Mixing
On the solution of Volterra integral equation
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
I got stuck with some strange point, solving Volterra integral equation:
$$
int_0^t (t-s)f(s) ds =sqrtt.
$$
The solution can be obtained by ssuccessive differnetiation
$$
int_0^t f(s)ds=frac12sqrtt, quad mboxand then
$$
$$
f(t)=-frac14tsqrtt
$$
But, when I substitute this solution to the original equation
$$
int_0^t(t-s)left[-frac14ssqrtsright]ds=left[fracs+t2sqrtsright]_0^t=sqrtt-infty
$$
I can't figure out where I was wrong. Please explain.
calculus real-analysis integration integral-equations
edited Mar 9 '16 at 14:41
Harry Peter
5,47311438
asked Mar 8 '16 at 5:10
user155214
15512
add a comment |Â
up vote
2
down vote
favorite
I got stuck with some strange point, solving Volterra integral equation:
$$
int_0^t (t-s)f(s) ds =sqrtt.
$$
The solution can be obtained by ssuccessive differnetiation
$$
int_0^t f(s)ds=frac12sqrtt, quad mboxand then
$$
$$
f(t)=-frac14tsqrtt
$$
But, when I substitute this solution to the original equation
$$
int_0^t(t-s)left[-frac14ssqrtsright]ds=left[fracs+t2sqrtsright]_0^t=sqrtt-infty
$$
I can't figure out where I was wrong. Please explain.
calculus real-analysis integration integral-equations
edited Mar 9 '16 at 14:41
Harry Peter
5,47311438
asked Mar 8 '16 at 5:10
user155214
15512
@Chip I already considered your concern. The differntiation with the upper limit part is cancled out due to (t-s) term.
– user155214
Mar 8 '16 at 6:01
@Chip I can't get your point. You mean $$fracddt(frac12sqrtt)neq -frac14tsqrtt$$?
– user155214
Mar 8 '16 at 6:59
sorry, you are correct. I misread because of bad fonts / LaTeX rendering.
– Chip
Mar 8 '16 at 8:24
Where is your Volterra operator defined? On which algebra of functions are you seeking a solution to this integral equation?
– TheOscillator
Apr 27 '16 at 11:19
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I got stuck with some strange point, solving Volterra integral equation:
$$
int_0^t (t-s)f(s) ds =sqrtt.
$$
The solution can be obtained by ssuccessive differnetiation
$$
int_0^t f(s)ds=frac12sqrtt, quad mboxand then
$$
$$
f(t)=-frac14tsqrtt
$$
But, when I substitute this solution to the original equation
$$
int_0^t(t-s)left[-frac14ssqrtsright]ds=left[fracs+t2sqrtsright]_0^t=sqrtt-infty
$$
I can't figure out where I was wrong. Please explain.
calculus real-analysis integration integral-equations
edited Mar 9 '16 at 14:41
Harry Peter
5,47311438
asked Mar 8 '16 at 5:10
user155214
15512
I got stuck with some strange point, solving Volterra integral equation:
$$
int_0^t (t-s)f(s) ds =sqrtt.
$$
The solution can be obtained by ssuccessive differnetiation
$$
int_0^t f(s)ds=frac12sqrtt, quad mboxand then
$$
$$
f(t)=-frac14tsqrtt
$$
But, when I substitute this solution to the original equation
$$
int_0^t(t-s)left[-frac14ssqrtsright]ds=left[fracs+t2sqrtsright]_0^t=sqrtt-infty
$$
I can't figure out where I was wrong. Please explain.
calculus real-analysis integration integral-equations
edited Mar 9 '16 at 14:41
Harry Peter
5,47311438
asked Mar 8 '16 at 5:10
user155214
15512
edited Mar 9 '16 at 14:41
Harry Peter
5,47311438
edited Mar 9 '16 at 14:41
Harry Peter
5,47311438
edited Mar 9 '16 at 14:41
Harry Peter
5,47311438
5,47311438
asked Mar 8 '16 at 5:10
user155214
15512
asked Mar 8 '16 at 5:10
user155214
15512
asked Mar 8 '16 at 5:10
user155214
15512
15512
@Chip I already considered your concern. The differntiation with the upper limit part is cancled out due to (t-s) term.
– user155214
Mar 8 '16 at 6:01
@Chip I can't get your point. You mean $$fracddt(frac12sqrtt)neq -frac14tsqrtt$$?
– user155214
Mar 8 '16 at 6:59
sorry, you are correct. I misread because of bad fonts / LaTeX rendering.
– Chip
Mar 8 '16 at 8:24
Where is your Volterra operator defined? On which algebra of functions are you seeking a solution to this integral equation?
– TheOscillator
Apr 27 '16 at 11:19
add a comment |Â
@Chip I already considered your concern. The differntiation with the upper limit part is cancled out due to (t-s) term.
– user155214
Mar 8 '16 at 6:01
@Chip I can't get your point. You mean $$fracddt(frac12sqrtt)neq -frac14tsqrtt$$?
– user155214
Mar 8 '16 at 6:59
sorry, you are correct. I misread because of bad fonts / LaTeX rendering.
– Chip
Mar 8 '16 at 8:24
Where is your Volterra operator defined? On which algebra of functions are you seeking a solution to this integral equation?
– TheOscillator
Apr 27 '16 at 11:19
@Chip I already considered your concern. The differntiation with the upper limit part is cancled out due to (t-s) term.
– user155214
Mar 8 '16 at 6:01
@Chip I already considered your concern. The differntiation with the upper limit part is cancled out due to (t-s) term.
– user155214
Mar 8 '16 at 6:01
@Chip I can't get your point. You mean $$fracddt(frac12sqrtt)neq -frac14tsqrtt$$?
– user155214
Mar 8 '16 at 6:59
@Chip I can't get your point. You mean $$fracddt(frac12sqrtt)neq -frac14tsqrtt$$?
– user155214
Mar 8 '16 at 6:59
sorry, you are correct. I misread because of bad fonts / LaTeX rendering.
– Chip
Mar 8 '16 at 8:24
sorry, you are correct. I misread because of bad fonts / LaTeX rendering.
– Chip
Mar 8 '16 at 8:24
Where is your Volterra operator defined? On which algebra of functions are you seeking a solution to this integral equation?
– TheOscillator
Apr 27 '16 at 11:19
Where is your Volterra operator defined? On which algebra of functions are you seeking a solution to this integral equation?
– TheOscillator
Apr 27 '16 at 11:19
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
Notice that integration by part gives $$int_0^t(t-s),f(s) , ds = left[(t-s)int_0^sf(u)du right]_s=0^s=t +int_0^tint_0^sf(u),du,ds = V^2(f)(t)$$
where $V^2(f)$ is understood as the composition of Volterra operatorer $$V(f)(t):= int_0^tf(u) , du$$ with itself.
Now for the Volterra operator to make sense, we must at least have that $f$ is locally integrable in a neighbourhood of $0$, say $fin L^1(0,1))$. But then $V(f)$ lands into the space of continuous functions on $[0,1]$ (This is a straightforward appliction of Dominated convergence theorem).
Consequently composing with $V$ again we obtain $V^2(f)in C^1([0,1])$.
Going back to your initial equation, we wished to solve $$V^2(f)(t) = sqrtt$$ The previous argument shows that we cannot even find an $L^1(0,1)$-solution $f$ to this equation, since $trightarrow sqrtt$ fails to be continuously differentiable on $[0,1]$. What I meant to explain was that if you cannot even find a solution with minimal to no restrictions, then you are pretty much out of luck.
answered Apr 27 '16 at 11:49
TheOscillator
2,0441716
add a comment |Â
up vote
0
down vote
I am not an expert on integral equations or the like, but the origin of the trouble is clear -- the differentiation process loses sense of a constant. The infinity comes because your function is not integrable near zero. Being only mildly acquainted with integral equations, I would try two other approaches -- Neumann series and Laplace transform -- and see if they give something reasonable and get some intuition from the result.
answered Apr 27 '16 at 8:51
kstar
465
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Notice that integration by part gives $$int_0^t(t-s),f(s) , ds = left[(t-s)int_0^sf(u)du right]_s=0^s=t +int_0^tint_0^sf(u),du,ds = V^2(f)(t)$$
where $V^2(f)$ is understood as the composition of Volterra operatorer $$V(f)(t):= int_0^tf(u) , du$$ with itself.
Now for the Volterra operator to make sense, we must at least have that $f$ is locally integrable in a neighbourhood of $0$, say $fin L^1(0,1))$. But then $V(f)$ lands into the space of continuous functions on $[0,1]$ (This is a straightforward appliction of Dominated convergence theorem).
Consequently composing with $V$ again we obtain $V^2(f)in C^1([0,1])$.
Going back to your initial equation, we wished to solve $$V^2(f)(t) = sqrtt$$ The previous argument shows that we cannot even find an $L^1(0,1)$-solution $f$ to this equation, since $trightarrow sqrtt$ fails to be continuously differentiable on $[0,1]$. What I meant to explain was that if you cannot even find a solution with minimal to no restrictions, then you are pretty much out of luck.
answered Apr 27 '16 at 11:49
TheOscillator
2,0441716
add a comment |Â
up vote
1
down vote
Notice that integration by part gives $$int_0^t(t-s),f(s) , ds = left[(t-s)int_0^sf(u)du right]_s=0^s=t +int_0^tint_0^sf(u),du,ds = V^2(f)(t)$$
where $V^2(f)$ is understood as the composition of Volterra operatorer $$V(f)(t):= int_0^tf(u) , du$$ with itself.
Now for the Volterra operator to make sense, we must at least have that $f$ is locally integrable in a neighbourhood of $0$, say $fin L^1(0,1))$. But then $V(f)$ lands into the space of continuous functions on $[0,1]$ (This is a straightforward appliction of Dominated convergence theorem).
Consequently composing with $V$ again we obtain $V^2(f)in C^1([0,1])$.
Going back to your initial equation, we wished to solve $$V^2(f)(t) = sqrtt$$ The previous argument shows that we cannot even find an $L^1(0,1)$-solution $f$ to this equation, since $trightarrow sqrtt$ fails to be continuously differentiable on $[0,1]$. What I meant to explain was that if you cannot even find a solution with minimal to no restrictions, then you are pretty much out of luck.
answered Apr 27 '16 at 11:49
TheOscillator
2,0441716
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Notice that integration by part gives $$int_0^t(t-s),f(s) , ds = left[(t-s)int_0^sf(u)du right]_s=0^s=t +int_0^tint_0^sf(u),du,ds = V^2(f)(t)$$
where $V^2(f)$ is understood as the composition of Volterra operatorer $$V(f)(t):= int_0^tf(u) , du$$ with itself.
Now for the Volterra operator to make sense, we must at least have that $f$ is locally integrable in a neighbourhood of $0$, say $fin L^1(0,1))$. But then $V(f)$ lands into the space of continuous functions on $[0,1]$ (This is a straightforward appliction of Dominated convergence theorem).
Consequently composing with $V$ again we obtain $V^2(f)in C^1([0,1])$.
Going back to your initial equation, we wished to solve $$V^2(f)(t) = sqrtt$$ The previous argument shows that we cannot even find an $L^1(0,1)$-solution $f$ to this equation, since $trightarrow sqrtt$ fails to be continuously differentiable on $[0,1]$. What I meant to explain was that if you cannot even find a solution with minimal to no restrictions, then you are pretty much out of luck.
answered Apr 27 '16 at 11:49
TheOscillator
2,0441716
Notice that integration by part gives $$int_0^t(t-s),f(s) , ds = left[(t-s)int_0^sf(u)du right]_s=0^s=t +int_0^tint_0^sf(u),du,ds = V^2(f)(t)$$
where $V^2(f)$ is understood as the composition of Volterra operatorer $$V(f)(t):= int_0^tf(u) , du$$ with itself.
Now for the Volterra operator to make sense, we must at least have that $f$ is locally integrable in a neighbourhood of $0$, say $fin L^1(0,1))$. But then $V(f)$ lands into the space of continuous functions on $[0,1]$ (This is a straightforward appliction of Dominated convergence theorem).
Consequently composing with $V$ again we obtain $V^2(f)in C^1([0,1])$.
Going back to your initial equation, we wished to solve $$V^2(f)(t) = sqrtt$$ The previous argument shows that we cannot even find an $L^1(0,1)$-solution $f$ to this equation, since $trightarrow sqrtt$ fails to be continuously differentiable on $[0,1]$. What I meant to explain was that if you cannot even find a solution with minimal to no restrictions, then you are pretty much out of luck.
answered Apr 27 '16 at 11:49
TheOscillator
2,0441716
edited Aug 6 at 20:11
answered Apr 27 '16 at 11:49
TheOscillator
2,0441716
answered Apr 27 '16 at 11:49
TheOscillator
2,0441716
answered Apr 27 '16 at 11:49
TheOscillator
2,0441716
2,0441716
add a comment |Â
add a comment |Â
up vote
0
down vote
I am not an expert on integral equations or the like, but the origin of the trouble is clear -- the differentiation process loses sense of a constant. The infinity comes because your function is not integrable near zero. Being only mildly acquainted with integral equations, I would try two other approaches -- Neumann series and Laplace transform -- and see if they give something reasonable and get some intuition from the result.
answered Apr 27 '16 at 8:51
kstar
465
add a comment |Â
up vote
0
down vote
I am not an expert on integral equations or the like, but the origin of the trouble is clear -- the differentiation process loses sense of a constant. The infinity comes because your function is not integrable near zero. Being only mildly acquainted with integral equations, I would try two other approaches -- Neumann series and Laplace transform -- and see if they give something reasonable and get some intuition from the result.
answered Apr 27 '16 at 8:51
kstar
465
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I am not an expert on integral equations or the like, but the origin of the trouble is clear -- the differentiation process loses sense of a constant. The infinity comes because your function is not integrable near zero. Being only mildly acquainted with integral equations, I would try two other approaches -- Neumann series and Laplace transform -- and see if they give something reasonable and get some intuition from the result.
answered Apr 27 '16 at 8:51
kstar
465
I am not an expert on integral equations or the like, but the origin of the trouble is clear -- the differentiation process loses sense of a constant. The infinity comes because your function is not integrable near zero. Being only mildly acquainted with integral equations, I would try two other approaches -- Neumann series and Laplace transform -- and see if they give something reasonable and get some intuition from the result.
answered Apr 27 '16 at 8:51
kstar
465
answered Apr 27 '16 at 8:51
kstar
465
answered Apr 27 '16 at 8:51
kstar
465
answered Apr 27 '16 at 8:51
kstar
465
465
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1688018%2fon-the-solution-of-volterra-integral-equation%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
@Chip I already considered your concern. The differntiation with the upper limit part is cancled out due to (t-s) term.
– user155214
Mar 8 '16 at 6:01
@Chip I can't get your point. You mean $$fracddt(frac12sqrtt)neq -frac14tsqrtt$$?
– user155214
Mar 8 '16 at 6:59
sorry, you are correct. I misread because of bad fonts / LaTeX rendering.
– Chip
Mar 8 '16 at 8:24
Where is your Volterra operator defined? On which algebra of functions are you seeking a solution to this integral equation?
– TheOscillator
Apr 27 '16 at 11:19