I want to get continuous point between two equations.

My equations are:

To[x_] := 2 c1 Cosh [x Sqrt[((f0 - b g0)/q)]] + g0/(f0 - b g0);

Tu[x_] := 10 Exp[-m x];

I try to find c1 and m constants which are in terms of g0.

I have boundary conditions which are;

T = Tu   @ x = +L/2

D^2 k T'[x] = Du^2  ku Tu'[x]     @ x = + L/2

So, I have two uknown "c1,m" and two boundary conditions for single point. It should be continuous at L/2. Actually boundary conditions for +- L/2 ( because cosh is symmetric, but I assume m is positive , so just for positive part of boundary)

Like that plot,for example, he had found Tu= 10.3 e^-598(x-0.005), it is continuous at 0.5 mm with cosh function.

enter image description here

In the same manner, it should be obtained thanks to boundary conditions.

Then I applied boundary conditions in order to find "c1,m" to be continuous at L/2.

Solve[g0/(f0 - b g0) + 2 c1 Cosh[1/2 L Sqrt[(f0 - b g0)/q]] - 
   10 E^(-((L m)/2)) == 0, c1]

{{c1 -> -((
    E^(-((L m)/2)) (-10 f0 + 10 b g0 + E^((L m)/2) g0) Sech[
      1/2 L Sqrt[(f0 - b g0)/q]])/(2 (f0 - b g0)))}}

c1 is okay, then for finding m I applied derivative boundary condition,

Solve[2 c1 Sqrt[(f0 - b g0)/q]
     Sinh[1/2 L Sqrt[(f0 - b g0)/q]] dia^2 k - (-10 E^(-((L m)/2))
      m Du^2 ku) == 0, m]

And program gave error as:

S`olve::nsmet: This system cannot be solved with the methods available to Solve.`

How can I get c1 and m in terms of g0 ?

When I obtain c1 and m in terms of g0, constant temperature condition will be used for g0:


  • 1
    $\begingroup$ I get a warning that it is using inverse functions (so may not find all the answers) but your final Solve does give an answer for m. Maybe quit the kernel and try again. $\endgroup$ – bill s Feb 27 '14 at 14:03
  • $\begingroup$ Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the faq! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! $\endgroup$ – Dr. belisarius Feb 27 '14 at 21:02

Let's do it with simplified variable names:

To[x_] := 2 c1 Cosh[x a1] + a2
Tu[x_] := 10 Exp[-m x]
bcs = {To[L/2] == Tu[L/2], To'[L/2] == a4 Tu'[L/2]};
s = First@Solve[bcs[[1]], c1]
Quiet@Solve[bcs[[2]] /. s[[1]], m]

Mathematica graphics

  • $\begingroup$ I was thinking that if I solve for positive part, also the problem will be done for negative part. Because cosh is symmetric. But Tu must be zero at +- Infinity. Because of this reason,I assumed that Tu=10 Exp[-mx], when x is positive it goes zero at +infinity and when x is negative it goes zero at -infinity. I applied your code, it is okay for @ x=L/2 but it not satisfied for @ x=-L/2. Can you help me? $\endgroup$ – CanYusuf Feb 27 '14 at 20:54
  • $\begingroup$ Why don't use another function to the left (the symmetric one) $\endgroup$ – Dr. belisarius Feb 27 '14 at 21:02
  • $\begingroup$ When I put negative number for x, it ise mirror of positive part, isn't it ? $\endgroup$ – CanYusuf Feb 27 '14 at 21:06

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