2次元の偏微分方程式の解き方

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Honoka Kobayashi
Honoka Kobayashi on 1 Oct 2023
Commented: Honoka Kobayashi on 17 Oct 2023
以下の各初期値を用いて、式1より"u(x,y)"を解きたいのですが、MATLABで解く方法がわかりません。もし詳しい方いらっしゃいましたら教えていただきたいです。
%初期値
alpha=0.03;
beta=4;
nx=8;
ny=8;
ita=0.03;
u=zeros(nx,ny);
e=zeros(nx,ny);
m_0=[0 0 0 0.5 0.5 0.1 0.7 0.4; %m_0は各要素が[0,1]を満たす、nx × nyのランダムな行列
0.2 0.2 0.2 0 0 0 0 0;
0.1 0 0 0 0.3 0.5 0.9 0;
0.3 0.1 0.2 0 0 0.5 0 0;
0 0 0 0.5 0.5 0.1 0.7 0.4;
0.2 0.2 0.2 0 0 0 0 0;
0.1 0 0 0 0.3 0.5 0.9 0;
0.3 0.1 0.2 0 0 0.5 0 0];
%以下の式を解きたい
alpha*laplacian(u,[x y])=beta*u-(m_0) %式1
  4 Comments
Dyuman Joshi
Dyuman Joshi on 12 Oct 2023
"u that I want to solve with equation 1 is a function"
Then please provide the definition of u.
Honoka Kobayashi
Honoka Kobayashi on 17 Oct 2023
u is not clear function, but it's a function that is derived by equation 1.

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