Mixing
Solve advection equation $v_t + v_x = 1$ numerically with Matlab
$begingroup$
Consider the advection equation
$$ v_t + v_x = 1 $$
with initial condition
$$ v(x,0) = begin{cases} sin^2 pi (x-1), & x in [1,2] \ 0, & text{otherwise} end{cases}$$
Clearly, we know that for any $F$, the general solution is
$$ v(x,t) = F(x - s t) $$ and $v(x,0) = F(x) = sin^2 pi (x-1)$. Therefore, the solution we are looking for is
$$ v(x,t) = sin^2 pi (x-1-st) $$
where $s$ is constant.
My question is how do we implement the solution numerically in matlab? Numerically, we can discretize the PDE using the following scheme Lax
$$ frac{ u_j^{n+1} - frac{1}{2}( u_{j+1}^n + u_{j-1}^n) }{Delta t} + frac{ u_{j+1}^n - u_{j-1}^n }{2 Delta x} =0 $$
say for $x in [0,6]$ and $t in [0,4]$
pde numerical-methods matlab hyperbolic-equations
edited Jan 13 at 16:32
Harry49
6,40231132
asked Jan 12 at 22:37
Jimmy SabaterJimmy Sabater
2,605322
$endgroup$
add a comment |
$begingroup$
Consider the advection equation
$$ v_t + v_x = 1 $$
with initial condition
$$ v(x,0) = begin{cases} sin^2 pi (x-1), & x in [1,2] \ 0, & text{otherwise} end{cases}$$
Clearly, we know that for any $F$, the general solution is
$$ v(x,t) = F(x - s t) $$ and $v(x,0) = F(x) = sin^2 pi (x-1)$. Therefore, the solution we are looking for is
$$ v(x,t) = sin^2 pi (x-1-st) $$
where $s$ is constant.
My question is how do we implement the solution numerically in matlab? Numerically, we can discretize the PDE using the following scheme Lax
$$ frac{ u_j^{n+1} - frac{1}{2}( u_{j+1}^n + u_{j-1}^n) }{Delta t} + frac{ u_{j+1}^n - u_{j-1}^n }{2 Delta x} =0 $$
say for $x in [0,6]$ and $t in [0,4]$
pde numerical-methods matlab hyperbolic-equations
edited Jan 13 at 16:32
Harry49
6,40231132
asked Jan 12 at 22:37
Jimmy SabaterJimmy Sabater
2,605322
$endgroup$
$begingroup$
What is the domain of the problem? $xin cdots$
$endgroup$
– caverac
Jan 12 at 23:43
add a comment |
$begingroup$
Consider the advection equation
$$ v_t + v_x = 1 $$
with initial condition
$$ v(x,0) = begin{cases} sin^2 pi (x-1), & x in [1,2] \ 0, & text{otherwise} end{cases}$$
Clearly, we know that for any $F$, the general solution is
$$ v(x,t) = F(x - s t) $$ and $v(x,0) = F(x) = sin^2 pi (x-1)$. Therefore, the solution we are looking for is
$$ v(x,t) = sin^2 pi (x-1-st) $$
where $s$ is constant.
My question is how do we implement the solution numerically in matlab? Numerically, we can discretize the PDE using the following scheme Lax
$$ frac{ u_j^{n+1} - frac{1}{2}( u_{j+1}^n + u_{j-1}^n) }{Delta t} + frac{ u_{j+1}^n - u_{j-1}^n }{2 Delta x} =0 $$
say for $x in [0,6]$ and $t in [0,4]$
pde numerical-methods matlab hyperbolic-equations
edited Jan 13 at 16:32
Harry49
6,40231132
asked Jan 12 at 22:37
Jimmy SabaterJimmy Sabater
2,605322
$endgroup$
Consider the advection equation
$$ v_t + v_x = 1 $$
with initial condition
$$ v(x,0) = begin{cases} sin^2 pi (x-1), & x in [1,2] \ 0, & text{otherwise} end{cases}$$
Clearly, we know that for any $F$, the general solution is
$$ v(x,t) = F(x - s t) $$ and $v(x,0) = F(x) = sin^2 pi (x-1)$. Therefore, the solution we are looking for is
$$ v(x,t) = sin^2 pi (x-1-st) $$
where $s$ is constant.
My question is how do we implement the solution numerically in matlab? Numerically, we can discretize the PDE using the following scheme Lax
$$ frac{ u_j^{n+1} - frac{1}{2}( u_{j+1}^n + u_{j-1}^n) }{Delta t} + frac{ u_{j+1}^n - u_{j-1}^n }{2 Delta x} =0 $$
say for $x in [0,6]$ and $t in [0,4]$
pde numerical-methods matlab hyperbolic-equations
pde numerical-methods matlab hyperbolic-equations
edited Jan 13 at 16:32
Harry49
6,40231132
asked Jan 12 at 22:37
Jimmy SabaterJimmy Sabater
2,605322
edited Jan 13 at 16:32
Harry49
6,40231132
asked Jan 12 at 22:37
Jimmy SabaterJimmy Sabater
2,605322
edited Jan 13 at 16:32
Harry49
6,40231132
edited Jan 13 at 16:32
Harry49
6,40231132
edited Jan 13 at 16:32
Harry49
6,40231132
6,40231132
asked Jan 12 at 22:37
Jimmy SabaterJimmy Sabater
2,605322
asked Jan 12 at 22:37
Jimmy SabaterJimmy Sabater
2,605322
asked Jan 12 at 22:37
Jimmy SabaterJimmy Sabater
2,605322
2,605322
$begingroup$
What is the domain of the problem? $xin cdots$
$endgroup$
– caverac
Jan 12 at 23:43
add a comment |
$begingroup$
What is the domain of the problem? $xin cdots$
$endgroup$
– caverac
Jan 12 at 23:43
$begingroup$
What is the domain of the problem? $xin cdots$
$endgroup$
– caverac
Jan 12 at 23:43
$begingroup$
What is the domain of the problem? $xin cdots$
$endgroup$
– caverac
Jan 12 at 23:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For $v_t+v_x=1$, the solution to the Cauchy problem $v(x,0)=F(x)$ obtained with the method of characteristics is
$$
v(x,t) = F(x-t) +t .
$$
The Lax-Friedrichs method reads
$$
frac{v_i^{n+1}-frac{1}{2}(v_{i-1}^{n}+v_{i+1}^{n})}{Delta t} + frac{v_{i+1}^{n}-v_{i-1}^{n}}{2 Delta x} = 1
$$
where $v_i^n simeq v(iDelta x, nDelta t)$. This method is stable for small time steps according to the Courant-Friedrichs-Lewy condition $Delta t < Delta x$.
Now, we only need to translate the previous algorithm into MATLAB syntax.
%% Initialisation
F = @(x) sin(pi*(x-1)).^2 .* (1<x).*(x<2);
n = 100;
x = linspace(0,6,n);
dx = 6/(n-1);
t = 0;
dt = 0.95*dx;
v = F(x);
figure;
plot(x,v,'k-');
%% Scheme iterations
while t<4
v(2:n-1) = 0.5*((1+dt/dx)*v(1:n-2) + (1-dt/dx)*v(3:n)) + dt;
v(1) = v(2);
v(n) = v(n-1);
t = t + dt;
end
%% Output
plot(x,v,'bo');
hold on
plot(x,F(x-t)+t,'k-');
answered Jan 13 at 12:13
Harry49Harry49
6,40231132
$endgroup$
$begingroup$
It would be more idiomatic to replace the inner loop with a vectorized operationvtemp(2:n-1) = 0.5*(v(1:n-2)+v(3:n)) - 0.5*dt/dx*(v(3:n)-v(1:n-2)) + dt;
$endgroup$
– LutzL
Jan 13 at 12:56
$begingroup$
@LutzL Thanks for the suggestion. We can even factorize to avoid multiple vector extractions and remove the temporary data vectorvtemp.
$endgroup$
– Harry49
Jan 13 at 16:31
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 $v_t+v_x=1$, the solution to the Cauchy problem $v(x,0)=F(x)$ obtained with the method of characteristics is
$$
v(x,t) = F(x-t) +t .
$$
The Lax-Friedrichs method reads
$$
frac{v_i^{n+1}-frac{1}{2}(v_{i-1}^{n}+v_{i+1}^{n})}{Delta t} + frac{v_{i+1}^{n}-v_{i-1}^{n}}{2 Delta x} = 1
$$
where $v_i^n simeq v(iDelta x, nDelta t)$. This method is stable for small time steps according to the Courant-Friedrichs-Lewy condition $Delta t < Delta x$.
Now, we only need to translate the previous algorithm into MATLAB syntax.
%% Initialisation
F = @(x) sin(pi*(x-1)).^2 .* (1<x).*(x<2);
n = 100;
x = linspace(0,6,n);
dx = 6/(n-1);
t = 0;
dt = 0.95*dx;
v = F(x);
figure;
plot(x,v,'k-');
%% Scheme iterations
while t<4
v(2:n-1) = 0.5*((1+dt/dx)*v(1:n-2) + (1-dt/dx)*v(3:n)) + dt;
v(1) = v(2);
v(n) = v(n-1);
t = t + dt;
end
%% Output
plot(x,v,'bo');
hold on
plot(x,F(x-t)+t,'k-');
answered Jan 13 at 12:13
Harry49Harry49
6,40231132
$endgroup$
$begingroup$
It would be more idiomatic to replace the inner loop with a vectorized operationvtemp(2:n-1) = 0.5*(v(1:n-2)+v(3:n)) - 0.5*dt/dx*(v(3:n)-v(1:n-2)) + dt;
$endgroup$
– LutzL
Jan 13 at 12:56
$begingroup$
@LutzL Thanks for the suggestion. We can even factorize to avoid multiple vector extractions and remove the temporary data vectorvtemp.
$endgroup$
– Harry49
Jan 13 at 16:31
add a comment |
$begingroup$
For $v_t+v_x=1$, the solution to the Cauchy problem $v(x,0)=F(x)$ obtained with the method of characteristics is
$$
v(x,t) = F(x-t) +t .
$$
The Lax-Friedrichs method reads
$$
frac{v_i^{n+1}-frac{1}{2}(v_{i-1}^{n}+v_{i+1}^{n})}{Delta t} + frac{v_{i+1}^{n}-v_{i-1}^{n}}{2 Delta x} = 1
$$
where $v_i^n simeq v(iDelta x, nDelta t)$. This method is stable for small time steps according to the Courant-Friedrichs-Lewy condition $Delta t < Delta x$.
Now, we only need to translate the previous algorithm into MATLAB syntax.
%% Initialisation
F = @(x) sin(pi*(x-1)).^2 .* (1<x).*(x<2);
n = 100;
x = linspace(0,6,n);
dx = 6/(n-1);
t = 0;
dt = 0.95*dx;
v = F(x);
figure;
plot(x,v,'k-');
%% Scheme iterations
while t<4
v(2:n-1) = 0.5*((1+dt/dx)*v(1:n-2) + (1-dt/dx)*v(3:n)) + dt;
v(1) = v(2);
v(n) = v(n-1);
t = t + dt;
end
%% Output
plot(x,v,'bo');
hold on
plot(x,F(x-t)+t,'k-');
answered Jan 13 at 12:13
Harry49Harry49
6,40231132
$endgroup$
$begingroup$
It would be more idiomatic to replace the inner loop with a vectorized operationvtemp(2:n-1) = 0.5*(v(1:n-2)+v(3:n)) - 0.5*dt/dx*(v(3:n)-v(1:n-2)) + dt;
$endgroup$
– LutzL
Jan 13 at 12:56
$begingroup$
@LutzL Thanks for the suggestion. We can even factorize to avoid multiple vector extractions and remove the temporary data vectorvtemp.
$endgroup$
– Harry49
Jan 13 at 16:31
add a comment |
$begingroup$
For $v_t+v_x=1$, the solution to the Cauchy problem $v(x,0)=F(x)$ obtained with the method of characteristics is
$$
v(x,t) = F(x-t) +t .
$$
The Lax-Friedrichs method reads
$$
frac{v_i^{n+1}-frac{1}{2}(v_{i-1}^{n}+v_{i+1}^{n})}{Delta t} + frac{v_{i+1}^{n}-v_{i-1}^{n}}{2 Delta x} = 1
$$
where $v_i^n simeq v(iDelta x, nDelta t)$. This method is stable for small time steps according to the Courant-Friedrichs-Lewy condition $Delta t < Delta x$.
Now, we only need to translate the previous algorithm into MATLAB syntax.
%% Initialisation
F = @(x) sin(pi*(x-1)).^2 .* (1<x).*(x<2);
n = 100;
x = linspace(0,6,n);
dx = 6/(n-1);
t = 0;
dt = 0.95*dx;
v = F(x);
figure;
plot(x,v,'k-');
%% Scheme iterations
while t<4
v(2:n-1) = 0.5*((1+dt/dx)*v(1:n-2) + (1-dt/dx)*v(3:n)) + dt;
v(1) = v(2);
v(n) = v(n-1);
t = t + dt;
end
%% Output
plot(x,v,'bo');
hold on
plot(x,F(x-t)+t,'k-');
answered Jan 13 at 12:13
Harry49Harry49
6,40231132
$endgroup$
For $v_t+v_x=1$, the solution to the Cauchy problem $v(x,0)=F(x)$ obtained with the method of characteristics is
$$
v(x,t) = F(x-t) +t .
$$
The Lax-Friedrichs method reads
$$
frac{v_i^{n+1}-frac{1}{2}(v_{i-1}^{n}+v_{i+1}^{n})}{Delta t} + frac{v_{i+1}^{n}-v_{i-1}^{n}}{2 Delta x} = 1
$$
where $v_i^n simeq v(iDelta x, nDelta t)$. This method is stable for small time steps according to the Courant-Friedrichs-Lewy condition $Delta t < Delta x$.
Now, we only need to translate the previous algorithm into MATLAB syntax.
%% Initialisation
F = @(x) sin(pi*(x-1)).^2 .* (1<x).*(x<2);
n = 100;
x = linspace(0,6,n);
dx = 6/(n-1);
t = 0;
dt = 0.95*dx;
v = F(x);
figure;
plot(x,v,'k-');
%% Scheme iterations
while t<4
v(2:n-1) = 0.5*((1+dt/dx)*v(1:n-2) + (1-dt/dx)*v(3:n)) + dt;
v(1) = v(2);
v(n) = v(n-1);
t = t + dt;
end
%% Output
plot(x,v,'bo');
hold on
plot(x,F(x-t)+t,'k-');
answered Jan 13 at 12:13
Harry49Harry49
6,40231132
edited Jan 13 at 16:38
answered Jan 13 at 12:13
Harry49Harry49
6,40231132
answered Jan 13 at 12:13
Harry49Harry49
6,40231132
answered Jan 13 at 12:13
Harry49Harry49
6,40231132
6,40231132
$begingroup$
It would be more idiomatic to replace the inner loop with a vectorized operationvtemp(2:n-1) = 0.5*(v(1:n-2)+v(3:n)) - 0.5*dt/dx*(v(3:n)-v(1:n-2)) + dt;
$endgroup$
– LutzL
Jan 13 at 12:56
$begingroup$
@LutzL Thanks for the suggestion. We can even factorize to avoid multiple vector extractions and remove the temporary data vectorvtemp.
$endgroup$
– Harry49
Jan 13 at 16:31
add a comment |
$begingroup$
It would be more idiomatic to replace the inner loop with a vectorized operationvtemp(2:n-1) = 0.5*(v(1:n-2)+v(3:n)) - 0.5*dt/dx*(v(3:n)-v(1:n-2)) + dt;
$endgroup$
– LutzL
Jan 13 at 12:56
$begingroup$
@LutzL Thanks for the suggestion. We can even factorize to avoid multiple vector extractions and remove the temporary data vectorvtemp.
$endgroup$
– Harry49
Jan 13 at 16:31
$begingroup$
It would be more idiomatic to replace the inner loop with a vectorized operation
vtemp(2:n-1) = 0.5*(v(1:n-2)+v(3:n)) - 0.5*dt/dx*(v(3:n)-v(1:n-2)) + dt;$endgroup$
– LutzL
Jan 13 at 12:56
$begingroup$
It would be more idiomatic to replace the inner loop with a vectorized operation
vtemp(2:n-1) = 0.5*(v(1:n-2)+v(3:n)) - 0.5*dt/dx*(v(3:n)-v(1:n-2)) + dt;$endgroup$
– LutzL
Jan 13 at 12:56
$begingroup$
@LutzL Thanks for the suggestion. We can even factorize to avoid multiple vector extractions and remove the temporary data vector
vtemp.$endgroup$
– Harry49
Jan 13 at 16:31
$begingroup$
@LutzL Thanks for the suggestion. We can even factorize to avoid multiple vector extractions and remove the temporary data vector
vtemp.$endgroup$
– Harry49
Jan 13 at 16:31
add a comment |
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$begingroup$
What is the domain of the problem? $xin cdots$
$endgroup$
– caverac
Jan 12 at 23:43