Symbolic solution for the energy of potential flow












5














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?










share|improve this question
























  • 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
    Dec 3 '18 at 1:12
















5














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?










share|improve this question
























  • 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
    Dec 3 '18 at 1:12














5












5








5


1





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?










share|improve this question















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






share|improve this question















share|improve this question













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share|improve this question








edited Dec 3 '18 at 0:17









chris

12.2k441109




12.2k441109










asked Dec 2 '18 at 16:27









Tom

412




412












  • 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
    Dec 3 '18 at 1:12


















  • 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
    Dec 3 '18 at 1:12
















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
Dec 3 '18 at 1:12




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
Dec 3 '18 at 1:12










2 Answers
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5














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}]


enter image description here



Decreasing eps corresponds to increasing values of x and t at the turning points.






share|improve this answer





























    5














    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"}]


    fig1
    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]))*)





    share|improve this answer























    • Thank you Alex! Helps a lot
      – Tom
      Dec 2 '18 at 18:44











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    2 Answers
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    2 Answers
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    5














    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}]


    enter image description here



    Decreasing eps corresponds to increasing values of x and t at the turning points.






    share|improve this answer


























      5














      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}]


      enter image description here



      Decreasing eps corresponds to increasing values of x and t at the turning points.






      share|improve this answer
























        5












        5








        5






        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}]


        enter image description here



        Decreasing eps corresponds to increasing values of x and t at the turning points.






        share|improve this answer












        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}]


        enter image description here



        Decreasing eps corresponds to increasing values of x and t at the turning points.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Dec 2 '18 at 22:55









        bbgodfrey

        44.3k958109




        44.3k958109























            5














            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"}]


            fig1
            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]))*)





            share|improve this answer























            • Thank you Alex! Helps a lot
              – Tom
              Dec 2 '18 at 18:44
















            5














            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"}]


            fig1
            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]))*)





            share|improve this answer























            • Thank you Alex! Helps a lot
              – Tom
              Dec 2 '18 at 18:44














            5












            5








            5






            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"}]


            fig1
            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]))*)





            share|improve this answer














            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"}]


            fig1
            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]))*)






            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Dec 3 '18 at 0:58

























            answered Dec 2 '18 at 17:07









            Alex Trounev

            6,1201419




            6,1201419












            • Thank you Alex! Helps a lot
              – Tom
              Dec 2 '18 at 18:44


















            • Thank you Alex! Helps a lot
              – Tom
              Dec 2 '18 at 18:44
















            Thank you Alex! Helps a lot
            – Tom
            Dec 2 '18 at 18:44




            Thank you Alex! Helps a lot
            – Tom
            Dec 2 '18 at 18:44


















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