An Approach to Mathematic Functions Basics

by Frank C. Fung ( 1st published in December, 2014. )

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Section XXXVI - Solving the Differential Equation for [ N = 3 ]

Summary for the Section :

• In this Section , we shall attempt to solve the Differential Equation
  • for the Central Force being a { Gravitation-Type Force varying as the [ inverse of R-Cube ] } .

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Setting up the Differential Equation for [ N = 3 ] :

• Let us now bring back , from Section XXXIV , the general Differential Equation
  • for the Central Force being a Gravitation-type Force varying as the [ inverse of R-raised-to-the-Nth-Power ] ,

  i.e. , this equation :

  • 

    with :

    • 

  Therefore for [ N = 3 ] , the Differential Equation is :

  • 

    yielding :

  • 

  On re-arranging terms , we have :

  • 

  And therefore this Differential Equation :

  • 

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The First Special Case of the Circular Orbit :

• Let us now look at the first Special Case with :
  • 

  Our original Differential Equation ,

  • i.e. , this equation :
    • 

    would now become :

    • 

• Let us now bring in the { Standardized Set of Initial Conditions } as detailed in Section XXIX ,
  • and the proposed format of our solution is then :
    • 

  We then have :

  • 

  And therefore :

  • 

  And the Final Solution here is then :

  • 

    and this is the equation for a Circular Orbit .

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The Second Special Case of the Straight-Line Orbit :

• Let us now permit the mass of the [ Mass at the Point-Of-Origin ] drop to [ zero ] .

  Consequently , the original Differential Equation ,

  • i.e. , this equation :
    • 

    would now become :

    • 

• Let us now bring in the { Standardized Set of Initial Conditions } as detailed in Section XXIX ,
  • and the proposed format of our solution is then :
    • 

  And :

  • the first-derivative thereof is then :
    • 

  • the second-derivative thereof is then :
    • 

  And we see here that our proposed format above does satisfy the Differential Equation :

  • 

    as required .

  Our final solution here is then :

  • 

  And consequently :

  • 

    and this is the equation of a Straight-Line .

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Breaking down the domain into 2 Regions :

• For the General Solution to the original Differention Equation ,
  • i.e. , this equation :
    • 

  let us now break down the domain for our analysis here into two (2) regions :

  • 

  • 

  And we take note here , of course , that the case of :

  • 

    has already been dealt with under the First Special Case of the Circular Orbit above .

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Solution for REGION ONE :

• For Region One defined by :
  • 

  let us now set up the value [ H-square ] where :

  • 

  And therefore :

  • 

  The original Differential Equation , i.e. this equation :

  • 

    may now be re-written in this format :

  • 

• Let us now bring in the { Standardized Set of Initial Conditions } as detailed in Section XXIX ,
  • and the proposed format of our solution is then :
    • 

  And :

  • the first-derivative thereof is then :
    • 

  • the second-derivative thereof is then :
    • 

  Substituting these into the Differential Equation above , i.e. :

  • 

    then yields us this :

  • 

  Consequently , we have :

  • 

• We see here that if we set :
  • 

  • 

    the Differential Equation would be satisfied .

  Accordingly , we can now set up :

  • 

  And the final solution here is then :

  • 

  And consequently , we have :

  • 

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Solution for REGION TWO :

• For Region TWO defined by :
  • 

  let us now set up the value [ H-square ] where :

  • 

  And therefore :

  • 

  The original Differential Equation , i.e. this equation :

  • 

    may now be re-written in this format :

  • 

• Let us now bring in the { Standardized Set of Initial Conditions } as detailed in Section XXIX ,
  • and the proposed format of our solution is then :
    • 

  And the first-derivative thereof is then :

  • 

  And the second-derivative thereof is then :

  • 

  Substituting these into the Differential Equation above , i.e. :

  • 

    then yields us this :

  • 

  On re-arranging terms , this would become :

  • 

  Consequently, we have :

  • 

• We see here that if we set :
  • 

  • 

    the Differential Equation would be satisfied .

  Accordingly , we can now set up :

  • 

  And the final solution here is then :

  • 

  And consequently , we have :

  • 

    and this the equation for a [ spiral ] .

• But this equation does not satisfy our { Standardized Set of Initial Conditions } from Section XXVI ,
  • which would have required that :
    • 

  And for the equation above , the first-derivative here is :

  • 

    yielding :

  • 

  and the value for the expression on the right-hand-side here can never be [ zero ] for all values of [ Theta ] ;

  • i.e. :
    • 

• Let us now try again , but with the help of a { Mathematical Handbook of Formulas and Tables } .

  And based on this , our proposal for the solution is :

  • 

  And :

  • the first-derivative thereof is then :
    • 

  • the second-derivative thereof is then :
    • 

• Let us now bring back the Differential Equation for REGION TWO from above ,
  • i.e. , this equation :
    • 

  and substitute therein the expression for [ z ] and its second-derivative ,

  • to arrive at this equation :
    • 

    consequently yielding :

    • 

  Consolidating terms then yields us this equation :

  • 

• And we see here that if we set up :
  • 

    the above equation would be satisied .

  Thus , the proposed solution we have now is then :

  • 

• Let us now take a quick look at the { first-derivative of [ R ] with respect to [ Theta ] } via :
  • 

  And we take note here that :

  • First-of-All :
    • The first-derivative of [ z ] with respect to [ Theta ] here is :
      • 

        based on our expression for [ z ] immediately above .

  • Secondly :
    • Since we have defined [ z ] as the reciprical of [ R ] , we have :
      • 

        and therefore :

      • 

  Let us now substitute these back into :

  • 

    to arrive at this equation :

  • 

• Let us now bring in the { Standardized Set of Initial Conditions } again ,
  • to help us determine the values of Constants [ C-sub-1 ] and [ C-sub-2 ] .

  Let us recall that , for our { Standardized Set of Initial Conditions } , we did set up :

  • 
  • 

  And these particular { Initial Conditions } do dictate that :

  • 

    arising from the { Velocity in the Radial Direction } being [ zero ] at [ t = 0 ] .

• Let us now do the evaluation for the { first-derivative of [ R ] with respect to [ Theta ] } at [ t = 0 ] ,
  • at which time the value of [ Theta ] is [ zero ] .

  And the equation for the said { first-derivative } , from above , is :

  • 

  Therefore ,

  • 

    yielding :

  • 

  Consequently :

  • 

  And we see here that the above expression can be [ turned-into-zero ]

  • if we set :
    • 

• We then set up the value [ C-sub-BAR ] such that :
  • 

  And the expression for [ z ] in terms of [ Theta ] can now be written as :

  • 

    yielding :

  • 

• Let us recall that :
  • 

    arising from the { Standiardized Set of Initial Conditions } .

  Thus , we can now write this equation below expressing [ z ] in terms of [ Theta ] :

  • 

  And finally , we have this equation expressing [ R ] in terms of [ Theta ] :

  • 

• Isn't that nice !

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go to the next section : Section XXXVII - A Graphic look at the Solutions for [ N = 3 ]

go to the last section : Section XXXV - Re-doing the Differential Equation for [ N = 2 ]

return to the MATH Functions Basics HomePage

Original dated 2014-12-07