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If I have a differential equation $ y'(t)=A y(t)$ where A is a constant square matrix that is not diagonalizable(although it is surely possible to calculate the eigenvalues) and no initial condition is given. And now I am interested in the fundamental matrix. Is there a general method to determine this matrix? I do not want to use the exponential function and the Jordan normal form, as this is quite exhausting. Maybe there is also an ansatz possible as it is for the special case, where this differential equation is equivalent to an n-th order ode. I saw a method where they calculated the eigenvalues of the matrix and depending on the multiplicity n of this eigenvalue they used an exponential term(with the eigenvalue) and in each component an n-th order polynomial as a possible ansatz. Though they only did this, when they were interested in a initial value problem, so with an initial condition and not for a general solution.

I was asked to deliver an example: so $y'(t)=\begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix} y(t)$ If somebody can construct a fundamental matrix for this system, than this should be sufficient

  • Do you have a specific differential equation? Maybe seeing it suggests some trick that is good for that specific one. – OR. Jul 08 '13 at 19:28
  • I will include one –  Jul 08 '13 at 19:30
  • I said it wrong. I meant to say that if you were interested in a specific differential equation it would be better to give it. If you are interested how to avoid in general exponential and Jordan it is not so important to give an example. – OR. Jul 08 '13 at 19:34
  • Here is a method. – Mhenni Benghorbal Jul 08 '13 at 19:39
  • The solution to $\vec{y}'_t = A \vec{y}_t$ is $y_t = e^{tA}$. Now you need to diagonalize the matrix to find what $e^{tA}$ really is. – gt6989b Jul 08 '13 at 19:50
  • that is the point $A$ is not necessarily diagonalizable!!! –  Jul 08 '13 at 20:02
  • so maybe there is somebody who knows how to calculate the fundamental matrix to the ODE that I have written down there. –  Jul 08 '13 at 20:03
  • @MhenniBenghorbal I don't know whether this method also works for non-diagonalizable matrices –  Jul 08 '13 at 20:19
  • @Lipschitz: You need only to find the eigenvalues $\lambda_1,\lambda_2$ and the corresponding eigenvectors $v_1,v_2$ and then construct the general solution $y(t)=c_1v_1e^{\lambda_1 t} + c_2 v_2e^{\lambda_2 t}$. – Mhenni Benghorbal Jul 08 '13 at 20:40
  • But A is not diagonalizable! Does really nobody know how to do this? Maybe it has something to do with generalized eigenspaces –  Jul 08 '13 at 20:41
  • @Lipschitz: Do you know how to find the Eigen values and Eigen vectors? You do not need to diagonalize your matrix for this process. – Mhenni Benghorbal Jul 08 '13 at 20:44
  • yeah, well not every matrix is diagonalizable. –  Jul 08 '13 at 20:51

2 Answers2

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We have many ways to proceed and this is only a $2x2$. We can choose from:

$$\tag 1 e^{At} = \left[I+ \sum_{k=1}^\infty \dfrac{(A-\lambda I)^k}{k!}t^k\right]e^{\lambda t}$$

For the matrix $A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$, we have:

$\det [A-\lambda I] = \det \begin{bmatrix} 3-\lambda & -4 \\ 1 & -1-\lambda \end{bmatrix} = 0 \rightarrow \lambda^2-2 \lambda+1 = 0 \rightarrow \lambda_{1,2} = 1,1$ (a double eigenvalue). From the eigenvalues, we derive the eigenvalue/eigenvector pairs:

  • $\lambda_1 = 1, v_1 = (2, 1)$
  • $\lambda_2 = 1, v_2 = (1, 0)$ (the second eigenvector is a generalized one)

Lets find the matrix exponential using two different methods.

Method 1

From $(1)$, we have:

$$e^{At} = \left[I + \dfrac{(A-\lambda I)^0}{1!}t^1 \right]e^{\lambda t} = \left[\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 2 & -4 \\ 1 & -2 \end{bmatrix}t\right] = e^{t}\begin{bmatrix} 1+2t & -4t \\ t & 1-2t \end{bmatrix} $$

Method 2

Use the Laplace Transform.

$$e^{At}=\mathcal{L}^{-1}\left((sI-A)^{-1}\right)= \mathcal{L}^{-1}\left(\begin{bmatrix}s-3 & 4 \\ -1 & s+1\end{bmatrix}^{-1}\right) = e^{t}\begin{bmatrix}1+2 t & -4 t \\ t & 1-2 t\end{bmatrix}$$

$\vdots$

Method n

Try other approaches discussed above!

Update: Method n+1

If you wanted to write $A$ using Jordan Normal Form, we would have:

$$A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} = PJP^{-1} = \begin{bmatrix} 2 & 1 \\ 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix} \cdot \begin{bmatrix} 0 & 1 \\1 & -2 \end{bmatrix}$$

To write the matrix exponential for this, we take advantage of the Jordan Block and have:

$$e^{At} = e^{PJP^{-1}t} = Pe^{Jt}P^{-1} = \begin{bmatrix} 2 & 1 \\ 1 & 0 \end{bmatrix} \cdot e^{\begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}t} \cdot \begin{bmatrix} 0 & 1 \\1 & -2 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} e^t & te^t \\ 0 & e^t\end{bmatrix} \cdot \begin{bmatrix} 0 & 1 \\1 & -2 \end{bmatrix} = e^{t}\begin{bmatrix}1+2 t & -4 t \\ t & 1-2 t\end{bmatrix}$$

Lastly, it is worth noting that sometimes the Fundamental Matrix is given as:

$$\phi(t, t_0) = \phi(t) \cdot \phi^{-1}(t_0)$$

Amzoti
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0

There are many methods for determining the matrix exponential, even for a non-diagonalizable matrix. One of the easiest is via the Laplace transform. You can check that

$$\mathcal{L}(e^{tA})(s) = (sI-A)^{-1}.$$

For your example,

$$\mathcal{L}(e^{tA})(s) = \begin{bmatrix}s-3 & 4 \\ -1 & s+1\end{bmatrix}^{-1} = \frac{1}{(s-1)^2}\begin{bmatrix}s+1 & -4 \\ 1 & s-3\end{bmatrix},$$

and a (component-wise) inverse Laplace transform gives $$ e^{tA} = \begin{bmatrix}(2t+1)e^t & -4te^t \\ te^t & -(2t-1)e^t\end{bmatrix}. $$

Formally, this is only valid for $t>0$, but since the elements in $e^{tA}$ are holomorphic, the identity theorem for holomorphic functions shows that the equality is valid for all $t$.

mrf
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