Differential difference equation












1














$y'(t)+y(t-frac{1}{2}pi)-y(t-pi)=cos t$



is the equation I am trying to solve. I want to find all solutions with period $2pi$. Now I have tried with assuming that $y(t)sim sum c_ne^{int}=frac{a_0}{2}+sum (a_ncos nt +b_nsin nt)$ and then differentiating and putting it back into the equation, but I do not know how to handle the second term $y(t-frac{1}{2}pi)$.



Let me show you: $y(t-frac{1}{2}pi)sim sum c_ne^{int-infrac{pi}{2}}=sum c_ne^{int}e^{-infrac{pi}{2}}$ ... from here I don't know how to convert the $e^{-infrac{pi}{2}}$ into something "nicer", for example $e^{-inpi}=(-1)^n$. Is there an analog for this term?



I hope I made myself clear. If someone would show a complete solution that would help, too! Best regards //






differential-equations recurrence-relations fourier-series







share|cite|improve this question



























    1














    $y'(t)+y(t-frac{1}{2}pi)-y(t-pi)=cos t$



    is the equation I am trying to solve. I want to find all solutions with period $2pi$. Now I have tried with assuming that $y(t)sim sum c_ne^{int}=frac{a_0}{2}+sum (a_ncos nt +b_nsin nt)$ and then differentiating and putting it back into the equation, but I do not know how to handle the second term $y(t-frac{1}{2}pi)$.



    Let me show you: $y(t-frac{1}{2}pi)sim sum c_ne^{int-infrac{pi}{2}}=sum c_ne^{int}e^{-infrac{pi}{2}}$ ... from here I don't know how to convert the $e^{-infrac{pi}{2}}$ into something "nicer", for example $e^{-inpi}=(-1)^n$. Is there an analog for this term?



    I hope I made myself clear. If someone would show a complete solution that would help, too! Best regards //






    differential-equations recurrence-relations fourier-series







    share|cite|improve this question

























      1












      1








      1


      2





      $y'(t)+y(t-frac{1}{2}pi)-y(t-pi)=cos t$



      is the equation I am trying to solve. I want to find all solutions with period $2pi$. Now I have tried with assuming that $y(t)sim sum c_ne^{int}=frac{a_0}{2}+sum (a_ncos nt +b_nsin nt)$ and then differentiating and putting it back into the equation, but I do not know how to handle the second term $y(t-frac{1}{2}pi)$.



      Let me show you: $y(t-frac{1}{2}pi)sim sum c_ne^{int-infrac{pi}{2}}=sum c_ne^{int}e^{-infrac{pi}{2}}$ ... from here I don't know how to convert the $e^{-infrac{pi}{2}}$ into something "nicer", for example $e^{-inpi}=(-1)^n$. Is there an analog for this term?



      I hope I made myself clear. If someone would show a complete solution that would help, too! Best regards //






      differential-equations recurrence-relations fourier-series







      share|cite|improve this question













      $y'(t)+y(t-frac{1}{2}pi)-y(t-pi)=cos t$



      is the equation I am trying to solve. I want to find all solutions with period $2pi$. Now I have tried with assuming that $y(t)sim sum c_ne^{int}=frac{a_0}{2}+sum (a_ncos nt +b_nsin nt)$ and then differentiating and putting it back into the equation, but I do not know how to handle the second term $y(t-frac{1}{2}pi)$.



      Let me show you: $y(t-frac{1}{2}pi)sim sum c_ne^{int-infrac{pi}{2}}=sum c_ne^{int}e^{-infrac{pi}{2}}$ ... from here I don't know how to convert the $e^{-infrac{pi}{2}}$ into something "nicer", for example $e^{-inpi}=(-1)^n$. Is there an analog for this term?



      I hope I made myself clear. If someone would show a complete solution that would help, too! Best regards //






      differential-equations recurrence-relations fourier-series




      differential-equations recurrence-relations fourier-series






      share|cite|improve this question













      share|cite|improve this question











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      asked Jan 5 at 22:27









      SimpleProgrammer SimpleProgrammer

      326




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          1 Answer
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          To the complex numbers question: $e^{-ifracpi2}=-i$ so $e^{-ifracpi2}=(-i)^n=i^{-n}$.




          Consider first $y_k(t)=y(t-kfracpi2)$ as independent $2pi$ periodic functions. At the end, the integration constant will have to be adapted so that they all are the same function again, if at all possible.



          Then these $4$ functions satisfy the first order system
          $$
          pmatrix{y_0'\y_1'\y_2'\y_3'}
          +
          pmatrix{0&1&-1&0\0&0&1&-1\-1&0&0&1\1&-1&0&0}
          pmatrix{y_0\y_1\y_2\y_3}
          =
          pmatrix{cos(t)\sin(t)\-cos(t)\-sin(t)}
          $$
          The circulant matrix can be written as $A=Q-Q^2$ where $Q^4=I$, so $A$ has eigenvalues $lambda_k=q^k-(q^k)^2=0$, $1+i$, $-2$, $1-i$ for the eigenvalues $q^k=i^k$ and eigenvectors $v^k_j=i^{jk}$, $j=0,1,2,3$, $k=0,1,2,3$, of $Q$. Of these components of the homogeneous solution, only the first one as a constant is again $2pi$ periodic, the others have exponentially increasing or decreasing factors.



          So the qualitative conclusion is that a periodic solution for $y$ can only have the components $$y(t)=A+Bcos(t)+Csin(t).$$ Inserting this gives
          $$
          (-Bsin t+Ccos t)+(A+Bsin t-Ccos t)-(A-Bcos t-Csin t)=cos t
          \
          impliesleft{begin{aligned}
          B&=1\C&=0
          end{aligned}\right.
          $$




          One could also do the first part slightly different by considering linear combinations of the shifted functions,
          $$u_k(t)=sum_{k=0}^3i^ky(t-kfracpi2),~~k=0,1,2,3, ~~~~ (i^4=1, ~~ y(t-2pi)=y(t)). $$ Then
          begin{align}
          u_k'(t)+sum_{m=0}^3i^{km}y(t-(m+1)fracpi2)-sum_{m=0}^3i^{km}y(t-(m+2)fracpi2)
          &=sum_{k=0}^3i^{km}cos(t-mfracpi2)
          \~\
          u_k'(t)+(i^k-(-1)^k)u(t)&=(1-(-1)^k)cos(t)+i^k(1-(-1)^k)sin(t)
          end{align}
          This is a linear ODE with constant coefficients for each of the 4 values of $k$, and the solution can be reconstructed from them by solving the corresponding linear system or a là the inverse Fourier transform.






          share|cite|improve this answer























          • I appreciate your answer, although I am yet not familiar with the Fourier transform or its inverse. Is there another way to solve it? I don't understand what the q represents, and how do you factor the second term? I don't follow.
            – SimpleProgrammer
            2 days ago










          • This is just 4-dimensional linear algebra. Essentially it shows that the solution has to be a combination of $sin t$ and $cos t$ as you have to exclude the exponential terms for a periodic solution. Note that $sin(t+kfracpi2)$ is $sin t,cos t, -sin t, -cos t, sin t,...$ for $k=0,1,2,3,4,..$
            – LutzL
            2 days ago











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          1 Answer
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          1 Answer
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          active

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          1














          To the complex numbers question: $e^{-ifracpi2}=-i$ so $e^{-ifracpi2}=(-i)^n=i^{-n}$.




          Consider first $y_k(t)=y(t-kfracpi2)$ as independent $2pi$ periodic functions. At the end, the integration constant will have to be adapted so that they all are the same function again, if at all possible.



          Then these $4$ functions satisfy the first order system
          $$
          pmatrix{y_0'\y_1'\y_2'\y_3'}
          +
          pmatrix{0&1&-1&0\0&0&1&-1\-1&0&0&1\1&-1&0&0}
          pmatrix{y_0\y_1\y_2\y_3}
          =
          pmatrix{cos(t)\sin(t)\-cos(t)\-sin(t)}
          $$
          The circulant matrix can be written as $A=Q-Q^2$ where $Q^4=I$, so $A$ has eigenvalues $lambda_k=q^k-(q^k)^2=0$, $1+i$, $-2$, $1-i$ for the eigenvalues $q^k=i^k$ and eigenvectors $v^k_j=i^{jk}$, $j=0,1,2,3$, $k=0,1,2,3$, of $Q$. Of these components of the homogeneous solution, only the first one as a constant is again $2pi$ periodic, the others have exponentially increasing or decreasing factors.



          So the qualitative conclusion is that a periodic solution for $y$ can only have the components $$y(t)=A+Bcos(t)+Csin(t).$$ Inserting this gives
          $$
          (-Bsin t+Ccos t)+(A+Bsin t-Ccos t)-(A-Bcos t-Csin t)=cos t
          \
          impliesleft{begin{aligned}
          B&=1\C&=0
          end{aligned}\right.
          $$




          One could also do the first part slightly different by considering linear combinations of the shifted functions,
          $$u_k(t)=sum_{k=0}^3i^ky(t-kfracpi2),~~k=0,1,2,3, ~~~~ (i^4=1, ~~ y(t-2pi)=y(t)). $$ Then
          begin{align}
          u_k'(t)+sum_{m=0}^3i^{km}y(t-(m+1)fracpi2)-sum_{m=0}^3i^{km}y(t-(m+2)fracpi2)
          &=sum_{k=0}^3i^{km}cos(t-mfracpi2)
          \~\
          u_k'(t)+(i^k-(-1)^k)u(t)&=(1-(-1)^k)cos(t)+i^k(1-(-1)^k)sin(t)
          end{align}
          This is a linear ODE with constant coefficients for each of the 4 values of $k$, and the solution can be reconstructed from them by solving the corresponding linear system or a là the inverse Fourier transform.






          share|cite|improve this answer























          • I appreciate your answer, although I am yet not familiar with the Fourier transform or its inverse. Is there another way to solve it? I don't understand what the q represents, and how do you factor the second term? I don't follow.
            – SimpleProgrammer
            2 days ago










          • This is just 4-dimensional linear algebra. Essentially it shows that the solution has to be a combination of $sin t$ and $cos t$ as you have to exclude the exponential terms for a periodic solution. Note that $sin(t+kfracpi2)$ is $sin t,cos t, -sin t, -cos t, sin t,...$ for $k=0,1,2,3,4,..$
            – LutzL
            2 days ago
















          1














          To the complex numbers question: $e^{-ifracpi2}=-i$ so $e^{-ifracpi2}=(-i)^n=i^{-n}$.




          Consider first $y_k(t)=y(t-kfracpi2)$ as independent $2pi$ periodic functions. At the end, the integration constant will have to be adapted so that they all are the same function again, if at all possible.



          Then these $4$ functions satisfy the first order system
          $$
          pmatrix{y_0'\y_1'\y_2'\y_3'}
          +
          pmatrix{0&1&-1&0\0&0&1&-1\-1&0&0&1\1&-1&0&0}
          pmatrix{y_0\y_1\y_2\y_3}
          =
          pmatrix{cos(t)\sin(t)\-cos(t)\-sin(t)}
          $$
          The circulant matrix can be written as $A=Q-Q^2$ where $Q^4=I$, so $A$ has eigenvalues $lambda_k=q^k-(q^k)^2=0$, $1+i$, $-2$, $1-i$ for the eigenvalues $q^k=i^k$ and eigenvectors $v^k_j=i^{jk}$, $j=0,1,2,3$, $k=0,1,2,3$, of $Q$. Of these components of the homogeneous solution, only the first one as a constant is again $2pi$ periodic, the others have exponentially increasing or decreasing factors.



          So the qualitative conclusion is that a periodic solution for $y$ can only have the components $$y(t)=A+Bcos(t)+Csin(t).$$ Inserting this gives
          $$
          (-Bsin t+Ccos t)+(A+Bsin t-Ccos t)-(A-Bcos t-Csin t)=cos t
          \
          impliesleft{begin{aligned}
          B&=1\C&=0
          end{aligned}\right.
          $$




          One could also do the first part slightly different by considering linear combinations of the shifted functions,
          $$u_k(t)=sum_{k=0}^3i^ky(t-kfracpi2),~~k=0,1,2,3, ~~~~ (i^4=1, ~~ y(t-2pi)=y(t)). $$ Then
          begin{align}
          u_k'(t)+sum_{m=0}^3i^{km}y(t-(m+1)fracpi2)-sum_{m=0}^3i^{km}y(t-(m+2)fracpi2)
          &=sum_{k=0}^3i^{km}cos(t-mfracpi2)
          \~\
          u_k'(t)+(i^k-(-1)^k)u(t)&=(1-(-1)^k)cos(t)+i^k(1-(-1)^k)sin(t)
          end{align}
          This is a linear ODE with constant coefficients for each of the 4 values of $k$, and the solution can be reconstructed from them by solving the corresponding linear system or a là the inverse Fourier transform.






          share|cite|improve this answer























          • I appreciate your answer, although I am yet not familiar with the Fourier transform or its inverse. Is there another way to solve it? I don't understand what the q represents, and how do you factor the second term? I don't follow.
            – SimpleProgrammer
            2 days ago










          • This is just 4-dimensional linear algebra. Essentially it shows that the solution has to be a combination of $sin t$ and $cos t$ as you have to exclude the exponential terms for a periodic solution. Note that $sin(t+kfracpi2)$ is $sin t,cos t, -sin t, -cos t, sin t,...$ for $k=0,1,2,3,4,..$
            – LutzL
            2 days ago














          1












          1








          1






          To the complex numbers question: $e^{-ifracpi2}=-i$ so $e^{-ifracpi2}=(-i)^n=i^{-n}$.




          Consider first $y_k(t)=y(t-kfracpi2)$ as independent $2pi$ periodic functions. At the end, the integration constant will have to be adapted so that they all are the same function again, if at all possible.



          Then these $4$ functions satisfy the first order system
          $$
          pmatrix{y_0'\y_1'\y_2'\y_3'}
          +
          pmatrix{0&1&-1&0\0&0&1&-1\-1&0&0&1\1&-1&0&0}
          pmatrix{y_0\y_1\y_2\y_3}
          =
          pmatrix{cos(t)\sin(t)\-cos(t)\-sin(t)}
          $$
          The circulant matrix can be written as $A=Q-Q^2$ where $Q^4=I$, so $A$ has eigenvalues $lambda_k=q^k-(q^k)^2=0$, $1+i$, $-2$, $1-i$ for the eigenvalues $q^k=i^k$ and eigenvectors $v^k_j=i^{jk}$, $j=0,1,2,3$, $k=0,1,2,3$, of $Q$. Of these components of the homogeneous solution, only the first one as a constant is again $2pi$ periodic, the others have exponentially increasing or decreasing factors.



          So the qualitative conclusion is that a periodic solution for $y$ can only have the components $$y(t)=A+Bcos(t)+Csin(t).$$ Inserting this gives
          $$
          (-Bsin t+Ccos t)+(A+Bsin t-Ccos t)-(A-Bcos t-Csin t)=cos t
          \
          impliesleft{begin{aligned}
          B&=1\C&=0
          end{aligned}\right.
          $$




          One could also do the first part slightly different by considering linear combinations of the shifted functions,
          $$u_k(t)=sum_{k=0}^3i^ky(t-kfracpi2),~~k=0,1,2,3, ~~~~ (i^4=1, ~~ y(t-2pi)=y(t)). $$ Then
          begin{align}
          u_k'(t)+sum_{m=0}^3i^{km}y(t-(m+1)fracpi2)-sum_{m=0}^3i^{km}y(t-(m+2)fracpi2)
          &=sum_{k=0}^3i^{km}cos(t-mfracpi2)
          \~\
          u_k'(t)+(i^k-(-1)^k)u(t)&=(1-(-1)^k)cos(t)+i^k(1-(-1)^k)sin(t)
          end{align}
          This is a linear ODE with constant coefficients for each of the 4 values of $k$, and the solution can be reconstructed from them by solving the corresponding linear system or a là the inverse Fourier transform.






          share|cite|improve this answer














          To the complex numbers question: $e^{-ifracpi2}=-i$ so $e^{-ifracpi2}=(-i)^n=i^{-n}$.




          Consider first $y_k(t)=y(t-kfracpi2)$ as independent $2pi$ periodic functions. At the end, the integration constant will have to be adapted so that they all are the same function again, if at all possible.



          Then these $4$ functions satisfy the first order system
          $$
          pmatrix{y_0'\y_1'\y_2'\y_3'}
          +
          pmatrix{0&1&-1&0\0&0&1&-1\-1&0&0&1\1&-1&0&0}
          pmatrix{y_0\y_1\y_2\y_3}
          =
          pmatrix{cos(t)\sin(t)\-cos(t)\-sin(t)}
          $$
          The circulant matrix can be written as $A=Q-Q^2$ where $Q^4=I$, so $A$ has eigenvalues $lambda_k=q^k-(q^k)^2=0$, $1+i$, $-2$, $1-i$ for the eigenvalues $q^k=i^k$ and eigenvectors $v^k_j=i^{jk}$, $j=0,1,2,3$, $k=0,1,2,3$, of $Q$. Of these components of the homogeneous solution, only the first one as a constant is again $2pi$ periodic, the others have exponentially increasing or decreasing factors.



          So the qualitative conclusion is that a periodic solution for $y$ can only have the components $$y(t)=A+Bcos(t)+Csin(t).$$ Inserting this gives
          $$
          (-Bsin t+Ccos t)+(A+Bsin t-Ccos t)-(A-Bcos t-Csin t)=cos t
          \
          impliesleft{begin{aligned}
          B&=1\C&=0
          end{aligned}\right.
          $$




          One could also do the first part slightly different by considering linear combinations of the shifted functions,
          $$u_k(t)=sum_{k=0}^3i^ky(t-kfracpi2),~~k=0,1,2,3, ~~~~ (i^4=1, ~~ y(t-2pi)=y(t)). $$ Then
          begin{align}
          u_k'(t)+sum_{m=0}^3i^{km}y(t-(m+1)fracpi2)-sum_{m=0}^3i^{km}y(t-(m+2)fracpi2)
          &=sum_{k=0}^3i^{km}cos(t-mfracpi2)
          \~\
          u_k'(t)+(i^k-(-1)^k)u(t)&=(1-(-1)^k)cos(t)+i^k(1-(-1)^k)sin(t)
          end{align}
          This is a linear ODE with constant coefficients for each of the 4 values of $k$, and the solution can be reconstructed from them by solving the corresponding linear system or a là the inverse Fourier transform.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 days ago

























          answered Jan 6 at 0:02









          LutzLLutzL

          56.5k42054




          56.5k42054












          • I appreciate your answer, although I am yet not familiar with the Fourier transform or its inverse. Is there another way to solve it? I don't understand what the q represents, and how do you factor the second term? I don't follow.
            – SimpleProgrammer
            2 days ago










          • This is just 4-dimensional linear algebra. Essentially it shows that the solution has to be a combination of $sin t$ and $cos t$ as you have to exclude the exponential terms for a periodic solution. Note that $sin(t+kfracpi2)$ is $sin t,cos t, -sin t, -cos t, sin t,...$ for $k=0,1,2,3,4,..$
            – LutzL
            2 days ago


















          • I appreciate your answer, although I am yet not familiar with the Fourier transform or its inverse. Is there another way to solve it? I don't understand what the q represents, and how do you factor the second term? I don't follow.
            – SimpleProgrammer
            2 days ago










          • This is just 4-dimensional linear algebra. Essentially it shows that the solution has to be a combination of $sin t$ and $cos t$ as you have to exclude the exponential terms for a periodic solution. Note that $sin(t+kfracpi2)$ is $sin t,cos t, -sin t, -cos t, sin t,...$ for $k=0,1,2,3,4,..$
            – LutzL
            2 days ago
















          I appreciate your answer, although I am yet not familiar with the Fourier transform or its inverse. Is there another way to solve it? I don't understand what the q represents, and how do you factor the second term? I don't follow.
          – SimpleProgrammer
          2 days ago




          I appreciate your answer, although I am yet not familiar with the Fourier transform or its inverse. Is there another way to solve it? I don't understand what the q represents, and how do you factor the second term? I don't follow.
          – SimpleProgrammer
          2 days ago












          This is just 4-dimensional linear algebra. Essentially it shows that the solution has to be a combination of $sin t$ and $cos t$ as you have to exclude the exponential terms for a periodic solution. Note that $sin(t+kfracpi2)$ is $sin t,cos t, -sin t, -cos t, sin t,...$ for $k=0,1,2,3,4,..$
          – LutzL
          2 days ago




          This is just 4-dimensional linear algebra. Essentially it shows that the solution has to be a combination of $sin t$ and $cos t$ as you have to exclude the exponential terms for a periodic solution. Note that $sin(t+kfracpi2)$ is $sin t,cos t, -sin t, -cos t, sin t,...$ for $k=0,1,2,3,4,..$
          – LutzL
          2 days ago


















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          Antonio Litta Visconti Arese

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          Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
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