D[Tu1[x], {x, 2}] qu - Tu1[x] (fu0 + I w p) == Tou[x] f1

    D[T1[x], {x, 2}] q - T1[x] (f0 - b g0 + I w p)== 
 To[x] (f1 - b g1) - g1

For T1 , Range[0, L/2]

For Tu1, Range[L/2, La/2]

Here, there are two differential equations with coupled boundary conditions;

Tu1[La/2] == 0
Tu1[L/2] == T1[L/2]
a4 Tu1'[L/2] == T1'[L/2]

And also,

Tou[x_] := E^(a3 x) c2 + E^(-a3 x) c3
To[x_] := 2 c1 Cosh[x a1] + a2

I know all constants in the equations.

I tried following code, but I coud not get answer, " error: not the same shape "

{T1, Tu1} = {T1, Tu1} /. 
  DSolve[{D[T1[x], {x, 2}] q - T1[x] (f0 - b g0 + I w p) - 
      To[x] (f1 - b g1) + g1 == 0, 
    D[Tu1[x], {x, 2}] qu - Tu1[x] (fu0 + I w p) - Tu[x] f1 == 0, 
    T1'[0] == 0, Tu1[L/2] == T1[L/2], a4 Tu1'[L/2] == T1'[L/2], 
    Tu1[La/2] == 0}, {T1, Tu1}, x]

How can I find solution of differential equations?

  • 1
    $\begingroup$ have you tried ?DSolve $\endgroup$
    – chris
    Apr 8, 2014 at 8:00
  • $\begingroup$ I have tried solving with Green's function technique, but failed $\endgroup$
    – CanYusuf
    Apr 8, 2014 at 8:01
  • 1
    $\begingroup$ what is Tou in the first equation? (which should have == instead of =) $\endgroup$
    – chris
    Apr 8, 2014 at 8:07
  • $\begingroup$ Sorry mate, I fixed now. $\endgroup$
    – CanYusuf
    Apr 8, 2014 at 8:12
  • $\begingroup$ no you haven't :-) $\endgroup$
    – chris
    Apr 8, 2014 at 8:17

1 Answer 1


This is not an answer to your question but a copy and paste from the documentation which you might find inspiring:

Boundary Value Problems with Parameters

In many of the applications where boundary value problems arise, there may be undetermined parameters, such as eigenvalues, in the problem itself that may be a part of the desired solution. By introducing the parameters as dependent variables, the problem can often be written as a boundary value problem in standard form.

For example, the flow in a channel can be modeled by

$$f'''-R\,((f')^2-f f'')+R\, a==0$$

where $R$ (the Reynolds number) is given, but a is to be determined.

To find the solution $f$ and the value of a, just add the equation a'==0.

This solves the flow problem with R==1 for f and a, plots the solution f and returns the value of a.

 Block[{R = 1}, 
   sol = NDSolve[{f'''[t] - R ((f'[t])^2 - f[t] f''[t]) + R a[t] == 0, 
                  a'[t] == 0, f[0] == f'[0] == f'[1] == 0, f[1] == 1}, {f, a}, t];
   Column[{Plot[f[t] /. First[sol], {t, 0, 1}],
           a[0] /. First[sol]}]]

enter image description here


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