Symbolic solution for the energy of potential flow
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I have a question to a physical task in Mathematica.
We have this equation of motion:
$$mcdotddot{x} = -m(omega_0^2cdot x+(epsilon x^3))=-frac{d}{dx}V(x)$$
For energy of masspoint there is the condition :
$$epsilon Ell momega_0^4$$
I have to write a procedure that uses the law of the conservation of energy for the potential $V(x)$ to calculate $t(x_1) - t(x_0)$ when there are given two points $x_0$ and $x_1.
How could I do this in Mathematica?
differential-equations equation-solving
New contributor
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up vote
4
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favorite
I have a question to a physical task in Mathematica.
We have this equation of motion:
$$mcdotddot{x} = -m(omega_0^2cdot x+(epsilon x^3))=-frac{d}{dx}V(x)$$
For energy of masspoint there is the condition :
$$epsilon Ell momega_0^4$$
I have to write a procedure that uses the law of the conservation of energy for the potential $V(x)$ to calculate $t(x_1) - t(x_0)$ when there are given two points $x_0$ and $x_1.
How could I do this in Mathematica?
differential-equations equation-solving
New contributor
Welcome to Mathematica.SE! I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise.
– bbgodfrey
2 hours ago
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4
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up vote
4
down vote
favorite
I have a question to a physical task in Mathematica.
We have this equation of motion:
$$mcdotddot{x} = -m(omega_0^2cdot x+(epsilon x^3))=-frac{d}{dx}V(x)$$
For energy of masspoint there is the condition :
$$epsilon Ell momega_0^4$$
I have to write a procedure that uses the law of the conservation of energy for the potential $V(x)$ to calculate $t(x_1) - t(x_0)$ when there are given two points $x_0$ and $x_1.
How could I do this in Mathematica?
differential-equations equation-solving
New contributor
I have a question to a physical task in Mathematica.
We have this equation of motion:
$$mcdotddot{x} = -m(omega_0^2cdot x+(epsilon x^3))=-frac{d}{dx}V(x)$$
For energy of masspoint there is the condition :
$$epsilon Ell momega_0^4$$
I have to write a procedure that uses the law of the conservation of energy for the potential $V(x)$ to calculate $t(x_1) - t(x_0)$ when there are given two points $x_0$ and $x_1.
How could I do this in Mathematica?
differential-equations equation-solving
differential-equations equation-solving
New contributor
New contributor
edited 3 hours ago
chris
12.2k440108
12.2k440108
New contributor
asked 11 hours ago
Tom
211
211
New contributor
New contributor
Welcome to Mathematica.SE! I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise.
– bbgodfrey
2 hours ago
add a comment |
Welcome to Mathematica.SE! I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise.
– bbgodfrey
2 hours ago
Welcome to Mathematica.SE! I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise.
– bbgodfrey
2 hours ago
Welcome to Mathematica.SE! I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise.
– bbgodfrey
2 hours ago
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2 Answers
2
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up vote
4
down vote
In a numerical model, energy is conserved with some accuracy; in this example, the deviation from the initial value is about 1.5*10^-12
m = 1; omega0 = 1; eps = 1/100; v0 = 1;
eq = m*x''[t] == -m*(omega0^2*x[t] + eps*x[t]^3);
ic = {x[0] == 0, x'[0] == v0};
X = NDSolveValue[{eq, ic}, x, {t, 0, 10}, WorkingPrecision -> 30];
Plot[m/2*X'[t]^2 + m/2*omega0^2*X[t]^2 + m/4*eps*X[t]^4 -
m/2*v0^2, {t, 0, 10},AxesLabel -> {"t", "E-E0"}]
Using the law of conservation of energy, we express $x'(t) $ and then time as a function of $x$
t=Integrate[1/Sqrt[v0^2 - omega0^2*x^2 - eps/2*x^4], x]
(*-((I Sqrt[2 + (2 eps x^2)/(omega0^2 - Sqrt[omega0^4 + 2 eps v0^2])]
Sqrt[1 + (eps x^2)/(omega0^2 + Sqrt[omega0^4 + 2 eps v0^2])]
EllipticF[
I ArcSinh[Sqrt[eps/(omega0^2 - Sqrt[omega0^4 + 2 eps v0^2])] x], (
omega0^2 - Sqrt[omega0^4 + 2 eps v0^2])/(
omega0^2 + Sqrt[omega0^4 + 2 eps v0^2])])/(
Sqrt[eps/(omega0^2 - Sqrt[omega0^4 + 2 eps v0^2])] Sqrt[
2 v0^2 - 2 omega0^2 x^2 - eps x^4]))*)
Thank you Alex! Helps a lot
– Tom
8 hours ago
add a comment |
up vote
3
down vote
This problem can be solved symbolically as follows. Multiply the expression (m (omega0^2 x[t] + eps x[t]^3) + m x''[t])
by x'[t]
and integrate to obtain an expression for the energy of this nonlinear oscillator.
eq = Integrate[(m (omega0^2 x[t] + eps x[t]^3) + m x''[t]) x'[t], t]
(* 1/2 m omega0^2 x[t]^2 + 1/4 eps m x[t]^4 + 1/2 m x'[t]^2 *)
with constant of integration v0
, the conserved energy. Then, apply DSolve
.
s = DSolve[eq == v0, x[t], t] // Last
(* {x[t] -> InverseFunction[-((I EllipticF[I ArcSinh[Sqrt[(eps m)/(m omega0^2 -
Sqrt[m (m omega0^4 + 4 eps v0)])] #1], (m omega0^2 - Sqrt[m (m omega0^4 +
4 eps v0)])/(m omega0^2 + Sqrt[m (m omega0^4 + 4 eps v0)])] Sqrt[1 + (eps m #1^2)
/(m omega0^2 - Sqrt[m (m omega0^4 + 4 eps v0)])] Sqrt[1 + (eps m #1^2)
/(m omega0^2 + Sqrt[m (m omega0^4 + 4 eps v0)])])/(Sqrt[(eps m)/(m omega0^2 -
Sqrt[m (m omega0^4 + 4 eps v0)])] Sqrt[4 v0 - m #1^2 (2 omega0^2 + eps #1^2)])) &]
[t/(Sqrt[2] Sqrt[m]) + C[1]]} *)
(The other solution is the negative of the first.) Since the question requests t
as a function of x
, s
must be inverted. In the absence of a Mathematica command to accomplish this, we use the following ungainly expression.
st = Rule[(s[[1, 2, 1]] /. C[1] -> 0) Sqrt[2] Sqrt[m],
Head[s[[1, 2]]][[1]][x[t]] Sqrt[2] Sqrt[m]]
(* t -> -((I Sqrt[2] Sqrt[m] EllipticF[I ArcSinh[
Sqrt[(eps m)/(m omega0^2 - Sqrt[m (m omega0^4 + 4 eps v0)])] x[t]],
(m omega0^2 - Sqrt[m (m omega0^4 + 4 eps v0)])/
(m omega0^2 + Sqrt[m (m omega0^4 + 4 eps v0)])]
Sqrt[1 + (eps m x[t]^2)/(m omega0^2 - Sqrt[m (m omega0^4 + 4 eps v0)])]
Sqrt[1 + (eps m x[t]^2)/(m omega0^2 + Sqrt[m (m omega0^4 + 4 eps v0)])])/(
Sqrt[(eps m)/(m omega0^2 - Sqrt[m (m omega0^4 + 4 eps v0)])]
Sqrt[4 v0 - m x[t]^2 (2 omega0^2 + eps x[t]^2)])) *)
This result for various values of eps
can be plotted as
st /. {m -> 1, omega0 -> 1, v0 -> 1};
Plot[Evaluate@Table[Last[%], {eps, {10^1, 1, 10^-1, 10^-10}}], {x[t], -2, 2},
AxesLabel -> {x, t}, AspectRatio -> 1, ImageSize -> Large,
LabelStyle -> {Bold, Black, 15}]
Decreasing eps
corresponds to increasing values of x
and t
at the turning points.
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
In a numerical model, energy is conserved with some accuracy; in this example, the deviation from the initial value is about 1.5*10^-12
m = 1; omega0 = 1; eps = 1/100; v0 = 1;
eq = m*x''[t] == -m*(omega0^2*x[t] + eps*x[t]^3);
ic = {x[0] == 0, x'[0] == v0};
X = NDSolveValue[{eq, ic}, x, {t, 0, 10}, WorkingPrecision -> 30];
Plot[m/2*X'[t]^2 + m/2*omega0^2*X[t]^2 + m/4*eps*X[t]^4 -
m/2*v0^2, {t, 0, 10},AxesLabel -> {"t", "E-E0"}]
Using the law of conservation of energy, we express $x'(t) $ and then time as a function of $x$
t=Integrate[1/Sqrt[v0^2 - omega0^2*x^2 - eps/2*x^4], x]
(*-((I Sqrt[2 + (2 eps x^2)/(omega0^2 - Sqrt[omega0^4 + 2 eps v0^2])]
Sqrt[1 + (eps x^2)/(omega0^2 + Sqrt[omega0^4 + 2 eps v0^2])]
EllipticF[
I ArcSinh[Sqrt[eps/(omega0^2 - Sqrt[omega0^4 + 2 eps v0^2])] x], (
omega0^2 - Sqrt[omega0^4 + 2 eps v0^2])/(
omega0^2 + Sqrt[omega0^4 + 2 eps v0^2])])/(
Sqrt[eps/(omega0^2 - Sqrt[omega0^4 + 2 eps v0^2])] Sqrt[
2 v0^2 - 2 omega0^2 x^2 - eps x^4]))*)
Thank you Alex! Helps a lot
– Tom
8 hours ago
add a comment |
up vote
4
down vote
In a numerical model, energy is conserved with some accuracy; in this example, the deviation from the initial value is about 1.5*10^-12
m = 1; omega0 = 1; eps = 1/100; v0 = 1;
eq = m*x''[t] == -m*(omega0^2*x[t] + eps*x[t]^3);
ic = {x[0] == 0, x'[0] == v0};
X = NDSolveValue[{eq, ic}, x, {t, 0, 10}, WorkingPrecision -> 30];
Plot[m/2*X'[t]^2 + m/2*omega0^2*X[t]^2 + m/4*eps*X[t]^4 -
m/2*v0^2, {t, 0, 10},AxesLabel -> {"t", "E-E0"}]
Using the law of conservation of energy, we express $x'(t) $ and then time as a function of $x$
t=Integrate[1/Sqrt[v0^2 - omega0^2*x^2 - eps/2*x^4], x]
(*-((I Sqrt[2 + (2 eps x^2)/(omega0^2 - Sqrt[omega0^4 + 2 eps v0^2])]
Sqrt[1 + (eps x^2)/(omega0^2 + Sqrt[omega0^4 + 2 eps v0^2])]
EllipticF[
I ArcSinh[Sqrt[eps/(omega0^2 - Sqrt[omega0^4 + 2 eps v0^2])] x], (
omega0^2 - Sqrt[omega0^4 + 2 eps v0^2])/(
omega0^2 + Sqrt[omega0^4 + 2 eps v0^2])])/(
Sqrt[eps/(omega0^2 - Sqrt[omega0^4 + 2 eps v0^2])] Sqrt[
2 v0^2 - 2 omega0^2 x^2 - eps x^4]))*)
Thank you Alex! Helps a lot
– Tom
8 hours ago
add a comment |
up vote
4
down vote
up vote
4
down vote
In a numerical model, energy is conserved with some accuracy; in this example, the deviation from the initial value is about 1.5*10^-12
m = 1; omega0 = 1; eps = 1/100; v0 = 1;
eq = m*x''[t] == -m*(omega0^2*x[t] + eps*x[t]^3);
ic = {x[0] == 0, x'[0] == v0};
X = NDSolveValue[{eq, ic}, x, {t, 0, 10}, WorkingPrecision -> 30];
Plot[m/2*X'[t]^2 + m/2*omega0^2*X[t]^2 + m/4*eps*X[t]^4 -
m/2*v0^2, {t, 0, 10},AxesLabel -> {"t", "E-E0"}]
Using the law of conservation of energy, we express $x'(t) $ and then time as a function of $x$
t=Integrate[1/Sqrt[v0^2 - omega0^2*x^2 - eps/2*x^4], x]
(*-((I Sqrt[2 + (2 eps x^2)/(omega0^2 - Sqrt[omega0^4 + 2 eps v0^2])]
Sqrt[1 + (eps x^2)/(omega0^2 + Sqrt[omega0^4 + 2 eps v0^2])]
EllipticF[
I ArcSinh[Sqrt[eps/(omega0^2 - Sqrt[omega0^4 + 2 eps v0^2])] x], (
omega0^2 - Sqrt[omega0^4 + 2 eps v0^2])/(
omega0^2 + Sqrt[omega0^4 + 2 eps v0^2])])/(
Sqrt[eps/(omega0^2 - Sqrt[omega0^4 + 2 eps v0^2])] Sqrt[
2 v0^2 - 2 omega0^2 x^2 - eps x^4]))*)
In a numerical model, energy is conserved with some accuracy; in this example, the deviation from the initial value is about 1.5*10^-12
m = 1; omega0 = 1; eps = 1/100; v0 = 1;
eq = m*x''[t] == -m*(omega0^2*x[t] + eps*x[t]^3);
ic = {x[0] == 0, x'[0] == v0};
X = NDSolveValue[{eq, ic}, x, {t, 0, 10}, WorkingPrecision -> 30];
Plot[m/2*X'[t]^2 + m/2*omega0^2*X[t]^2 + m/4*eps*X[t]^4 -
m/2*v0^2, {t, 0, 10},AxesLabel -> {"t", "E-E0"}]
Using the law of conservation of energy, we express $x'(t) $ and then time as a function of $x$
t=Integrate[1/Sqrt[v0^2 - omega0^2*x^2 - eps/2*x^4], x]
(*-((I Sqrt[2 + (2 eps x^2)/(omega0^2 - Sqrt[omega0^4 + 2 eps v0^2])]
Sqrt[1 + (eps x^2)/(omega0^2 + Sqrt[omega0^4 + 2 eps v0^2])]
EllipticF[
I ArcSinh[Sqrt[eps/(omega0^2 - Sqrt[omega0^4 + 2 eps v0^2])] x], (
omega0^2 - Sqrt[omega0^4 + 2 eps v0^2])/(
omega0^2 + Sqrt[omega0^4 + 2 eps v0^2])])/(
Sqrt[eps/(omega0^2 - Sqrt[omega0^4 + 2 eps v0^2])] Sqrt[
2 v0^2 - 2 omega0^2 x^2 - eps x^4]))*)
edited 2 hours ago
answered 10 hours ago
Alex Trounev
5,0901418
5,0901418
Thank you Alex! Helps a lot
– Tom
8 hours ago
add a comment |
Thank you Alex! Helps a lot
– Tom
8 hours ago
Thank you Alex! Helps a lot
– Tom
8 hours ago
Thank you Alex! Helps a lot
– Tom
8 hours ago
add a comment |
up vote
3
down vote
This problem can be solved symbolically as follows. Multiply the expression (m (omega0^2 x[t] + eps x[t]^3) + m x''[t])
by x'[t]
and integrate to obtain an expression for the energy of this nonlinear oscillator.
eq = Integrate[(m (omega0^2 x[t] + eps x[t]^3) + m x''[t]) x'[t], t]
(* 1/2 m omega0^2 x[t]^2 + 1/4 eps m x[t]^4 + 1/2 m x'[t]^2 *)
with constant of integration v0
, the conserved energy. Then, apply DSolve
.
s = DSolve[eq == v0, x[t], t] // Last
(* {x[t] -> InverseFunction[-((I EllipticF[I ArcSinh[Sqrt[(eps m)/(m omega0^2 -
Sqrt[m (m omega0^4 + 4 eps v0)])] #1], (m omega0^2 - Sqrt[m (m omega0^4 +
4 eps v0)])/(m omega0^2 + Sqrt[m (m omega0^4 + 4 eps v0)])] Sqrt[1 + (eps m #1^2)
/(m omega0^2 - Sqrt[m (m omega0^4 + 4 eps v0)])] Sqrt[1 + (eps m #1^2)
/(m omega0^2 + Sqrt[m (m omega0^4 + 4 eps v0)])])/(Sqrt[(eps m)/(m omega0^2 -
Sqrt[m (m omega0^4 + 4 eps v0)])] Sqrt[4 v0 - m #1^2 (2 omega0^2 + eps #1^2)])) &]
[t/(Sqrt[2] Sqrt[m]) + C[1]]} *)
(The other solution is the negative of the first.) Since the question requests t
as a function of x
, s
must be inverted. In the absence of a Mathematica command to accomplish this, we use the following ungainly expression.
st = Rule[(s[[1, 2, 1]] /. C[1] -> 0) Sqrt[2] Sqrt[m],
Head[s[[1, 2]]][[1]][x[t]] Sqrt[2] Sqrt[m]]
(* t -> -((I Sqrt[2] Sqrt[m] EllipticF[I ArcSinh[
Sqrt[(eps m)/(m omega0^2 - Sqrt[m (m omega0^4 + 4 eps v0)])] x[t]],
(m omega0^2 - Sqrt[m (m omega0^4 + 4 eps v0)])/
(m omega0^2 + Sqrt[m (m omega0^4 + 4 eps v0)])]
Sqrt[1 + (eps m x[t]^2)/(m omega0^2 - Sqrt[m (m omega0^4 + 4 eps v0)])]
Sqrt[1 + (eps m x[t]^2)/(m omega0^2 + Sqrt[m (m omega0^4 + 4 eps v0)])])/(
Sqrt[(eps m)/(m omega0^2 - Sqrt[m (m omega0^4 + 4 eps v0)])]
Sqrt[4 v0 - m x[t]^2 (2 omega0^2 + eps x[t]^2)])) *)
This result for various values of eps
can be plotted as
st /. {m -> 1, omega0 -> 1, v0 -> 1};
Plot[Evaluate@Table[Last[%], {eps, {10^1, 1, 10^-1, 10^-10}}], {x[t], -2, 2},
AxesLabel -> {x, t}, AspectRatio -> 1, ImageSize -> Large,
LabelStyle -> {Bold, Black, 15}]
Decreasing eps
corresponds to increasing values of x
and t
at the turning points.
add a comment |
up vote
3
down vote
This problem can be solved symbolically as follows. Multiply the expression (m (omega0^2 x[t] + eps x[t]^3) + m x''[t])
by x'[t]
and integrate to obtain an expression for the energy of this nonlinear oscillator.
eq = Integrate[(m (omega0^2 x[t] + eps x[t]^3) + m x''[t]) x'[t], t]
(* 1/2 m omega0^2 x[t]^2 + 1/4 eps m x[t]^4 + 1/2 m x'[t]^2 *)
with constant of integration v0
, the conserved energy. Then, apply DSolve
.
s = DSolve[eq == v0, x[t], t] // Last
(* {x[t] -> InverseFunction[-((I EllipticF[I ArcSinh[Sqrt[(eps m)/(m omega0^2 -
Sqrt[m (m omega0^4 + 4 eps v0)])] #1], (m omega0^2 - Sqrt[m (m omega0^4 +
4 eps v0)])/(m omega0^2 + Sqrt[m (m omega0^4 + 4 eps v0)])] Sqrt[1 + (eps m #1^2)
/(m omega0^2 - Sqrt[m (m omega0^4 + 4 eps v0)])] Sqrt[1 + (eps m #1^2)
/(m omega0^2 + Sqrt[m (m omega0^4 + 4 eps v0)])])/(Sqrt[(eps m)/(m omega0^2 -
Sqrt[m (m omega0^4 + 4 eps v0)])] Sqrt[4 v0 - m #1^2 (2 omega0^2 + eps #1^2)])) &]
[t/(Sqrt[2] Sqrt[m]) + C[1]]} *)
(The other solution is the negative of the first.) Since the question requests t
as a function of x
, s
must be inverted. In the absence of a Mathematica command to accomplish this, we use the following ungainly expression.
st = Rule[(s[[1, 2, 1]] /. C[1] -> 0) Sqrt[2] Sqrt[m],
Head[s[[1, 2]]][[1]][x[t]] Sqrt[2] Sqrt[m]]
(* t -> -((I Sqrt[2] Sqrt[m] EllipticF[I ArcSinh[
Sqrt[(eps m)/(m omega0^2 - Sqrt[m (m omega0^4 + 4 eps v0)])] x[t]],
(m omega0^2 - Sqrt[m (m omega0^4 + 4 eps v0)])/
(m omega0^2 + Sqrt[m (m omega0^4 + 4 eps v0)])]
Sqrt[1 + (eps m x[t]^2)/(m omega0^2 - Sqrt[m (m omega0^4 + 4 eps v0)])]
Sqrt[1 + (eps m x[t]^2)/(m omega0^2 + Sqrt[m (m omega0^4 + 4 eps v0)])])/(
Sqrt[(eps m)/(m omega0^2 - Sqrt[m (m omega0^4 + 4 eps v0)])]
Sqrt[4 v0 - m x[t]^2 (2 omega0^2 + eps x[t]^2)])) *)
This result for various values of eps
can be plotted as
st /. {m -> 1, omega0 -> 1, v0 -> 1};
Plot[Evaluate@Table[Last[%], {eps, {10^1, 1, 10^-1, 10^-10}}], {x[t], -2, 2},
AxesLabel -> {x, t}, AspectRatio -> 1, ImageSize -> Large,
LabelStyle -> {Bold, Black, 15}]
Decreasing eps
corresponds to increasing values of x
and t
at the turning points.
add a comment |
up vote
3
down vote
up vote
3
down vote
This problem can be solved symbolically as follows. Multiply the expression (m (omega0^2 x[t] + eps x[t]^3) + m x''[t])
by x'[t]
and integrate to obtain an expression for the energy of this nonlinear oscillator.
eq = Integrate[(m (omega0^2 x[t] + eps x[t]^3) + m x''[t]) x'[t], t]
(* 1/2 m omega0^2 x[t]^2 + 1/4 eps m x[t]^4 + 1/2 m x'[t]^2 *)
with constant of integration v0
, the conserved energy. Then, apply DSolve
.
s = DSolve[eq == v0, x[t], t] // Last
(* {x[t] -> InverseFunction[-((I EllipticF[I ArcSinh[Sqrt[(eps m)/(m omega0^2 -
Sqrt[m (m omega0^4 + 4 eps v0)])] #1], (m omega0^2 - Sqrt[m (m omega0^4 +
4 eps v0)])/(m omega0^2 + Sqrt[m (m omega0^4 + 4 eps v0)])] Sqrt[1 + (eps m #1^2)
/(m omega0^2 - Sqrt[m (m omega0^4 + 4 eps v0)])] Sqrt[1 + (eps m #1^2)
/(m omega0^2 + Sqrt[m (m omega0^4 + 4 eps v0)])])/(Sqrt[(eps m)/(m omega0^2 -
Sqrt[m (m omega0^4 + 4 eps v0)])] Sqrt[4 v0 - m #1^2 (2 omega0^2 + eps #1^2)])) &]
[t/(Sqrt[2] Sqrt[m]) + C[1]]} *)
(The other solution is the negative of the first.) Since the question requests t
as a function of x
, s
must be inverted. In the absence of a Mathematica command to accomplish this, we use the following ungainly expression.
st = Rule[(s[[1, 2, 1]] /. C[1] -> 0) Sqrt[2] Sqrt[m],
Head[s[[1, 2]]][[1]][x[t]] Sqrt[2] Sqrt[m]]
(* t -> -((I Sqrt[2] Sqrt[m] EllipticF[I ArcSinh[
Sqrt[(eps m)/(m omega0^2 - Sqrt[m (m omega0^4 + 4 eps v0)])] x[t]],
(m omega0^2 - Sqrt[m (m omega0^4 + 4 eps v0)])/
(m omega0^2 + Sqrt[m (m omega0^4 + 4 eps v0)])]
Sqrt[1 + (eps m x[t]^2)/(m omega0^2 - Sqrt[m (m omega0^4 + 4 eps v0)])]
Sqrt[1 + (eps m x[t]^2)/(m omega0^2 + Sqrt[m (m omega0^4 + 4 eps v0)])])/(
Sqrt[(eps m)/(m omega0^2 - Sqrt[m (m omega0^4 + 4 eps v0)])]
Sqrt[4 v0 - m x[t]^2 (2 omega0^2 + eps x[t]^2)])) *)
This result for various values of eps
can be plotted as
st /. {m -> 1, omega0 -> 1, v0 -> 1};
Plot[Evaluate@Table[Last[%], {eps, {10^1, 1, 10^-1, 10^-10}}], {x[t], -2, 2},
AxesLabel -> {x, t}, AspectRatio -> 1, ImageSize -> Large,
LabelStyle -> {Bold, Black, 15}]
Decreasing eps
corresponds to increasing values of x
and t
at the turning points.
This problem can be solved symbolically as follows. Multiply the expression (m (omega0^2 x[t] + eps x[t]^3) + m x''[t])
by x'[t]
and integrate to obtain an expression for the energy of this nonlinear oscillator.
eq = Integrate[(m (omega0^2 x[t] + eps x[t]^3) + m x''[t]) x'[t], t]
(* 1/2 m omega0^2 x[t]^2 + 1/4 eps m x[t]^4 + 1/2 m x'[t]^2 *)
with constant of integration v0
, the conserved energy. Then, apply DSolve
.
s = DSolve[eq == v0, x[t], t] // Last
(* {x[t] -> InverseFunction[-((I EllipticF[I ArcSinh[Sqrt[(eps m)/(m omega0^2 -
Sqrt[m (m omega0^4 + 4 eps v0)])] #1], (m omega0^2 - Sqrt[m (m omega0^4 +
4 eps v0)])/(m omega0^2 + Sqrt[m (m omega0^4 + 4 eps v0)])] Sqrt[1 + (eps m #1^2)
/(m omega0^2 - Sqrt[m (m omega0^4 + 4 eps v0)])] Sqrt[1 + (eps m #1^2)
/(m omega0^2 + Sqrt[m (m omega0^4 + 4 eps v0)])])/(Sqrt[(eps m)/(m omega0^2 -
Sqrt[m (m omega0^4 + 4 eps v0)])] Sqrt[4 v0 - m #1^2 (2 omega0^2 + eps #1^2)])) &]
[t/(Sqrt[2] Sqrt[m]) + C[1]]} *)
(The other solution is the negative of the first.) Since the question requests t
as a function of x
, s
must be inverted. In the absence of a Mathematica command to accomplish this, we use the following ungainly expression.
st = Rule[(s[[1, 2, 1]] /. C[1] -> 0) Sqrt[2] Sqrt[m],
Head[s[[1, 2]]][[1]][x[t]] Sqrt[2] Sqrt[m]]
(* t -> -((I Sqrt[2] Sqrt[m] EllipticF[I ArcSinh[
Sqrt[(eps m)/(m omega0^2 - Sqrt[m (m omega0^4 + 4 eps v0)])] x[t]],
(m omega0^2 - Sqrt[m (m omega0^4 + 4 eps v0)])/
(m omega0^2 + Sqrt[m (m omega0^4 + 4 eps v0)])]
Sqrt[1 + (eps m x[t]^2)/(m omega0^2 - Sqrt[m (m omega0^4 + 4 eps v0)])]
Sqrt[1 + (eps m x[t]^2)/(m omega0^2 + Sqrt[m (m omega0^4 + 4 eps v0)])])/(
Sqrt[(eps m)/(m omega0^2 - Sqrt[m (m omega0^4 + 4 eps v0)])]
Sqrt[4 v0 - m x[t]^2 (2 omega0^2 + eps x[t]^2)])) *)
This result for various values of eps
can be plotted as
st /. {m -> 1, omega0 -> 1, v0 -> 1};
Plot[Evaluate@Table[Last[%], {eps, {10^1, 1, 10^-1, 10^-10}}], {x[t], -2, 2},
AxesLabel -> {x, t}, AspectRatio -> 1, ImageSize -> Large,
LabelStyle -> {Bold, Black, 15}]
Decreasing eps
corresponds to increasing values of x
and t
at the turning points.
answered 4 hours ago
bbgodfrey
43.8k857107
43.8k857107
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