Is it possible to determine the infinite sum coefficients $a_n$ for the following identity:
$$1=\displaystyle \sum_{n=1}^{+\infty}\Big[a_n\Big(\cosh(\lambda_nx)-\frac{h}{k\lambda_n}\sinh(\lambda_nx)\Big)\Big]$$ Where $\lambda_n$ are the roots of the following transcendental equation: $$\lambda_n=\frac{h}{k}\coth \Big(\frac{\lambda_nL}{2}\Big)$$
In addition:
$0 \leq x \leq L$.
$h$ and $k$ are Real and positive constants.
Background:
$$X_n=a_n\cosh(\lambda_nx)+b_n\sinh(\lambda_nx)$$
This is the solution to a DE with boundary conditions: $$-kX'(0)=hX(0)$$ And: $$X_n'(L/2)=0$$ So: $$X_n'=\lambda_n a_n\sinh(\lambda_nx)+\lambda_n b_n\cosh(\lambda_nx)$$ $$-k\lambda b_n=ha_n$$ $$b_n=-\frac{ha_n}{k\lambda_n}$$ $$X_n=a_n\Big(\cosh(\lambda_nx)-\frac{h}{k\lambda_n}\sinh(\lambda_nx)\Big)$$ And: $$X_n'(L/2)=a_n\Big(\lambda_n\sinh(\lambda_nL/2)-\frac{h}{k}\cosh(\lambda_nL/2)\Big)=0$$ $$\lambda_n=\frac{h}{k}\coth(\lambda_nL/2)$$ With superposition: $$X(x)=\displaystyle \sum_{n=1}^{+\infty}\Big[a_n\Big(\cosh(\lambda_nx)-\frac{h}{k\lambda_n}\sinh(\lambda_nx)\Big)\Big]$$ In addition, from an initial condition: $$1=\displaystyle \sum_{n=1}^{+\infty}\Big[a_n\Big(\cosh(\lambda_nx)-\frac{h}{k\lambda_n}\sinh(\lambda_nx)\Big)\Big]$$
More background: I'm trying to solve the PDE (convective cooling of a straight, uniform rod, quasi-1D, transient regime):
$$u_t=\kappa u_{xx}-\frac{ph}{A\rho c}u$$
Where $u(x,t)=T(x,t)-T_{\infty}$. ($T_{\infty}$ is a constant)
Domain as above.
Initial condition: $u(x,0)=T_0-T_{\infty}$
Boundaries: $-ku_x(0)=hu(0)$ and $-ku_x(L)=hu(L)$
Symmetry: $u_x(L/2)=0$
Note: $\kappa=\frac{k}{\rho c}$ (where $\rho$ and $c$ are Real, positive constants).
Ansatz: $u(x,t)=X(x)\Gamma(t)$
Separation of variables is easy (hint: keep the source term with the $\Gamma(t)$ side of things!)
I find an easy solution with slightly different boundary conditions: $u_x(0,t)=u_x(L,t)=0$ because $X(x)$ is then purely trigonometric with simple eigenvalues.
However, using the boundary conditions I'm really interested in, I find hyperbolic $X(x)$ is needed. I could be wrong on this.
Further edit:
Like Georg, yesterday I also established that the hyperbolic solution for $X(x)$ is not a solution because $c_1=c_2=0$.
Now I've tried the separation constant $-m^2$ (having moved $\kappa$ to the $\Gamma$ side first):
$$X''+m^2X=0$$ $$X=a\cos mx+b\sin mx$$ $$X'=-ma\sin mx+mb\cos mx$$ Boundary condition: $$-kX'(0)=hX(0)$$ $$-kmb=ha\implies b=-\frac{ha}{km}$$ $$X=a\cos mx-\frac{ha}{km}\sin mx$$ $$X=a(\cos mx-\frac{h}{km}\sin mx)$$ $$X'=a(-m\sin mx-\frac{h}{k}\cos mx)$$ Symmetry condition: $$X'(L/2)=a(-m\sin mL/2-\frac{h}{k}\cos mL/2)=0$$ $$-m\sin mL/2=\frac{h}{k}\cos mL/2$$ So the eigenvalues are the roots of the following trigonometric equation: $$m_n=-\frac{h}{k}\cot\Big(\frac{m_nL}{2}\big)$$ Superposition: $$\Large{X(x)=\displaystyle \sum_{n=1}^{+\infty}a_n\big[\cos(m_nx)-\frac{h}{km_n}\sin(m_nx)\big]}$$
With the initial condition we get: $$1=\displaystyle \sum_{n=1}^{+\infty}a_n\big[\cos(m_nx)-\frac{h}{km_n}\sin(m_nx)\big]$$
So the coefficients $a_n$ still need determining.