Partial differential equations , rearranging and spotting

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Discussion Overview

The discussion revolves around the manipulation of a specific partial differential equation of the form r^2 (d^2R/dr^2) + 2r(dR/dr) - lambda*R = 0, with the assumption that R can be expressed as R ~ r^β. Participants are attempting to derive a related equation involving β and are seeking clarification on the substitution process and the implications of constants in the context of this equation.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose substituting R = r^β directly into the differential equation to derive the equation β(β − 1) rβ + 2β rβ − λ rβ = 0.
  • Others mention that the equation is an "Euler-Lagrange" type, noting that the substitution t = ln(r) transforms it into a "constant coefficients" problem, suggesting that solutions of the form e^{βt} could be explored.
  • One participant suggests setting R = constant * r^β, indicating that the derivatives will also involve constants, which they believe will cancel out.
  • Another participant questions the necessity of the constant in the expression, leading to a discussion about the meaning of the tilde symbol in this context, which is interpreted as indicating linear proportionality.

Areas of Agreement / Disagreement

Participants express differing views on the role of constants in the substitution process and the interpretation of the tilde symbol. There is no consensus on the necessity or implications of these constants, and the discussion remains unresolved regarding the best approach to the problem.

Contextual Notes

Participants have not fully resolved the implications of the substitution and the role of constants in the context of the differential equation. There are also varying interpretations of the notation used, which may affect the clarity of the discussion.

mohsin031211
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I have been given an equation : r^2 ( d^2R/dr^2) + 2r(dR/dr) - lambda*R = 0


It says to assume R~ r^β



Then i can't seem to spot how from that information we can produce this equation:

β(β − 1) rβ + 2β rβ − λ rβ = 0


Any help would be appreciated, thanks.
 
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mohsin031211 said:
I have been given an equation : r^2 ( d^2R/dr^2) + 2r(dR/dr) - lambda*R = 0It says to assume R~ r^β
Then i can't seem to spot how from that information we can produce this equation:

β(β − 1) rβ + 2β rβ − λ rβ = 0Any help would be appreciated, thanks.
Simply substitute R=r^\beta into the differential equation.
 
That is, by the way, an "Euler-Lagrange" type equation. Each derivative is multiplied by a power of x equal to the order of the derivative. The substitution t= ln(r) changes it to a "constant coefficients" problem. You should remember that for such an equation we "try" a solution of the form e^{\beta t} (although we then find that there are other solutions). With t= ln r, that becomes e^{\beta ln(r)}= e^{ln r^\beta}= r^\beta.
 
mohsin031211 said:
I have been given an equation : r^2 ( d^2R/dr^2) + 2r(dR/dr) - lambda*R = 0


It says to assume R~ r^β



Then i can't seem to spot how from that information we can produce this equation:

β(β − 1) rβ + 2β rβ − λ rβ = 0


Any help would be appreciated, thanks.

Set R= constant * r**beta so dR=constant * whatever

They are proportional
R=k*r--beta , all k's ( constants cancel )
 
stallionx said:
Set R= constant * r**beta so dR=constant * whatever

They are proportional
R=k*r--beta , all k's ( constants cancel )
Why the constant?
 
Hootenanny said:
Why the constant?

Well because I thought tilde is for (constant) linear proportionality.
 
I have been given an equation
 

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