Integral equation

**Riesz** · January 9th 2011, 03:56 PM

How to solve equation

?

,

real matrix;

,

real matrix;

, real column vector, composed by

elements;

, real constant;

The unknown is

and it should result equal to

.

**snowtea** · January 9th 2011, 08:48 PM

What does exponentiating a matrix mean (i.e. $e^A$ )?
Please excuse my ignorance.

[Edit: never mind just found out]

**Opalg** · January 10th 2011, 02:33 AM

Originally Posted by Riesz

How to solve equation

?

,

real matrix;

,

real matrix;

, real column vector, composed by

elements;

, real constant;

The unknown is

and it should result equal to

.

Two adjustments to the final line above: first, if B is an nxm matrix then u must be a function from $\mathbb{R}$ to $\mathbb{R}^m$ , not $\mathbb{R}^n$ . Second, in the formula $u(t) = B^{\textsc t}e^{A^{\textsc t}(\overline{t}-t)} \Bigl\{ \int_0^{\overline{t}}e^{At}BB^{\textsc t}e^{A^{\textsc t}t}dt\Bigr\}^{-1}\overline{x}$ , it is bad form to use the same letter t for both the variable in the function u and the dummy variable in the definite integral. It makes the formula clearer if you write it as $u(t) = B^{\textsc t}e^{A^{\textsc t}(\overline{t}-t)} \Bigl\{ \int_0^{\overline{t}}e^{As}BB^{\textsc t}e^{A^{\textsc t}s}ds\Bigr\}^{-1}\overline{x}$ .

In order to make sense of this result, it is helpful to try to separate the analytic and algebraic aspects of the formula. One way to do that is to consider the simple special case that occurs when $m=n=1.$ Then the matrices become scalars. If we write them as $a$ and $b$ instead of $A$ and $B$ , then the given equation becomes

. . . . . . $\int_0^{\overline{t}}e^{a(\overline{t}-t)}bu(t)\,dt = \overline{x}$ ,. . . .(1),

and the solution is given as

. . . . . . $u(t) = \dfrac{be^{a(\overline{t}-t)}\overline{x}}{\int_0^{\overline{t}}e^{as}b^2e^{ as}ds}.$ . . . .(2)

You can see straight away that there is something odd about that result. The given information (1) only tells you the single number $\overline{x}$ (given as the value of a definite integral), whereas the solution (2) consists of a function defined on the whole interval $[0,\overline{t}].$ Of course, there is no way that a single number can determine a whole function. So you have to conclude that the formula (2) just consists of one among infinitely many possible functions giving the value $\overline{x}$ when plugged into the definite integral (1).

The matrix-valued formula in the original problem arises in the same way. There is no way that the given equation $\int_0^{\overline{t}}e^{A(\overline{t}-t)}Bu(t)\,dt = \overline{x}$ can determine the vector-valued function $u(t)$ proposed as the "solution". All you can say is that the given formula for $u(t)$ is one possible solution, and the only way to establish that is to plug it into the given equation and check that it satisfies it.

I suppose that the reason for choosing the formula (2) is that the denominator is visibly positive and so cannot be zero. Similarly, in the solution proposed for the matrix-valued problem, the bracketed matrix is positive definite and therefore invertible.

**Riesz** · January 10th 2011, 12:50 PM

Everything right.

And if $u$ has also to satisfy $2\int_0^{\overline{t}}u(t)dt - c^T\int_0^{\overline{t}} e^{A(\overline{t}-t)}B\:dt=0$ ?

$c$ , real vector with $n$ elements;

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