Linear ordinary differential equations and matrix exponentiation

Question

I feel the solution to this problem is simple - but I am not entirely clear on what the question actually wants.

Given the following:

$$A = \left[ \begin{array}{ccc} -0.1005 & -0.266\\ -0.1498 & 0.2005 \end{array} \right] $$

$$\frac{dx}{dt} = Ax; $$ $$x(0) = x0 = [1; −2]$$

Show, in general, that the following form solves $\frac{dx}{dt} = Ax$

i)

$$ x(t) = e^{At}x_0$$

ii)

$$x(t) = \sum_{i}a_i e_i e^{\lambda_it}$$

Usually when I come across a problem I don'tt even understand, I try to solve simpler analogous problems. I can't quite find one comparable to this. Can someone point out the first couple of steps for me? Maybe for i), for example?

Answer 1

Given $x(t) = e^{At} x_0$, we have

$$\begin{align} x'(t) &= A ~ e^{At} x_0 \\ &= A~x(t) ~\Large\color{\green}{\unicode{10004}}\end{align}$$

From eigenvalues / eigenvectors, we know $\lambda_1 v_1 = Av_1, \lambda_2 v_2 = A v_2$ (this is in your book).

Given $\displaystyle x(t) = \sum_{i}c_i e^{\lambda_it} v_i$, we have

$$\begin{align} x'(t) &= \left(\displaystyle \sum_{i}c_i e^{\lambda_it} v_i\right)' \\ & = (c_1 e^{\lambda_1 t}v_1 + c_2 e^{\lambda_2 t} v_2)' \\ & = c_1 e^{\lambda_1 t} \lambda_1 v_1 + c_2 e^{\lambda_2 t} \lambda_2 v_2 \\ & = c_1 e^{\lambda_1 t} A v_1 + c_2 e^{\lambda_2 t} A v_2 \\ & = A(c_1 e^{\lambda_1 t}v_1 + c_2 e^{\lambda_2 t} v_2)\\ & = A~x(t) ~\Large\color{\green}{\unicode{10004}} \end{align}$$

For this particular problem, we have

$$A = \begin{bmatrix} -0.1005 & -0.266\\ -0.1498 & 0.2005 \end{bmatrix}, x(0) = x_0 = \begin{bmatrix} 1\\ -2 \end{bmatrix} $$

The solution is given by

$$x(t) = e^{At} x_0 = \left( \begin{array}{c} 1.26302 e^{0.299994 t}-0.263018 e^{-0.199994 t} \\ -0.0983787 e^{-0.199994 t}-1.90162 e^{0.299994 t} \\ \end{array} \right)$$

Note $A = P D P^{-1}, e^{At} = Pe^{Dt} P^{-1}$, can you find the eigenvalues / eigenvectors, matrix exponential and that $x(t)$?

Answer 2

Here is a brief outline. First the characteristic polynomial of $A$ is $$\lambda ^2-tr(A)\lambda +det(A)=0$$ Note that det$(A)$ is negative so the roots of the characteristic polynomial, i.e. the eigenvalues, are real. Moreover, the discriminant of the characteristic polynomial is negative, so the roots are distinct, i.e. there are two distinct real eigenvalues, say $\lambda _1$ and $\lambda _2.$ Thus we can find a real non-singular matrix $P$ such that $$P^{-1}AP=\begin{bmatrix}\lambda_1&0\\0&\lambda _2 \end{bmatrix}$$. The notation $\frac{dx}{dt}=Ax$ means that there are two funcions of $t$ , $x_1$ and $x_2$ and that $$\begin{bmatrix}\dot{x_1}\\\dot{x_2}\end{bmatrix}=A\begin{bmatrix} x_1\\x_2\end{bmatrix}(1)$$. We then make a giant leap of faith/proof by wishful thinking and wild use of symbolism and say that since the solution of $\dot x=cx$ is $x=ce^t$ the solution of our equation (1) above should be $$\begin{bmatrix}x_1\\x_2\end{bmatrix}=e^{At}\begin{bmatrix}c_1\\c_2\end{bmatrix}$$. If we put $t=0$ then $c_1=x_1(0)=1 $ and $c_2=x_2(0)=-2 $We can legitimize this formalism in several ways. We can define the exponential function of a matrix by the power series and prove that this series converges for any matrix. Note that the calculation of $e^{At}$ is not too hard because $$e^{At}=P\begin{bmatrix}e^{\lambda _1t}&0\\0&e^{\lambda_2t}\end{bmatrix}P^{-1}$$. Furthermore, we can obtain the solution legitimately by making the change of variables $$\begin{bmatrix}x_1\\x_2\end{bmatrix}=P\begin{bmatrix}u_1\\u_2\end{bmatrix}$$ which gives $\dot {u_1}=\lambda _1,\dot {u_2}=\lambda _2,$ whose solution we know. Then we can recover $x_1$ and $x_2$ from $u_1$ and $u_2$ and show that this legitimately obtained solution agrees with the one obtained by formal manipulation of symbols.