I'm trying to solve the following differential equation:

DSolve[{D[χ[β, βc, t],t] == -Γ (β D[χ[β, βc,t], β] + βc D[χ[β, βc, t], βc])}, χ, {β, βc, t}]

and this is the output

{* {χ -> Function[{β, βc, t}, C[1][βc/β, (t Γ - Log[β])/Γ]]} *}


  1. I don't understand why the result is written as a function of two arguments
  2. What is the meaning of (t Γ - Log[β])/Γ?
  • $\begingroup$ First, the solution is not a function of two arguments but of three {\[Beta], \[Beta]c, t}. Your solution is given in the form of PureFunction. For details about pure function see this link. Regarding last part of your question, t[CapitalGamma]-Log[[Beta]])/[CapitalGamma] is part of the solution (again see link for pure function). $\endgroup$ – ercegovac Feb 6 '17 at 19:15
eqn = D[χ[β, βc, t], 
    t] == -Γ (β D[χ[β, βc, 
         t], β] + βc D[χ[β, βc, t], βc]);

soln = DSolve[eqn, χ, {β, βc, t}][[1]]

{χ -> Function[{β, βc, t}, 
   C[1][βc/β, (t Γ - Log[β])/Γ]]}

Verifying the solution

eqn /. soln // Simplify

(*  True  *)

C[1] is an arbitrary function of its two shown parameters. For example. let C[1] be the undefined function f

soln2 = soln /. C[1] -> f

(*  {χ -> Function[{β, βc, t}, 
   f[βc/β, (t Γ - Log[β])/Γ]]}  *)

As expected, this solution also satisfies the equation

eqn /. soln2 // Simplify

(*  True  *)
  • $\begingroup$ I tried to impose χ[β, βc, 0]==E^(-(β βc)/2 but i didn't get anything $\endgroup$ – Kowalski Feb 6 '17 at 20:50
  • $\begingroup$ @Kowalski - your condition appears to be inconsistent with the general solution. \[Chi][\[Beta], \[Beta]c, 0] must be expressible as a function of the arguments {\[Beta]c / \[Beta], -Log[\[Beta]] / \[CapitalGamma]} $\endgroup$ – Bob Hanlon Feb 6 '17 at 23:59
  • $\begingroup$ But I found that also [Chi] -> Function[{β, βc, t}, C[1][β E^(-Γt), βc E^(-Γt)]], is a solution, so my condition is not inconsistent, i'm right? $\endgroup$ – Kowalski Feb 7 '17 at 8:28
  • $\begingroup$ @Kowalski - I'm not a mathematician and I don't know how to force Mathematica to provide a result using your constraint. $\endgroup$ – Bob Hanlon Feb 7 '17 at 16:26

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