I have obtained the following eigenfunctions as solutions to a PDE problem:


I also obtained the eigenfunctions to the adjoint problem:


Here, the $\mu_n$ are eigenvalues which satisfy a transcendental equation (which cannot be solved exactly):

$$ \mu_n(\tan{\mu_n} - \tanh{\mu_n}) = 2d $$

where $d>0$.

I have successfully proved (after a lot of algebra), using the equation above, that

$$\int_{0}^{1} u_n(y)v_m(y){\rm d}y = 0 \text{ when }m\neq n.$$

I would like to get Mathematica to do this in a similar way, but I'm not sure how.

I have used the following code:

eigEqnN = \[Mu][n] (Tanh[\[Mu][n]] - Tan[\[Mu][n]]) + 2 d == 0;
eigEqnM = \[Mu][m] (Tanh[\[Mu][m]] - Tan[\[Mu][m]]) + 2 d == 0;
u[n_, y_] = 
  Cos[\[Mu][n]] Cosh[\[Mu][n] y] - Cosh[\[Mu][n]] Cos[\[Mu][n] y];
v[n_, y_] = 
  Cos[\[Mu][n]] Cosh[\[Mu][n] y] + Cosh[\[Mu][n]] Cos[\[Mu][n] y];
Integrate[u[n, y] v[m, y], {y, 0, 1}, 
 Assumptions -> {eigEqnN, eigEqnM}]

but Mathematica hasn't simplified the integral fully to get zero.

I am unsure how to implement this properly. Does anyone know how I can use the transcendental equation in Mathematica to simplify the result?

  • $\begingroup$ Please post code as text, not as pictures. Could you explain more clearly what you want to accomplish with MMA here? Do you want to reproduce the condition you found? That sounds difficult to do. $\endgroup$
    – MarcoB
    Jul 13, 2022 at 19:04
  • 3
    $\begingroup$ You can use Solve[] to get Root[] expressions for your eigenvalues, after a bit of reformulation, and restriction to a finite range: With[{d = 3/2}, Solve[μ (Sin[μ] Cosh[μ] - Sinh[μ] Cos[μ]) - 2 d Cosh[μ] Cos[μ] == 0 && 0 < μ < 30, μ, Reals]] $\endgroup$ Jul 13, 2022 at 19:59


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