The orthogonality for Bessel functions is given by $\int_0 ^1 rJ_n(k_1r)J_n(k_2r) dr=0,\ (k_1 \neq k_2)\\ \neq 0, (k_1=k_2,\ J_n(k_1)=J_n(k_2)=0\ \mbox{or}\ J'_n(k_1)=J'_n(k_2)=0)$

This suggests a particular condition at the boundary $r=1$ for this orthogonality to hold. In case we have a different boundary condition for which $J_n(k_1,k_2) \neq 0\ \mbox{and}\ J_n'(k_1,k_2) \neq 0$, how do we establish an orthogonality condition for this case? How can we approach this problem?


$y=J_n(kr)$ is a solution of Bessel´s differential equation $r^2y'' +ry'+(r^2k^2-n^2)y=0$, which can be rewritten as $(ry')'+(rk^2-n^2/r)y = 0$.

If $u=J_n(ar)$ and $v=J_n(br)$, then they fulfill the equations

$$(ru')'+(ra^2-n^2/r)u = 0$$ $$(rv')'+(rb^2-n^2/r)v = 0$$

Multiply the first by $v$, the second by $u$ and substract them, and you get

$$(b^2-a^2)ruv =u(rv')'-v(ru')'=(vru'-urv')'$$

Integrating this, you get that

$$(b^2-a^2)\int_0^1ruvdr = \left.(vru'-urv')\right|_0^1=v(1)u'(1)-u(1)v'(1)$$

So if you want the left hand side to be $0$, then the right hand side must be $0$ as well, so you must have $aJ_n(b)J_n'(a)=bJ_n(a)J_n'(b)$.

This is fulfilled if $J_n(a)=J_n(b)=0$, or $J_n'(a)=J_n'(b)=0$, but also if $a J_n'(a)/J_n(a)=b J_n'(b)/J_n(b)$. So the boundary condition $y' = C y$ at $r=1$ will also work.


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.