homework

[Anonymous No.5932538]

Sup guise!

How does one find solutions to a system of infinite many homogeneous differential equations with variable coefficients?

[Anonymous No.5932565]

Linear (Can you write it as v'=Dv?)?
And do you know the maximal order of the involved operators?

[Anonymous No.5932575]

>>5932565
(dx/dt)(t) = x(t)A(t)
x is vector range n, A is matrix range nxn.
I can do it if A is constant.

[Anonymous No.5932578]

>>5932565
Derivatives are all 1st order. It is also homogeneous.

[Anonymous No.5932599]

You obviously have to use a hawking paradox to violate the space time continume first(I assume you've allready done this) after which, you use a collapsing microscope to view the alpha center of a cell!

[Anonymous No.5932632]

>>5932575
In one dimension, you have
<span class="math">x(t):=x_0\mathrm{exp}(\int^t A(s)ds)[/spoiler]

<span class="math">\frac{d}{dt}x(t)=\left(\frac{d}{dt}(\int^t A(s)ds)\right)x(t)=A(t)x(t)[/spoiler]

This can be extended to relatively tame A-operators as well. If A doesn't contain more differential operators, then the situation is easier. Still, if A is an infinite "matrix", you'll have to be more specific about it - I don't know the recipe of how to interpret the exponetial map It'll be some tricky version of the general series. Unless of course, you know how to diagonalize A.
You can find the standard results of the theory within the first 30 pages of
"Partial Differential Equations I: Basic Theory - Front Cover Michael Eugene Taylor"

[Anonymous No.5932648]

>>5932632
I've been looking for some good books on PDE's for a while, that are theoretically rigourous, would you recommend this?

[Anonymous No.5932717]

>>5932648
yes, it's a series of books, actually.

[Anonymous No.5932741]

>>5932632
I call bullshit. Even for the 2x2 A case its not clear how to this extends.

[Anonymous No.5932854]

>>5932741
For a finite dimensional matrix A whose coefficients are continous in t,

<span class="math">x(t)=S(t)x(0)[/spoiler]

with

<span class="math">x(t)=\lim_{n\to\infty}\ \{\mathrm{exp}\left( \frac{t}{n}A((n-1)\frac{t}{n}) \right \cdots \mathrm{exp}\left( \frac{t}{n}A(0) \right \}[/spoiler]

For specific simpler A's one can make "the integration" more specific.
In the case of constant A, the exp-function is just the power series on the ring of matrices. Mathematica will solve you the differential equations in one line. For example here the harmonic oscillator, viewed as problem in 2-dimensional phase space:

http://www.wolframalpha.com/input/?i=MatrixExp[Omega*{{0%2C-1}%2C{1%2C0}}t].{x%2Cp}

But people also go in the other extreme. E.g. the Laplaceoperator, not depnending on time, lets you solve the diffusion equation
<span class="math">\frac{\partial}{\partial t}\phi(x,t)=c\ \Delta \phi(x,t)[/spoiler]
via
<span class="math">S(t)\sim \mathrm{exp}(t\cdot c\ \Delta)[/spoiler]

But as soon as things go infinite dimensional, people through with zeta functions alla
http://en.wikipedia.org/wiki/Minakshisundaram%E2%80%93Pleijel_zeta_function

[Anonymous No.5932858]

>>5932741
For a finite dimensional matrix A whose coefficients are continous in t,

<span class="math">x(t)=S(t)x(0)[/spoiler]

with

<span class="math">x(t)=\lim_{n\to\infty}\ \{\mathrm{exp}\left( \frac{t}{n}A((n-1)\frac{t}{n}) \right \cdots \mathrm{exp}\left( \frac{t}{n}A(0) \right) \}[/spoiler]

For specific simpler A's one can make "the integration" more specific.
In the case of constant A, the exp-function is just the power series on the ring of matrices. Mathematica will solve you the differential equations in one line. For example here the harmonic oscillator, viewed as problem in 2-dimensional phase space:

http://www.wolframalpha.com/input/?i=MatrixExp[Omega*{{0%2C-1}%2C{1%2C0}}t].{x%2Cp}

But people also go in the other extreme. E.g. the Laplaceoperator, not depnending on time, lets you solve the diffusion equation

<span class="math">\frac{\partial}{\partial t}\phi(x,t)=c\ \Delta \phi(x,t)[/spoiler]

via

<span class="math">S(t)\sim \mathrm{exp}(t\cdot c\ \Delta)[/spoiler]

But as soon as things go infinite dimensional, people through with zeta functions alla

http://en.wikipedia.org/wiki/Minakshisundaram%E2%80%93Pleijel_zeta_function

[Anonymous No.5932863]

>>5932741
For a finite dimensional matrix A whose coefficients are continous in t,

<span class="math">x(t)=S(t)x(0)[/spoiler]

with

<span class="math">x(t)=\lim_{n\to\infty}\ \{\mathrm{exp}\left( \frac{t}{n}A((n-1)\frac{t}{n}) \right) \cdots \mathrm{exp}\left( \frac{t}{n}A(0) \right) \}[/spoiler]

For specific simpler A's one can make "the integration" more specific.
In the case of constant A, the exp-function is just the power series on the ring of matrices. Mathematica will solve you the differential equations in one line. For example here the harmonic oscillator, viewed as problem in 2-dimensional phase space:

http://www.wolframalpha.com/input/?i=MatrixExp[Omega*{{0%2C-1}%2C{1%2C0}}t].{x%2Cp}

But people also go in the other extreme. E.g. the Laplaceoperator, not depnending on time, lets you solve the diffusion equation

<span class="math">\frac{\partial}{\partial t}\phi(x,t)=c\ \Delta \phi(x,t)[/spoiler]

via

<span class="math">S(t)\sim \mathrm{exp}(t\cdot c\ \Delta)[/spoiler]

But as soon as things go infinite dimensional, people through with zeta functions alla

http://en.wikipedia.org/wiki/Minakshisundaram%E2%80%93Pleijel_zeta_function

[Anonymous No.5932873]

>>5932741
For a finite dimensional matrix A whose coefficients are continous in t,

<span class="math">x(t)=S(t)x(0)[/spoiler]

with

<span class="math">S(t)=\lim_{n\to\infty}\ \{\mathrm{exp}\left( \frac{t}{n}A((n-1)\frac{t}{n}) \right) \cdots \mathrm{exp}\left( \frac{t}{n}A(0) \right) \}[/spoiler]

This is from the book I posted, page 25.
For specific simpler A's one can make "the integration" more specific.
In the case of constant A, the exp-function is just the power series on the ring of matrices. Mathematica will solve you the differential equations in one line. For example here the harmonic oscillator, viewed as problem in 2-dimensional phase space:

http://www.wolframalpha.com/input/?i=MatrixExp[Omega*{{0%2C-1}%2C{1%2C0}}t].{x%2Cp}

But people also go in the other extreme. E.g. the Laplaceoperator, not depnending on time, lets you solve the diffusion equation

<span class="math">\frac{\partial}{\partial t}\phi(x,t)=c\ \Delta \phi(x,t)[/spoiler]

via

<span class="math">S(t)=\mathrm{exp}(t\cdot c\ \Delta)[/spoiler]

But as soon as things go infinite dimensional, people through with zeta functions alla

http://en.wikipedia.org/wiki/Minakshisundaram%E2%80%93Pleijel_zeta_function

[Anonymous No.5932990]

>>5932873
Calling the formula for S(t) an extension of the 1-D case is a stretch. The exponential formula extends for constant A to N-D, but for A=A(t) it doesn't really. In fact, the formula inside the curly braces is simply the repeated use of the formula for constant A, where for large enough n, the "time step" t/n is so small that A can be considered constant. It's essentially an Euler scheme for solving the differential equation. As the time step gets smaller and smaller, the expression in the curly braces converges to the solution.

As for the Laplacian, this is already "infinite dimensional" in the sense of this post. For example, on a finite interval, the operator can be diagonalized using the standard eigenfunctions exp(i n x), n=0,1,2,...,\infty

[Anonymous No.5933029]

>>5932990
Yeah. It's already infinite dimension - the zeta function posted directly relates to the example.
A stretch...well I think it's the right way of calling it. The formula reduces to the matrix exponential in the A=const case. But that's semantics anyway, it's the way to solve it if A isn't further specified.