Solve for f in the integral equation $f(t) = sin(t) + int f(s)ds $












2












$begingroup$


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







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 29 '18 at 17:22









      Joshua Mundinger

      2,5141026




      2,5141026










      asked Nov 29 '18 at 1:32









      SarahSarah

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          1 Answer
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          active

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          1












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$









          • 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











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          1 Answer
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          active

          oldest

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          1












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$









          • 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
















          1












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$









          • 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














          1












          1








          1





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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














          • 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


















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