Generalizing separation technique

[jostpuur]

<br /> \left[\begin{array}{c}<br /> y&#039;_1(x) \\ \vdots \\ y&#039;_n(x) \\<br /> \end{array}\right]<br /> =<br /> \left[\begin{array}{ccc}<br /> f_{11}(y) &amp; \cdots &amp; f_{1n}(y) \\<br /> \vdots &amp; &amp; \vdots \\<br /> f_{n1}(y) &amp; \cdots &amp; f_{nn}(y) \\<br /> \end{array}\right]<br /> \left[\begin{array}{c}<br /> g_1(x) \\ \vdots \\ g_n(x) \\<br /> \end{array}\right]<br />

If n=1, then this can be solved with the separation technique. Suppose n>1 and that f is invertible. Could the separation technique be generalized to give some explicit formula for solution y(x)? I tried without success. Anyone dealt with problems like this ever?

 

[lurflurf]

As written I cannot say I do not know what y on the right hand side means.

If you mean, or have use for
\left[\begin{array}{c}y&#039;_1(x) \\ \vdots \\ y&#039;_n(x) \\\end{array}\right]=\left[\begin{array}{ccc}f_{11}(x) &amp; \cdots &amp; f_{1n}(x) \\\vdots &amp; &amp; \vdots \\f_{n1}(x) &amp; \cdots &amp; f_{nn}(x) \\\end{array}\right]\left[\begin{array}{c}y_1(x) \\ \vdots \\ y_n(x) \\\end{array}\right]

Then yes this is called the method of Peano-Baker.
A good reference is Ince Ordinary differential equations.
The idea is
y'=Ay
A and y being functions of x
A a linear operator y a vector
let I be integration say from 0 to x
suppose when x=0 y=c
y=c+IAy
y'=Ac+AIAy
y=c+IAc+IAIAy
y=c+IAc+IAIAc+IAIAIAy
leting the sum go to infinity we have a geometric series
y=((IA)^0+(IA)^1+(IA)^2+...)c
or
y={[(IA)^0-(IA)^1]^-1}c
we can prove this works.
It can be used in principle, but is more usefule to prove existence-uniqueness since it is highly impractical.
we may think of this as a generalization of
y'=Ay,A'=0,y(0)=c->y=exp(Ax)c
ie
y'=Ay,y(0)=c->y={[(IA)^0-(IA)^1]^-1}c
in both cases the "explicit" formula as written is not enough we often want to know more

 

[jostpuur]

I see. The equation

<br /> y&#039;(x) = A(x)y(x)<br />

is equivalent with

<br /> y(x) = y(0) + \int\limits_0^x du\; A(u)y(u),<br />

so the iteration attempt

<br /> y_0(x) = y(0)<br />

<br /> y_{n+1}(x) = y(0) + \int\limits_0^x du\; A(u)y_n(u)<br />

seems natural. If the iterations converge towards the solution, it follows that the solution is given by the series

<br /> y(x) \;=\; \Big(1 \;+\; \int\limits_0^x du\; A(u) \;+\; \int\limits_0^x du\; \int\limits_0^u du&#039;\; A(u)A(u&#039;) \;+\; \int\limits_0^x du\; \int\limits_0^u du&#039;\; \int\limits_0^{u&#039;} du&#039;&#039;\; A(u) A(u&#039;) A(u&#039;&#039;) \;+\; \cdots\Big)y(0). <br />

But I didn't quite get the geometric series part. From equation

<br /> S = 1 + q + q^2 + q^3 + \cdots<br />

follows

<br /> qS = q \;+\; q^2 \;+\; q^3 \;+\; \cdots = S - 1\quad\implies\quad S = \frac{1}{1-q},<br />

but how do you do the same with the iterated integrals? If we define the time evolution operator to be

<br /> U(x) \;=\; 1 \;+\; \int\limits_0^x du\; A(u) \;+\; \int\limits_0^x du\;\int\limits_0^u du&#039;\; A(u) A(u&#039;) \;+\; \cdots,<br />

then the calculation

<br /> \int\limits_0^{x&#039;} dx\; A(x)U(x) \;=\; \int\limits_0^{x&#039;} dx\; A(x) \;+\; \int\limits_0^{x&#039;} dx\;\int\limits_0^x du\; A(x) A(u) \;+\; \cdots \;=\; U(x&#039;) \;-\; 1<br />

doesn't lead anywhere.