Solving a PDE w/ given boundary and initial conditions

  • #1
367
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Firstly, my main question boils down to speaking about the initial conditions and boundary conditions.

I was given:

$$ u(0,y,t) = u(\pi,y,t) = u(x,0,t) = u(x,\pi,t) = 0 $$

but then the initial condition was:

$$ u(x,y,0) = 1 $$

Aren't the initial and boundary conditions inconsistent in such a case? For example, what is the value of ##u(0,0,0)##? Based on the boundary conditions, it should be 0, but based on the initial conditions, it should be 1, no? Why exactly is this okay, if it is okay?
 

Answers and Replies

  • #2
pasmith
Homework Helper
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(a) A discontinuous initial condition of that sort poses no mathematical difficulty. The values of points on the boundary are given by the boundary condition, not the initial condition.

(b) This is a mathematical abstraction of the physically realistic situation where there is a very thin boundary layer in which [itex]u[/itex] reduces smoothly from 1 to 0 and which we aren't interested in modelling.
 
  • #3
367
13
(a) A discontinuous initial condition of that sort poses no mathematical difficulty. The values of points on the boundary are given by the boundary condition, not the initial condition.

(b) This is a mathematical abstraction of the physically realistic situation where there is a very thin boundary layer in which [itex]u[/itex] reduces smoothly from 1 to 0 and which we aren't interested in modelling.
Thank you for the response.


I still do not quite understand how it poses no mathematics difficulty. Based on the boundary conditions, if ## u(0,y,t) = 0 ## then ##u(0,0,0) = 0##, too, right? But based on the initial condition ## u(x,y,0) = 1 ##, then ## u(0,0,0) = 1 ##, no? Don't these boundary conditions hold for all time, and the initial conditions for all space? If so, then how can both the initial conditions and boundary conditions be satisfied by a function? In such a case, u would have to have two values for the coordinate (0,0,0) and therefore no longer a function?

The answer derived for u is:

$$ u(x,y,t) = \frac {16}{\pi ^2} \sum _{j=0}^\infty \sum _{k=0}^\infty \frac {1}{(2j+1)(2k+1)} e^{-[(2j+1)^2 + (2k+1)^2]t} \sin[(2j+1)x] \sin[(2k+1)y] $$

And in such a case, ## u(0,0,0) = 0 ## which violates the initial condition?

It seems like I am definitely missing something here, since although I agree that the solutions may converge to different values at different points, they should not be equal to two separate values at the same point.

Are you saying the initial conditions are valid everywhere except the boundary? In which case, why isn't this usually explicitly stated?
 
  • #4
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You have a square region in which the value of u is u=1 throughout. At time t = 0, you suddenly drop the value of u at the boundary to zero. Does that make physical sense to you?
 
  • #5
367
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You have a square region in which the value of u is u=1 throughout. At time t = 0, you suddenly drop the value of u at the boundary to zero. Does that make physical sense to you?
Yes, I can visualize that happening. But to reflect this situation, it would only be equal to 1 at t = 0, no? Although my concern still remains in how that's accurately represented by these boundary/initial conditions above and the solution I put forth.
 
  • #6
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Yes, I can visualize that happening. But to reflect this situation, it would only be equal to 1 at t = 0, no? Although my concern still remains in how that's accurately represented by these boundary/initial conditions above and the solution I put forth.
The solution you put down matches the initial conditions everywhere, except at the boundaries where it is zero.
 

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