Linear differential equations systems with constant coefficients.

Example: suppose we want to solve the following system:

$$\left\{\begin{array}{lllllll} y_1 ' (x) &= & y_1(x) &- & y_2(... ..._2 ' (x) &=& 2 y_1(x)& +& 4 y_2(x)& +& e^x \end{array} \right.$$

with initial data y₁(0)=y₂(0)=0.

For scalar linear differential equations with constant coefficients, it is a well-known procedure to use Laplace transform, e.g. to solve

a y'' + b y' + c y = f(x)

one would perform the ${\cal L}$-transform and get:

$$(a s^2 + b s + c) {\cal L}(y) = {\cal L}(f) (s) + a (s f(0) + f'(0)) + b f(0)$$

where f(0) and f'(0) are the initial consitions at x=0. If ${\cal L}(f)$ is a rational fraction, ILAP allows you to recover y by inverse Laplace transform of

$$\frac{ {\cal L}(f) (s) + a (s f(0) + f'(0)) + b f(0)}{ a s^2 + b s + c}$$

But this method may be applied to systems of linear differential equations as well. The aim of LDEC is to help you solving 1st order systems (note that higher order systems may always be rewritten as 1st order systems). Let y=(y₁, ..., y_n) be a vector of functions of x and suppose that we want to solve:

y' = A y + b

where A is a $n \times n$ constant matrix and b a vector of n functions of x. Denote by ${\cal L}(b)$ the vector of the n Laplace transform of the n functions of b , and by y(0) the vector of n initial data at x=0 . Then

$$(s I - A ) {\cal L}(y) = {\cal L}(b) + y(0)$$

(I denotes the identity matrix), hence:

$${\cal L}(y) = ( s I - A )^{-1} ( {\cal L}(b) + y(0) )$$

and:

$$y={\cal L}^{-1} \left[ ( s I - A )^{-1} ( {\cal L}(b) ) \right] + {\cal L}^{-1}\left[ (sI-A)^{-1} y(0) \right]$$ (1)

The purpose of LDEC is to compute

$$y={\cal L}^{-1} ( ( s I - A )^{-1} ( {\cal L}(b) ) )$$

the solution with initial data y(0)=0, given A and b. The stack should be prepared as:
stk2: b
stk1: A,
then type LDEC and you will get y at stack level 1. Stack level 3 is the comatrix, stack level 2 is the determinant of (sI-A).

Let's do it for the example above:
First put b on the stack: { 1 'EXP(X)' } (here VX is set as usual to 'X'). Now put the matrix A on the stack:
{ { 1 -1 } { 2 4 } }
After calling LDEC, we get:

$$\left\{ \begin{array}{rcl} y_1 &=& -1/2 e^x-2/3+2 e^{2x}-5/6 ... ...x} \\ y_2 &=& 1/3 - 2 e^{2x} + 5/3 e^{3x} \end{array} \right.$$

At level 2, we have $\det(sI-A)$ (with s=X):

$$\det(sI-A)=(2-s)(3-s)$$

and level 3 is the comatrix of sI-A:

$$\left(\begin{array}{cc} -4+s & -1\\ 2 & -1+s \end{array}\right)$$

Level 2 and 3 will be useful if we want to compute the solution of the LDEC with non-zero initial data , e.g. y₁(0)=1, y₂(0)=2. If we look at Equation (1), we see that we have to add ${\cal L}^{-1}\left[ (sI-A)^{-1} y(0) \right]$ to the previous solution. To do this, type { 1 2 }, multiply by the comatrix from level 3, multiply by the inverse of the determinant from level 2 (you can note divide a vector, you must first invert the determinant and then multiply):

$$(sI-A)^{-1}(1,2)=(\frac{-6+s}{6-5s+s^2},\frac{2s}{6-5s+s^2})$$

call ILAP:

(4e^2x-3e^3x, -4e^2x+6e^3x )

and add the result to the previous solution:

$$\left\{ \begin{array}{rcl} y_1 &=& -1/2 e^x-2/3+6 e^{2x}-23/6... ...} \\ y_2 &=& 1/3 - 6 e^{2x} + 23/3 e^{3x} \end{array} \right.$$