[Numerical] System of first order ordinary diff equations with given asymptotic

ala
Messages
21
Reaction score
0
I have system of first order ordinary diff equations, indipendent variable is x cordinate. I know asymptotic solution in left and right region (i.e. when x->-infinity or x->infinity, e.g. when abs(x)>1000), it's const plus exponentially falling function. I want to find numerical solution in middle region, witch will have given asymptotic in left and right region.

If I give initial value at left (i.e. at x=-1000) numerical solution blow up at right (I also have exponentially growing functions on right).

How to do this?
 
Physics news on Phys.org
Okay, bellow is my attempt:

Let:

y(x)=c(1-w(x)exp(-kx))

Then

<br /> w(x)=(1-y/c)exp(kt)
<br /> \dot{w}=-\frac{\dot{y}}{c}exp(-kx)+k \ (1-y/c)exp(-kx)<br />

Now time to make some substitutions
<br /> \dot{w}=-\frac{f(y)}{c}exp(kt)+k \ (1-\left(c(1-w \ exp(-kx)) \right)/c)exp(kt)
=-\frac{f(y)}{c}exp(kx)-w \ \left(1-\frac{k}{c}\right)exp(kx)+k \ w \
where y is given above as:

y=c (1- w \ exp(-kx))

and f(y) is the original differential equation.

edit: The above only seems useful if \frac{1}{x} is much bigger then k.
 
Last edited:
For that x, I have asymptotic solution. I want to find numerical solution in the middle, but don't know how. (I don't have 1 ODE, I have system of ODE)
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top