フィルターのクリア

Solving a PDE symbolically

24 ビュー (過去 30 日間)
richard
richard 2022 年 10 月 18 日
コメント済み: richard 2022 年 10 月 19 日
Greetings! Background: I'm studying "Interconnection and damping assignmemnt passivity based control" (abbreviated as "IDA-PBC"). The general challenge is to solve the matching equations. I am using symbollic math to enter systems of differential equations.
Example: I have a system of differential equations with two state variables like so (this is only an example from my head, just to illustrate what I'm trying to do):
x1_prime = x2 + u*x1 + dH/dx1 <--- dH/dx1 is the partial deriv of H wrt x1
x2_prime = x1 + dH/dx2 <--- dH/dx2 is the partial deriv of H wrt x2
H is the energy function that I'm trying to find
--> I know the partial derivatives of the function H, but I don't know what H itself is. Is there a matlab function that will help me find H symbollically?

採用された回答

Torsten
Torsten 2022 年 10 月 18 日
編集済み: Torsten 2022 年 10 月 18 日
First check whether (dH/dx1,dH/dx2) is conservative.
Then you can reconstruct H by using
H(x1,x2) = H(x10,x20) + integral_{z=x10}^{z=x1} dH/dx1(z,x20) dz + integral_{z=x20}^{z=x2} dH/dx2(x1,z) dz
where H(x10,x20) must be given (thus H must be fixed at a single point).
Following the example under
syms x y x1 y1
dHdx = y*cos(x) + y^2;
dHdy = sin(x)+2*x*y-2*y;
x0 = 0;
y0 = 0;
H0 = 0;
H = H0 + int(subs(dHdx,y,y0),x,x0,x1) + int(subs(dHdy,x,x1),y,y0,y1);
H = subs(H,[x1,y1],[x,y])
H = 
  3 件のコメント
Torsten
Torsten 2022 年 10 月 18 日
[dH/dx1,dH/dx2] is the vector field of H.
Read my answer more carefully.
richard
richard 2022 年 10 月 19 日
Thank you again Torsten - my previous response shows just how much I need to learn!
--> I certainly APPRECIATE your example code that you gave!

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

Help Center および File ExchangeBoundary Conditions についてさらに検索

タグ

製品


リリース

R2021b

Community Treasure Hunt

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

Start Hunting!

Translated by