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.



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.

  • $\begingroup$ Only a remark: a) Put $x=0$. You get $\displaystyle 1=\sum_{n\geq 1} a_n$. b) Derive your expression with respect to $x$(as we do not know the $a_n$, perhaps we cannot do that). We get $$0=\sum_{n\geq 1}a_n(\lambda_n \sinh(\lambda_n x)-\frac{h}{k}\cosh(\lambda_nx))$$ and put $x=0$ again; If $h\not =0$, you get $\sum_{n\geq 1} a_n=0$. Hence there is probably a problem in the question. $\endgroup$ – Kelenner Aug 13 '16 at 15:35

I'm not sure if this will help or if you've already thought of it: $$\tanh\left(\frac {\lambda_n L}2\right)=\frac{h}{\lambda_nk}$$

$$1=\sum_n\left[a_n\left(\cosh(\lambda_nx)-\tanh(\frac {\lambda_n L}2)\sinh(\lambda_nx)\right)\right]$$$$=\sum_n\left[\frac{a_n}{\cosh(\frac {\lambda_n L}2)}\left(\cosh(\lambda_nx)\cosh(\frac {\lambda_n L}2)-\sinh(\frac {\lambda_n L}2)\sinh(\lambda_nx)\right)\right]$$ $$=\sum_n a_n\frac{\cosh(\lambda_n(x-\frac L2))}{\cosh\left(\frac {\lambda_n L}2\right)}$$

I don't think there's a single coefficient $a_n$ that satisfies the identity. Maybe some more context could clarify the question.

$$\boxed{\textbf{EDIT}}$$ I felt that $h$ and $k$ were for convection and conduction! I believe that this is what you should do:

$$X(x)\dot \Gamma(t)=\kappa X''(x)\Gamma(t)- CX(x)\Gamma(t)$$ $$\frac {\dot \Gamma(t)}{\Gamma(t)}+C=\frac{\kappa X''(x)}{X(x)}=-A$$

$$X''(x)+\lambda X(x)=0$$ where $\lambda=A/\kappa$

If $\lambda<0$ then $$X(x)=c_1\cosh(\sqrt{-\lambda}x)+c_2\sinh(\sqrt{-\lambda}x)$$

Since there's thermal symmetry $X(0)=X(L)$ and $X'(0)=X'(L)$ $$c_1=c_1\cosh(\sqrt{-\lambda}L)+c_2\sinh(\sqrt{-\lambda}L)$$ $$c_1(1-\cosh(\sqrt{-\lambda}L))-c_2\sinh(\sqrt{-\lambda}L)=0$$ and $$c_2(1-\cosh(\sqrt{-\lambda}L))-c_1\sinh(\sqrt{-\lambda}L)=0$$

This becomes: $$\frac{c_2}{c_1}=\frac{c_1}{c_2}$$ $$c_1^2=c_2^2$$ $$c_1=c_2\neq0 \quad\text{or}\quad c_1=-c_2\neq 0 \quad\text{or}\quad c_1=c_2=0$$

Replacing the first two solutions in either one of the equations yields $$\frac{\sinh(\sqrt{-\lambda}L)}{1-\cosh(\sqrt{-\lambda}L)}=1 \quad\text{or}\quad \frac{\sinh(\sqrt{-\lambda}L)}{1-\cosh(\sqrt{-\lambda}L)}=-1$$

Both of which have no solution for $L$ in the real domain which means that $c_1=c_2=0$ then both hyperbolic terms vanish! For the other cases ($\lambda>0$ and $\lambda=0$), you solve using Fourier! $$\boxed{\textbf{Further EDIT}}$$

$$1=\sum_n a_n\cos(m_n x)+\sum_n b_n \sin(m_n x)$$ $$\cos(m_k x)=\sum_n a_n\cos(m_n x)\cos(m_k x)+\sum_n b_n \sin(m_n x)\cos(m_k x)$$ $$\int_T \cos(m_k x)dx=\int_T \sum_n a_n\cos(m_n x)\cos(m_k x)dx+\int_T \sum_n b_n \sin(m_n x)\cos(m_k x)dx$$ $$\int_T \cos(m_k x)dx= \sum_n a_n\int_T\cos(m_n x)\cos(m_k x)dx+ \sum_n b_n \int_T\sin(m_n x)\cos(m_k x)dx$$

The last term disappears. The cosine function is orthogonal, that is you can easily prove that $$\frac{1}{T}\int_T\cos(m_n x)\cos(m_k x)=0 \text{ if } n\neq k$$ $$\frac{1}{T}\int_T\cos(m_n x)\cos(m_k x)=1 \text{ if } n= k$$

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  • $\begingroup$ Thanks you, Georg. I've added some background. $\endgroup$ – Gert Aug 13 '16 at 20:59
  • $\begingroup$ @Gert can you also include the actual differential equation and physical context? If we were dealing with trigonometric functions, extracting $a_n$ would've been easy, but the hyperbolic functions complicate things. There also seems to be something wrong as Kelenner pointed out. $\endgroup$ – GeorgSaliba Aug 13 '16 at 22:29
  • $\begingroup$ Added. Thanks so much for your continued interest. I've a feeling my mistake is a 'silly' one! $\endgroup$ – Gert Aug 13 '16 at 23:31
  • $\begingroup$ @Gert I think I found your problem $\endgroup$ – GeorgSaliba Aug 14 '16 at 6:59
  • $\begingroup$ Yes.I realised my mistake too. But the coefficients $a_n$ for a trigonometric solution still need determining. See my further edit. Any ideas on that? Solution accepted, BTW. Many thanks for your help! $\endgroup$ – Gert Aug 14 '16 at 12:53

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