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Mikael Hakman

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These are answers submitted by Mikael Hakman

Corrected...

Mikael Hakman 15
June 20 2016
0 1

Yes, I had few errors. In the PDEs there should be a minus after the equal sign in both equations. In expression for dJ/dx there should be a ɸ before ε. Also some of ε (unicode 03B5) were ϵ (unicode 03F5). I'm attaching the correct file. It checks for the consistency now.

 

restart``

 

Original PDE system:

 

pdesys := {diff(C(x, t), t) = -phi*varepsilon*J(x, t)*C(x, t), diff(J(x, t), x) = -phi*varepsilon*J(x, t)*C(x, t)}:

``

C(x,t) is concentration of a reactant, J(x,t) is light intensity, epsilon is absorption coefficient, phi is quatum yield, x is distance within reactant volume along light path, and t is time.

 

Maple pdsolve gives:

 

pdsolve(pdesys, [J, C])

{C(x, t) = 0, J(x, t) = _F1(t)}, {C(x, t) = _F1(x)*_C1, J(x, t) = 0}

 (1)

``

Above is a trivial 0 solution and it isn't what we want. I don't need a tool to tell me that anything multiplied by 0 is 0, and that df/dx is 0 if f doesn't depend on x.

 

The real solution was derived in 1977 by people at Institute of Mathematics, Uppsala University, Sweden and it is:

 

J := proc (x, t) options operator, arrow; J__0*exp(J__0*varepsilon*phi*t)/(exp(J__0*varepsilon*phi*t)+exp(C__0*varepsilon*phi*x)-1) end proc

proc (x, t) options operator, arrow; J__0*exp(J__0*varepsilon*phi*t)/(exp(J__0*varepsilon*phi*t)+exp(C__0*varepsilon*phi*x)-1) end proc

 (2)

``C := proc (x, t) options operator, arrow; C__0*exp(C__0*varepsilon*phi*x)/(exp(J__0*varepsilon*phi*t)+exp(C__0*varepsilon*phi*x)-1) end proc

proc (x, t) options operator, arrow; C__0*exp(C__0*varepsilon*phi*x)/(exp(J__0*varepsilon*phi*t)+exp(C__0*varepsilon*phi*x)-1) end proc

 (3)

Check if solution is the right one:

 

(diff(J(x, t), x))/(-phi*varepsilon*J(x, t)*C(x, t))

1

 (4)

(diff(C(x, t), t))/(-phi*varepsilon*J(x, t)*C(x, t))

1

 (5)

``

 

Download Photo2.mw

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