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How exactly is the Wronskian tied with Cramer's rule in a system of ODEs?



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1












$begingroup$


If we have a nonhomogeneous system of linear differential equations:
$$frac{dvec{y}}{dx}=Avec{y}+vec{b}$$
where:
$$frac{dvec{y}}{dx}=begin{bmatrix}y_1'\y_2'\y_3'end{bmatrix},vec{y}=begin{bmatrix}y_1\y_2\y_3end{bmatrix},A=begin{bmatrix}a_{11}&a_{12}&a_{13}\a_{21}&a_{22}&a_{23}\a_{31}&a_{32}&a_{33}end{bmatrix}, vec{b}=begin{bmatrix}b_1(x)\b_2(x)\b_3(x)end{bmatrix}$$
the particular solution involving these three functions could be written:
$$y_{particular}=c_1(x)e^{lambda_1x}vec{u_1}+c_2(x)e^{lambda_2x}vec{u_2}+c_3(x)e^{lambda_3x}vec{u_3}$$
where $lambda_i, vec{u_i}$ are the eigenvalues and eigenvectors of the constant matrix $A$ (I chose it constant to make things simpler.)



When trying to evaluate these $c_i(x)$ functions, I was told to evaluate them by:
$$c_i(x)=intfrac{W_i}{W}dx$$
Where $W$ is the Wronskian determinant whose columns are the elements of the vectors $c_i(x)e^{lambda_3i}vec{u_i}$, and $W_i$ is the Wronskian determinant whose $i$th column was replaced by the contents of the $vec{b}$ vector.



My questions are: How does the integral show up? Why does the process look like Cramer's rule? How did we lead up to that result? Where did this 'Wronskian' determinant come from?



If I have missed something, please point it out and I'll add it to the post.



Thanks in advance.










share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    If we have a nonhomogeneous system of linear differential equations:
    $$frac{dvec{y}}{dx}=Avec{y}+vec{b}$$
    where:
    $$frac{dvec{y}}{dx}=begin{bmatrix}y_1'\y_2'\y_3'end{bmatrix},vec{y}=begin{bmatrix}y_1\y_2\y_3end{bmatrix},A=begin{bmatrix}a_{11}&a_{12}&a_{13}\a_{21}&a_{22}&a_{23}\a_{31}&a_{32}&a_{33}end{bmatrix}, vec{b}=begin{bmatrix}b_1(x)\b_2(x)\b_3(x)end{bmatrix}$$
    the particular solution involving these three functions could be written:
    $$y_{particular}=c_1(x)e^{lambda_1x}vec{u_1}+c_2(x)e^{lambda_2x}vec{u_2}+c_3(x)e^{lambda_3x}vec{u_3}$$
    where $lambda_i, vec{u_i}$ are the eigenvalues and eigenvectors of the constant matrix $A$ (I chose it constant to make things simpler.)



    When trying to evaluate these $c_i(x)$ functions, I was told to evaluate them by:
    $$c_i(x)=intfrac{W_i}{W}dx$$
    Where $W$ is the Wronskian determinant whose columns are the elements of the vectors $c_i(x)e^{lambda_3i}vec{u_i}$, and $W_i$ is the Wronskian determinant whose $i$th column was replaced by the contents of the $vec{b}$ vector.



    My questions are: How does the integral show up? Why does the process look like Cramer's rule? How did we lead up to that result? Where did this 'Wronskian' determinant come from?



    If I have missed something, please point it out and I'll add it to the post.



    Thanks in advance.










    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      If we have a nonhomogeneous system of linear differential equations:
      $$frac{dvec{y}}{dx}=Avec{y}+vec{b}$$
      where:
      $$frac{dvec{y}}{dx}=begin{bmatrix}y_1'\y_2'\y_3'end{bmatrix},vec{y}=begin{bmatrix}y_1\y_2\y_3end{bmatrix},A=begin{bmatrix}a_{11}&a_{12}&a_{13}\a_{21}&a_{22}&a_{23}\a_{31}&a_{32}&a_{33}end{bmatrix}, vec{b}=begin{bmatrix}b_1(x)\b_2(x)\b_3(x)end{bmatrix}$$
      the particular solution involving these three functions could be written:
      $$y_{particular}=c_1(x)e^{lambda_1x}vec{u_1}+c_2(x)e^{lambda_2x}vec{u_2}+c_3(x)e^{lambda_3x}vec{u_3}$$
      where $lambda_i, vec{u_i}$ are the eigenvalues and eigenvectors of the constant matrix $A$ (I chose it constant to make things simpler.)



      When trying to evaluate these $c_i(x)$ functions, I was told to evaluate them by:
      $$c_i(x)=intfrac{W_i}{W}dx$$
      Where $W$ is the Wronskian determinant whose columns are the elements of the vectors $c_i(x)e^{lambda_3i}vec{u_i}$, and $W_i$ is the Wronskian determinant whose $i$th column was replaced by the contents of the $vec{b}$ vector.



      My questions are: How does the integral show up? Why does the process look like Cramer's rule? How did we lead up to that result? Where did this 'Wronskian' determinant come from?



      If I have missed something, please point it out and I'll add it to the post.



      Thanks in advance.










      share|cite|improve this question











      $endgroup$




      If we have a nonhomogeneous system of linear differential equations:
      $$frac{dvec{y}}{dx}=Avec{y}+vec{b}$$
      where:
      $$frac{dvec{y}}{dx}=begin{bmatrix}y_1'\y_2'\y_3'end{bmatrix},vec{y}=begin{bmatrix}y_1\y_2\y_3end{bmatrix},A=begin{bmatrix}a_{11}&a_{12}&a_{13}\a_{21}&a_{22}&a_{23}\a_{31}&a_{32}&a_{33}end{bmatrix}, vec{b}=begin{bmatrix}b_1(x)\b_2(x)\b_3(x)end{bmatrix}$$
      the particular solution involving these three functions could be written:
      $$y_{particular}=c_1(x)e^{lambda_1x}vec{u_1}+c_2(x)e^{lambda_2x}vec{u_2}+c_3(x)e^{lambda_3x}vec{u_3}$$
      where $lambda_i, vec{u_i}$ are the eigenvalues and eigenvectors of the constant matrix $A$ (I chose it constant to make things simpler.)



      When trying to evaluate these $c_i(x)$ functions, I was told to evaluate them by:
      $$c_i(x)=intfrac{W_i}{W}dx$$
      Where $W$ is the Wronskian determinant whose columns are the elements of the vectors $c_i(x)e^{lambda_3i}vec{u_i}$, and $W_i$ is the Wronskian determinant whose $i$th column was replaced by the contents of the $vec{b}$ vector.



      My questions are: How does the integral show up? Why does the process look like Cramer's rule? How did we lead up to that result? Where did this 'Wronskian' determinant come from?



      If I have missed something, please point it out and I'll add it to the post.



      Thanks in advance.







      linear-algebra ordinary-differential-equations systems-of-equations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 17 at 17:13









      J. W. Tanner

      4,1961320




      4,1961320










      asked Mar 17 at 17:03









      Gradient EntropyGradient Entropy

      204




      204






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Write the general solution of the homogeneous linear ODE
          $$
          tag{1}
          frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y}
          $$

          in the matrix form as
          $$
          Phi(x; x_0) vec{c},
          $$

          where $vec{c} = mathrm{col}(c_1, dots, c_n)$ is a constant matrix (that is, independent of $x$), and $Phi$ is the state-transition matrix such that $Phi(x_0, x_0) = I$ (the identity matrix).



          Now we are looking for a general solution of the nonhomogeneous linear ODE
          $$
          tag{2}
          frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y} + vec{b}(x)
          $$

          in the form
          $$
          Phi(x, x_0) vec{c}(x).
          $$

          Incidentally, here is the source of the term "variation of constants". We look for conditions for the above function of $x$ to be a solution of $(2)$, that is, to have
          $$
          frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = A(x) (Phi(x, x_0) vec{c}(x)) + vec{b}(x).
          $$

          But
          $$
          frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) cdot vec{c}(x) + Phi(x, x_0) cdot vec{c}'(x)
          $$

          and
          $$
          frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) = A(x) Phi(x, x_0),
          $$

          so after cancellation we obtain
          $$
          Phi(x, x_0) cdot vec{c}'(x) = vec{b}(x),
          $$

          which is, for each fixed $x$, a Cramer (because $det{Phi(x, x_0)} = W(x, x_0) ne 0$) system of $n$ linear equations with $n$ unknowns, $c'_1(x), dots, c'_n(x)$. The unknowns are given by the Cramer rule. We have thus obtained the derivative of the matrix function $vec{c}(x)$, so we have to integrate that derivative. And that is the source of the integral.



          The "Wrońskian" determinant was named after a nineteenth-century Polish philosopher and mathematician, Józef Maria Hoene-Wroński.



          EDIT: I changed slightly the terminology and provided a reference to a Wikipedia entry.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Thank you for your answer, but I am having difficulty in understanding what this ' fundamental matrix ' is, as well as its notation. What is that semicolon's purpose?
            $endgroup$
            – Gradient Entropy
            Mar 17 at 21:53










          • $begingroup$
            I replaced the semicolons with commas, to keep the notation in line with the Wikipedia entry.
            $endgroup$
            – user539887
            Mar 17 at 22:00










          • $begingroup$
            One last question. I'm quite unfamiliar with control theory, and I don't know how this state transition matrix is composed of. I assume it's a 3x3 matrix (according to my example) and it's invertible, because the wronskian determinant is non zero. This means we can write: $vec{c}'=Phi^{-1}(x,x_0)vec{b}$, then integrate to find the vector c. Is this correct? If so, how is that done? That is why I'm asking about the elements of the state transition matrix.
            $endgroup$
            – Gradient Entropy
            Mar 17 at 23:01












          • $begingroup$
            It is only the name "state-transition matrix" that is used in Control Theory. It is one of the most fundamental concepts in the theory of linear systems of ODEs: it is the fundamental matrix that takes value $I$ at $x_0$. In other words, if $Psi(x)$ is any fundamental matrix of $(1)$ then the state-transition matrix of $(1)$ equals $Phi(x,x_0)=Psi(x)Psi^{-1}(x_0)$. Another name (when $x_0$ is fixed): principal fundamental matrix, see, e.g., Linear Differential System.
            $endgroup$
            – user539887
            Mar 18 at 7:18












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          active

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          active

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          1












          $begingroup$

          Write the general solution of the homogeneous linear ODE
          $$
          tag{1}
          frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y}
          $$

          in the matrix form as
          $$
          Phi(x; x_0) vec{c},
          $$

          where $vec{c} = mathrm{col}(c_1, dots, c_n)$ is a constant matrix (that is, independent of $x$), and $Phi$ is the state-transition matrix such that $Phi(x_0, x_0) = I$ (the identity matrix).



          Now we are looking for a general solution of the nonhomogeneous linear ODE
          $$
          tag{2}
          frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y} + vec{b}(x)
          $$

          in the form
          $$
          Phi(x, x_0) vec{c}(x).
          $$

          Incidentally, here is the source of the term "variation of constants". We look for conditions for the above function of $x$ to be a solution of $(2)$, that is, to have
          $$
          frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = A(x) (Phi(x, x_0) vec{c}(x)) + vec{b}(x).
          $$

          But
          $$
          frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) cdot vec{c}(x) + Phi(x, x_0) cdot vec{c}'(x)
          $$

          and
          $$
          frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) = A(x) Phi(x, x_0),
          $$

          so after cancellation we obtain
          $$
          Phi(x, x_0) cdot vec{c}'(x) = vec{b}(x),
          $$

          which is, for each fixed $x$, a Cramer (because $det{Phi(x, x_0)} = W(x, x_0) ne 0$) system of $n$ linear equations with $n$ unknowns, $c'_1(x), dots, c'_n(x)$. The unknowns are given by the Cramer rule. We have thus obtained the derivative of the matrix function $vec{c}(x)$, so we have to integrate that derivative. And that is the source of the integral.



          The "Wrońskian" determinant was named after a nineteenth-century Polish philosopher and mathematician, Józef Maria Hoene-Wroński.



          EDIT: I changed slightly the terminology and provided a reference to a Wikipedia entry.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Thank you for your answer, but I am having difficulty in understanding what this ' fundamental matrix ' is, as well as its notation. What is that semicolon's purpose?
            $endgroup$
            – Gradient Entropy
            Mar 17 at 21:53










          • $begingroup$
            I replaced the semicolons with commas, to keep the notation in line with the Wikipedia entry.
            $endgroup$
            – user539887
            Mar 17 at 22:00










          • $begingroup$
            One last question. I'm quite unfamiliar with control theory, and I don't know how this state transition matrix is composed of. I assume it's a 3x3 matrix (according to my example) and it's invertible, because the wronskian determinant is non zero. This means we can write: $vec{c}'=Phi^{-1}(x,x_0)vec{b}$, then integrate to find the vector c. Is this correct? If so, how is that done? That is why I'm asking about the elements of the state transition matrix.
            $endgroup$
            – Gradient Entropy
            Mar 17 at 23:01












          • $begingroup$
            It is only the name "state-transition matrix" that is used in Control Theory. It is one of the most fundamental concepts in the theory of linear systems of ODEs: it is the fundamental matrix that takes value $I$ at $x_0$. In other words, if $Psi(x)$ is any fundamental matrix of $(1)$ then the state-transition matrix of $(1)$ equals $Phi(x,x_0)=Psi(x)Psi^{-1}(x_0)$. Another name (when $x_0$ is fixed): principal fundamental matrix, see, e.g., Linear Differential System.
            $endgroup$
            – user539887
            Mar 18 at 7:18
















          1












          $begingroup$

          Write the general solution of the homogeneous linear ODE
          $$
          tag{1}
          frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y}
          $$

          in the matrix form as
          $$
          Phi(x; x_0) vec{c},
          $$

          where $vec{c} = mathrm{col}(c_1, dots, c_n)$ is a constant matrix (that is, independent of $x$), and $Phi$ is the state-transition matrix such that $Phi(x_0, x_0) = I$ (the identity matrix).



          Now we are looking for a general solution of the nonhomogeneous linear ODE
          $$
          tag{2}
          frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y} + vec{b}(x)
          $$

          in the form
          $$
          Phi(x, x_0) vec{c}(x).
          $$

          Incidentally, here is the source of the term "variation of constants". We look for conditions for the above function of $x$ to be a solution of $(2)$, that is, to have
          $$
          frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = A(x) (Phi(x, x_0) vec{c}(x)) + vec{b}(x).
          $$

          But
          $$
          frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) cdot vec{c}(x) + Phi(x, x_0) cdot vec{c}'(x)
          $$

          and
          $$
          frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) = A(x) Phi(x, x_0),
          $$

          so after cancellation we obtain
          $$
          Phi(x, x_0) cdot vec{c}'(x) = vec{b}(x),
          $$

          which is, for each fixed $x$, a Cramer (because $det{Phi(x, x_0)} = W(x, x_0) ne 0$) system of $n$ linear equations with $n$ unknowns, $c'_1(x), dots, c'_n(x)$. The unknowns are given by the Cramer rule. We have thus obtained the derivative of the matrix function $vec{c}(x)$, so we have to integrate that derivative. And that is the source of the integral.



          The "Wrońskian" determinant was named after a nineteenth-century Polish philosopher and mathematician, Józef Maria Hoene-Wroński.



          EDIT: I changed slightly the terminology and provided a reference to a Wikipedia entry.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Thank you for your answer, but I am having difficulty in understanding what this ' fundamental matrix ' is, as well as its notation. What is that semicolon's purpose?
            $endgroup$
            – Gradient Entropy
            Mar 17 at 21:53










          • $begingroup$
            I replaced the semicolons with commas, to keep the notation in line with the Wikipedia entry.
            $endgroup$
            – user539887
            Mar 17 at 22:00










          • $begingroup$
            One last question. I'm quite unfamiliar with control theory, and I don't know how this state transition matrix is composed of. I assume it's a 3x3 matrix (according to my example) and it's invertible, because the wronskian determinant is non zero. This means we can write: $vec{c}'=Phi^{-1}(x,x_0)vec{b}$, then integrate to find the vector c. Is this correct? If so, how is that done? That is why I'm asking about the elements of the state transition matrix.
            $endgroup$
            – Gradient Entropy
            Mar 17 at 23:01












          • $begingroup$
            It is only the name "state-transition matrix" that is used in Control Theory. It is one of the most fundamental concepts in the theory of linear systems of ODEs: it is the fundamental matrix that takes value $I$ at $x_0$. In other words, if $Psi(x)$ is any fundamental matrix of $(1)$ then the state-transition matrix of $(1)$ equals $Phi(x,x_0)=Psi(x)Psi^{-1}(x_0)$. Another name (when $x_0$ is fixed): principal fundamental matrix, see, e.g., Linear Differential System.
            $endgroup$
            – user539887
            Mar 18 at 7:18














          1












          1








          1





          $begingroup$

          Write the general solution of the homogeneous linear ODE
          $$
          tag{1}
          frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y}
          $$

          in the matrix form as
          $$
          Phi(x; x_0) vec{c},
          $$

          where $vec{c} = mathrm{col}(c_1, dots, c_n)$ is a constant matrix (that is, independent of $x$), and $Phi$ is the state-transition matrix such that $Phi(x_0, x_0) = I$ (the identity matrix).



          Now we are looking for a general solution of the nonhomogeneous linear ODE
          $$
          tag{2}
          frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y} + vec{b}(x)
          $$

          in the form
          $$
          Phi(x, x_0) vec{c}(x).
          $$

          Incidentally, here is the source of the term "variation of constants". We look for conditions for the above function of $x$ to be a solution of $(2)$, that is, to have
          $$
          frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = A(x) (Phi(x, x_0) vec{c}(x)) + vec{b}(x).
          $$

          But
          $$
          frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) cdot vec{c}(x) + Phi(x, x_0) cdot vec{c}'(x)
          $$

          and
          $$
          frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) = A(x) Phi(x, x_0),
          $$

          so after cancellation we obtain
          $$
          Phi(x, x_0) cdot vec{c}'(x) = vec{b}(x),
          $$

          which is, for each fixed $x$, a Cramer (because $det{Phi(x, x_0)} = W(x, x_0) ne 0$) system of $n$ linear equations with $n$ unknowns, $c'_1(x), dots, c'_n(x)$. The unknowns are given by the Cramer rule. We have thus obtained the derivative of the matrix function $vec{c}(x)$, so we have to integrate that derivative. And that is the source of the integral.



          The "Wrońskian" determinant was named after a nineteenth-century Polish philosopher and mathematician, Józef Maria Hoene-Wroński.



          EDIT: I changed slightly the terminology and provided a reference to a Wikipedia entry.






          share|cite|improve this answer











          $endgroup$



          Write the general solution of the homogeneous linear ODE
          $$
          tag{1}
          frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y}
          $$

          in the matrix form as
          $$
          Phi(x; x_0) vec{c},
          $$

          where $vec{c} = mathrm{col}(c_1, dots, c_n)$ is a constant matrix (that is, independent of $x$), and $Phi$ is the state-transition matrix such that $Phi(x_0, x_0) = I$ (the identity matrix).



          Now we are looking for a general solution of the nonhomogeneous linear ODE
          $$
          tag{2}
          frac{mathrm{d}vec{y}}{mathrm{d}x} = A(x) vec{y} + vec{b}(x)
          $$

          in the form
          $$
          Phi(x, x_0) vec{c}(x).
          $$

          Incidentally, here is the source of the term "variation of constants". We look for conditions for the above function of $x$ to be a solution of $(2)$, that is, to have
          $$
          frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = A(x) (Phi(x, x_0) vec{c}(x)) + vec{b}(x).
          $$

          But
          $$
          frac{mathrm{d}}{mathrm{d}x}(Phi(x, x_0) vec{c}(x)) = frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) cdot vec{c}(x) + Phi(x, x_0) cdot vec{c}'(x)
          $$

          and
          $$
          frac{mathrm{d}}{mathrm{d}x} Phi(x, x_0) = A(x) Phi(x, x_0),
          $$

          so after cancellation we obtain
          $$
          Phi(x, x_0) cdot vec{c}'(x) = vec{b}(x),
          $$

          which is, for each fixed $x$, a Cramer (because $det{Phi(x, x_0)} = W(x, x_0) ne 0$) system of $n$ linear equations with $n$ unknowns, $c'_1(x), dots, c'_n(x)$. The unknowns are given by the Cramer rule. We have thus obtained the derivative of the matrix function $vec{c}(x)$, so we have to integrate that derivative. And that is the source of the integral.



          The "Wrońskian" determinant was named after a nineteenth-century Polish philosopher and mathematician, Józef Maria Hoene-Wroński.



          EDIT: I changed slightly the terminology and provided a reference to a Wikipedia entry.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 17 at 21:59

























          answered Mar 17 at 21:17









          user539887user539887

          1,7811414




          1,7811414








          • 1




            $begingroup$
            Thank you for your answer, but I am having difficulty in understanding what this ' fundamental matrix ' is, as well as its notation. What is that semicolon's purpose?
            $endgroup$
            – Gradient Entropy
            Mar 17 at 21:53










          • $begingroup$
            I replaced the semicolons with commas, to keep the notation in line with the Wikipedia entry.
            $endgroup$
            – user539887
            Mar 17 at 22:00










          • $begingroup$
            One last question. I'm quite unfamiliar with control theory, and I don't know how this state transition matrix is composed of. I assume it's a 3x3 matrix (according to my example) and it's invertible, because the wronskian determinant is non zero. This means we can write: $vec{c}'=Phi^{-1}(x,x_0)vec{b}$, then integrate to find the vector c. Is this correct? If so, how is that done? That is why I'm asking about the elements of the state transition matrix.
            $endgroup$
            – Gradient Entropy
            Mar 17 at 23:01












          • $begingroup$
            It is only the name "state-transition matrix" that is used in Control Theory. It is one of the most fundamental concepts in the theory of linear systems of ODEs: it is the fundamental matrix that takes value $I$ at $x_0$. In other words, if $Psi(x)$ is any fundamental matrix of $(1)$ then the state-transition matrix of $(1)$ equals $Phi(x,x_0)=Psi(x)Psi^{-1}(x_0)$. Another name (when $x_0$ is fixed): principal fundamental matrix, see, e.g., Linear Differential System.
            $endgroup$
            – user539887
            Mar 18 at 7:18














          • 1




            $begingroup$
            Thank you for your answer, but I am having difficulty in understanding what this ' fundamental matrix ' is, as well as its notation. What is that semicolon's purpose?
            $endgroup$
            – Gradient Entropy
            Mar 17 at 21:53










          • $begingroup$
            I replaced the semicolons with commas, to keep the notation in line with the Wikipedia entry.
            $endgroup$
            – user539887
            Mar 17 at 22:00










          • $begingroup$
            One last question. I'm quite unfamiliar with control theory, and I don't know how this state transition matrix is composed of. I assume it's a 3x3 matrix (according to my example) and it's invertible, because the wronskian determinant is non zero. This means we can write: $vec{c}'=Phi^{-1}(x,x_0)vec{b}$, then integrate to find the vector c. Is this correct? If so, how is that done? That is why I'm asking about the elements of the state transition matrix.
            $endgroup$
            – Gradient Entropy
            Mar 17 at 23:01












          • $begingroup$
            It is only the name "state-transition matrix" that is used in Control Theory. It is one of the most fundamental concepts in the theory of linear systems of ODEs: it is the fundamental matrix that takes value $I$ at $x_0$. In other words, if $Psi(x)$ is any fundamental matrix of $(1)$ then the state-transition matrix of $(1)$ equals $Phi(x,x_0)=Psi(x)Psi^{-1}(x_0)$. Another name (when $x_0$ is fixed): principal fundamental matrix, see, e.g., Linear Differential System.
            $endgroup$
            – user539887
            Mar 18 at 7:18








          1




          1




          $begingroup$
          Thank you for your answer, but I am having difficulty in understanding what this ' fundamental matrix ' is, as well as its notation. What is that semicolon's purpose?
          $endgroup$
          – Gradient Entropy
          Mar 17 at 21:53




          $begingroup$
          Thank you for your answer, but I am having difficulty in understanding what this ' fundamental matrix ' is, as well as its notation. What is that semicolon's purpose?
          $endgroup$
          – Gradient Entropy
          Mar 17 at 21:53












          $begingroup$
          I replaced the semicolons with commas, to keep the notation in line with the Wikipedia entry.
          $endgroup$
          – user539887
          Mar 17 at 22:00




          $begingroup$
          I replaced the semicolons with commas, to keep the notation in line with the Wikipedia entry.
          $endgroup$
          – user539887
          Mar 17 at 22:00












          $begingroup$
          One last question. I'm quite unfamiliar with control theory, and I don't know how this state transition matrix is composed of. I assume it's a 3x3 matrix (according to my example) and it's invertible, because the wronskian determinant is non zero. This means we can write: $vec{c}'=Phi^{-1}(x,x_0)vec{b}$, then integrate to find the vector c. Is this correct? If so, how is that done? That is why I'm asking about the elements of the state transition matrix.
          $endgroup$
          – Gradient Entropy
          Mar 17 at 23:01






          $begingroup$
          One last question. I'm quite unfamiliar with control theory, and I don't know how this state transition matrix is composed of. I assume it's a 3x3 matrix (according to my example) and it's invertible, because the wronskian determinant is non zero. This means we can write: $vec{c}'=Phi^{-1}(x,x_0)vec{b}$, then integrate to find the vector c. Is this correct? If so, how is that done? That is why I'm asking about the elements of the state transition matrix.
          $endgroup$
          – Gradient Entropy
          Mar 17 at 23:01














          $begingroup$
          It is only the name "state-transition matrix" that is used in Control Theory. It is one of the most fundamental concepts in the theory of linear systems of ODEs: it is the fundamental matrix that takes value $I$ at $x_0$. In other words, if $Psi(x)$ is any fundamental matrix of $(1)$ then the state-transition matrix of $(1)$ equals $Phi(x,x_0)=Psi(x)Psi^{-1}(x_0)$. Another name (when $x_0$ is fixed): principal fundamental matrix, see, e.g., Linear Differential System.
          $endgroup$
          – user539887
          Mar 18 at 7:18




          $begingroup$
          It is only the name "state-transition matrix" that is used in Control Theory. It is one of the most fundamental concepts in the theory of linear systems of ODEs: it is the fundamental matrix that takes value $I$ at $x_0$. In other words, if $Psi(x)$ is any fundamental matrix of $(1)$ then the state-transition matrix of $(1)$ equals $Phi(x,x_0)=Psi(x)Psi^{-1}(x_0)$. Another name (when $x_0$ is fixed): principal fundamental matrix, see, e.g., Linear Differential System.
          $endgroup$
          – user539887
          Mar 18 at 7:18


















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