Why aren't the essential boundary conditions fulfilled?

I made a Poisson-solver based on Legendre-spectral method and I would like to test it. I used the pdetool, created the square region [-1,1]x[-1,1] and specified the PDE to be an elliptic one with parameters c=-1, a=0, f=x.*y. The boundary conditions: left:y, right: -y, top:-x, bottom:x. I run the task with different mesh density and obtained the interesting result that at (-1,-1) the obtained value is not -1 as it should be in my opinion. For an elliptic PDE, these Dirichlet boundary conditions are essential boundary conditions therefore they are taken into account when applying the weak form. Why aren't these boundary condition satisfied?
Thanks, Zoli

 採用された回答

Bill Greene
Bill Greene 2014 年 6 月 6 日

0 投票

I ran this example in R2014a of MATLAB and I do get -1 at the lower left corner (and upper right)
I created the example in pdetool with the following steps: 1. Create the square. 2. After selecting Boundary Conditions, click on each edge and define r to be the values you list above (y,-y,-x,x). 3. Click the = icon.
What version of MATLAB are you running?
If you want to post your code, I'm happy to take a look.
Bill

5 件のコメント

Zoltán Csáti
Zoltán Csáti 2014 年 6 月 7 日
I did the same steps as you. My MATLAB version is R2011a.
Bill Greene
Bill Greene 2014 年 6 月 7 日
I tried this in R2011a on Windows and also get good answers in that version. I have attached my script if you want to try it.
If you haven't already done this, I would suggest exporting the mesh and solution into the MATLAB workspace. The four corner nodes should have numbers 1:4. You can confirm this and find which index corresponds to which corner by examining the p-matrix. Then you can display u(1:4) to see exactly what the solution values are.
Bill
Zoltán Csáti
Zoltán Csáti 2014 年 6 月 8 日
編集済み: Zoltán Csáti 2014 年 6 月 8 日
I examined the p and u matrices and u(1:4) gives the correct values. Then why isn't it so on the attached screenshot? It is a good lesson for me to take the returned numerical values into account and not the graphics. But why doesn't the figure show the correct values?
Bill Greene
Bill Greene 2014 年 6 月 8 日
This plotting problem was fixed in R2013a.
Bill
Zoltán Csáti
Zoltán Csáti 2014 年 6 月 8 日
Thank you.

サインインしてコメントする。

その他の回答 (0 件)

カテゴリ

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by