphard --- I suspect it is possible, but I'm still not 100% sure of exactly what you want. (There is no theta in your code, and I don't know where the sqrt comes from in the work you did by hand.) Perhaps I can give a better answer if you upload a screenshot from Maple and highlight the terms that you want to isolate.
– Ian Thompson
Jan 17 at 9:19
I've added screenshot. Sorry for the lack of info, I've thought Maple automatically solves it as was needed.
– Kelly Shepphard
Jan 17 at 13:40
add a comment |
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2 Answers
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2 Answers
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It does not make sense to expect Maple to represent the result in terms of either sin(theta)
or cos(theta)
, using some formulas for those terms that appear nowhere in the problem specification and are entirely introduced by you.
The following is obtained using a radical (square root) in each of cond1
and cond2
.
restart;
de := diff(x(t),t,t)+2*lambda*diff(x(t),t)+omega^2*x(t)=F*cos(gamma1*t):
cond := x(0)=0, D(x)(0)=0:
sol := dsolve({cond, de}):
E,T := selectremove(hastype,rhs(sol),specfunc(anything,exp)):
lprint(T);
F*(2*sin(gamma1*t)*gamma1*lambda-cos(gamma1*t)*gamma1^2
+cos(gamma1*t)*omega^2)/(gamma1^4+4*gamma1^2*lambda^2-
2*gamma1^2*omega^2+omega^4)
cond1 := cos(theta) = (-gamma1^2+omega^2)
/((-gamma1^2+omega^2)^2+4*lambda^2*gamma1^2)^(1/2):
cond2 := sin(theta) = (-2*lambda*gamma1)
/((-gamma1^2+omega^2)^2+4*lambda^2*gamma1^2)^(1/2):
frontend(algsubs, [numer(rhs(cond1))=lhs(cond1)*denom(rhs(cond1)),
numer(T)],
[{`+`,`*`,`=`},{}]):
frontend(algsubs, [numer(rhs(cond2))=lhs(cond2)*denom(rhs(cond2)),
%],
[{`+`,`*`,`=`},{}]):
ans := collect(combine(%, trig),cos)/denom(T):
lprint(ans);
F*cos(gamma1*t+theta)/(gamma1^4+4*gamma1^2*lambda^2-
2*gamma1^2*omega^2+omega^4)^(1/2)
subsans := eval(eval(ans,[lambda=beta/(2*m),omega=sqrt(k/m)]),
[F=2,m=1,k=4]):
lprint(subsans);
2*cos(gamma1*t+theta)
/(beta^2*gamma1^2+gamma1^4-8*gamma1^2+16)^(1/2)
add a comment |
It does not make sense to expect Maple to represent the result in terms of either sin(theta)
or cos(theta)
, using some formulas for those terms that appear nowhere in the problem specification and are entirely introduced by you.
The following is obtained using a radical (square root) in each of cond1
and cond2
.
restart;
de := diff(x(t),t,t)+2*lambda*diff(x(t),t)+omega^2*x(t)=F*cos(gamma1*t):
cond := x(0)=0, D(x)(0)=0:
sol := dsolve({cond, de}):
E,T := selectremove(hastype,rhs(sol),specfunc(anything,exp)):
lprint(T);
F*(2*sin(gamma1*t)*gamma1*lambda-cos(gamma1*t)*gamma1^2
+cos(gamma1*t)*omega^2)/(gamma1^4+4*gamma1^2*lambda^2-
2*gamma1^2*omega^2+omega^4)
cond1 := cos(theta) = (-gamma1^2+omega^2)
/((-gamma1^2+omega^2)^2+4*lambda^2*gamma1^2)^(1/2):
cond2 := sin(theta) = (-2*lambda*gamma1)
/((-gamma1^2+omega^2)^2+4*lambda^2*gamma1^2)^(1/2):
frontend(algsubs, [numer(rhs(cond1))=lhs(cond1)*denom(rhs(cond1)),
numer(T)],
[{`+`,`*`,`=`},{}]):
frontend(algsubs, [numer(rhs(cond2))=lhs(cond2)*denom(rhs(cond2)),
%],
[{`+`,`*`,`=`},{}]):
ans := collect(combine(%, trig),cos)/denom(T):
lprint(ans);
F*cos(gamma1*t+theta)/(gamma1^4+4*gamma1^2*lambda^2-
2*gamma1^2*omega^2+omega^4)^(1/2)
subsans := eval(eval(ans,[lambda=beta/(2*m),omega=sqrt(k/m)]),
[F=2,m=1,k=4]):
lprint(subsans);
2*cos(gamma1*t+theta)
/(beta^2*gamma1^2+gamma1^4-8*gamma1^2+16)^(1/2)
add a comment |
It does not make sense to expect Maple to represent the result in terms of either sin(theta)
or cos(theta)
, using some formulas for those terms that appear nowhere in the problem specification and are entirely introduced by you.
The following is obtained using a radical (square root) in each of cond1
and cond2
.
restart;
de := diff(x(t),t,t)+2*lambda*diff(x(t),t)+omega^2*x(t)=F*cos(gamma1*t):
cond := x(0)=0, D(x)(0)=0:
sol := dsolve({cond, de}):
E,T := selectremove(hastype,rhs(sol),specfunc(anything,exp)):
lprint(T);
F*(2*sin(gamma1*t)*gamma1*lambda-cos(gamma1*t)*gamma1^2
+cos(gamma1*t)*omega^2)/(gamma1^4+4*gamma1^2*lambda^2-
2*gamma1^2*omega^2+omega^4)
cond1 := cos(theta) = (-gamma1^2+omega^2)
/((-gamma1^2+omega^2)^2+4*lambda^2*gamma1^2)^(1/2):
cond2 := sin(theta) = (-2*lambda*gamma1)
/((-gamma1^2+omega^2)^2+4*lambda^2*gamma1^2)^(1/2):
frontend(algsubs, [numer(rhs(cond1))=lhs(cond1)*denom(rhs(cond1)),
numer(T)],
[{`+`,`*`,`=`},{}]):
frontend(algsubs, [numer(rhs(cond2))=lhs(cond2)*denom(rhs(cond2)),
%],
[{`+`,`*`,`=`},{}]):
ans := collect(combine(%, trig),cos)/denom(T):
lprint(ans);
F*cos(gamma1*t+theta)/(gamma1^4+4*gamma1^2*lambda^2-
2*gamma1^2*omega^2+omega^4)^(1/2)
subsans := eval(eval(ans,[lambda=beta/(2*m),omega=sqrt(k/m)]),
[F=2,m=1,k=4]):
lprint(subsans);
2*cos(gamma1*t+theta)
/(beta^2*gamma1^2+gamma1^4-8*gamma1^2+16)^(1/2)
It does not make sense to expect Maple to represent the result in terms of either sin(theta)
or cos(theta)
, using some formulas for those terms that appear nowhere in the problem specification and are entirely introduced by you.
The following is obtained using a radical (square root) in each of cond1
and cond2
.
restart;
de := diff(x(t),t,t)+2*lambda*diff(x(t),t)+omega^2*x(t)=F*cos(gamma1*t):
cond := x(0)=0, D(x)(0)=0:
sol := dsolve({cond, de}):
E,T := selectremove(hastype,rhs(sol),specfunc(anything,exp)):
lprint(T);
F*(2*sin(gamma1*t)*gamma1*lambda-cos(gamma1*t)*gamma1^2
+cos(gamma1*t)*omega^2)/(gamma1^4+4*gamma1^2*lambda^2-
2*gamma1^2*omega^2+omega^4)
cond1 := cos(theta) = (-gamma1^2+omega^2)
/((-gamma1^2+omega^2)^2+4*lambda^2*gamma1^2)^(1/2):
cond2 := sin(theta) = (-2*lambda*gamma1)
/((-gamma1^2+omega^2)^2+4*lambda^2*gamma1^2)^(1/2):
frontend(algsubs, [numer(rhs(cond1))=lhs(cond1)*denom(rhs(cond1)),
numer(T)],
[{`+`,`*`,`=`},{}]):
frontend(algsubs, [numer(rhs(cond2))=lhs(cond2)*denom(rhs(cond2)),
%],
[{`+`,`*`,`=`},{}]):
ans := collect(combine(%, trig),cos)/denom(T):
lprint(ans);
F*cos(gamma1*t+theta)/(gamma1^4+4*gamma1^2*lambda^2-
2*gamma1^2*omega^2+omega^4)^(1/2)
subsans := eval(eval(ans,[lambda=beta/(2*m),omega=sqrt(k/m)]),
[F=2,m=1,k=4]):
lprint(subsans);
2*cos(gamma1*t+theta)
/(beta^2*gamma1^2+gamma1^4-8*gamma1^2+16)^(1/2)
answered Jan 31 at 10:12
aceracer
5,0071015
5,0071015
add a comment |
add a comment |
I'm not sure I fully understand your question, but when I run your code I get some terms with exponentials, some with sines and some with cosines. You can grab the coefficient of the sine terms with
coeff( collect( rhs( sol ) , sin( gamma1 * t ) ) , sin( gamma1 * t ) , 1 )
Thank you for the answer, but I need to find coeff only for the last summand (which doesn't contain exp(-t*...)) and term will be like f(gamma1)*sin(gamma1*t+theta), where theta is also a function of lambda,omega and gamma1. I've tried to collect this, but it seems that it won't work (collecting gives the same 3rd term as before). At least, now I know that it's impossible. +1 for the help, thank you. This idea was very useful for understanding.
– Kelly Shepphard
Jan 17 at 6:43
@KellyShepphard --- I suspect it is possible, but I'm still not 100% sure of exactly what you want. (There is no theta in your code, and I don't know where the sqrt comes from in the work you did by hand.) Perhaps I can give a better answer if you upload a screenshot from Maple and highlight the terms that you want to isolate.
– Ian Thompson
Jan 17 at 9:19
I've added screenshot. Sorry for the lack of info, I've thought Maple automatically solves it as was needed.
– Kelly Shepphard
Jan 17 at 13:40
add a comment |
I'm not sure I fully understand your question, but when I run your code I get some terms with exponentials, some with sines and some with cosines. You can grab the coefficient of the sine terms with
coeff( collect( rhs( sol ) , sin( gamma1 * t ) ) , sin( gamma1 * t ) , 1 )
Thank you for the answer, but I need to find coeff only for the last summand (which doesn't contain exp(-t*...)) and term will be like f(gamma1)*sin(gamma1*t+theta), where theta is also a function of lambda,omega and gamma1. I've tried to collect this, but it seems that it won't work (collecting gives the same 3rd term as before). At least, now I know that it's impossible. +1 for the help, thank you. This idea was very useful for understanding.
– Kelly Shepphard
Jan 17 at 6:43
@KellyShepphard --- I suspect it is possible, but I'm still not 100% sure of exactly what you want. (There is no theta in your code, and I don't know where the sqrt comes from in the work you did by hand.) Perhaps I can give a better answer if you upload a screenshot from Maple and highlight the terms that you want to isolate.
– Ian Thompson
Jan 17 at 9:19
I've added screenshot. Sorry for the lack of info, I've thought Maple automatically solves it as was needed.
– Kelly Shepphard
Jan 17 at 13:40
add a comment |
I'm not sure I fully understand your question, but when I run your code I get some terms with exponentials, some with sines and some with cosines. You can grab the coefficient of the sine terms with
coeff( collect( rhs( sol ) , sin( gamma1 * t ) ) , sin( gamma1 * t ) , 1 )
I'm not sure I fully understand your question, but when I run your code I get some terms with exponentials, some with sines and some with cosines. You can grab the coefficient of the sine terms with
coeff( collect( rhs( sol ) , sin( gamma1 * t ) ) , sin( gamma1 * t ) , 1 )
answered Jan 16 at 22:34
Ian ThompsonIan Thompson
1464
1464
Thank you for the answer, but I need to find coeff only for the last summand (which doesn't contain exp(-t*...)) and term will be like f(gamma1)*sin(gamma1*t+theta), where theta is also a function of lambda,omega and gamma1. I've tried to collect this, but it seems that it won't work (collecting gives the same 3rd term as before). At least, now I know that it's impossible. +1 for the help, thank you. This idea was very useful for understanding.
– Kelly Shepphard
Jan 17 at 6:43
@KellyShepphard --- I suspect it is possible, but I'm still not 100% sure of exactly what you want. (There is no theta in your code, and I don't know where the sqrt comes from in the work you did by hand.) Perhaps I can give a better answer if you upload a screenshot from Maple and highlight the terms that you want to isolate.
– Ian Thompson
Jan 17 at 9:19
I've added screenshot. Sorry for the lack of info, I've thought Maple automatically solves it as was needed.
– Kelly Shepphard
Jan 17 at 13:40
add a comment |
Thank you for the answer, but I need to find coeff only for the last summand (which doesn't contain exp(-t*...)) and term will be like f(gamma1)*sin(gamma1*t+theta), where theta is also a function of lambda,omega and gamma1. I've tried to collect this, but it seems that it won't work (collecting gives the same 3rd term as before). At least, now I know that it's impossible. +1 for the help, thank you. This idea was very useful for understanding.
– Kelly Shepphard
Jan 17 at 6:43
@KellyShepphard --- I suspect it is possible, but I'm still not 100% sure of exactly what you want. (There is no theta in your code, and I don't know where the sqrt comes from in the work you did by hand.) Perhaps I can give a better answer if you upload a screenshot from Maple and highlight the terms that you want to isolate.
– Ian Thompson
Jan 17 at 9:19
I've added screenshot. Sorry for the lack of info, I've thought Maple automatically solves it as was needed.
– Kelly Shepphard
Jan 17 at 13:40
Thank you for the answer, but I need to find coeff only for the last summand (which doesn't contain exp(-t*...)) and term will be like f(gamma1)*sin(gamma1*t+theta), where theta is also a function of lambda,omega and gamma1. I've tried to collect this, but it seems that it won't work (collecting gives the same 3rd term as before). At least, now I know that it's impossible. +1 for the help, thank you. This idea was very useful for understanding.
– Kelly Shepphard
Jan 17 at 6:43
Thank you for the answer, but I need to find coeff only for the last summand (which doesn't contain exp(-t*...)) and term will be like f(gamma1)*sin(gamma1*t+theta), where theta is also a function of lambda,omega and gamma1. I've tried to collect this, but it seems that it won't work (collecting gives the same 3rd term as before). At least, now I know that it's impossible. +1 for the help, thank you. This idea was very useful for understanding.
– Kelly Shepphard
Jan 17 at 6:43
@KellyShepphard --- I suspect it is possible, but I'm still not 100% sure of exactly what you want. (There is no theta in your code, and I don't know where the sqrt comes from in the work you did by hand.) Perhaps I can give a better answer if you upload a screenshot from Maple and highlight the terms that you want to isolate.
– Ian Thompson
Jan 17 at 9:19
@KellyShepphard --- I suspect it is possible, but I'm still not 100% sure of exactly what you want. (There is no theta in your code, and I don't know where the sqrt comes from in the work you did by hand.) Perhaps I can give a better answer if you upload a screenshot from Maple and highlight the terms that you want to isolate.
– Ian Thompson
Jan 17 at 9:19
I've added screenshot. Sorry for the lack of info, I've thought Maple automatically solves it as was needed.
– Kelly Shepphard
Jan 17 at 13:40
I've added screenshot. Sorry for the lack of info, I've thought Maple automatically solves it as was needed.
– Kelly Shepphard
Jan 17 at 13:40
add a comment |
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