Unsolvable first order nonlinear differential equation? $frac{dx(t)}{dt} + sin(x(t)) = sin(omega t)$
I do want to solve the following differential equation analytically:
$$frac{dx(t)}{dt} + sin(x(t)) = sin(omega t)$$
I tried several methods to solve this equation, unfortunately without any success. In the last week, I have read a lot of papers related to that kind of prototype and have looked into all the books that deal with first order differential equations. It still seems to be impossible. Neither Wolfram-alpha, nor Matlab's symbolic toolbox can give me a solution for it. Because of that, I am wondering if there is actually a solution for that kind of differential equation?
Thanks for your help!
calculus differential-equations
|
show 3 more comments
I do want to solve the following differential equation analytically:
$$frac{dx(t)}{dt} + sin(x(t)) = sin(omega t)$$
I tried several methods to solve this equation, unfortunately without any success. In the last week, I have read a lot of papers related to that kind of prototype and have looked into all the books that deal with first order differential equations. It still seems to be impossible. Neither Wolfram-alpha, nor Matlab's symbolic toolbox can give me a solution for it. Because of that, I am wondering if there is actually a solution for that kind of differential equation?
Thanks for your help!
calculus differential-equations
2
No CAS being able to solve is a very good sign that you can't neither. Due to the non-linearity (sine), I doubt that there is a closed-form expression. For small angles, use $sin xapprox x$.
– Yves Daoust
Nov 20 at 17:21
What about assumptions? Could that help you to find a solution? Actually I can not believe that such an equation is not solvable.
– R. Caloudis
Nov 20 at 17:24
3
It's very likely that there are no closed-form solutions to this differential equation. Most differential equations are like that. Of course there are solutions. You can solve the differential equation numerically, or find arbitrarily many terms of a series.
– Robert Israel
Nov 20 at 17:24
Thanks. However, I cannot use the small-angle approximation, because I do need that term in order to describe a certain physical phenomena. Any more guesses?
– R. Caloudis
Nov 20 at 17:26
The problem is the following: You do not have saturation nor harmonics without the sinus term. If you excite with a sinusoid at high amplitude the response would contain only the fundamental frequency without the term.
– R. Caloudis
Nov 20 at 17:30
|
show 3 more comments
I do want to solve the following differential equation analytically:
$$frac{dx(t)}{dt} + sin(x(t)) = sin(omega t)$$
I tried several methods to solve this equation, unfortunately without any success. In the last week, I have read a lot of papers related to that kind of prototype and have looked into all the books that deal with first order differential equations. It still seems to be impossible. Neither Wolfram-alpha, nor Matlab's symbolic toolbox can give me a solution for it. Because of that, I am wondering if there is actually a solution for that kind of differential equation?
Thanks for your help!
calculus differential-equations
I do want to solve the following differential equation analytically:
$$frac{dx(t)}{dt} + sin(x(t)) = sin(omega t)$$
I tried several methods to solve this equation, unfortunately without any success. In the last week, I have read a lot of papers related to that kind of prototype and have looked into all the books that deal with first order differential equations. It still seems to be impossible. Neither Wolfram-alpha, nor Matlab's symbolic toolbox can give me a solution for it. Because of that, I am wondering if there is actually a solution for that kind of differential equation?
Thanks for your help!
calculus differential-equations
calculus differential-equations
edited Nov 20 at 18:10
asked Nov 20 at 17:09
R. Caloudis
444
444
2
No CAS being able to solve is a very good sign that you can't neither. Due to the non-linearity (sine), I doubt that there is a closed-form expression. For small angles, use $sin xapprox x$.
– Yves Daoust
Nov 20 at 17:21
What about assumptions? Could that help you to find a solution? Actually I can not believe that such an equation is not solvable.
– R. Caloudis
Nov 20 at 17:24
3
It's very likely that there are no closed-form solutions to this differential equation. Most differential equations are like that. Of course there are solutions. You can solve the differential equation numerically, or find arbitrarily many terms of a series.
– Robert Israel
Nov 20 at 17:24
Thanks. However, I cannot use the small-angle approximation, because I do need that term in order to describe a certain physical phenomena. Any more guesses?
– R. Caloudis
Nov 20 at 17:26
The problem is the following: You do not have saturation nor harmonics without the sinus term. If you excite with a sinusoid at high amplitude the response would contain only the fundamental frequency without the term.
– R. Caloudis
Nov 20 at 17:30
|
show 3 more comments
2
No CAS being able to solve is a very good sign that you can't neither. Due to the non-linearity (sine), I doubt that there is a closed-form expression. For small angles, use $sin xapprox x$.
– Yves Daoust
Nov 20 at 17:21
What about assumptions? Could that help you to find a solution? Actually I can not believe that such an equation is not solvable.
– R. Caloudis
Nov 20 at 17:24
3
It's very likely that there are no closed-form solutions to this differential equation. Most differential equations are like that. Of course there are solutions. You can solve the differential equation numerically, or find arbitrarily many terms of a series.
– Robert Israel
Nov 20 at 17:24
Thanks. However, I cannot use the small-angle approximation, because I do need that term in order to describe a certain physical phenomena. Any more guesses?
– R. Caloudis
Nov 20 at 17:26
The problem is the following: You do not have saturation nor harmonics without the sinus term. If you excite with a sinusoid at high amplitude the response would contain only the fundamental frequency without the term.
– R. Caloudis
Nov 20 at 17:30
2
2
No CAS being able to solve is a very good sign that you can't neither. Due to the non-linearity (sine), I doubt that there is a closed-form expression. For small angles, use $sin xapprox x$.
– Yves Daoust
Nov 20 at 17:21
No CAS being able to solve is a very good sign that you can't neither. Due to the non-linearity (sine), I doubt that there is a closed-form expression. For small angles, use $sin xapprox x$.
– Yves Daoust
Nov 20 at 17:21
What about assumptions? Could that help you to find a solution? Actually I can not believe that such an equation is not solvable.
– R. Caloudis
Nov 20 at 17:24
What about assumptions? Could that help you to find a solution? Actually I can not believe that such an equation is not solvable.
– R. Caloudis
Nov 20 at 17:24
3
3
It's very likely that there are no closed-form solutions to this differential equation. Most differential equations are like that. Of course there are solutions. You can solve the differential equation numerically, or find arbitrarily many terms of a series.
– Robert Israel
Nov 20 at 17:24
It's very likely that there are no closed-form solutions to this differential equation. Most differential equations are like that. Of course there are solutions. You can solve the differential equation numerically, or find arbitrarily many terms of a series.
– Robert Israel
Nov 20 at 17:24
Thanks. However, I cannot use the small-angle approximation, because I do need that term in order to describe a certain physical phenomena. Any more guesses?
– R. Caloudis
Nov 20 at 17:26
Thanks. However, I cannot use the small-angle approximation, because I do need that term in order to describe a certain physical phenomena. Any more guesses?
– R. Caloudis
Nov 20 at 17:26
The problem is the following: You do not have saturation nor harmonics without the sinus term. If you excite with a sinusoid at high amplitude the response would contain only the fundamental frequency without the term.
– R. Caloudis
Nov 20 at 17:30
The problem is the following: You do not have saturation nor harmonics without the sinus term. If you excite with a sinusoid at high amplitude the response would contain only the fundamental frequency without the term.
– R. Caloudis
Nov 20 at 17:30
|
show 3 more comments
1 Answer
1
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oldest
votes
Of course solutions do exist, we might just be unable to find closed form representations for them. This doesn't mean we can't compute them numerically: here are some IVP trajectories with $omega=1$
You can plot just a fundamental patch $tin[0,2pi]$, $xin[-pi,pi]$ thanks to the periodicity (I've highlighted in red the stable periodic solution and in blue the unstable periodic solution):
Alternative visualization:
Edit. Code for the plots, in Mathematica:
sol = ParametricNDSolve[{
x'[t] + Sin[x[t]] == Sin[t],
x[0] == x0},
x, {t, 0, 4 [Pi]}, {{x0, -3 [Pi], 3 [Pi]}}];
Plot[Evaluate@Table[
x[x0][t] /. sol, {x0, -3 [Pi], 2.5 [Pi], .25}], {t, 0, 3.5 [Pi]}]
per = NDSolve[{
x'[t] + Sin[x[t]] == Sin[t],
x[0] == x[2 [Pi]]},
x, {t, 0, 2 [Pi]}];
Show[StreamPlot[{1, Sin[t] - Sin[x]}, {t, 0, 2 [Pi]}, {x, -[Pi], [Pi]}],
Plot[Evaluate[x[t] /. per], {t, 0, 2 [Pi]}, PlotStyle -> Red]]
Thanks! What exactly does the fundamental patch tell me? It does not seem to be just another representation of the solution x(t) with respect to t.
– R. Caloudis
Nov 20 at 18:01
1
Well, you know that the entire plane is filled with this repeating pattern, so this picture gives a pretty complete idea of the behavior of the solutions.
– Federico
Nov 20 at 18:03
I've now added a plot of the periodic solution overlaid on top of the stream plot. From the first picture you can see that this periodic solutions are attractive. You might be interested in studying this phenomenon, related to the stability of the system
– Federico
Nov 20 at 18:05
1
A bit harder to see in that plot, but there's also another (unstable) periodic solution with $y(0) approx 2.604965594$.
– Robert Israel
Nov 20 at 19:48
@RobertIsrael Updated to the plot, thanks
– Federico
Nov 21 at 15:24
add a comment |
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Of course solutions do exist, we might just be unable to find closed form representations for them. This doesn't mean we can't compute them numerically: here are some IVP trajectories with $omega=1$
You can plot just a fundamental patch $tin[0,2pi]$, $xin[-pi,pi]$ thanks to the periodicity (I've highlighted in red the stable periodic solution and in blue the unstable periodic solution):
Alternative visualization:
Edit. Code for the plots, in Mathematica:
sol = ParametricNDSolve[{
x'[t] + Sin[x[t]] == Sin[t],
x[0] == x0},
x, {t, 0, 4 [Pi]}, {{x0, -3 [Pi], 3 [Pi]}}];
Plot[Evaluate@Table[
x[x0][t] /. sol, {x0, -3 [Pi], 2.5 [Pi], .25}], {t, 0, 3.5 [Pi]}]
per = NDSolve[{
x'[t] + Sin[x[t]] == Sin[t],
x[0] == x[2 [Pi]]},
x, {t, 0, 2 [Pi]}];
Show[StreamPlot[{1, Sin[t] - Sin[x]}, {t, 0, 2 [Pi]}, {x, -[Pi], [Pi]}],
Plot[Evaluate[x[t] /. per], {t, 0, 2 [Pi]}, PlotStyle -> Red]]
Thanks! What exactly does the fundamental patch tell me? It does not seem to be just another representation of the solution x(t) with respect to t.
– R. Caloudis
Nov 20 at 18:01
1
Well, you know that the entire plane is filled with this repeating pattern, so this picture gives a pretty complete idea of the behavior of the solutions.
– Federico
Nov 20 at 18:03
I've now added a plot of the periodic solution overlaid on top of the stream plot. From the first picture you can see that this periodic solutions are attractive. You might be interested in studying this phenomenon, related to the stability of the system
– Federico
Nov 20 at 18:05
1
A bit harder to see in that plot, but there's also another (unstable) periodic solution with $y(0) approx 2.604965594$.
– Robert Israel
Nov 20 at 19:48
@RobertIsrael Updated to the plot, thanks
– Federico
Nov 21 at 15:24
add a comment |
Of course solutions do exist, we might just be unable to find closed form representations for them. This doesn't mean we can't compute them numerically: here are some IVP trajectories with $omega=1$
You can plot just a fundamental patch $tin[0,2pi]$, $xin[-pi,pi]$ thanks to the periodicity (I've highlighted in red the stable periodic solution and in blue the unstable periodic solution):
Alternative visualization:
Edit. Code for the plots, in Mathematica:
sol = ParametricNDSolve[{
x'[t] + Sin[x[t]] == Sin[t],
x[0] == x0},
x, {t, 0, 4 [Pi]}, {{x0, -3 [Pi], 3 [Pi]}}];
Plot[Evaluate@Table[
x[x0][t] /. sol, {x0, -3 [Pi], 2.5 [Pi], .25}], {t, 0, 3.5 [Pi]}]
per = NDSolve[{
x'[t] + Sin[x[t]] == Sin[t],
x[0] == x[2 [Pi]]},
x, {t, 0, 2 [Pi]}];
Show[StreamPlot[{1, Sin[t] - Sin[x]}, {t, 0, 2 [Pi]}, {x, -[Pi], [Pi]}],
Plot[Evaluate[x[t] /. per], {t, 0, 2 [Pi]}, PlotStyle -> Red]]
Thanks! What exactly does the fundamental patch tell me? It does not seem to be just another representation of the solution x(t) with respect to t.
– R. Caloudis
Nov 20 at 18:01
1
Well, you know that the entire plane is filled with this repeating pattern, so this picture gives a pretty complete idea of the behavior of the solutions.
– Federico
Nov 20 at 18:03
I've now added a plot of the periodic solution overlaid on top of the stream plot. From the first picture you can see that this periodic solutions are attractive. You might be interested in studying this phenomenon, related to the stability of the system
– Federico
Nov 20 at 18:05
1
A bit harder to see in that plot, but there's also another (unstable) periodic solution with $y(0) approx 2.604965594$.
– Robert Israel
Nov 20 at 19:48
@RobertIsrael Updated to the plot, thanks
– Federico
Nov 21 at 15:24
add a comment |
Of course solutions do exist, we might just be unable to find closed form representations for them. This doesn't mean we can't compute them numerically: here are some IVP trajectories with $omega=1$
You can plot just a fundamental patch $tin[0,2pi]$, $xin[-pi,pi]$ thanks to the periodicity (I've highlighted in red the stable periodic solution and in blue the unstable periodic solution):
Alternative visualization:
Edit. Code for the plots, in Mathematica:
sol = ParametricNDSolve[{
x'[t] + Sin[x[t]] == Sin[t],
x[0] == x0},
x, {t, 0, 4 [Pi]}, {{x0, -3 [Pi], 3 [Pi]}}];
Plot[Evaluate@Table[
x[x0][t] /. sol, {x0, -3 [Pi], 2.5 [Pi], .25}], {t, 0, 3.5 [Pi]}]
per = NDSolve[{
x'[t] + Sin[x[t]] == Sin[t],
x[0] == x[2 [Pi]]},
x, {t, 0, 2 [Pi]}];
Show[StreamPlot[{1, Sin[t] - Sin[x]}, {t, 0, 2 [Pi]}, {x, -[Pi], [Pi]}],
Plot[Evaluate[x[t] /. per], {t, 0, 2 [Pi]}, PlotStyle -> Red]]
Of course solutions do exist, we might just be unable to find closed form representations for them. This doesn't mean we can't compute them numerically: here are some IVP trajectories with $omega=1$
You can plot just a fundamental patch $tin[0,2pi]$, $xin[-pi,pi]$ thanks to the periodicity (I've highlighted in red the stable periodic solution and in blue the unstable periodic solution):
Alternative visualization:
Edit. Code for the plots, in Mathematica:
sol = ParametricNDSolve[{
x'[t] + Sin[x[t]] == Sin[t],
x[0] == x0},
x, {t, 0, 4 [Pi]}, {{x0, -3 [Pi], 3 [Pi]}}];
Plot[Evaluate@Table[
x[x0][t] /. sol, {x0, -3 [Pi], 2.5 [Pi], .25}], {t, 0, 3.5 [Pi]}]
per = NDSolve[{
x'[t] + Sin[x[t]] == Sin[t],
x[0] == x[2 [Pi]]},
x, {t, 0, 2 [Pi]}];
Show[StreamPlot[{1, Sin[t] - Sin[x]}, {t, 0, 2 [Pi]}, {x, -[Pi], [Pi]}],
Plot[Evaluate[x[t] /. per], {t, 0, 2 [Pi]}, PlotStyle -> Red]]
edited Nov 21 at 15:33
answered Nov 20 at 17:39
Federico
4,649514
4,649514
Thanks! What exactly does the fundamental patch tell me? It does not seem to be just another representation of the solution x(t) with respect to t.
– R. Caloudis
Nov 20 at 18:01
1
Well, you know that the entire plane is filled with this repeating pattern, so this picture gives a pretty complete idea of the behavior of the solutions.
– Federico
Nov 20 at 18:03
I've now added a plot of the periodic solution overlaid on top of the stream plot. From the first picture you can see that this periodic solutions are attractive. You might be interested in studying this phenomenon, related to the stability of the system
– Federico
Nov 20 at 18:05
1
A bit harder to see in that plot, but there's also another (unstable) periodic solution with $y(0) approx 2.604965594$.
– Robert Israel
Nov 20 at 19:48
@RobertIsrael Updated to the plot, thanks
– Federico
Nov 21 at 15:24
add a comment |
Thanks! What exactly does the fundamental patch tell me? It does not seem to be just another representation of the solution x(t) with respect to t.
– R. Caloudis
Nov 20 at 18:01
1
Well, you know that the entire plane is filled with this repeating pattern, so this picture gives a pretty complete idea of the behavior of the solutions.
– Federico
Nov 20 at 18:03
I've now added a plot of the periodic solution overlaid on top of the stream plot. From the first picture you can see that this periodic solutions are attractive. You might be interested in studying this phenomenon, related to the stability of the system
– Federico
Nov 20 at 18:05
1
A bit harder to see in that plot, but there's also another (unstable) periodic solution with $y(0) approx 2.604965594$.
– Robert Israel
Nov 20 at 19:48
@RobertIsrael Updated to the plot, thanks
– Federico
Nov 21 at 15:24
Thanks! What exactly does the fundamental patch tell me? It does not seem to be just another representation of the solution x(t) with respect to t.
– R. Caloudis
Nov 20 at 18:01
Thanks! What exactly does the fundamental patch tell me? It does not seem to be just another representation of the solution x(t) with respect to t.
– R. Caloudis
Nov 20 at 18:01
1
1
Well, you know that the entire plane is filled with this repeating pattern, so this picture gives a pretty complete idea of the behavior of the solutions.
– Federico
Nov 20 at 18:03
Well, you know that the entire plane is filled with this repeating pattern, so this picture gives a pretty complete idea of the behavior of the solutions.
– Federico
Nov 20 at 18:03
I've now added a plot of the periodic solution overlaid on top of the stream plot. From the first picture you can see that this periodic solutions are attractive. You might be interested in studying this phenomenon, related to the stability of the system
– Federico
Nov 20 at 18:05
I've now added a plot of the periodic solution overlaid on top of the stream plot. From the first picture you can see that this periodic solutions are attractive. You might be interested in studying this phenomenon, related to the stability of the system
– Federico
Nov 20 at 18:05
1
1
A bit harder to see in that plot, but there's also another (unstable) periodic solution with $y(0) approx 2.604965594$.
– Robert Israel
Nov 20 at 19:48
A bit harder to see in that plot, but there's also another (unstable) periodic solution with $y(0) approx 2.604965594$.
– Robert Israel
Nov 20 at 19:48
@RobertIsrael Updated to the plot, thanks
– Federico
Nov 21 at 15:24
@RobertIsrael Updated to the plot, thanks
– Federico
Nov 21 at 15:24
add a comment |
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2
No CAS being able to solve is a very good sign that you can't neither. Due to the non-linearity (sine), I doubt that there is a closed-form expression. For small angles, use $sin xapprox x$.
– Yves Daoust
Nov 20 at 17:21
What about assumptions? Could that help you to find a solution? Actually I can not believe that such an equation is not solvable.
– R. Caloudis
Nov 20 at 17:24
3
It's very likely that there are no closed-form solutions to this differential equation. Most differential equations are like that. Of course there are solutions. You can solve the differential equation numerically, or find arbitrarily many terms of a series.
– Robert Israel
Nov 20 at 17:24
Thanks. However, I cannot use the small-angle approximation, because I do need that term in order to describe a certain physical phenomena. Any more guesses?
– R. Caloudis
Nov 20 at 17:26
The problem is the following: You do not have saturation nor harmonics without the sinus term. If you excite with a sinusoid at high amplitude the response would contain only the fundamental frequency without the term.
– R. Caloudis
Nov 20 at 17:30