d x d t = ω A − x 2 ,where A , ω &gt; 0 and x = x 0 at t = 0.It is to be solved from t = 0 to t = 50.0. It has analytical solution x ( t ) = A sin ⁡ ( ω t + φ ) ,where sin ⁡ ( φ ) = x 0 A if x 0 ≤ A .The question I am trying to solve is the followingRewrite the differential equation you have been given at the start of this document in the correct form for applying the Euler and Euler-Cauchy numerical schemes. Write down an appropriate 1 Euler method recursive scheme to solve this differential equation for the following values of the parameters and initial conditions: ω = 3.1, A = 10.0, x 0 = 2.0.

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d      x              d      t        =      ω        A    −          x      2        ,where   A  ,      ω    &amp;gt;  0 and   x  =      x    0   at   t  =  0.It is to be solved from   t  =  0 to   t  =  50.0. It has analytical solution  x  (  t  )  =      A    sin  ⁡  (      ω    t  +      φ    )  ,where   sin  ⁡  (      φ    )  =                    x        0                    A             if       x    0    ≤      A  .The question I am trying to solve is the followingRewrite the differential equation you have been given at the start of this document in the correct form for applying the Euler and Euler-Cauchy numerical schemes. Write down an appropriate 1 Euler method recursive scheme to solve this differential equation for the following values of the parameters and initial conditions:      ω    =  3.1,   A  =  10.0,       x    0    =  2.0.

mainzollbtt · Accepted Answer

The right side of the ODE is always positive, so that the given function is only a solution as long as it is growing. After reaching the maximum it will stay constant at this maximum value       A  .The Euler method computes       Δ    x  =  f  (  t  ,  x  )      Δ    t    ⟹        x          n      +      1        =      x    n    +  f  (      t    n    ,      x    n    )      Δ    tNote that this formula does not detect and handle domain boundaries. However, the domain of the given equation is bounded by       |    x      |    ≤      A  . Just applying the method without further preparations will run into errors in computing square roots of negative numbers.There are some natural continuations of the square root that remove this domain limitation, like            x      ˙        =      ω        max    (    0    ,    A    −          x      2        )           or                   x      ˙        =      ω          s    i    g    n    (  A  −      x    2    )            |        A    −          x      2              |        .The numerical solutions by the Euler method will reach this fixed point   x  =      A   and step over it. In the first variant the solution will then stay constant, in the second variant it will oscillate around the fixed point, possibly quite chaotically due to the vertical tangent at   x  =      A  .This can be repaiblack to some degree by replacing the vertical tangent by a segment with a steep but finite slope,            x      ˙        =      ω              A      −              x        2                    max      (      ε      ,              |            A      −              x        2                    |            )        .

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