Mixing
Solve a system of differential equations where argument of one function depend at another argument
$begingroup$
I need to solve a system of differential equations using Mathcad.
Moreover, the condition argument of one function depends on $t$, and delated $y = t - t_0$.
How make a substitution argument in the function $ N(y) $?
Editor's comment: Class of coupled / delayed differential equations.
ordinary-differential-equations
edited Mar 8 '15 at 16:51
Narasimham
20.6k52158
asked Mar 8 '15 at 15:32
timyrik20timyrik20
1
$endgroup$
add a comment |
$begingroup$
I need to solve a system of differential equations using Mathcad.
Moreover, the condition argument of one function depends on $t$, and delated $y = t - t_0$.
How make a substitution argument in the function $ N(y) $?
Editor's comment: Class of coupled / delayed differential equations.
ordinary-differential-equations
edited Mar 8 '15 at 16:51
Narasimham
20.6k52158
asked Mar 8 '15 at 15:32
timyrik20timyrik20
1
$endgroup$
$begingroup$
In your cited image it is $N(t)$?
$endgroup$
– LutzL
Mar 8 '15 at 16:46
$begingroup$
It's N(y) where y=t-t0
$endgroup$
– timyrik20
Mar 8 '15 at 16:52
add a comment |
$begingroup$
I need to solve a system of differential equations using Mathcad.
Moreover, the condition argument of one function depends on $t$, and delated $y = t - t_0$.
How make a substitution argument in the function $ N(y) $?
Editor's comment: Class of coupled / delayed differential equations.
ordinary-differential-equations
edited Mar 8 '15 at 16:51
Narasimham
20.6k52158
asked Mar 8 '15 at 15:32
timyrik20timyrik20
1
$endgroup$
I need to solve a system of differential equations using Mathcad.
Moreover, the condition argument of one function depends on $t$, and delated $y = t - t_0$.
How make a substitution argument in the function $ N(y) $?
Editor's comment: Class of coupled / delayed differential equations.
ordinary-differential-equations
ordinary-differential-equations
edited Mar 8 '15 at 16:51
Narasimham
20.6k52158
asked Mar 8 '15 at 15:32
timyrik20timyrik20
1
edited Mar 8 '15 at 16:51
Narasimham
20.6k52158
asked Mar 8 '15 at 15:32
timyrik20timyrik20
1
edited Mar 8 '15 at 16:51
Narasimham
20.6k52158
edited Mar 8 '15 at 16:51
Narasimham
20.6k52158
edited Mar 8 '15 at 16:51
Narasimham
20.6k52158
20.6k52158
asked Mar 8 '15 at 15:32
timyrik20timyrik20
1
asked Mar 8 '15 at 15:32
timyrik20timyrik20
1
asked Mar 8 '15 at 15:32
timyrik20timyrik20
1
1
$begingroup$
In your cited image it is $N(t)$?
$endgroup$
– LutzL
Mar 8 '15 at 16:46
$begingroup$
It's N(y) where y=t-t0
$endgroup$
– timyrik20
Mar 8 '15 at 16:52
add a comment |
$begingroup$
In your cited image it is $N(t)$?
$endgroup$
– LutzL
Mar 8 '15 at 16:46
$begingroup$
It's N(y) where y=t-t0
$endgroup$
– timyrik20
Mar 8 '15 at 16:52
$begingroup$
In your cited image it is $N(t)$?
$endgroup$
– LutzL
Mar 8 '15 at 16:46
$begingroup$
In your cited image it is $N(t)$?
$endgroup$
– LutzL
Mar 8 '15 at 16:46
$begingroup$
It's N(y) where y=t-t0
$endgroup$
– timyrik20
Mar 8 '15 at 16:52
$begingroup$
It's N(y) where y=t-t0
$endgroup$
– timyrik20
Mar 8 '15 at 16:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Set $M(t)=N(t-t_0)$, then you have a boundary value problems with linear boundary conditions at $t=0$ and $t=t_0$.
Resulting system
a*R'(t)+b*R''(t) = R(t)*(c-d*R(t)+e*M(t)), R(0) = R0, R'(0) = DR0,
a*M'(t)+b*M''(t) = M(t)*(f-g*M(t)+h*R(t)), M(t0) = N0, M'(t0) = DN0.
or with the constants as supplied by timyrik20
Now you can try to use ODEsolve in some versions of MathCad (15 but not Prime? by topics in the community forum). Single or multiple shooting strategies are usually employed in this case. Basically, the initial conditions of $M$ at $t=0$ are introduced as variables, the integration process can be reduced to a function to the values of the other boundary, where the boundary conditions form a non-linear system. This can be solved via Newton or derivative free (adjoint) secant methods. This single shooting approach seems to be used by MathCad, or you can emulate it by explicitly employing the solver on a parametrized solution. Other answers in the forum employ the MathCad command line with additional specialized procedures,...
answered Mar 8 '15 at 16:47
LutzLLutzL
57.1k42054
$endgroup$
$begingroup$
Whether I have understood you, when t0=3 i.stack.imgur.com/nl5H8.png
$endgroup$
– timyrik20
Mar 8 '15 at 19:39
$begingroup$
Yes, like that, apart from the arguments of Odesolve, check the documentation, the second argument should bet0or a somewhat larger constant, the last can be omitted, 10 subdivisions could be too small.
$endgroup$
– LutzL
Jan 3 at 10:17
add a comment |
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1 Answer
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$begingroup$
Set $M(t)=N(t-t_0)$, then you have a boundary value problems with linear boundary conditions at $t=0$ and $t=t_0$.
Resulting system
a*R'(t)+b*R''(t) = R(t)*(c-d*R(t)+e*M(t)), R(0) = R0, R'(0) = DR0,
a*M'(t)+b*M''(t) = M(t)*(f-g*M(t)+h*R(t)), M(t0) = N0, M'(t0) = DN0.
or with the constants as supplied by timyrik20
Now you can try to use ODEsolve in some versions of MathCad (15 but not Prime? by topics in the community forum). Single or multiple shooting strategies are usually employed in this case. Basically, the initial conditions of $M$ at $t=0$ are introduced as variables, the integration process can be reduced to a function to the values of the other boundary, where the boundary conditions form a non-linear system. This can be solved via Newton or derivative free (adjoint) secant methods. This single shooting approach seems to be used by MathCad, or you can emulate it by explicitly employing the solver on a parametrized solution. Other answers in the forum employ the MathCad command line with additional specialized procedures,...
answered Mar 8 '15 at 16:47
LutzLLutzL
57.1k42054
$endgroup$
$begingroup$
Whether I have understood you, when t0=3 i.stack.imgur.com/nl5H8.png
$endgroup$
– timyrik20
Mar 8 '15 at 19:39
$begingroup$
Yes, like that, apart from the arguments of Odesolve, check the documentation, the second argument should bet0or a somewhat larger constant, the last can be omitted, 10 subdivisions could be too small.
$endgroup$
– LutzL
Jan 3 at 10:17
add a comment |
$begingroup$
Set $M(t)=N(t-t_0)$, then you have a boundary value problems with linear boundary conditions at $t=0$ and $t=t_0$.
Resulting system
a*R'(t)+b*R''(t) = R(t)*(c-d*R(t)+e*M(t)), R(0) = R0, R'(0) = DR0,
a*M'(t)+b*M''(t) = M(t)*(f-g*M(t)+h*R(t)), M(t0) = N0, M'(t0) = DN0.
or with the constants as supplied by timyrik20
Now you can try to use ODEsolve in some versions of MathCad (15 but not Prime? by topics in the community forum). Single or multiple shooting strategies are usually employed in this case. Basically, the initial conditions of $M$ at $t=0$ are introduced as variables, the integration process can be reduced to a function to the values of the other boundary, where the boundary conditions form a non-linear system. This can be solved via Newton or derivative free (adjoint) secant methods. This single shooting approach seems to be used by MathCad, or you can emulate it by explicitly employing the solver on a parametrized solution. Other answers in the forum employ the MathCad command line with additional specialized procedures,...
answered Mar 8 '15 at 16:47
LutzLLutzL
57.1k42054
$endgroup$
$begingroup$
Whether I have understood you, when t0=3 i.stack.imgur.com/nl5H8.png
$endgroup$
– timyrik20
Mar 8 '15 at 19:39
$begingroup$
Yes, like that, apart from the arguments of Odesolve, check the documentation, the second argument should bet0or a somewhat larger constant, the last can be omitted, 10 subdivisions could be too small.
$endgroup$
– LutzL
Jan 3 at 10:17
add a comment |
$begingroup$
Set $M(t)=N(t-t_0)$, then you have a boundary value problems with linear boundary conditions at $t=0$ and $t=t_0$.
Resulting system
a*R'(t)+b*R''(t) = R(t)*(c-d*R(t)+e*M(t)), R(0) = R0, R'(0) = DR0,
a*M'(t)+b*M''(t) = M(t)*(f-g*M(t)+h*R(t)), M(t0) = N0, M'(t0) = DN0.
or with the constants as supplied by timyrik20
Now you can try to use ODEsolve in some versions of MathCad (15 but not Prime? by topics in the community forum). Single or multiple shooting strategies are usually employed in this case. Basically, the initial conditions of $M$ at $t=0$ are introduced as variables, the integration process can be reduced to a function to the values of the other boundary, where the boundary conditions form a non-linear system. This can be solved via Newton or derivative free (adjoint) secant methods. This single shooting approach seems to be used by MathCad, or you can emulate it by explicitly employing the solver on a parametrized solution. Other answers in the forum employ the MathCad command line with additional specialized procedures,...
answered Mar 8 '15 at 16:47
LutzLLutzL
57.1k42054
$endgroup$
Set $M(t)=N(t-t_0)$, then you have a boundary value problems with linear boundary conditions at $t=0$ and $t=t_0$.
Resulting system
a*R'(t)+b*R''(t) = R(t)*(c-d*R(t)+e*M(t)), R(0) = R0, R'(0) = DR0,
a*M'(t)+b*M''(t) = M(t)*(f-g*M(t)+h*R(t)), M(t0) = N0, M'(t0) = DN0.
or with the constants as supplied by timyrik20
Now you can try to use ODEsolve in some versions of MathCad (15 but not Prime? by topics in the community forum). Single or multiple shooting strategies are usually employed in this case. Basically, the initial conditions of $M$ at $t=0$ are introduced as variables, the integration process can be reduced to a function to the values of the other boundary, where the boundary conditions form a non-linear system. This can be solved via Newton or derivative free (adjoint) secant methods. This single shooting approach seems to be used by MathCad, or you can emulate it by explicitly employing the solver on a parametrized solution. Other answers in the forum employ the MathCad command line with additional specialized procedures,...
answered Mar 8 '15 at 16:47
LutzLLutzL
57.1k42054
edited Jan 3 at 10:13
answered Mar 8 '15 at 16:47
LutzLLutzL
57.1k42054
answered Mar 8 '15 at 16:47
LutzLLutzL
57.1k42054
answered Mar 8 '15 at 16:47
LutzLLutzL
57.1k42054
57.1k42054
$begingroup$
Whether I have understood you, when t0=3 i.stack.imgur.com/nl5H8.png
$endgroup$
– timyrik20
Mar 8 '15 at 19:39
$begingroup$
Yes, like that, apart from the arguments of Odesolve, check the documentation, the second argument should bet0or a somewhat larger constant, the last can be omitted, 10 subdivisions could be too small.
$endgroup$
– LutzL
Jan 3 at 10:17
add a comment |
$begingroup$
Whether I have understood you, when t0=3 i.stack.imgur.com/nl5H8.png
$endgroup$
– timyrik20
Mar 8 '15 at 19:39
$begingroup$
Yes, like that, apart from the arguments of Odesolve, check the documentation, the second argument should bet0or a somewhat larger constant, the last can be omitted, 10 subdivisions could be too small.
$endgroup$
– LutzL
Jan 3 at 10:17
$begingroup$
Whether I have understood you, when t0=3 i.stack.imgur.com/nl5H8.png
$endgroup$
– timyrik20
Mar 8 '15 at 19:39
$begingroup$
Whether I have understood you, when t0=3 i.stack.imgur.com/nl5H8.png
$endgroup$
– timyrik20
Mar 8 '15 at 19:39
$begingroup$
Yes, like that, apart from the arguments of Odesolve, check the documentation, the second argument should be
t0 or a somewhat larger constant, the last can be omitted, 10 subdivisions could be too small.$endgroup$
– LutzL
Jan 3 at 10:17
$begingroup$
Yes, like that, apart from the arguments of Odesolve, check the documentation, the second argument should be
t0 or a somewhat larger constant, the last can be omitted, 10 subdivisions could be too small.$endgroup$
– LutzL
Jan 3 at 10:17
add a comment |
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$begingroup$
In your cited image it is $N(t)$?
$endgroup$
– LutzL
Mar 8 '15 at 16:46
$begingroup$
It's N(y) where y=t-t0
$endgroup$
– timyrik20
Mar 8 '15 at 16:52