Differential difference equation












1












$begingroup$


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



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 //






ordinary-differential-equations recurrence-relations fourier-series







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


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



    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 //






    ordinary-differential-equations recurrence-relations fourier-series







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      2



      $begingroup$


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



      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 //






      ordinary-differential-equations recurrence-relations fourier-series







      share|cite|improve this question









      $endgroup$




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



      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 //






      ordinary-differential-equations recurrence-relations fourier-series




      ordinary-differential-equations recurrence-relations fourier-series






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Jan 5 at 22:27









      SimpleProgrammer SimpleProgrammer

      719




      719






















          1 Answer
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          active

          oldest

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          2












          $begingroup$

          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











          $endgroup$













          • $begingroup$
            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.
            $endgroup$
            – SimpleProgrammer
            Jan 6 at 10:13






          • 1




            $begingroup$
            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 $k=0,1,2,3,4,..$
            $endgroup$
            – LutzL
            Jan 6 at 10:17












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          1 Answer
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          1 Answer
          1






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          active

          oldest

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          2












          $begingroup$

          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











          $endgroup$













          • $begingroup$
            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.
            $endgroup$
            – SimpleProgrammer
            Jan 6 at 10:13






          • 1




            $begingroup$
            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 $k=0,1,2,3,4,..$
            $endgroup$
            – LutzL
            Jan 6 at 10:17
















          2












          $begingroup$

          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











          $endgroup$













          • $begingroup$
            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.
            $endgroup$
            – SimpleProgrammer
            Jan 6 at 10:13






          • 1




            $begingroup$
            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 $k=0,1,2,3,4,..$
            $endgroup$
            – LutzL
            Jan 6 at 10:17














          2












          2








          2





          $begingroup$

          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











          $endgroup$



          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 Jan 6 at 11:19

























          answered Jan 6 at 0:02









          LutzLLutzL

          60.3k42057




          60.3k42057












          • $begingroup$
            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.
            $endgroup$
            – SimpleProgrammer
            Jan 6 at 10:13






          • 1




            $begingroup$
            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 $k=0,1,2,3,4,..$
            $endgroup$
            – LutzL
            Jan 6 at 10:17


















          • $begingroup$
            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.
            $endgroup$
            – SimpleProgrammer
            Jan 6 at 10:13






          • 1




            $begingroup$
            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 $k=0,1,2,3,4,..$
            $endgroup$
            – LutzL
            Jan 6 at 10:17
















          $begingroup$
          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.
          $endgroup$
          – SimpleProgrammer
          Jan 6 at 10:13




          $begingroup$
          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.
          $endgroup$
          – SimpleProgrammer
          Jan 6 at 10:13




          1




          1




          $begingroup$
          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 $k=0,1,2,3,4,..$
          $endgroup$
          – LutzL
          Jan 6 at 10:17




          $begingroup$
          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 $k=0,1,2,3,4,..$
          $endgroup$
          – LutzL
          Jan 6 at 10:17


















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          0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
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          Aardman Animations

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          Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
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          Are they similar matrix

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          0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
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