Using alternating-direction implicit method to solve nonzero boundary value problem (transform into zero boundary value problem)Initial-boundary value problem for PDEHeat Equation: Initial value boundary value problemHow to solve this initial-boundary value problem for a PDERunge-Kutta method for PDESolve Burgers' equation after shock formsFinite difference method and division by zero problem with no flux boundary conditionUsing implicit method to solve system analytically and finding errorSolve parabolic PDE with non-zero boundary and initial conditionsHow to Solve This PDE and Deal with Boundary Value Using Method of Characteristics. Work Shown.How to implement boundary value in Python using spectral methods
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Using alternating-direction implicit method to solve nonzero boundary value problem (transform into zero boundary value problem)
Initial-boundary value problem for PDEHeat Equation: Initial value boundary value problemHow to solve this initial-boundary value problem for a PDERunge-Kutta method for PDESolve Burgers' equation after shock formsFinite difference method and division by zero problem with no flux boundary conditionUsing implicit method to solve system analytically and finding errorSolve parabolic PDE with non-zero boundary and initial conditionsHow to Solve This PDE and Deal with Boundary Value Using Method of Characteristics. Work Shown.How to implement boundary value in Python using spectral methods
$begingroup$
$$u_t = 0.25(u_xx+u_yy) -8x(x^2+y^2), qquad 0 leq x,y leq 1, qquad 0 leq t leq 0.5,$$
$$u(0,y,t) = y^2 - y, qquad u(1,y,t) = y^2 + y + 1,$$
$$u(x,0,t) = x, qquad u(x,1,t) = 3x,$$
$$u(x,y,0) = y^" + 2xy + x -y.$$
I don't know how to implement the nonzero boundary conditions or transform the problem into a problem with zero boundary conditions.
Here is my attempt:
function [u,x,y,t]=q2(T,K,N,M)
% solves the 2D heat equation u_t=K u_xx+ f(x,y,t) in the unit square
% and on the interval [0,T] with zero boundary condition by ADI method.
%K is the (heat conduction) coefficient in the heat equation.
%N is the number of subintervals in [0,1].
%M is the number of time steps.
h=1/N;
tau=T/M;
gamma=tau*K/h^2;
u=zeros(N+1,N+1,M+1);
u1=zeros(N+1,N+1);
alpha=zeros(N,1);
beta=zeros(N,1);
% grid points
x=(0:N)*h;
y=x;
t=(0:M)*tau;
%initial condition
for k=2:N
for j=2:N
u(k,j,1)=gg(x(k),y(j));
end
end
% ADI method using the double-sweep method
for k=1:N-1
alpha(k+1)=gamma/(2*(1+gamma)-gamma*alpha(k));
end
for n=2:M+1
for j=2:N
for k=2:N
F=-0.5*gamma*(u(k,j-1,n-1)+u(k,j+1,n-1))-(1-gamma)*u(k,j,n-1)...
-0.25*tau*(ff(x(k),y(j),t(n))+ff(x(k),y(j),t(n-1)));
beta(k)=(beta(k-1)*gamma-2*F)/(2*(1+gamma)-gamma*alpha(k-1));
end
for k=N:(-1):2
u1(k,j)=alpha(k)*u1(k+1,j)+beta(k);
end
end
for k=2:N
for j=2:N
F=-0.5*gamma*(u1(k-1,j)+u1(k+1,j))-(1-gamma)*u1(k,j)...
-0.25*tau*(ff(x(k),y(j),t(n))+ff(x(k),y(j),t(n-1)));
beta(j)=(beta(j-1)*gamma-2*F)/(2*(1+gamma)-gamma*alpha(j-1));
end
for j=N:(-1):2
u(k,j,n)=alpha(j)*u(k,j+1,n)+beta(j);
end
end
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function f(x,y,t) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function f=ff(x,y,t)
f=-8*x*(x^2+y^2);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function g(x,y) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function g=gg(x,y)
g=y^2+2*x*y+x-y;
pde matlab
$endgroup$
add a comment |
$begingroup$
$$u_t = 0.25(u_xx+u_yy) -8x(x^2+y^2), qquad 0 leq x,y leq 1, qquad 0 leq t leq 0.5,$$
$$u(0,y,t) = y^2 - y, qquad u(1,y,t) = y^2 + y + 1,$$
$$u(x,0,t) = x, qquad u(x,1,t) = 3x,$$
$$u(x,y,0) = y^" + 2xy + x -y.$$
I don't know how to implement the nonzero boundary conditions or transform the problem into a problem with zero boundary conditions.
Here is my attempt:
function [u,x,y,t]=q2(T,K,N,M)
% solves the 2D heat equation u_t=K u_xx+ f(x,y,t) in the unit square
% and on the interval [0,T] with zero boundary condition by ADI method.
%K is the (heat conduction) coefficient in the heat equation.
%N is the number of subintervals in [0,1].
%M is the number of time steps.
h=1/N;
tau=T/M;
gamma=tau*K/h^2;
u=zeros(N+1,N+1,M+1);
u1=zeros(N+1,N+1);
alpha=zeros(N,1);
beta=zeros(N,1);
% grid points
x=(0:N)*h;
y=x;
t=(0:M)*tau;
%initial condition
for k=2:N
for j=2:N
u(k,j,1)=gg(x(k),y(j));
end
end
% ADI method using the double-sweep method
for k=1:N-1
alpha(k+1)=gamma/(2*(1+gamma)-gamma*alpha(k));
end
for n=2:M+1
for j=2:N
for k=2:N
F=-0.5*gamma*(u(k,j-1,n-1)+u(k,j+1,n-1))-(1-gamma)*u(k,j,n-1)...
-0.25*tau*(ff(x(k),y(j),t(n))+ff(x(k),y(j),t(n-1)));
beta(k)=(beta(k-1)*gamma-2*F)/(2*(1+gamma)-gamma*alpha(k-1));
end
for k=N:(-1):2
u1(k,j)=alpha(k)*u1(k+1,j)+beta(k);
end
end
for k=2:N
for j=2:N
F=-0.5*gamma*(u1(k-1,j)+u1(k+1,j))-(1-gamma)*u1(k,j)...
-0.25*tau*(ff(x(k),y(j),t(n))+ff(x(k),y(j),t(n-1)));
beta(j)=(beta(j-1)*gamma-2*F)/(2*(1+gamma)-gamma*alpha(j-1));
end
for j=N:(-1):2
u(k,j,n)=alpha(j)*u(k,j+1,n)+beta(j);
end
end
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function f(x,y,t) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function f=ff(x,y,t)
f=-8*x*(x^2+y^2);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function g(x,y) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function g=gg(x,y)
g=y^2+2*x*y+x-y;
pde matlab
$endgroup$
$begingroup$
Why is there a $y''$ in the initial condition?
$endgroup$
– Dylan
Mar 18 at 14:19
add a comment |
$begingroup$
$$u_t = 0.25(u_xx+u_yy) -8x(x^2+y^2), qquad 0 leq x,y leq 1, qquad 0 leq t leq 0.5,$$
$$u(0,y,t) = y^2 - y, qquad u(1,y,t) = y^2 + y + 1,$$
$$u(x,0,t) = x, qquad u(x,1,t) = 3x,$$
$$u(x,y,0) = y^" + 2xy + x -y.$$
I don't know how to implement the nonzero boundary conditions or transform the problem into a problem with zero boundary conditions.
Here is my attempt:
function [u,x,y,t]=q2(T,K,N,M)
% solves the 2D heat equation u_t=K u_xx+ f(x,y,t) in the unit square
% and on the interval [0,T] with zero boundary condition by ADI method.
%K is the (heat conduction) coefficient in the heat equation.
%N is the number of subintervals in [0,1].
%M is the number of time steps.
h=1/N;
tau=T/M;
gamma=tau*K/h^2;
u=zeros(N+1,N+1,M+1);
u1=zeros(N+1,N+1);
alpha=zeros(N,1);
beta=zeros(N,1);
% grid points
x=(0:N)*h;
y=x;
t=(0:M)*tau;
%initial condition
for k=2:N
for j=2:N
u(k,j,1)=gg(x(k),y(j));
end
end
% ADI method using the double-sweep method
for k=1:N-1
alpha(k+1)=gamma/(2*(1+gamma)-gamma*alpha(k));
end
for n=2:M+1
for j=2:N
for k=2:N
F=-0.5*gamma*(u(k,j-1,n-1)+u(k,j+1,n-1))-(1-gamma)*u(k,j,n-1)...
-0.25*tau*(ff(x(k),y(j),t(n))+ff(x(k),y(j),t(n-1)));
beta(k)=(beta(k-1)*gamma-2*F)/(2*(1+gamma)-gamma*alpha(k-1));
end
for k=N:(-1):2
u1(k,j)=alpha(k)*u1(k+1,j)+beta(k);
end
end
for k=2:N
for j=2:N
F=-0.5*gamma*(u1(k-1,j)+u1(k+1,j))-(1-gamma)*u1(k,j)...
-0.25*tau*(ff(x(k),y(j),t(n))+ff(x(k),y(j),t(n-1)));
beta(j)=(beta(j-1)*gamma-2*F)/(2*(1+gamma)-gamma*alpha(j-1));
end
for j=N:(-1):2
u(k,j,n)=alpha(j)*u(k,j+1,n)+beta(j);
end
end
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function f(x,y,t) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function f=ff(x,y,t)
f=-8*x*(x^2+y^2);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function g(x,y) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function g=gg(x,y)
g=y^2+2*x*y+x-y;
pde matlab
$endgroup$
$$u_t = 0.25(u_xx+u_yy) -8x(x^2+y^2), qquad 0 leq x,y leq 1, qquad 0 leq t leq 0.5,$$
$$u(0,y,t) = y^2 - y, qquad u(1,y,t) = y^2 + y + 1,$$
$$u(x,0,t) = x, qquad u(x,1,t) = 3x,$$
$$u(x,y,0) = y^" + 2xy + x -y.$$
I don't know how to implement the nonzero boundary conditions or transform the problem into a problem with zero boundary conditions.
Here is my attempt:
function [u,x,y,t]=q2(T,K,N,M)
% solves the 2D heat equation u_t=K u_xx+ f(x,y,t) in the unit square
% and on the interval [0,T] with zero boundary condition by ADI method.
%K is the (heat conduction) coefficient in the heat equation.
%N is the number of subintervals in [0,1].
%M is the number of time steps.
h=1/N;
tau=T/M;
gamma=tau*K/h^2;
u=zeros(N+1,N+1,M+1);
u1=zeros(N+1,N+1);
alpha=zeros(N,1);
beta=zeros(N,1);
% grid points
x=(0:N)*h;
y=x;
t=(0:M)*tau;
%initial condition
for k=2:N
for j=2:N
u(k,j,1)=gg(x(k),y(j));
end
end
% ADI method using the double-sweep method
for k=1:N-1
alpha(k+1)=gamma/(2*(1+gamma)-gamma*alpha(k));
end
for n=2:M+1
for j=2:N
for k=2:N
F=-0.5*gamma*(u(k,j-1,n-1)+u(k,j+1,n-1))-(1-gamma)*u(k,j,n-1)...
-0.25*tau*(ff(x(k),y(j),t(n))+ff(x(k),y(j),t(n-1)));
beta(k)=(beta(k-1)*gamma-2*F)/(2*(1+gamma)-gamma*alpha(k-1));
end
for k=N:(-1):2
u1(k,j)=alpha(k)*u1(k+1,j)+beta(k);
end
end
for k=2:N
for j=2:N
F=-0.5*gamma*(u1(k-1,j)+u1(k+1,j))-(1-gamma)*u1(k,j)...
-0.25*tau*(ff(x(k),y(j),t(n))+ff(x(k),y(j),t(n-1)));
beta(j)=(beta(j-1)*gamma-2*F)/(2*(1+gamma)-gamma*alpha(j-1));
end
for j=N:(-1):2
u(k,j,n)=alpha(j)*u(k,j+1,n)+beta(j);
end
end
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function f(x,y,t) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function f=ff(x,y,t)
f=-8*x*(x^2+y^2);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% function g(x,y) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function g=gg(x,y)
g=y^2+2*x*y+x-y;
pde matlab
pde matlab
edited Mar 17 at 13:30
user572780
asked Mar 17 at 13:23
user572780user572780
335
335
$begingroup$
Why is there a $y''$ in the initial condition?
$endgroup$
– Dylan
Mar 18 at 14:19
add a comment |
$begingroup$
Why is there a $y''$ in the initial condition?
$endgroup$
– Dylan
Mar 18 at 14:19
$begingroup$
Why is there a $y''$ in the initial condition?
$endgroup$
– Dylan
Mar 18 at 14:19
$begingroup$
Why is there a $y''$ in the initial condition?
$endgroup$
– Dylan
Mar 18 at 14:19
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For the general boundary conditions
beginalign
u(0,y,t) &= a_1(y) \
u(1,y,t) &= b_1(y) \
u(x,0,t) &= a_2(x) \
u(x,1,t) &= b_2(x)
endalign
You can write $u(x,y,t) = w_1(x,y) + w_2(x,y) + v(x,y,t)$ such that $v(x,y,t)$ is homogeneous and
beginarrayll
beginaligned
w_1(0,y) &= a_1(y) \
w_1(1,y) &= a_2(y)
endaligned &&&
beginaligned
w_2(x,0) &= a_2(x) - w_1(x,0) \
w_2(x,1) &= b_2(x) - w_1(x,1)
endaligned
endarray
The standard approach is to set
beginalign
w_1(x,y) &= (1-x)a_1(y) + xb_1(y) \
w_2(x,y) &= (1-y)[a_2(x) - (1-x)a_1(0) - xb_1(0)] + y[b_2(x) - (1-x)a_1(1) - xb_1(1)]
endalign
For this specific boundary problem, the simplified solution is
$$ w(x,y) = w_1(x,y) + w_2(x,y) = y^2 - y + x + 2xy $$
Note that this approach works only when the boundary is continuous at the "sharp points", i.e. $a_1(0)=a_2(0)$, $b_1(0)=a_2(1)$, $a_1(1)=b_2(0)$, $b_1(1)=b_2(1)$
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For the general boundary conditions
beginalign
u(0,y,t) &= a_1(y) \
u(1,y,t) &= b_1(y) \
u(x,0,t) &= a_2(x) \
u(x,1,t) &= b_2(x)
endalign
You can write $u(x,y,t) = w_1(x,y) + w_2(x,y) + v(x,y,t)$ such that $v(x,y,t)$ is homogeneous and
beginarrayll
beginaligned
w_1(0,y) &= a_1(y) \
w_1(1,y) &= a_2(y)
endaligned &&&
beginaligned
w_2(x,0) &= a_2(x) - w_1(x,0) \
w_2(x,1) &= b_2(x) - w_1(x,1)
endaligned
endarray
The standard approach is to set
beginalign
w_1(x,y) &= (1-x)a_1(y) + xb_1(y) \
w_2(x,y) &= (1-y)[a_2(x) - (1-x)a_1(0) - xb_1(0)] + y[b_2(x) - (1-x)a_1(1) - xb_1(1)]
endalign
For this specific boundary problem, the simplified solution is
$$ w(x,y) = w_1(x,y) + w_2(x,y) = y^2 - y + x + 2xy $$
Note that this approach works only when the boundary is continuous at the "sharp points", i.e. $a_1(0)=a_2(0)$, $b_1(0)=a_2(1)$, $a_1(1)=b_2(0)$, $b_1(1)=b_2(1)$
$endgroup$
add a comment |
$begingroup$
For the general boundary conditions
beginalign
u(0,y,t) &= a_1(y) \
u(1,y,t) &= b_1(y) \
u(x,0,t) &= a_2(x) \
u(x,1,t) &= b_2(x)
endalign
You can write $u(x,y,t) = w_1(x,y) + w_2(x,y) + v(x,y,t)$ such that $v(x,y,t)$ is homogeneous and
beginarrayll
beginaligned
w_1(0,y) &= a_1(y) \
w_1(1,y) &= a_2(y)
endaligned &&&
beginaligned
w_2(x,0) &= a_2(x) - w_1(x,0) \
w_2(x,1) &= b_2(x) - w_1(x,1)
endaligned
endarray
The standard approach is to set
beginalign
w_1(x,y) &= (1-x)a_1(y) + xb_1(y) \
w_2(x,y) &= (1-y)[a_2(x) - (1-x)a_1(0) - xb_1(0)] + y[b_2(x) - (1-x)a_1(1) - xb_1(1)]
endalign
For this specific boundary problem, the simplified solution is
$$ w(x,y) = w_1(x,y) + w_2(x,y) = y^2 - y + x + 2xy $$
Note that this approach works only when the boundary is continuous at the "sharp points", i.e. $a_1(0)=a_2(0)$, $b_1(0)=a_2(1)$, $a_1(1)=b_2(0)$, $b_1(1)=b_2(1)$
$endgroup$
add a comment |
$begingroup$
For the general boundary conditions
beginalign
u(0,y,t) &= a_1(y) \
u(1,y,t) &= b_1(y) \
u(x,0,t) &= a_2(x) \
u(x,1,t) &= b_2(x)
endalign
You can write $u(x,y,t) = w_1(x,y) + w_2(x,y) + v(x,y,t)$ such that $v(x,y,t)$ is homogeneous and
beginarrayll
beginaligned
w_1(0,y) &= a_1(y) \
w_1(1,y) &= a_2(y)
endaligned &&&
beginaligned
w_2(x,0) &= a_2(x) - w_1(x,0) \
w_2(x,1) &= b_2(x) - w_1(x,1)
endaligned
endarray
The standard approach is to set
beginalign
w_1(x,y) &= (1-x)a_1(y) + xb_1(y) \
w_2(x,y) &= (1-y)[a_2(x) - (1-x)a_1(0) - xb_1(0)] + y[b_2(x) - (1-x)a_1(1) - xb_1(1)]
endalign
For this specific boundary problem, the simplified solution is
$$ w(x,y) = w_1(x,y) + w_2(x,y) = y^2 - y + x + 2xy $$
Note that this approach works only when the boundary is continuous at the "sharp points", i.e. $a_1(0)=a_2(0)$, $b_1(0)=a_2(1)$, $a_1(1)=b_2(0)$, $b_1(1)=b_2(1)$
$endgroup$
For the general boundary conditions
beginalign
u(0,y,t) &= a_1(y) \
u(1,y,t) &= b_1(y) \
u(x,0,t) &= a_2(x) \
u(x,1,t) &= b_2(x)
endalign
You can write $u(x,y,t) = w_1(x,y) + w_2(x,y) + v(x,y,t)$ such that $v(x,y,t)$ is homogeneous and
beginarrayll
beginaligned
w_1(0,y) &= a_1(y) \
w_1(1,y) &= a_2(y)
endaligned &&&
beginaligned
w_2(x,0) &= a_2(x) - w_1(x,0) \
w_2(x,1) &= b_2(x) - w_1(x,1)
endaligned
endarray
The standard approach is to set
beginalign
w_1(x,y) &= (1-x)a_1(y) + xb_1(y) \
w_2(x,y) &= (1-y)[a_2(x) - (1-x)a_1(0) - xb_1(0)] + y[b_2(x) - (1-x)a_1(1) - xb_1(1)]
endalign
For this specific boundary problem, the simplified solution is
$$ w(x,y) = w_1(x,y) + w_2(x,y) = y^2 - y + x + 2xy $$
Note that this approach works only when the boundary is continuous at the "sharp points", i.e. $a_1(0)=a_2(0)$, $b_1(0)=a_2(1)$, $a_1(1)=b_2(0)$, $b_1(1)=b_2(1)$
answered Mar 19 at 9:07
DylanDylan
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$begingroup$
Why is there a $y''$ in the initial condition?
$endgroup$
– Dylan
Mar 18 at 14:19