2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $$\frac{1}{T}\int_T\cos(m_n x)\cos(m_k x)=0 \text{ if } n\neq k$$ $$\frac{1}{T}\int_T\cos(m_n x)\cos(m_k x)=1 \text{ if } n= k$$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.