0
$\begingroup$

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]]} *}

Now:

  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?
$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$
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  *)
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.