I am trying to model transient cooling of two concentric cylinders sharing an interface along the length. Heat will flow in radial direction only.
Radius of inner cylinder (or inner radius of outer cylinder) $= a$
Outer Radius of outer cylinder $= b$
Interface is at $r =a$
I am trying to solve one dimensional heat equation:
Equation for outer cylinder 2 is:
$$ \dfrac {\partial T_2^2}{\partial r^2} +\dfrac{1}{r}\frac{\partial T_2}{\partial r}-\dfrac{1}{\alpha}\dfrac{\partial T_2}{\partial t} =0 .$$
Initial conditions: $ T_2(r,0) = A$.
Boundary Condition: $ T_2 = 0 $ at $r=b$.
Two other conditions on the interface: $T_2 = T_1$ at $r =a$, and $k_2\dfrac {\partial T_2}{\partial r}=k_1\dfrac {\partial T_1}{\partial r}$ at $r = a$
I have applied separation of variables to above problem and I have reached the following step:
$$ T_2 = \sum_{m=1}^ \infty c_m e^{-\alpha_2\lambda_m^2t}\left(\dfrac{-Y_0(\lambda_mb)J_0(\lambda_m r)}{J_0(\lambda_mb)}+Y_0(\lambda_mr)\right)$$
I need to find $c_m$ which is usually found using orthogonality condition. This is done by multiplying both sides with $rJ_0(\lambda_nr)$ and integrating both sides with respect to r. But here in above equation I can't apply the orthogonality condition due to presence of $Y_0(\lambda_mr)$ on right side of equation.
How do I apply orthogonality condition in this case?
I just want to find out the coefficient $c_m$ and somehow get rid of the summation sign.
Please help!
$\Large \textbf {UPDATE:}$
Note: I have replaced n with m and vice versa in the below discussion.
I managed to get Gradshteyn and Ryzhik book today. The book is really useful.
Now that we have got the formula.....
$\begin{split} \int x X_p(\alpha x) Z_p(\beta x) & = \frac{x}{\alpha^2-\beta^2} \left( \alpha X_{p+1}(\alpha x) Z_p(\beta x) -\beta X_{p}(\alpha x) Z_{p+1}(\beta x) \right) \\ & = \frac{x}{\alpha^2-\beta^2} \left( \beta X_{p}(\alpha x) Z_{p-1}(\beta x) -\alpha X_{p-1}(\alpha x) Z_{p}(\beta x) \right) .\end{split}$
...... I would like to evaluate the constant $c_n$ for the following equation:
$ T = \sum^{∞}_{n=1} c_n e^{-\alpha \lambda^{2}_{n}t}(\frac{-Y_0(\lambda_nb)J_0(\lambda_nr)}{J_0(\lambda_nb)} +Y_0(\lambda_nr))$
Applying initial condition leads to: $T(r,0) = X $
$ X = \sum^{∞}_{n=1} c_n (\frac{-Y_0(\lambda_nb)J_0(\lambda_nr)}{J_0(\lambda_nb)} +Y_0(\lambda_nr))$
Multiply both sides by $rJ_0(\lambda_mr)$ and integrate from r=a to r=b.
$ ∫^{b}_{a}rJ_0(\lambda_mr)Xdr = \sum^{∞}_{n=1} c_n (∫^{b}_{a}\frac{-Y_0(\lambda_nb)rJ_0(\lambda_mr)J_0(\lambda_nr)}{J_0(\lambda_nb)}dr +∫^{b}_{a}rJ_0(\lambda_mr)Y_0(\lambda_nr)dr)$
The first term on the right side of equation will be taken care of by using the orthogonality condition:
$\int_a^b xJ_0(\lambda_nx)J_0(\lambda_mx)dx = 0$ for $n\not=m$ So for first term on right side all integrals of the series would vanish except for the case n =m.
The second term on right side will be evaluated using the formula that you mentioned in your post.
Since for our case, the below equation is not equal to zero:
$\begin{split} \int x X_p(\alpha x) Z_p(\beta x) & = \frac{x}{\alpha^2-\beta^2} \left( \alpha X_{p+1}(\alpha x) Z_p(\beta x) -\beta X_{p}(\alpha x) Z_{p+1}(\beta x) \right) \\ & = \frac{x}{\alpha^2-\beta^2} \left( \beta X_{p}(\alpha x) Z_{p-1}(\beta x) -\alpha X_{p-1}(\alpha x) Z_{p}(\beta x) \right) .\end{split}\not=0$
.... the summation sign would stay intact and $c_n$ cannot be evaluated.
Would you please help me get rid of the summation sign. I want to evaluate $c_n$.