Orthogonality condition for disimilar Bessel functions

[itisali]

As per orthogonality condition this equation is valid:

\int_0^b xJ_0(\lambda_nx)J_0(\lambda_mx)dx = 0 for m\not=n

I want to know the outcome of the following:

\int_0^b xJ_0(\lambda_nx)Y_0(\lambda_mx)dx = 0

for two cases:
m\not=n
m=n

 

[fzero]

I didn't have access to Gradshteyn and Ryzhik earlier. Integral 5.54 claims that, for $X,Z$ any of the Bessel functions ($J,Y,\ldots$),

$$\begin{split} \int x X_p(\alpha x) Z_p(\beta x) & = \frac{x}{\alpha^2-\beta^2} \left( \alpha X_{p+1}(\alpha x) Z_p(\beta x) -\beta X_{p}(\alpha x) Z_{p+1}(\beta x) \right) \\
& = \frac{x}{\alpha^2-\beta^2} \left( \beta X_{p}(\alpha x) Z_{p-1}(\beta x) -\alpha X_{p-1}(\alpha x) Z_{p}(\beta x) \right) .\end{split} $$

You should be able to work out your integral by setting the appropriate integration limits and using other identities as necessary.

 

[itisali]

Thanks alot!

I can now solve the problem. But there is still one more thing left. Is it safe to assume that above integral will be equal to zero in case


\int x X_p(\alpha x) Z_q(\beta x) = ? <br /> <br /> \\<br /> <br /> when p\not = q

 

[fzero]

Thanks alot!

I can now solve the problem. But there is still one more thing left. Is it safe to assume that above integral will be equal to zero in case


\int x X_p(\alpha x) Z_q(\beta x) = ? <br /> <br /> \\<br /> <br /> when p\not = q

I am unable to find precisely that integral. The integral I gave above can be derived from the expressions for the derivative of the Bessel functions (which involves specific factors of $x$), using integration by parts. The intermediate result that applies to $p\neq q$ is

So the part involving the integral you want involves $x^{-1}$ instead of $x$.

However, the universality of the formula seems to agree with my suggestion that the orthogonality relations of $J$ with $Y$ are precisely analogous to $J$ with $J$. I urge you to try to get your hands on a copy of Gradshteyn and Rhyzik, at least from a library, since you might note a useful result that I have overlooked.

 

[itisali]

I managed to get Gradshteyn and Ryzhik 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) &amp; = \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) \\ <br /> &amp; = \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) &amp; = \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) \\ <br /> &amp; = \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.