7
$\begingroup$

Consider the following dual integral equations

\begin{align} \int_0^\infty q^3 f_0(q) J_0 (qr) \, \mathrm{d} q &= g(r) \qquad\qquad\quad (0<r<1) , \\ \int_0^\infty f_0(q) J_0 (qr) \, \mathrm{d} q &= 0 \,\quad\qquad\qquad\quad (r>1) \, , \end{align} where $$ g(r) = \frac{9}{4\pi} \frac{16h^6-72h^4 r^2 + 18h^2 r^4 +r^6}{(h^2+r^2)^{11/2}} \, . $$

We search a solution of the integral form \begin{equation} f_0 (q) = \int_0^1 \lambda(t) \sin (qt) \, \mathrm{d} t \, , \label{integralWithHeaviside_sin} \end{equation} which clearly satisfies the integral equation for $r>1$ by making use of the relation \begin{equation} \int_0^\infty J_0 (qr) \sin(qt) \, \mathrm{d} q = \frac{H(t-r)}{(t^2-r^2)^{1/2}} \, , \end{equation} where $H(\cdot)$ denotes Heaviside function.

For $0<r<1$, it follows from three successive integration by parts that \begin{equation}\label{longEqBending} \begin{split} \int_0^\infty & J_0(qr) \, \mathrm{d} q \int_0^1 q^3 \lambda(t)\sin(qt) \, \mathrm{d} t \\ &= \int_0^\infty J_0(qr) \, \mathrm{d} q \bigg( \left(\lambda''(1)-q^2 \lambda(1)\right) \cos(q) + q \lambda'(1) \sin(q) -\lambda''(0) + q^2 \lambda(0) \\ &\qquad- \int_0^1 \lambda'''(t) \cos(qt) \, \mathrm{d} t \bigg) \, . \end{split} \end{equation}

For the integral on the right-hand side of the latter equation to be convergent, we require that $\lambda(0)=\lambda(1) =\lambda'(1)=0$. Thus, the latter equation becomes \begin{equation}\label{secondTermBending} \int_0^\infty J_0(qr) \, \mathrm{d} q \int_0^1 q^3 \lambda(t)\sin(qt) = - \frac{\lambda''(0)}{r} - \int_0^r \frac{\lambda'''(t) \, \mathrm{d} t}{(r^2-t^2)^{1/2}} \, , \end{equation} after using identity \begin{equation} \int_0^\infty J_0 (qr) \cos(qt) \, \mathrm{d} q = \frac{H(r-t)}{(r^2-t^2)^{1/2}} \, . \label{integralWithHeaviside_cos} \end{equation}

Thus, the integral equation can be simplified as \begin{equation} \frac{\lambda''(0)}{r} + \int_0^r \frac{\lambda'''(t) \, \mathrm{d} t}{(r^2-t^2)^{1/2}} = -g(r) \, , \end{equation}

By multiplying both members of the latter equation by $r/(s^2-r^2)^{1/2}$ and integrating with respect to $r$ from 0 to $s$, the resulting equation reads \begin{equation} \lambda''(s) = - \frac{24 h^3}{\pi^2 } \frac{s(3h^2 - 5s^2)}{(s^2+h^2)^5} \, . \end{equation}

The latter equation can now be easily solved. But the problem is that we have required that $\lambda(0)=\lambda(1)=0$ but also $\lambda'(1)=0$. As the final equation is a second order ODE, only 2 boundary conditions are in principal required. I would be grateful if someone here could be of help and clarify how this could be explained.

Thank you!

$\endgroup$
  • 1
    $\begingroup$ The first thing I would do is to find the function $\lambda$ and then check if the conditions are satisfied (or better yet, that it makes your intermediate expression convergent, which is the whole reason you introduced them). $\endgroup$ – Yuriy S Nov 29 '18 at 13:59
  • $\begingroup$ @YuriyS Apparently the expression of λ as calculated from the last math step with λ(0)=λ(1)=0 does not satisfy the condition λ′(1)=0. Accordingly, one intermediate step would in principle NOT lead to a convergence of the overall integral. Maybe the first derivative should just be by construction discontinuous? Any inputs are highly desirable $\endgroup$ – Daddy Nov 29 '18 at 16:43

Your Answer

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

Browse other questions tagged or ask your own question.