Mixing
How to definine a function, when it contains the integral of the function itself?
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My question is basically if it is possible to define the following function. I tried in GeoGebra which solely says it is a circular definition, which I understand, however I do strongly believe that there most be one way or another ot calculate this. It is for a model of how the in and outflow of the Mediterranean behave - I a very thankful for any help!
$V(t)=int{sqrt{sin{t}-V(t)}dt}$
(full equation: $V(t)_{c}=A_{cs}int{sqrt{2g(a sin{(t)}+c-frac{V(t)_{c}+V_{M}}{A_{M}})}dt}$
Where this is basically water flowing from one bucket to another and back when one of which's height changes sinusoidally and the other in reaction to the first one.)
I am sorry for the most probably very fundamental nature of the question, however I couldn't find anything related to that in the forum.
Thanks for any answers in advance!
integration
asked 2 days ago
xyaang
133
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up vote
2
down vote
favorite
My question is basically if it is possible to define the following function. I tried in GeoGebra which solely says it is a circular definition, which I understand, however I do strongly believe that there most be one way or another ot calculate this. It is for a model of how the in and outflow of the Mediterranean behave - I a very thankful for any help!
$V(t)=int{sqrt{sin{t}-V(t)}dt}$
(full equation: $V(t)_{c}=A_{cs}int{sqrt{2g(a sin{(t)}+c-frac{V(t)_{c}+V_{M}}{A_{M}})}dt}$
Where this is basically water flowing from one bucket to another and back when one of which's height changes sinusoidally and the other in reaction to the first one.)
I am sorry for the most probably very fundamental nature of the question, however I couldn't find anything related to that in the forum.
Thanks for any answers in advance!
integration
asked 2 days ago
xyaang
133
New contributor
xyaang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What you have there is a specific integral equation, which is, more-or-less, a differential equation in reverse. There are techniques for solving such equations, though it might be helpful to know where you encountered this.
– Xander Henderson
yesterday
Thank you very much for your quick response! It was about a simple approach in tackling the tides in the Mediterranean. I left out all the constants, but it comes down to the fact that the speed of the flow is dependent on the difference in height and hence on how much water flowed up to a certain point in time.
– xyaang
yesterday
If you derivate your equation, you find $frac{dV}{dt}=sqrt{sin t-V}$ which is a non-circular differential equation for $V$. But a non-linear one, so good luck with that
– thedude
yesterday
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
My question is basically if it is possible to define the following function. I tried in GeoGebra which solely says it is a circular definition, which I understand, however I do strongly believe that there most be one way or another ot calculate this. It is for a model of how the in and outflow of the Mediterranean behave - I a very thankful for any help!
$V(t)=int{sqrt{sin{t}-V(t)}dt}$
(full equation: $V(t)_{c}=A_{cs}int{sqrt{2g(a sin{(t)}+c-frac{V(t)_{c}+V_{M}}{A_{M}})}dt}$
Where this is basically water flowing from one bucket to another and back when one of which's height changes sinusoidally and the other in reaction to the first one.)
I am sorry for the most probably very fundamental nature of the question, however I couldn't find anything related to that in the forum.
Thanks for any answers in advance!
integration
asked 2 days ago
xyaang
133
New contributor
xyaang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
My question is basically if it is possible to define the following function. I tried in GeoGebra which solely says it is a circular definition, which I understand, however I do strongly believe that there most be one way or another ot calculate this. It is for a model of how the in and outflow of the Mediterranean behave - I a very thankful for any help!
$V(t)=int{sqrt{sin{t}-V(t)}dt}$
(full equation: $V(t)_{c}=A_{cs}int{sqrt{2g(a sin{(t)}+c-frac{V(t)_{c}+V_{M}}{A_{M}})}dt}$
Where this is basically water flowing from one bucket to another and back when one of which's height changes sinusoidally and the other in reaction to the first one.)
I am sorry for the most probably very fundamental nature of the question, however I couldn't find anything related to that in the forum.
Thanks for any answers in advance!
integration
integration
asked 2 days ago
xyaang
133
New contributor
xyaang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
xyaang
133
New contributor
xyaang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
asked 2 days ago
xyaang
133
New contributor
xyaang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
xyaang
133
asked 2 days ago
xyaang
133
133
New contributor
xyaang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
xyaang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
xyaang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What you have there is a specific integral equation, which is, more-or-less, a differential equation in reverse. There are techniques for solving such equations, though it might be helpful to know where you encountered this.
– Xander Henderson
yesterday
Thank you very much for your quick response! It was about a simple approach in tackling the tides in the Mediterranean. I left out all the constants, but it comes down to the fact that the speed of the flow is dependent on the difference in height and hence on how much water flowed up to a certain point in time.
– xyaang
yesterday
If you derivate your equation, you find $frac{dV}{dt}=sqrt{sin t-V}$ which is a non-circular differential equation for $V$. But a non-linear one, so good luck with that
– thedude
yesterday
add a comment |
What you have there is a specific integral equation, which is, more-or-less, a differential equation in reverse. There are techniques for solving such equations, though it might be helpful to know where you encountered this.
– Xander Henderson
yesterday
Thank you very much for your quick response! It was about a simple approach in tackling the tides in the Mediterranean. I left out all the constants, but it comes down to the fact that the speed of the flow is dependent on the difference in height and hence on how much water flowed up to a certain point in time.
– xyaang
yesterday
If you derivate your equation, you find $frac{dV}{dt}=sqrt{sin t-V}$ which is a non-circular differential equation for $V$. But a non-linear one, so good luck with that
– thedude
yesterday
What you have there is a specific integral equation, which is, more-or-less, a differential equation in reverse. There are techniques for solving such equations, though it might be helpful to know where you encountered this.
– Xander Henderson
yesterday
What you have there is a specific integral equation, which is, more-or-less, a differential equation in reverse. There are techniques for solving such equations, though it might be helpful to know where you encountered this.
– Xander Henderson
yesterday
Thank you very much for your quick response! It was about a simple approach in tackling the tides in the Mediterranean. I left out all the constants, but it comes down to the fact that the speed of the flow is dependent on the difference in height and hence on how much water flowed up to a certain point in time.
– xyaang
yesterday
Thank you very much for your quick response! It was about a simple approach in tackling the tides in the Mediterranean. I left out all the constants, but it comes down to the fact that the speed of the flow is dependent on the difference in height and hence on how much water flowed up to a certain point in time.
– xyaang
yesterday
If you derivate your equation, you find $frac{dV}{dt}=sqrt{sin t-V}$ which is a non-circular differential equation for $V$. But a non-linear one, so good luck with that
– thedude
yesterday
If you derivate your equation, you find $frac{dV}{dt}=sqrt{sin t-V}$ which is a non-circular differential equation for $V$. But a non-linear one, so good luck with that
– thedude
yesterday
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
What you write is a integral equation, which can be easily be turned to an ordinary differential equation by differentiation.
$$V'(x)=sqrt{sin t-V(t)}$$ or
$$V'^2(t)+V(t)=sin t.$$
(We have many more techniques for solving ODEs than integral equations.)
Unfortunately, this one resists to resolution.
answered yesterday
Yves Daoust
121k668218
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
What you write is a integral equation, which can be easily be turned to an ordinary differential equation by differentiation.
$$V'(x)=sqrt{sin t-V(t)}$$ or
$$V'^2(t)+V(t)=sin t.$$
(We have many more techniques for solving ODEs than integral equations.)
Unfortunately, this one resists to resolution.
answered yesterday
Yves Daoust
121k668218
add a comment |
up vote
1
down vote
accepted
What you write is a integral equation, which can be easily be turned to an ordinary differential equation by differentiation.
$$V'(x)=sqrt{sin t-V(t)}$$ or
$$V'^2(t)+V(t)=sin t.$$
(We have many more techniques for solving ODEs than integral equations.)
Unfortunately, this one resists to resolution.
answered yesterday
Yves Daoust
121k668218
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
What you write is a integral equation, which can be easily be turned to an ordinary differential equation by differentiation.
$$V'(x)=sqrt{sin t-V(t)}$$ or
$$V'^2(t)+V(t)=sin t.$$
(We have many more techniques for solving ODEs than integral equations.)
Unfortunately, this one resists to resolution.
answered yesterday
Yves Daoust
121k668218
What you write is a integral equation, which can be easily be turned to an ordinary differential equation by differentiation.
$$V'(x)=sqrt{sin t-V(t)}$$ or
$$V'^2(t)+V(t)=sin t.$$
(We have many more techniques for solving ODEs than integral equations.)
Unfortunately, this one resists to resolution.
answered yesterday
Yves Daoust
121k668218
answered yesterday
Yves Daoust
121k668218
answered yesterday
Yves Daoust
121k668218
answered yesterday
Yves Daoust
121k668218
121k668218
add a comment |
add a comment |
xyaang is a new contributor. Be nice, and check out our Code of Conduct.
xyaang is a new contributor. Be nice, and check out our Code of Conduct.
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What you have there is a specific integral equation, which is, more-or-less, a differential equation in reverse. There are techniques for solving such equations, though it might be helpful to know where you encountered this.
– Xander Henderson
yesterday
Thank you very much for your quick response! It was about a simple approach in tackling the tides in the Mediterranean. I left out all the constants, but it comes down to the fact that the speed of the flow is dependent on the difference in height and hence on how much water flowed up to a certain point in time.
– xyaang
yesterday
If you derivate your equation, you find $frac{dV}{dt}=sqrt{sin t-V}$ which is a non-circular differential equation for $V$. But a non-linear one, so good luck with that
– thedude
yesterday