Message Boards Message Boards

0
|
10295 Views
|
6 Replies
|
5 Total Likes
View groups...
Share
Share this post:

Solution of differential equations with boundary conditions contain "Limit"

Robert Pham, Education Department
Posted 11 years ago

Dear Friends,

I am going to solve the following differential equation:

$y''[x] = {k^2}y[x]$

with boundary conditions is $y'[0] - hy[0] = 0$ and $\mathop {Lim}\limits_{x \to \infty } [\left| {y[x]} \right| < \infty $ I tried to solve it in Mathematica by

DSolve[{y''[x] == {k^2}y[x],y'[0] - hy[0] == 0,Limit[Abs[y[x]], x -> \[Infinity] < \[Infinity]]},y[x],x]

But I get the following error:

Equation or list of equations expected instead of Limit[Abs[y[x]],x->False] in the first argument.

I can use the boundary conditions contain "Limit" how to correct?

I would highly appreciate it if somebody can help me.

Best regards.
POSTED BY: Robert Pham
6 Replies
Sort By:

This is what I expected!! I really appreciate your help with my problem! Thank you so much!!
POSTED BY: Robert Pham

Thank you very much for helping me! But I still exist two problems:

  1. The boundary conditions of the problem is: $\mathop {Lim}\limits_{x \to \infty } [\left| {y[x]} \right| < \infty$. (y [x] is always finite).
  2. I can not use Plot command for $y[x]$ because the answer is still one constant C.

Could I eliminate constant C by the second condition? how can i do that?

Please help me one more!! I sincerely thank you very much!
POSTED BY: Robert Pham

For a linear DE with linear BCs, there is always an arbitrary multiplicative constant in the solution. The constant is often determined by a normalization condition on f(x).
POSTED BY: S M Blinder

Well, if C[1] is not equal zero, then the solution will diverge. So if you want the limit to be finite, i.e. $<\infty$, then the only thing that works is C[1]=0. So using that condition of yours, the only choice for the constant that works seems to be C[1]=0.

Cheers,

M.
POSTED BY: Marco Thiel

Syntax errors. Try following:

In[1]:= DSolve[{y''[x] == k^2 y[x], y'[0] - h y[0] == 0}, y[x], x]

Out[1]= {{y[x] -> ( E^(-k x) (-h + E^(2 k x) h + k + E^(2 k x) k) C[1])/(h + k)}}

Clearly then Lim x->Infinity, gives y[x]->Infinity
POSTED BY: S M Blinder

Unless C[1]=0, in which case the solution is constant zero, which appears to be a valid solution.

M.
POSTED BY: Marco Thiel
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Tag limit exceeded
Note: Only the first five people you tag will receive an email notification; the other tagged names will appear as links to their profiles.
Reply Preview
Attachments
Remove
or Discard

Group Abstract Group Abstract