Mixing
How to solve this ordinary differential equation from Van Long Dynamic games?
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I have to ask because I did the linear algebra/ode course 2 years ago so I don't remember. It doesn't look like a complicated problem but I can't recall anything.It's from Ngo Van Long's A survey in dynamic games.
Here it is here
It gets you both S(hat) and E(hat). I can do 1.13/1.14, how to proceed further?
δ, c and ρ are constants.
linear-algebra ordinary-differential-equations
asked Jan 26 at 6:16
SidSid
82
$endgroup$
|
show 1 more comment
$begingroup$
I have to ask because I did the linear algebra/ode course 2 years ago so I don't remember. It doesn't look like a complicated problem but I can't recall anything.It's from Ngo Van Long's A survey in dynamic games.
Here it is here
It gets you both S(hat) and E(hat). I can do 1.13/1.14, how to proceed further?
δ, c and ρ are constants.
linear-algebra ordinary-differential-equations
asked Jan 26 at 6:16
SidSid
82
$endgroup$
$begingroup$
E(hat)=δ*S(hat)/2
$endgroup$
– Sid
Jan 26 at 6:18
$begingroup$
What are $E$ and $S$? I see $S^*$ in the second equation. Does this mean that $E$ and $S$ are complex square matrices, and $S^*$ is the adjoint (conjugate transpose)? If not, then what is the $*$ operator?
$endgroup$
– Paul Sinclair
Jan 26 at 15:57
$begingroup$
@PaulSinclair E and S are functions of t. E(t) and S(t). S* is the rate of change of S, no it is not the adjoint. It's an economics problem so the star is to indicate the optimal value for the variable. They are not matrices.
$endgroup$
– Sid
Jan 29 at 10:38
$begingroup$
Make up your mind: Does $*$ mean "rate of change" or does it mean "optimal value for the variable"? The dot over $E$ and $S$ usually denotes rate of change, so I assume that $*$ is the latter, but in that case, how are you defining "optimal value"?
$endgroup$
– Paul Sinclair
Jan 29 at 17:45
$begingroup$
S(dot) implies rate of change whereas * implies the optimal value of the rate of change, or the optimal rate of change of variable.
$endgroup$
– Sid
Jan 30 at 18:21
|
show 1 more comment
$begingroup$
I have to ask because I did the linear algebra/ode course 2 years ago so I don't remember. It doesn't look like a complicated problem but I can't recall anything.It's from Ngo Van Long's A survey in dynamic games.
Here it is here
It gets you both S(hat) and E(hat). I can do 1.13/1.14, how to proceed further?
δ, c and ρ are constants.
linear-algebra ordinary-differential-equations
asked Jan 26 at 6:16
SidSid
82
$endgroup$
I have to ask because I did the linear algebra/ode course 2 years ago so I don't remember. It doesn't look like a complicated problem but I can't recall anything.It's from Ngo Van Long's A survey in dynamic games.
Here it is here
It gets you both S(hat) and E(hat). I can do 1.13/1.14, how to proceed further?
δ, c and ρ are constants.
linear-algebra ordinary-differential-equations
linear-algebra ordinary-differential-equations
asked Jan 26 at 6:16
SidSid
82
asked Jan 26 at 6:16
SidSid
82
asked Jan 26 at 6:16
SidSid
82
asked Jan 26 at 6:16
SidSid
82
asked Jan 26 at 6:16
SidSid
82
82
$begingroup$
E(hat)=δ*S(hat)/2
$endgroup$
– Sid
Jan 26 at 6:18
$begingroup$
What are $E$ and $S$? I see $S^*$ in the second equation. Does this mean that $E$ and $S$ are complex square matrices, and $S^*$ is the adjoint (conjugate transpose)? If not, then what is the $*$ operator?
$endgroup$
– Paul Sinclair
Jan 26 at 15:57
$begingroup$
@PaulSinclair E and S are functions of t. E(t) and S(t). S* is the rate of change of S, no it is not the adjoint. It's an economics problem so the star is to indicate the optimal value for the variable. They are not matrices.
$endgroup$
– Sid
Jan 29 at 10:38
$begingroup$
Make up your mind: Does $*$ mean "rate of change" or does it mean "optimal value for the variable"? The dot over $E$ and $S$ usually denotes rate of change, so I assume that $*$ is the latter, but in that case, how are you defining "optimal value"?
$endgroup$
– Paul Sinclair
Jan 29 at 17:45
$begingroup$
S(dot) implies rate of change whereas * implies the optimal value of the rate of change, or the optimal rate of change of variable.
$endgroup$
– Sid
Jan 30 at 18:21
|
show 1 more comment
$begingroup$
E(hat)=δ*S(hat)/2
$endgroup$
– Sid
Jan 26 at 6:18
$begingroup$
What are $E$ and $S$? I see $S^*$ in the second equation. Does this mean that $E$ and $S$ are complex square matrices, and $S^*$ is the adjoint (conjugate transpose)? If not, then what is the $*$ operator?
$endgroup$
– Paul Sinclair
Jan 26 at 15:57
$begingroup$
@PaulSinclair E and S are functions of t. E(t) and S(t). S* is the rate of change of S, no it is not the adjoint. It's an economics problem so the star is to indicate the optimal value for the variable. They are not matrices.
$endgroup$
– Sid
Jan 29 at 10:38
$begingroup$
Make up your mind: Does $*$ mean "rate of change" or does it mean "optimal value for the variable"? The dot over $E$ and $S$ usually denotes rate of change, so I assume that $*$ is the latter, but in that case, how are you defining "optimal value"?
$endgroup$
– Paul Sinclair
Jan 29 at 17:45
$begingroup$
S(dot) implies rate of change whereas * implies the optimal value of the rate of change, or the optimal rate of change of variable.
$endgroup$
– Sid
Jan 30 at 18:21
$begingroup$
E(hat)=δ*S(hat)/2
$endgroup$
– Sid
Jan 26 at 6:18
$begingroup$
E(hat)=δ*S(hat)/2
$endgroup$
– Sid
Jan 26 at 6:18
$begingroup$
What are $E$ and $S$? I see $S^*$ in the second equation. Does this mean that $E$ and $S$ are complex square matrices, and $S^*$ is the adjoint (conjugate transpose)? If not, then what is the $*$ operator?
$endgroup$
– Paul Sinclair
Jan 26 at 15:57
$begingroup$
What are $E$ and $S$? I see $S^*$ in the second equation. Does this mean that $E$ and $S$ are complex square matrices, and $S^*$ is the adjoint (conjugate transpose)? If not, then what is the $*$ operator?
$endgroup$
– Paul Sinclair
Jan 26 at 15:57
$begingroup$
@PaulSinclair E and S are functions of t. E(t) and S(t). S* is the rate of change of S, no it is not the adjoint. It's an economics problem so the star is to indicate the optimal value for the variable. They are not matrices.
$endgroup$
– Sid
Jan 29 at 10:38
$begingroup$
@PaulSinclair E and S are functions of t. E(t) and S(t). S* is the rate of change of S, no it is not the adjoint. It's an economics problem so the star is to indicate the optimal value for the variable. They are not matrices.
$endgroup$
– Sid
Jan 29 at 10:38
$begingroup$
Make up your mind: Does $*$ mean "rate of change" or does it mean "optimal value for the variable"? The dot over $E$ and $S$ usually denotes rate of change, so I assume that $*$ is the latter, but in that case, how are you defining "optimal value"?
$endgroup$
– Paul Sinclair
Jan 29 at 17:45
$begingroup$
Make up your mind: Does $*$ mean "rate of change" or does it mean "optimal value for the variable"? The dot over $E$ and $S$ usually denotes rate of change, so I assume that $*$ is the latter, but in that case, how are you defining "optimal value"?
$endgroup$
– Paul Sinclair
Jan 29 at 17:45
$begingroup$
S(dot) implies rate of change whereas * implies the optimal value of the rate of change, or the optimal rate of change of variable.
$endgroup$
– Sid
Jan 30 at 18:21
$begingroup$
S(dot) implies rate of change whereas * implies the optimal value of the rate of change, or the optimal rate of change of variable.
$endgroup$
– Sid
Jan 30 at 18:21
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Actually, it doesn't matter what you mean by "optimal value for the variable", provided that it is true that when $E$ and $S$ are constant, so is $S^*$.
The two equations in question are
$$dot E(t) = cS(t) + (rho+delta)(E(t) - A)tag{1.13}$$
$$dot {S^*}(t) = 2E(t) - delta S(t)tag{1.14}$$
Making the assumption above about $S^*$, there is only one other bit of information you need: $hat S$ and $hat E$ are steady-state values of $S$ and $E$. "Steady-state" means unchanging in time, so $dot E = 0$, and by the assumption, $dot {S^*} = 0$ as well. Plugging these in, we have
$$0 = chat S + (rho+delta)(hat E - A)$$
$$0 = 2hat E - delta hat Stag{1.14}$$
Now just solve for $hat S$ and $hat E$.
answered Jan 30 at 0:15
Paul SinclairPaul Sinclair
20.5k21543
$endgroup$
$begingroup$
thank you for the straightforward answer, I was thinking in the wrong direction
$endgroup$
– Sid
Jan 30 at 18:24
add a comment |
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$begingroup$
Actually, it doesn't matter what you mean by "optimal value for the variable", provided that it is true that when $E$ and $S$ are constant, so is $S^*$.
The two equations in question are
$$dot E(t) = cS(t) + (rho+delta)(E(t) - A)tag{1.13}$$
$$dot {S^*}(t) = 2E(t) - delta S(t)tag{1.14}$$
Making the assumption above about $S^*$, there is only one other bit of information you need: $hat S$ and $hat E$ are steady-state values of $S$ and $E$. "Steady-state" means unchanging in time, so $dot E = 0$, and by the assumption, $dot {S^*} = 0$ as well. Plugging these in, we have
$$0 = chat S + (rho+delta)(hat E - A)$$
$$0 = 2hat E - delta hat Stag{1.14}$$
Now just solve for $hat S$ and $hat E$.
answered Jan 30 at 0:15
Paul SinclairPaul Sinclair
20.5k21543
$endgroup$
$begingroup$
thank you for the straightforward answer, I was thinking in the wrong direction
$endgroup$
– Sid
Jan 30 at 18:24
add a comment |
$begingroup$
Actually, it doesn't matter what you mean by "optimal value for the variable", provided that it is true that when $E$ and $S$ are constant, so is $S^*$.
The two equations in question are
$$dot E(t) = cS(t) + (rho+delta)(E(t) - A)tag{1.13}$$
$$dot {S^*}(t) = 2E(t) - delta S(t)tag{1.14}$$
Making the assumption above about $S^*$, there is only one other bit of information you need: $hat S$ and $hat E$ are steady-state values of $S$ and $E$. "Steady-state" means unchanging in time, so $dot E = 0$, and by the assumption, $dot {S^*} = 0$ as well. Plugging these in, we have
$$0 = chat S + (rho+delta)(hat E - A)$$
$$0 = 2hat E - delta hat Stag{1.14}$$
Now just solve for $hat S$ and $hat E$.
answered Jan 30 at 0:15
Paul SinclairPaul Sinclair
20.5k21543
$endgroup$
$begingroup$
thank you for the straightforward answer, I was thinking in the wrong direction
$endgroup$
– Sid
Jan 30 at 18:24
add a comment |
$begingroup$
Actually, it doesn't matter what you mean by "optimal value for the variable", provided that it is true that when $E$ and $S$ are constant, so is $S^*$.
The two equations in question are
$$dot E(t) = cS(t) + (rho+delta)(E(t) - A)tag{1.13}$$
$$dot {S^*}(t) = 2E(t) - delta S(t)tag{1.14}$$
Making the assumption above about $S^*$, there is only one other bit of information you need: $hat S$ and $hat E$ are steady-state values of $S$ and $E$. "Steady-state" means unchanging in time, so $dot E = 0$, and by the assumption, $dot {S^*} = 0$ as well. Plugging these in, we have
$$0 = chat S + (rho+delta)(hat E - A)$$
$$0 = 2hat E - delta hat Stag{1.14}$$
Now just solve for $hat S$ and $hat E$.
answered Jan 30 at 0:15
Paul SinclairPaul Sinclair
20.5k21543
$endgroup$
Actually, it doesn't matter what you mean by "optimal value for the variable", provided that it is true that when $E$ and $S$ are constant, so is $S^*$.
The two equations in question are
$$dot E(t) = cS(t) + (rho+delta)(E(t) - A)tag{1.13}$$
$$dot {S^*}(t) = 2E(t) - delta S(t)tag{1.14}$$
Making the assumption above about $S^*$, there is only one other bit of information you need: $hat S$ and $hat E$ are steady-state values of $S$ and $E$. "Steady-state" means unchanging in time, so $dot E = 0$, and by the assumption, $dot {S^*} = 0$ as well. Plugging these in, we have
$$0 = chat S + (rho+delta)(hat E - A)$$
$$0 = 2hat E - delta hat Stag{1.14}$$
Now just solve for $hat S$ and $hat E$.
answered Jan 30 at 0:15
Paul SinclairPaul Sinclair
20.5k21543
answered Jan 30 at 0:15
Paul SinclairPaul Sinclair
20.5k21543
answered Jan 30 at 0:15
Paul SinclairPaul Sinclair
20.5k21543
answered Jan 30 at 0:15
Paul SinclairPaul Sinclair
20.5k21543
20.5k21543
$begingroup$
thank you for the straightforward answer, I was thinking in the wrong direction
$endgroup$
– Sid
Jan 30 at 18:24
add a comment |
$begingroup$
thank you for the straightforward answer, I was thinking in the wrong direction
$endgroup$
– Sid
Jan 30 at 18:24
$begingroup$
thank you for the straightforward answer, I was thinking in the wrong direction
$endgroup$
– Sid
Jan 30 at 18:24
$begingroup$
thank you for the straightforward answer, I was thinking in the wrong direction
$endgroup$
– Sid
Jan 30 at 18:24
add a comment |
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$begingroup$
E(hat)=δ*S(hat)/2
$endgroup$
– Sid
Jan 26 at 6:18
$begingroup$
What are $E$ and $S$? I see $S^*$ in the second equation. Does this mean that $E$ and $S$ are complex square matrices, and $S^*$ is the adjoint (conjugate transpose)? If not, then what is the $*$ operator?
$endgroup$
– Paul Sinclair
Jan 26 at 15:57
$begingroup$
@PaulSinclair E and S are functions of t. E(t) and S(t). S* is the rate of change of S, no it is not the adjoint. It's an economics problem so the star is to indicate the optimal value for the variable. They are not matrices.
$endgroup$
– Sid
Jan 29 at 10:38
$begingroup$
Make up your mind: Does $*$ mean "rate of change" or does it mean "optimal value for the variable"? The dot over $E$ and $S$ usually denotes rate of change, so I assume that $*$ is the latter, but in that case, how are you defining "optimal value"?
$endgroup$
– Paul Sinclair
Jan 29 at 17:45
$begingroup$
S(dot) implies rate of change whereas * implies the optimal value of the rate of change, or the optimal rate of change of variable.
$endgroup$
– Sid
Jan 30 at 18:21