Solving a dual integral equation involving a zeroth-order Bessel function
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Consider the following dual integral equations
begin{align}
int_0^infty q^3 f_0(q) J_0 (qr) , mathrm{d} q &= g(r) qquadqquadquad (0<r<1) , \
int_0^infty f_0(q) J_0 (qr) , mathrm{d} q &= 0 ,quadqquadqquadquad (r>1) , ,
end{align}
where
$$
g(r) = frac{9}{4pi} frac{16h^6-72h^4 r^2 + 18h^2 r^4 +r^6}{(h^2+r^2)^{11/2}} , .
$$
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!
real-analysis integration complex-analysis differential-equations improper-integrals
asked Nov 23 at 18:39
Math Student
58520
add a comment |
up vote
7
down vote
favorite
Consider the following dual integral equations
begin{align}
int_0^infty q^3 f_0(q) J_0 (qr) , mathrm{d} q &= g(r) qquadqquadquad (0<r<1) , \
int_0^infty f_0(q) J_0 (qr) , mathrm{d} q &= 0 ,quadqquadqquadquad (r>1) , ,
end{align}
where
$$
g(r) = frac{9}{4pi} frac{16h^6-72h^4 r^2 + 18h^2 r^4 +r^6}{(h^2+r^2)^{11/2}} , .
$$
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!
real-analysis integration complex-analysis differential-equations improper-integrals
asked Nov 23 at 18:39
Math Student
58520
1
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).
– Yuriy S
Nov 29 at 13:59
@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
– Math Student
Nov 29 at 16:43
add a comment |
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Consider the following dual integral equations
begin{align}
int_0^infty q^3 f_0(q) J_0 (qr) , mathrm{d} q &= g(r) qquadqquadquad (0<r<1) , \
int_0^infty f_0(q) J_0 (qr) , mathrm{d} q &= 0 ,quadqquadqquadquad (r>1) , ,
end{align}
where
$$
g(r) = frac{9}{4pi} frac{16h^6-72h^4 r^2 + 18h^2 r^4 +r^6}{(h^2+r^2)^{11/2}} , .
$$
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!
real-analysis integration complex-analysis differential-equations improper-integrals
asked Nov 23 at 18:39
Math Student
58520
Consider the following dual integral equations
begin{align}
int_0^infty q^3 f_0(q) J_0 (qr) , mathrm{d} q &= g(r) qquadqquadquad (0<r<1) , \
int_0^infty f_0(q) J_0 (qr) , mathrm{d} q &= 0 ,quadqquadqquadquad (r>1) , ,
end{align}
where
$$
g(r) = frac{9}{4pi} frac{16h^6-72h^4 r^2 + 18h^2 r^4 +r^6}{(h^2+r^2)^{11/2}} , .
$$
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!
real-analysis integration complex-analysis differential-equations improper-integrals
real-analysis integration complex-analysis differential-equations improper-integrals
asked Nov 23 at 18:39
Math Student
58520
asked Nov 23 at 18:39
Math Student
58520
edited Nov 29 at 8:18
asked Nov 23 at 18:39
Math Student
58520
asked Nov 23 at 18:39
Math Student
58520
asked Nov 23 at 18:39
Math Student
58520
58520
1
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).
– Yuriy S
Nov 29 at 13:59
@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
– Math Student
Nov 29 at 16:43
add a comment |
1
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).
– Yuriy S
Nov 29 at 13:59
@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
– Math Student
Nov 29 at 16:43
1
1
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).
– Yuriy S
Nov 29 at 13:59
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).
– Yuriy S
Nov 29 at 13:59
@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
– Math Student
Nov 29 at 16:43
@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
– Math Student
Nov 29 at 16:43
add a comment |
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1
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).
– Yuriy S
Nov 29 at 13:59
@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
– Math Student
Nov 29 at 16:43