Dirichlet Problem-Partial Differential Equations

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Homework Statement



Let [tex] K = \{ (x,y) | -1<x<1 , -1<y<1 \}[/tex] . Find the unique soloution of dirichlet problem:



[tex]\Delta u(x,y) =0 [/tex] , [tex] (x,y) \in K [/tex] ,
[tex] u(x,y) = |x+y| [/tex] , [tex](x,y) \in \partial K [/tex] .

Homework Equations


The Attempt at a Solution


We need to guess a soloution and not use seperation of variables!
in this particular homework assignment the next harominc functions appear:
[tex]1,x,y,xy, x^2-y^2 ,xye^{x^2-y^2-1} [/tex] ... maybe one of them should be in the soloution function...but none of them solve our equation...

Hope you'll be able to help me!

Thanks !
 

Answers and Replies

  • #2
AlephZero
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Draw a picture of your boundary conditions. You should see they are not a continuous function along each side of the square. In fact they consist of two straight line segments along each side of the square region.

Therefore, you won't be able to find a single function that satisfies the equation, unless it has some type of singularity at the points where the boundary conditions are not differentiable.

Or, you could look for a series solution, and guess the form of the terms of the series to fit the shape and symmetry of the boundary conditions.
 
  • #3
Dick
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Draw a picture of your boundary conditions. You should see they are not a continuous function along each side of the square. In fact they consist of two straight line segments along each side of the square region.

Therefore, you won't be able to find a single function that satisfies the equation, unless it has some type of singularity at the points where the boundary conditions are not differentiable.

Or, you could look for a series solution, and guess the form of the terms of the series to fit the shape and symmetry of the boundary conditions.

Now that's not right. The boundary condition is perfectly continuous along the boundary of K. And you can find a harmonic function that matches the boundary conditions. The OP should try a linear combination of the harmonic functions suggested.
 
  • #4
AlephZero
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Now that's not right. The boundary condition is perfectly continuous along the boundary of K.

Yes the boundary condition is continuous, but it is not differentiable at the four points (0,-1), (0,1), (-1, 0), (1,0) on the boundary.

Therefore the solution can not be a single "elementary function", like the examples the OP gave, which are all differentiable everywhere on the boundary of K, as well as everywhere inside K.

And you can find a harmonic function that matches the boundary conditions. The OP should try a linear combination of the harmonic functions suggested.

Sure, and that's the way I would do it. Each indvidual harmonic function is differentiable everywhere, but the limit of the series of functions is not differentiable at the four points on the boundary (and of course you don't want the limit to be differentiable at those four points, otherwise it would not match the boundary conditions!).

An alternative method would be to find a four solutions with the correct type of singularity to match the boundary conditions at the four points, but that approach is probably not what the OP is meant to be learning about right now. That is a good way to solve this type of problem numerically over a region with an arbitrary shaped boundary, but for the OP's problem a series solution is much easier.
 
  • #5
Dick
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Yes the boundary condition is continuous, but it is not differentiable at the four points (0,-1), (0,1), (-1, 0), (1,0) on the boundary.

Therefore the solution can not be a single "elementary function", like the examples the OP gave, which are all differentiable everywhere on the boundary of K, as well as everywhere inside K.



Sure, and that's the way I would do it. Each indvidual harmonic function is differentiable everywhere, but the limit of the series of functions is not differentiable at the four points on the boundary (and of course you don't want the limit to be differentiable at those four points, otherwise it would not match the boundary conditions!).

An alternative method would be to find a four solutions with the correct type of singularity to match the boundary conditions at the four points, but that approach is probably not what the OP is meant to be learning about right now. That is a good way to solve this type of problem numerically over a region with an arbitrary shaped boundary, but for the OP's problem a series solution is much easier.

It IS differentiable at (1,0) along the boundary curve. Look at it harder. The only places it could be not differentiable is where x+y=0. Trust me, the solution is perfectly differentiable defined in the plane and equals the given function on the boundary. Try and find it before you say it can't exist.
 
  • #6
AlephZero
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It IS differentiable at (1,0) along the boundary curve.

Oops. I misread the boundary conditions as

|x| + |y| on K,

not |x+y| on K

So the problem is a lot easier than I thought it was :blushing:
 

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