Solving PDEs with IC, BCs: Help from Kevin

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Discussion Overview

The discussion revolves around solving a partial differential equation (PDE) with initial and boundary conditions. The participants explore methods for addressing the equation, particularly through the use of Laplace transforms and separation of variables.

Discussion Character

  • Technical explanation, Mathematical reasoning, Homework-related

Main Points Raised

  • Kevin presents a PDE involving a delta function and specifies initial and boundary conditions, seeking assistance in solving it.
  • One participant questions whether the delta function is represented as δ(t), suggesting a similarity to a previous problem involving a fourth-order ordinary differential equation (ODE).
  • Kevin confirms the use of δ(t) and notes that applying the Laplace transform leads to a problem similar to the earlier one, but expresses difficulty in performing the inverse Laplace transform back to the time domain.
  • Kevin suggests that the problem may be separable, proposing a form Y(z,t) = Z(z)T(t) as a potential approach.

Areas of Agreement / Disagreement

Participants have not reached a consensus on the best method to solve the PDE, and multiple approaches are being considered, including the Laplace transform and separation of variables.

Contextual Notes

The discussion does not clarify the assumptions underlying the proposed methods or the specific steps required for the inverse Laplace transform, leaving some mathematical processes unresolved.

jimmygriffey
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Hi:

I have the following PDE:

ytzz=yzzzz+delta(t)

With I.C.: t=0, y=0; and B.C.s: z=0, y=0,yzz=0; z=-x,y=0,yz=0

Can someone show me how to solve it?

Kevin
 
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Yes, it is δ(t). The one is very similar to the previous one. After taking Laplace transform in t, this one will become the previous one. If I use the method pointed by other people, I have hard time to convert back (inverse Laplace transform) to time domain. That's why I ask the question again.

Kevin
 
jimmygriffey said:
Yes, it is δ(t). The one is very similar to the previous one. After taking Laplace transform in t, this one will become the previous one. If I use the method pointed by other people, I have hard time to convert back (inverse Laplace transform) to time domain. That's why I ask the question again.

Kevin
Yes - if one uses the Laplace Transform (which takes an equation from time domain to frequncy domain) then I believe it is the same problem. I'm expecting the problem is separable, i.e. Y(z,t) = Z(z)T(t). See if that works.
 

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