- #1

- 22

- 0

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?

- Thread starter ala
- Start date

- #1

- 22

- 0

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?

- #2

- 488

- 2

Okay, bellow is my attempt:

Let:

[tex]y(x)=c(1-w(x)exp(-kx))[/tex]

Then

[tex]

w(x)=(1-y/c)exp(kt)[/tex]

[tex]

\dot{w}=-\frac{\dot{y}}{c}exp(-kx)+k \ (1-y/c)exp(-kx)

[/tex]

Now time to make some substitutions

[tex]

\dot{w}=-\frac{f(y)}{c}exp(kt)+k \ (1-\left(c(1-w \ exp(-kx)) \right)/c)exp(kt)[/tex]

[tex]=-\frac{f(y)}{c}exp(kx)-w \ \left(1-\frac{k}{c}\right)exp(kx)+k \ w \ [/tex]

where [tex]y[/tex] is given above as:

[tex]y=c (1- w \ exp(-kx))[/tex]

and [tex]f(y)[/tex] is the original differential equation.

**edit:** The above only seems useful if [tex]\frac{1}{x}[/tex] is much bigger then [tex]k[/tex].

Let:

[tex]y(x)=c(1-w(x)exp(-kx))[/tex]

Then

[tex]

w(x)=(1-y/c)exp(kt)[/tex]

[tex]

\dot{w}=-\frac{\dot{y}}{c}exp(-kx)+k \ (1-y/c)exp(-kx)

[/tex]

Now time to make some substitutions

[tex]

\dot{w}=-\frac{f(y)}{c}exp(kt)+k \ (1-\left(c(1-w \ exp(-kx)) \right)/c)exp(kt)[/tex]

[tex]=-\frac{f(y)}{c}exp(kx)-w \ \left(1-\frac{k}{c}\right)exp(kx)+k \ w \ [/tex]

where [tex]y[/tex] is given above as:

[tex]y=c (1- w \ exp(-kx))[/tex]

and [tex]f(y)[/tex] is the original differential equation.

Last edited:

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