Solve for f in the integral equation $f(t) = sin(t) + int f(s)ds $
$begingroup$
Solve for f in the integral equation $$f(t) = sin t + int_{0}^{t} f(s)ds$$ using $ (V^nx)(t) =int_{0}^{t} frac{(t-s)^{n-1}}{(n-1)!}x(s)ds $ to to where V is the Volterra operator $V$ on $L^2(0,1)$ is an operator of integration such as the one shown below
$$(Vx)(t) = int_{0}^{t} x(s)ds, 0leq tleq 1$$
I tried integrating both sides from $0$ to $t$: $$int_{0}^{t}f(s)ds = int_{0}^{t} sin(s) ds + int_{0}^{t} int_{0}^{s} f(s) ds = -cos(t) + cos(0) + (V^2f)(t) = 1 - cos(0) + int_{0}^{t} (t-s)f(s)ds $$
I'm not sure where to go from here.
real-analysis integral-equations
edited Nov 29 '18 at 17:22
Joshua Mundinger
2,5141026
asked Nov 29 '18 at 1:32
SarahSarah
162
$endgroup$
add a comment |
$begingroup$
Solve for f in the integral equation $$f(t) = sin t + int_{0}^{t} f(s)ds$$ using $ (V^nx)(t) =int_{0}^{t} frac{(t-s)^{n-1}}{(n-1)!}x(s)ds $ to to where V is the Volterra operator $V$ on $L^2(0,1)$ is an operator of integration such as the one shown below
$$(Vx)(t) = int_{0}^{t} x(s)ds, 0leq tleq 1$$
I tried integrating both sides from $0$ to $t$: $$int_{0}^{t}f(s)ds = int_{0}^{t} sin(s) ds + int_{0}^{t} int_{0}^{s} f(s) ds = -cos(t) + cos(0) + (V^2f)(t) = 1 - cos(0) + int_{0}^{t} (t-s)f(s)ds $$
I'm not sure where to go from here.
real-analysis integral-equations
edited Nov 29 '18 at 17:22
Joshua Mundinger
2,5141026
asked Nov 29 '18 at 1:32
SarahSarah
162
$endgroup$
add a comment |
$begingroup$
Solve for f in the integral equation $$f(t) = sin t + int_{0}^{t} f(s)ds$$ using $ (V^nx)(t) =int_{0}^{t} frac{(t-s)^{n-1}}{(n-1)!}x(s)ds $ to to where V is the Volterra operator $V$ on $L^2(0,1)$ is an operator of integration such as the one shown below
$$(Vx)(t) = int_{0}^{t} x(s)ds, 0leq tleq 1$$
I tried integrating both sides from $0$ to $t$: $$int_{0}^{t}f(s)ds = int_{0}^{t} sin(s) ds + int_{0}^{t} int_{0}^{s} f(s) ds = -cos(t) + cos(0) + (V^2f)(t) = 1 - cos(0) + int_{0}^{t} (t-s)f(s)ds $$
I'm not sure where to go from here.
real-analysis integral-equations
edited Nov 29 '18 at 17:22
Joshua Mundinger
2,5141026
asked Nov 29 '18 at 1:32
SarahSarah
162
$endgroup$
Solve for f in the integral equation $$f(t) = sin t + int_{0}^{t} f(s)ds$$ using $ (V^nx)(t) =int_{0}^{t} frac{(t-s)^{n-1}}{(n-1)!}x(s)ds $ to to where V is the Volterra operator $V$ on $L^2(0,1)$ is an operator of integration such as the one shown below
$$(Vx)(t) = int_{0}^{t} x(s)ds, 0leq tleq 1$$
I tried integrating both sides from $0$ to $t$: $$int_{0}^{t}f(s)ds = int_{0}^{t} sin(s) ds + int_{0}^{t} int_{0}^{s} f(s) ds = -cos(t) + cos(0) + (V^2f)(t) = 1 - cos(0) + int_{0}^{t} (t-s)f(s)ds $$
I'm not sure where to go from here.
real-analysis integral-equations
real-analysis integral-equations
edited Nov 29 '18 at 17:22
Joshua Mundinger
2,5141026
asked Nov 29 '18 at 1:32
SarahSarah
162
edited Nov 29 '18 at 17:22
Joshua Mundinger
2,5141026
asked Nov 29 '18 at 1:32
SarahSarah
162
edited Nov 29 '18 at 17:22
Joshua Mundinger
2,5141026
edited Nov 29 '18 at 17:22
Joshua Mundinger
2,5141026
edited Nov 29 '18 at 17:22
Joshua Mundinger
2,5141026
2,5141026
asked Nov 29 '18 at 1:32
SarahSarah
162
asked Nov 29 '18 at 1:32
SarahSarah
162
asked Nov 29 '18 at 1:32
SarahSarah
162
162
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I'm not sure of the solution using the Volterra operator, but it should be fairly straightforward to just take the derivative. If you take there derivative, then you'll have:
$$ f'(t) = cos(t) + f(t) $$
which is the same as:
$$ f'(t) - f(t) = cos(t)$$
Use the integrating factor $e^{-t}$, and you'll find
$$e^{-t} f'(t) - e^{-t}f(t) = (e^{-t} f(t))'$$
So, you have:
$$ (e^{-t} f(t))' = e^{-t} cos(t)$$
If you integrate both sides, you'll be able to get the solution easily. For the Volterra approach, the right thing is probably not to integrate per se, but to use the fact $int_0^t f(s) ds = f(t) - sin(t)$ so simplify the expressions involving the Volterra operator.
answered Nov 29 '18 at 1:42
msmmsm
1,243515
$endgroup$
1
$begingroup$
Unless I've missed something, there is no guarantee that $f$ is differentiable.
$endgroup$
– Mattos
Nov 29 '18 at 1:50
$begingroup$
Ah, that is a very good point that I did not even consider. But, if it is then this should work. But, you are probably right, there needs to be another solution.
$endgroup$
– msm
Nov 29 '18 at 2:05
$begingroup$
It comes down to whether $int_0^t f(s) ds$ is differentiable.
$endgroup$
– J.G.
Nov 29 '18 at 17:27
$begingroup$
This kind of solution is consistent because it makes an assumption and then checks it. The thing to be worried about is not whether you've found a solution (you have) but whether the original problem without the additional assumption had a unique solution in the first place.
$endgroup$
– Ian
Nov 29 '18 at 17:49
$begingroup$
Right, so $f$ actually must be differentiable. Suppose $f$ is defined on the interval $[a,b]$. The function $F(t) = int_0^{t} f(s) ds $ is continuous, so $f$ must be continuous. But, since $f$ is continuous, the fundamental theorem of calculus tells us that $F$ is differentiable on $(a,b$, and so $f$ must be differentiable on $(a,b)$. So, the above analysis works on $(a,b)$.
$endgroup$
– msm
Nov 29 '18 at 20:59
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I'm not sure of the solution using the Volterra operator, but it should be fairly straightforward to just take the derivative. If you take there derivative, then you'll have:
$$ f'(t) = cos(t) + f(t) $$
which is the same as:
$$ f'(t) - f(t) = cos(t)$$
Use the integrating factor $e^{-t}$, and you'll find
$$e^{-t} f'(t) - e^{-t}f(t) = (e^{-t} f(t))'$$
So, you have:
$$ (e^{-t} f(t))' = e^{-t} cos(t)$$
If you integrate both sides, you'll be able to get the solution easily. For the Volterra approach, the right thing is probably not to integrate per se, but to use the fact $int_0^t f(s) ds = f(t) - sin(t)$ so simplify the expressions involving the Volterra operator.
answered Nov 29 '18 at 1:42
msmmsm
1,243515
$endgroup$
1
$begingroup$
Unless I've missed something, there is no guarantee that $f$ is differentiable.
$endgroup$
– Mattos
Nov 29 '18 at 1:50
$begingroup$
Ah, that is a very good point that I did not even consider. But, if it is then this should work. But, you are probably right, there needs to be another solution.
$endgroup$
– msm
Nov 29 '18 at 2:05
$begingroup$
It comes down to whether $int_0^t f(s) ds$ is differentiable.
$endgroup$
– J.G.
Nov 29 '18 at 17:27
$begingroup$
This kind of solution is consistent because it makes an assumption and then checks it. The thing to be worried about is not whether you've found a solution (you have) but whether the original problem without the additional assumption had a unique solution in the first place.
$endgroup$
– Ian
Nov 29 '18 at 17:49
$begingroup$
Right, so $f$ actually must be differentiable. Suppose $f$ is defined on the interval $[a,b]$. The function $F(t) = int_0^{t} f(s) ds $ is continuous, so $f$ must be continuous. But, since $f$ is continuous, the fundamental theorem of calculus tells us that $F$ is differentiable on $(a,b$, and so $f$ must be differentiable on $(a,b)$. So, the above analysis works on $(a,b)$.
$endgroup$
– msm
Nov 29 '18 at 20:59
add a comment |
$begingroup$
I'm not sure of the solution using the Volterra operator, but it should be fairly straightforward to just take the derivative. If you take there derivative, then you'll have:
$$ f'(t) = cos(t) + f(t) $$
which is the same as:
$$ f'(t) - f(t) = cos(t)$$
Use the integrating factor $e^{-t}$, and you'll find
$$e^{-t} f'(t) - e^{-t}f(t) = (e^{-t} f(t))'$$
So, you have:
$$ (e^{-t} f(t))' = e^{-t} cos(t)$$
If you integrate both sides, you'll be able to get the solution easily. For the Volterra approach, the right thing is probably not to integrate per se, but to use the fact $int_0^t f(s) ds = f(t) - sin(t)$ so simplify the expressions involving the Volterra operator.
answered Nov 29 '18 at 1:42
msmmsm
1,243515
$endgroup$
1
$begingroup$
Unless I've missed something, there is no guarantee that $f$ is differentiable.
$endgroup$
– Mattos
Nov 29 '18 at 1:50
$begingroup$
Ah, that is a very good point that I did not even consider. But, if it is then this should work. But, you are probably right, there needs to be another solution.
$endgroup$
– msm
Nov 29 '18 at 2:05
$begingroup$
It comes down to whether $int_0^t f(s) ds$ is differentiable.
$endgroup$
– J.G.
Nov 29 '18 at 17:27
$begingroup$
This kind of solution is consistent because it makes an assumption and then checks it. The thing to be worried about is not whether you've found a solution (you have) but whether the original problem without the additional assumption had a unique solution in the first place.
$endgroup$
– Ian
Nov 29 '18 at 17:49
$begingroup$
Right, so $f$ actually must be differentiable. Suppose $f$ is defined on the interval $[a,b]$. The function $F(t) = int_0^{t} f(s) ds $ is continuous, so $f$ must be continuous. But, since $f$ is continuous, the fundamental theorem of calculus tells us that $F$ is differentiable on $(a,b$, and so $f$ must be differentiable on $(a,b)$. So, the above analysis works on $(a,b)$.
$endgroup$
– msm
Nov 29 '18 at 20:59
add a comment |
$begingroup$
I'm not sure of the solution using the Volterra operator, but it should be fairly straightforward to just take the derivative. If you take there derivative, then you'll have:
$$ f'(t) = cos(t) + f(t) $$
which is the same as:
$$ f'(t) - f(t) = cos(t)$$
Use the integrating factor $e^{-t}$, and you'll find
$$e^{-t} f'(t) - e^{-t}f(t) = (e^{-t} f(t))'$$
So, you have:
$$ (e^{-t} f(t))' = e^{-t} cos(t)$$
If you integrate both sides, you'll be able to get the solution easily. For the Volterra approach, the right thing is probably not to integrate per se, but to use the fact $int_0^t f(s) ds = f(t) - sin(t)$ so simplify the expressions involving the Volterra operator.
answered Nov 29 '18 at 1:42
msmmsm
1,243515
$endgroup$
I'm not sure of the solution using the Volterra operator, but it should be fairly straightforward to just take the derivative. If you take there derivative, then you'll have:
$$ f'(t) = cos(t) + f(t) $$
which is the same as:
$$ f'(t) - f(t) = cos(t)$$
Use the integrating factor $e^{-t}$, and you'll find
$$e^{-t} f'(t) - e^{-t}f(t) = (e^{-t} f(t))'$$
So, you have:
$$ (e^{-t} f(t))' = e^{-t} cos(t)$$
If you integrate both sides, you'll be able to get the solution easily. For the Volterra approach, the right thing is probably not to integrate per se, but to use the fact $int_0^t f(s) ds = f(t) - sin(t)$ so simplify the expressions involving the Volterra operator.
answered Nov 29 '18 at 1:42
msmmsm
1,243515
answered Nov 29 '18 at 1:42
msmmsm
1,243515
answered Nov 29 '18 at 1:42
msmmsm
1,243515
answered Nov 29 '18 at 1:42
msmmsm
1,243515
1,243515
1
$begingroup$
Unless I've missed something, there is no guarantee that $f$ is differentiable.
$endgroup$
– Mattos
Nov 29 '18 at 1:50
$begingroup$
Ah, that is a very good point that I did not even consider. But, if it is then this should work. But, you are probably right, there needs to be another solution.
$endgroup$
– msm
Nov 29 '18 at 2:05
$begingroup$
It comes down to whether $int_0^t f(s) ds$ is differentiable.
$endgroup$
– J.G.
Nov 29 '18 at 17:27
$begingroup$
This kind of solution is consistent because it makes an assumption and then checks it. The thing to be worried about is not whether you've found a solution (you have) but whether the original problem without the additional assumption had a unique solution in the first place.
$endgroup$
– Ian
Nov 29 '18 at 17:49
$begingroup$
Right, so $f$ actually must be differentiable. Suppose $f$ is defined on the interval $[a,b]$. The function $F(t) = int_0^{t} f(s) ds $ is continuous, so $f$ must be continuous. But, since $f$ is continuous, the fundamental theorem of calculus tells us that $F$ is differentiable on $(a,b$, and so $f$ must be differentiable on $(a,b)$. So, the above analysis works on $(a,b)$.
$endgroup$
– msm
Nov 29 '18 at 20:59
add a comment |
1
$begingroup$
Unless I've missed something, there is no guarantee that $f$ is differentiable.
$endgroup$
– Mattos
Nov 29 '18 at 1:50
$begingroup$
Ah, that is a very good point that I did not even consider. But, if it is then this should work. But, you are probably right, there needs to be another solution.
$endgroup$
– msm
Nov 29 '18 at 2:05
$begingroup$
It comes down to whether $int_0^t f(s) ds$ is differentiable.
$endgroup$
– J.G.
Nov 29 '18 at 17:27
$begingroup$
This kind of solution is consistent because it makes an assumption and then checks it. The thing to be worried about is not whether you've found a solution (you have) but whether the original problem without the additional assumption had a unique solution in the first place.
$endgroup$
– Ian
Nov 29 '18 at 17:49
$begingroup$
Right, so $f$ actually must be differentiable. Suppose $f$ is defined on the interval $[a,b]$. The function $F(t) = int_0^{t} f(s) ds $ is continuous, so $f$ must be continuous. But, since $f$ is continuous, the fundamental theorem of calculus tells us that $F$ is differentiable on $(a,b$, and so $f$ must be differentiable on $(a,b)$. So, the above analysis works on $(a,b)$.
$endgroup$
– msm
Nov 29 '18 at 20:59
1
1
$begingroup$
Unless I've missed something, there is no guarantee that $f$ is differentiable.
$endgroup$
– Mattos
Nov 29 '18 at 1:50
$begingroup$
Unless I've missed something, there is no guarantee that $f$ is differentiable.
$endgroup$
– Mattos
Nov 29 '18 at 1:50
$begingroup$
Ah, that is a very good point that I did not even consider. But, if it is then this should work. But, you are probably right, there needs to be another solution.
$endgroup$
– msm
Nov 29 '18 at 2:05
$begingroup$
Ah, that is a very good point that I did not even consider. But, if it is then this should work. But, you are probably right, there needs to be another solution.
$endgroup$
– msm
Nov 29 '18 at 2:05
$begingroup$
It comes down to whether $int_0^t f(s) ds$ is differentiable.
$endgroup$
– J.G.
Nov 29 '18 at 17:27
$begingroup$
It comes down to whether $int_0^t f(s) ds$ is differentiable.
$endgroup$
– J.G.
Nov 29 '18 at 17:27
$begingroup$
This kind of solution is consistent because it makes an assumption and then checks it. The thing to be worried about is not whether you've found a solution (you have) but whether the original problem without the additional assumption had a unique solution in the first place.
$endgroup$
– Ian
Nov 29 '18 at 17:49
$begingroup$
This kind of solution is consistent because it makes an assumption and then checks it. The thing to be worried about is not whether you've found a solution (you have) but whether the original problem without the additional assumption had a unique solution in the first place.
$endgroup$
– Ian
Nov 29 '18 at 17:49
$begingroup$
Right, so $f$ actually must be differentiable. Suppose $f$ is defined on the interval $[a,b]$. The function $F(t) = int_0^{t} f(s) ds $ is continuous, so $f$ must be continuous. But, since $f$ is continuous, the fundamental theorem of calculus tells us that $F$ is differentiable on $(a,b$, and so $f$ must be differentiable on $(a,b)$. So, the above analysis works on $(a,b)$.
$endgroup$
– msm
Nov 29 '18 at 20:59
$begingroup$
Right, so $f$ actually must be differentiable. Suppose $f$ is defined on the interval $[a,b]$. The function $F(t) = int_0^{t} f(s) ds $ is continuous, so $f$ must be continuous. But, since $f$ is continuous, the fundamental theorem of calculus tells us that $F$ is differentiable on $(a,b$, and so $f$ must be differentiable on $(a,b)$. So, the above analysis works on $(a,b)$.
$endgroup$
– msm
Nov 29 '18 at 20:59
add a comment |
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