Dirichlet Problem-Partial Differential Equations

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Homework Help Overview

The original poster presents a Dirichlet problem involving the Laplace equation, where the goal is to find a unique solution within a specified square region K, given boundary conditions defined by the function |x+y| along the perimeter of K.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the nature of the boundary conditions, noting potential issues with continuity and differentiability. Some suggest exploring harmonic functions or series solutions, while others question the assumptions regarding the differentiability of the boundary conditions.

Discussion Status

The discussion is ongoing, with various interpretations of the boundary conditions being explored. Some participants offer guidance on potential approaches, such as using harmonic functions or series solutions, while others express differing views on the differentiability of the boundary conditions.

Contextual Notes

There is a noted concern regarding the differentiability of the boundary conditions at specific points, which may affect the type of solution that can be found. Participants are also considering the implications of these conditions on the methods to be used for finding a solution.

WannaBe22
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Homework Statement



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



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

Homework Equations


The Attempt at a Solution


We need to guess a soloution and not use separation of variables!
in this particular homework assignment the next harominc functions appear:
1,x,y,xy, x^2-y^2 ,xye^{x^2-y^2-1} ... 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 !
 
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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.
 
AlephZero said:
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.
 
Dick said:
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.
 
AlephZero said:
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.
 
Dick said:
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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