Solving Cauchy's problem with a discontinuous function












0














I have the Cauchy problem: $$begin{cases} f'=g(t)+2(f-5) \ f(0)=2end{cases}$$



Now $g(t)$ is a periodic function: $$g(t)=begin{cases} 0,tin(24k,24k+8)\
2,tnotin(24k,24k+8) end{cases}$$



for $k=1,2,3...$
How can I solve this Cauchy Problem? Since $g$ is not continuous I don't know what to do.



I am using sagemath to solve it.






differential-equations initial-value-problems sagemath cauchy-problem







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  • 1




    Use Laplace transform.
    – Jacky Chong
    Nov 24 at 0:47
















0














I have the Cauchy problem: $$begin{cases} f'=g(t)+2(f-5) \ f(0)=2end{cases}$$



Now $g(t)$ is a periodic function: $$g(t)=begin{cases} 0,tin(24k,24k+8)\
2,tnotin(24k,24k+8) end{cases}$$



for $k=1,2,3...$
How can I solve this Cauchy Problem? Since $g$ is not continuous I don't know what to do.



I am using sagemath to solve it.






differential-equations initial-value-problems sagemath cauchy-problem







share|cite|improve this question


















  • 1




    Use Laplace transform.
    – Jacky Chong
    Nov 24 at 0:47














0












0








0







I have the Cauchy problem: $$begin{cases} f'=g(t)+2(f-5) \ f(0)=2end{cases}$$



Now $g(t)$ is a periodic function: $$g(t)=begin{cases} 0,tin(24k,24k+8)\
2,tnotin(24k,24k+8) end{cases}$$



for $k=1,2,3...$
How can I solve this Cauchy Problem? Since $g$ is not continuous I don't know what to do.



I am using sagemath to solve it.






differential-equations initial-value-problems sagemath cauchy-problem







share|cite|improve this question













I have the Cauchy problem: $$begin{cases} f'=g(t)+2(f-5) \ f(0)=2end{cases}$$



Now $g(t)$ is a periodic function: $$g(t)=begin{cases} 0,tin(24k,24k+8)\
2,tnotin(24k,24k+8) end{cases}$$



for $k=1,2,3...$
How can I solve this Cauchy Problem? Since $g$ is not continuous I don't know what to do.



I am using sagemath to solve it.






differential-equations initial-value-problems sagemath cauchy-problem




differential-equations initial-value-problems sagemath cauchy-problem






share|cite|improve this question













share|cite|improve this question











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asked Nov 24 at 0:44









John Keeper

540313




540313








  • 1




    Use Laplace transform.
    – Jacky Chong
    Nov 24 at 0:47














  • 1




    Use Laplace transform.
    – Jacky Chong
    Nov 24 at 0:47








1




1




Use Laplace transform.
– Jacky Chong
Nov 24 at 0:47




Use Laplace transform.
– Jacky Chong
Nov 24 at 0:47










1 Answer
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Rewrite as



$$ f' - 2f = g - 10 $$



A Fourier method is appropriate here. Since $g$ has a period of $24$, we can assume a series solution of similar periodicity



$$ f(t) = f_0 + sum_{n=1}^infty left[f_ncosleft(frac{npi}{12}tright) + f_n^*sinleft(frac{npi}{12}tright) right]$$



Next, decompose $g$ to its Fourier series



$$ g(t) = g_0 + sum_{n=1}^infty left[g_ncosleft(frac{npi}{12}tright) + g_n^*sinleft(frac{npi}{12}tright) right] $$



From the usual definitions:



$$ g_0 = frac{1}{24}int_0^{24} g(t) dt = int_8^{24} 2 dt = frac{4}{3} $$
$$ g_n = frac{1}{12}int_ 8^{24} 2cosleft(frac{npi}{12}tright) dt = -frac{2}{npi}sin frac{2npi}{3} $$
$$ g_n^* = frac{1}{12}int_8^{24} 2sinleft(frac{npi}{12}tright)dt = -frac{2}{npi}left[1-cos frac{2npi}{3} right] $$



Plug in the 2 series and compare coefficients, we get:



$$ -2f_0 = g_0 - 10 $$
$$ -2f_n + frac{npi}{12}f_n^* = g_n $$
$$ -2f_n^* - frac{npi}{12}f_n = g_n^* $$



Solve the above system of the equations to find $f_n$ and $f_n^*$






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    0














    Rewrite as



    $$ f' - 2f = g - 10 $$



    A Fourier method is appropriate here. Since $g$ has a period of $24$, we can assume a series solution of similar periodicity



    $$ f(t) = f_0 + sum_{n=1}^infty left[f_ncosleft(frac{npi}{12}tright) + f_n^*sinleft(frac{npi}{12}tright) right]$$



    Next, decompose $g$ to its Fourier series



    $$ g(t) = g_0 + sum_{n=1}^infty left[g_ncosleft(frac{npi}{12}tright) + g_n^*sinleft(frac{npi}{12}tright) right] $$



    From the usual definitions:



    $$ g_0 = frac{1}{24}int_0^{24} g(t) dt = int_8^{24} 2 dt = frac{4}{3} $$
    $$ g_n = frac{1}{12}int_ 8^{24} 2cosleft(frac{npi}{12}tright) dt = -frac{2}{npi}sin frac{2npi}{3} $$
    $$ g_n^* = frac{1}{12}int_8^{24} 2sinleft(frac{npi}{12}tright)dt = -frac{2}{npi}left[1-cos frac{2npi}{3} right] $$



    Plug in the 2 series and compare coefficients, we get:



    $$ -2f_0 = g_0 - 10 $$
    $$ -2f_n + frac{npi}{12}f_n^* = g_n $$
    $$ -2f_n^* - frac{npi}{12}f_n = g_n^* $$



    Solve the above system of the equations to find $f_n$ and $f_n^*$






    share|cite|improve this answer




























      0














      Rewrite as



      $$ f' - 2f = g - 10 $$



      A Fourier method is appropriate here. Since $g$ has a period of $24$, we can assume a series solution of similar periodicity



      $$ f(t) = f_0 + sum_{n=1}^infty left[f_ncosleft(frac{npi}{12}tright) + f_n^*sinleft(frac{npi}{12}tright) right]$$



      Next, decompose $g$ to its Fourier series



      $$ g(t) = g_0 + sum_{n=1}^infty left[g_ncosleft(frac{npi}{12}tright) + g_n^*sinleft(frac{npi}{12}tright) right] $$



      From the usual definitions:



      $$ g_0 = frac{1}{24}int_0^{24} g(t) dt = int_8^{24} 2 dt = frac{4}{3} $$
      $$ g_n = frac{1}{12}int_ 8^{24} 2cosleft(frac{npi}{12}tright) dt = -frac{2}{npi}sin frac{2npi}{3} $$
      $$ g_n^* = frac{1}{12}int_8^{24} 2sinleft(frac{npi}{12}tright)dt = -frac{2}{npi}left[1-cos frac{2npi}{3} right] $$



      Plug in the 2 series and compare coefficients, we get:



      $$ -2f_0 = g_0 - 10 $$
      $$ -2f_n + frac{npi}{12}f_n^* = g_n $$
      $$ -2f_n^* - frac{npi}{12}f_n = g_n^* $$



      Solve the above system of the equations to find $f_n$ and $f_n^*$






      share|cite|improve this answer


























        0












        0








        0






        Rewrite as



        $$ f' - 2f = g - 10 $$



        A Fourier method is appropriate here. Since $g$ has a period of $24$, we can assume a series solution of similar periodicity



        $$ f(t) = f_0 + sum_{n=1}^infty left[f_ncosleft(frac{npi}{12}tright) + f_n^*sinleft(frac{npi}{12}tright) right]$$



        Next, decompose $g$ to its Fourier series



        $$ g(t) = g_0 + sum_{n=1}^infty left[g_ncosleft(frac{npi}{12}tright) + g_n^*sinleft(frac{npi}{12}tright) right] $$



        From the usual definitions:



        $$ g_0 = frac{1}{24}int_0^{24} g(t) dt = int_8^{24} 2 dt = frac{4}{3} $$
        $$ g_n = frac{1}{12}int_ 8^{24} 2cosleft(frac{npi}{12}tright) dt = -frac{2}{npi}sin frac{2npi}{3} $$
        $$ g_n^* = frac{1}{12}int_8^{24} 2sinleft(frac{npi}{12}tright)dt = -frac{2}{npi}left[1-cos frac{2npi}{3} right] $$



        Plug in the 2 series and compare coefficients, we get:



        $$ -2f_0 = g_0 - 10 $$
        $$ -2f_n + frac{npi}{12}f_n^* = g_n $$
        $$ -2f_n^* - frac{npi}{12}f_n = g_n^* $$



        Solve the above system of the equations to find $f_n$ and $f_n^*$






        share|cite|improve this answer














        Rewrite as



        $$ f' - 2f = g - 10 $$



        A Fourier method is appropriate here. Since $g$ has a period of $24$, we can assume a series solution of similar periodicity



        $$ f(t) = f_0 + sum_{n=1}^infty left[f_ncosleft(frac{npi}{12}tright) + f_n^*sinleft(frac{npi}{12}tright) right]$$



        Next, decompose $g$ to its Fourier series



        $$ g(t) = g_0 + sum_{n=1}^infty left[g_ncosleft(frac{npi}{12}tright) + g_n^*sinleft(frac{npi}{12}tright) right] $$



        From the usual definitions:



        $$ g_0 = frac{1}{24}int_0^{24} g(t) dt = int_8^{24} 2 dt = frac{4}{3} $$
        $$ g_n = frac{1}{12}int_ 8^{24} 2cosleft(frac{npi}{12}tright) dt = -frac{2}{npi}sin frac{2npi}{3} $$
        $$ g_n^* = frac{1}{12}int_8^{24} 2sinleft(frac{npi}{12}tright)dt = -frac{2}{npi}left[1-cos frac{2npi}{3} right] $$



        Plug in the 2 series and compare coefficients, we get:



        $$ -2f_0 = g_0 - 10 $$
        $$ -2f_n + frac{npi}{12}f_n^* = g_n $$
        $$ -2f_n^* - frac{npi}{12}f_n = g_n^* $$



        Solve the above system of the equations to find $f_n$ and $f_n^*$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 25 at 12:12

























        answered Nov 25 at 10:23









        Dylan

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        12.1k31026






























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