Generalizing separation technique

  • Thread starter jostpuur
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  • #1
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Main Question or Discussion Point

[tex]
\left[\begin{array}{c}
y'_1(x) \\ \vdots \\ y'_n(x) \\
\end{array}\right]
=
\left[\begin{array}{ccc}
f_{11}(y) & \cdots & f_{1n}(y) \\
\vdots & & \vdots \\
f_{n1}(y) & \cdots & f_{nn}(y) \\
\end{array}\right]
\left[\begin{array}{c}
g_1(x) \\ \vdots \\ g_n(x) \\
\end{array}\right]
[/tex]

If n=1, then this can be solved with the separation technique. Suppose n>1 and that [itex]f[/itex] 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?
 

Answers and Replies

  • #2
lurflurf
Homework Helper
2,423
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As written I cannot say I do not know what y on the right hand side means.

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

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 existance-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
 
  • #3
2,111
16
I see. The equation

[tex]
y'(x) = A(x)y(x)
[/tex]

is equivalent with

[tex]
y(x) = y(0) + \int\limits_0^x du\; A(u)y(u),
[/tex]

so the iteration attempt

[tex]
y_0(x) = y(0)
[/tex]

[tex]
y_{n+1}(x) = y(0) + \int\limits_0^x du\; A(u)y_n(u)
[/tex]

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

[tex]
y(x) \;=\; \Big(1 \;+\; \int\limits_0^x du\; A(u) \;+\; \int\limits_0^x du\; \int\limits_0^u du'\; A(u)A(u') \;+\; \int\limits_0^x du\; \int\limits_0^u du'\; \int\limits_0^{u'} du''\; A(u) A(u') A(u'') \;+\; \cdots\Big)y(0).
[/tex]

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

[tex]
S = 1 + q + q^2 + q^3 + \cdots
[/tex]

follows

[tex]
qS = q \;+\; q^2 \;+\; q^3 \;+\; \cdots = S - 1\quad\implies\quad S = \frac{1}{1-q},
[/tex]

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

[tex]
U(x) \;=\; 1 \;+\; \int\limits_0^x du\; A(u) \;+\; \int\limits_0^x du\;\int\limits_0^u du'\; A(u) A(u') \;+\; \cdots,
[/tex]

then the calculation

[tex]
\int\limits_0^{x'} dx\; A(x)U(x) \;=\; \int\limits_0^{x'} dx\; A(x) \;+\; \int\limits_0^{x'} dx\;\int\limits_0^x du\; A(x) A(u) \;+\; \cdots \;=\; U(x') \;-\; 1
[/tex]

doesn't lead anywhere.
 
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