Linear system of ODEs

Question

Given is the ODE system

$y'=\left(\begin{matrix}1\\1\\0\\ \end{matrix}\right)+\left(\begin{matrix}0&0&0\\0&k&0\\0&-k&k\\ \end{matrix} \right)y$

with boundary conditions $y(T)=0$.

Why does for $y$ hold that

$ \int _t^T \frac{d}{ds}(e^{K(T-s)}y(s))ds=\int_t^T e^{K(T-s)}\left(\begin{matrix}1\\1\\0\\ \end{matrix}\right)ds $

and using the boundary conditions

$ y(t)=-e^{-K(T-t)}\int_t^Te^{K(T-s)}\left(\begin{matrix}1\\1\\0\\ \end{matrix}\right)ds $

Would somebody explain please why those two formulas follow from the system of ODEs. Thanks a lot.

Answer 1

This is like a matrix version of integrating factor: let

$$K = \left(\begin{matrix}0&0&0\\0&k&0\\0&-k&k\\ \end{matrix} \right), b= \left(\begin{matrix}1\\1\\0\\ \end{matrix}\right), $$

then $y' = b+ Ky \Rightarrow e^{K(T-s)} y' - e^{K(T-s)}Ky = e^{K(T-s)} b$, which is

$$\big(e^{K(T-s)} y(s)\big)' = e^{K(T-s)} b\ .$$

Integrating both side,

$$ y(T) - e^{K(T-t)} y(t) = \int_t^T e^{K(T-s)} b ds \ .$$

But $y(T)=0$.

Answer 2

You do not say what $K$ is. Probably it is

$$\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & k & 0 \\ 0 & -k & k% \end{array}% \right) . $$

This matrix is not invertible and makes the integral $$ \int_{t}^{T}ds\exp [K(T-s)]\left( \begin{array}{c} 1 \\ 1 \\ 0% \end{array}% \right)$$

slightly delicate.

The usual procedure to solve problems of the present type is to use Duhamel's formula (http://wiki.math.toronto.edu/DispersiveWiki/index.php/Duhamel's_formula) $$ \partial _{t}f(t)=Lf(t)+g(t)\;\Rightarrow \;f(t)=\exp [Lt]f(0)+\int_{0}^{t}ds\exp [L(t-s)]g(s).$$

Now $$ \partial _{t}\mathbf{y}(t)=\left( \begin{array}{c} 1 \\ 1 \\ 0% \end{array}% \right) +\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & k & 0 \\ 0 & -k & k% \end{array}% \right) \mathbf{y}(t)=\mathbf{a}+B\mathbf{y}(t),\;\mathbf{y}(T)=0.$$

The equation of motion for $y_{1}(t)$ is decoupled from the rest and gives $$ y_{1}(t)=t-T$$

Duhamel's formula leads to

$$\mathbf{y}^{\perp }(t)=\exp [Ct]\mathbf{y}^{\perp }(0)+\int_{0}^{t}ds\exp [C(t-s)]\mathbf{a}^{\perp }=\exp [Ct]\mathbf{y}(0)-\frac{\exp [C(t-s)]}{C}% |_{s=0}^{t}\mathbf{a}^{\perp }=\exp [Ct]\mathbf{y}^{\perp }(0)-\frac{1-\exp [Ct]}{C}\mathbf{a}^{\perp },$$

where $\mathbf{y}^{\perp }$ denotes the remaining components of $\mathbf{y}$ and $$ C=\left( \begin{array}{cc} 0 & k \\ -k & k% \end{array}% \right) $$ is invertible. Next $$ \mathbf{y}^{\perp }(T)=\exp [CT]\mathbf{y}^{\perp }(0)-\frac{1-\exp [CT]}{C}% \mathbf{a}^{\perp }=0,$$

gives $$ \mathbf{y}^{\perp }(0)=-\frac{1-\exp [-CT]}{C}\mathbf{a}^{\perp }, $$ so $$ \mathbf{y}^{\perp }(t)=\frac{1}{C}\{\exp [C(t-T)]-1\}\mathbf{a}^{\perp }$$