Help solving very complex first order ODEs using ode45 - MATLAB - movement of water
I am trying to solve this complex set of first order ODEs using ode45 on MATLAB. They describe the change in volume of water in the inner layer ($X_I$) and outer layer ($X_O$) of the Venus Flytrap's upper leaf (trap). This movement of water is what allows the trap to snap shut. I am looking to solve the ODEs for time for different parts of the prey capture process (which will have different conditions - eg. for the closing process, where the trap moves from an open to semi-closed state, we have $u_a$ and $u_c$ equal to zero.) These u terms are water transport rates driven by various gradients, $alpha$ is the water supply rate from the roots and $mu$ is the water consumption rate.
$dX_O/dt = frac{alpha X_O^2}{X_O^2 + X_I^2} + u_h + u_a - u_c -mu X_O$
$dX_I/dt = frac{alpha X_I^2}{X_O^2 + X_I^2} - u_h - u_a + u_c -mu X_I $
Where:
$X_O + X_I = 1$
$k_t(s^-1) = 10$
$k_f(s^-1) = 2$
$k_d(s^-1) = 0.000045$
$alpha (h^-1) = 1$
$mu (h^-1) = 1$
and
$u_h = k_t max{X_I(t) - X_O(t),0}delta _t(t)$
$u_a =left{
begin{array}{@{}ll@{}}
k_f X_I(t) delta _t(t), & text{if} C geq 14 \
0, & text{if} C<14 end{array}right.$
where $delta_ t(t) = left{
begin{array}{@{}ll@{}}
1, & text{if} 0 leq t leq 0.3 \
0, & text{if} t>0.3 end{array}right.$
$u_c(t) = k_d delta _c(t)$
where $delta _c(t) =left{
begin{array}{@{}ll@{}}
1, & text{if} T_{start} leq t leq T_{start}+T_D (T_D=15 hours) \
0, & text{otherwise} end{array}right.$
So far I have attempted to create a script file which defines the variables:
a = 3600;
mu = 3600;
Xo = 1 - Xi;
if (0 <= t1) && (t1 <= 0.3)
Dt1 = 1;
else
Dt1 = 0;
end
Uh = 10*max((Xo - Xi),0)*d;
if C >= 14
Ua = 2*Xi*Dt1;
else
Ua = 0;
end
if (0 <= t2) && (t2 <= 54000)
Dc = 1;
else
Dc = 0;
end
Uc = 0.000045*Dc;
where a represents $alpha$ and Dt and Dc represent $delta _t$ and $delta _c$ respectively. Note that I converted the values of $alpha$ and $mu$ into seconds.
I created two times, t1 and t2 so that I could put $T_{start}$ as $0$ for $delta _c(t)$.
If anyone could help me I would be extremely grateful, I've spent about 15 hours on this now and am still getting nowhere, despite my numerous searches online for hints.
differential-equations matlab
add a comment |
I am trying to solve this complex set of first order ODEs using ode45 on MATLAB. They describe the change in volume of water in the inner layer ($X_I$) and outer layer ($X_O$) of the Venus Flytrap's upper leaf (trap). This movement of water is what allows the trap to snap shut. I am looking to solve the ODEs for time for different parts of the prey capture process (which will have different conditions - eg. for the closing process, where the trap moves from an open to semi-closed state, we have $u_a$ and $u_c$ equal to zero.) These u terms are water transport rates driven by various gradients, $alpha$ is the water supply rate from the roots and $mu$ is the water consumption rate.
$dX_O/dt = frac{alpha X_O^2}{X_O^2 + X_I^2} + u_h + u_a - u_c -mu X_O$
$dX_I/dt = frac{alpha X_I^2}{X_O^2 + X_I^2} - u_h - u_a + u_c -mu X_I $
Where:
$X_O + X_I = 1$
$k_t(s^-1) = 10$
$k_f(s^-1) = 2$
$k_d(s^-1) = 0.000045$
$alpha (h^-1) = 1$
$mu (h^-1) = 1$
and
$u_h = k_t max{X_I(t) - X_O(t),0}delta _t(t)$
$u_a =left{
begin{array}{@{}ll@{}}
k_f X_I(t) delta _t(t), & text{if} C geq 14 \
0, & text{if} C<14 end{array}right.$
where $delta_ t(t) = left{
begin{array}{@{}ll@{}}
1, & text{if} 0 leq t leq 0.3 \
0, & text{if} t>0.3 end{array}right.$
$u_c(t) = k_d delta _c(t)$
where $delta _c(t) =left{
begin{array}{@{}ll@{}}
1, & text{if} T_{start} leq t leq T_{start}+T_D (T_D=15 hours) \
0, & text{otherwise} end{array}right.$
So far I have attempted to create a script file which defines the variables:
a = 3600;
mu = 3600;
Xo = 1 - Xi;
if (0 <= t1) && (t1 <= 0.3)
Dt1 = 1;
else
Dt1 = 0;
end
Uh = 10*max((Xo - Xi),0)*d;
if C >= 14
Ua = 2*Xi*Dt1;
else
Ua = 0;
end
if (0 <= t2) && (t2 <= 54000)
Dc = 1;
else
Dc = 0;
end
Uc = 0.000045*Dc;
where a represents $alpha$ and Dt and Dc represent $delta _t$ and $delta _c$ respectively. Note that I converted the values of $alpha$ and $mu$ into seconds.
I created two times, t1 and t2 so that I could put $T_{start}$ as $0$ for $delta _c(t)$.
If anyone could help me I would be extremely grateful, I've spent about 15 hours on this now and am still getting nowhere, despite my numerous searches online for hints.
differential-equations matlab
What do you mean by $k_t(s^-1)$? Does it perhaps mean that the dimensions of $k_t$ are $s^{-1}$?
– Fabio Somenzi
Nov 28 at 18:18
It's some kind of rate so 'per second'
– maria1991
Dec 1 at 17:03
add a comment |
I am trying to solve this complex set of first order ODEs using ode45 on MATLAB. They describe the change in volume of water in the inner layer ($X_I$) and outer layer ($X_O$) of the Venus Flytrap's upper leaf (trap). This movement of water is what allows the trap to snap shut. I am looking to solve the ODEs for time for different parts of the prey capture process (which will have different conditions - eg. for the closing process, where the trap moves from an open to semi-closed state, we have $u_a$ and $u_c$ equal to zero.) These u terms are water transport rates driven by various gradients, $alpha$ is the water supply rate from the roots and $mu$ is the water consumption rate.
$dX_O/dt = frac{alpha X_O^2}{X_O^2 + X_I^2} + u_h + u_a - u_c -mu X_O$
$dX_I/dt = frac{alpha X_I^2}{X_O^2 + X_I^2} - u_h - u_a + u_c -mu X_I $
Where:
$X_O + X_I = 1$
$k_t(s^-1) = 10$
$k_f(s^-1) = 2$
$k_d(s^-1) = 0.000045$
$alpha (h^-1) = 1$
$mu (h^-1) = 1$
and
$u_h = k_t max{X_I(t) - X_O(t),0}delta _t(t)$
$u_a =left{
begin{array}{@{}ll@{}}
k_f X_I(t) delta _t(t), & text{if} C geq 14 \
0, & text{if} C<14 end{array}right.$
where $delta_ t(t) = left{
begin{array}{@{}ll@{}}
1, & text{if} 0 leq t leq 0.3 \
0, & text{if} t>0.3 end{array}right.$
$u_c(t) = k_d delta _c(t)$
where $delta _c(t) =left{
begin{array}{@{}ll@{}}
1, & text{if} T_{start} leq t leq T_{start}+T_D (T_D=15 hours) \
0, & text{otherwise} end{array}right.$
So far I have attempted to create a script file which defines the variables:
a = 3600;
mu = 3600;
Xo = 1 - Xi;
if (0 <= t1) && (t1 <= 0.3)
Dt1 = 1;
else
Dt1 = 0;
end
Uh = 10*max((Xo - Xi),0)*d;
if C >= 14
Ua = 2*Xi*Dt1;
else
Ua = 0;
end
if (0 <= t2) && (t2 <= 54000)
Dc = 1;
else
Dc = 0;
end
Uc = 0.000045*Dc;
where a represents $alpha$ and Dt and Dc represent $delta _t$ and $delta _c$ respectively. Note that I converted the values of $alpha$ and $mu$ into seconds.
I created two times, t1 and t2 so that I could put $T_{start}$ as $0$ for $delta _c(t)$.
If anyone could help me I would be extremely grateful, I've spent about 15 hours on this now and am still getting nowhere, despite my numerous searches online for hints.
differential-equations matlab
I am trying to solve this complex set of first order ODEs using ode45 on MATLAB. They describe the change in volume of water in the inner layer ($X_I$) and outer layer ($X_O$) of the Venus Flytrap's upper leaf (trap). This movement of water is what allows the trap to snap shut. I am looking to solve the ODEs for time for different parts of the prey capture process (which will have different conditions - eg. for the closing process, where the trap moves from an open to semi-closed state, we have $u_a$ and $u_c$ equal to zero.) These u terms are water transport rates driven by various gradients, $alpha$ is the water supply rate from the roots and $mu$ is the water consumption rate.
$dX_O/dt = frac{alpha X_O^2}{X_O^2 + X_I^2} + u_h + u_a - u_c -mu X_O$
$dX_I/dt = frac{alpha X_I^2}{X_O^2 + X_I^2} - u_h - u_a + u_c -mu X_I $
Where:
$X_O + X_I = 1$
$k_t(s^-1) = 10$
$k_f(s^-1) = 2$
$k_d(s^-1) = 0.000045$
$alpha (h^-1) = 1$
$mu (h^-1) = 1$
and
$u_h = k_t max{X_I(t) - X_O(t),0}delta _t(t)$
$u_a =left{
begin{array}{@{}ll@{}}
k_f X_I(t) delta _t(t), & text{if} C geq 14 \
0, & text{if} C<14 end{array}right.$
where $delta_ t(t) = left{
begin{array}{@{}ll@{}}
1, & text{if} 0 leq t leq 0.3 \
0, & text{if} t>0.3 end{array}right.$
$u_c(t) = k_d delta _c(t)$
where $delta _c(t) =left{
begin{array}{@{}ll@{}}
1, & text{if} T_{start} leq t leq T_{start}+T_D (T_D=15 hours) \
0, & text{otherwise} end{array}right.$
So far I have attempted to create a script file which defines the variables:
a = 3600;
mu = 3600;
Xo = 1 - Xi;
if (0 <= t1) && (t1 <= 0.3)
Dt1 = 1;
else
Dt1 = 0;
end
Uh = 10*max((Xo - Xi),0)*d;
if C >= 14
Ua = 2*Xi*Dt1;
else
Ua = 0;
end
if (0 <= t2) && (t2 <= 54000)
Dc = 1;
else
Dc = 0;
end
Uc = 0.000045*Dc;
where a represents $alpha$ and Dt and Dc represent $delta _t$ and $delta _c$ respectively. Note that I converted the values of $alpha$ and $mu$ into seconds.
I created two times, t1 and t2 so that I could put $T_{start}$ as $0$ for $delta _c(t)$.
If anyone could help me I would be extremely grateful, I've spent about 15 hours on this now and am still getting nowhere, despite my numerous searches online for hints.
differential-equations matlab
differential-equations matlab
edited Nov 28 at 17:53
asked Nov 17 at 16:11
maria1991
104
104
What do you mean by $k_t(s^-1)$? Does it perhaps mean that the dimensions of $k_t$ are $s^{-1}$?
– Fabio Somenzi
Nov 28 at 18:18
It's some kind of rate so 'per second'
– maria1991
Dec 1 at 17:03
add a comment |
What do you mean by $k_t(s^-1)$? Does it perhaps mean that the dimensions of $k_t$ are $s^{-1}$?
– Fabio Somenzi
Nov 28 at 18:18
It's some kind of rate so 'per second'
– maria1991
Dec 1 at 17:03
What do you mean by $k_t(s^-1)$? Does it perhaps mean that the dimensions of $k_t$ are $s^{-1}$?
– Fabio Somenzi
Nov 28 at 18:18
What do you mean by $k_t(s^-1)$? Does it perhaps mean that the dimensions of $k_t$ are $s^{-1}$?
– Fabio Somenzi
Nov 28 at 18:18
It's some kind of rate so 'per second'
– maria1991
Dec 1 at 17:03
It's some kind of rate so 'per second'
– maria1991
Dec 1 at 17:03
add a comment |
1 Answer
1
active
oldest
votes
If $X_I + X_O = 1$, then $frac{d}{dt}(X_I+X_O) = 0$, which implies $alpha = mu$. We can also simplify the problem by letting ode45
compute, say, $X_O(t)$ and then computing $X_I(t) = 1 - X_O(t)$.
MATLAB's ode45
and its siblings are passed a handle to a function of $t$ and $X_O$ that computes $frac{dX_O}{dt}$. That function may be built in two stages. First we write a function with all the parameters as inputs (besides $t$ and $X_O$). It may look like this:
function xdot = dxodt(t, x, kt, kf, kd, alpha, mu, C, Tstart, TD)
c = alpha * x^2 / (2*x^2 - 2*x + 1);
deltat = 1.0 * (0 <= t && t <= 0.3);
uh = kt*max(1-2*x,0)*deltat;
if C >= 14
ua = kf * (1-x) * deltat;
else
ua = 0;
end
deltac = 1.0 * (Tstart <= t && t <= Tstart + TD);
uc = kd * deltac;
xdot = c + uh + ua - uc - mu*x;
end
We then fix the values of the parameters and invoke the solver by some code like this:
function venusflytrap
%VENUSFLYTRAP simulate Venus Flytrap
kt = 10; % 1/s
kf = 2; % 1/s
kd = 0.000045; % 1/s
alpha = 3600; % 1/s
mu = alpha;
Tstart = 0;
TD = 15*3600;
% These values are uneducated guesses.
C = 1;
xinit = 0;
Tfin = 1;
options = odeset('stats','on');
fh = @(t,x) dxodt(t, x, kt, kf, kd, alpha, mu, C, Tstart, TD);
sol = ode45(fh, [0, Tfin], xinit, options);
plot(sol.x,sol.y)
end
If these two functions are placed in the same file, venusflytrap
should come first, and the file should be named venusflytrap.m
.
Thanks so much! I don't have access to MATLAB at the minute to give this ago but as soon as I do I'll let you know how I get on!
– maria1991
Dec 1 at 17:04
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
If $X_I + X_O = 1$, then $frac{d}{dt}(X_I+X_O) = 0$, which implies $alpha = mu$. We can also simplify the problem by letting ode45
compute, say, $X_O(t)$ and then computing $X_I(t) = 1 - X_O(t)$.
MATLAB's ode45
and its siblings are passed a handle to a function of $t$ and $X_O$ that computes $frac{dX_O}{dt}$. That function may be built in two stages. First we write a function with all the parameters as inputs (besides $t$ and $X_O$). It may look like this:
function xdot = dxodt(t, x, kt, kf, kd, alpha, mu, C, Tstart, TD)
c = alpha * x^2 / (2*x^2 - 2*x + 1);
deltat = 1.0 * (0 <= t && t <= 0.3);
uh = kt*max(1-2*x,0)*deltat;
if C >= 14
ua = kf * (1-x) * deltat;
else
ua = 0;
end
deltac = 1.0 * (Tstart <= t && t <= Tstart + TD);
uc = kd * deltac;
xdot = c + uh + ua - uc - mu*x;
end
We then fix the values of the parameters and invoke the solver by some code like this:
function venusflytrap
%VENUSFLYTRAP simulate Venus Flytrap
kt = 10; % 1/s
kf = 2; % 1/s
kd = 0.000045; % 1/s
alpha = 3600; % 1/s
mu = alpha;
Tstart = 0;
TD = 15*3600;
% These values are uneducated guesses.
C = 1;
xinit = 0;
Tfin = 1;
options = odeset('stats','on');
fh = @(t,x) dxodt(t, x, kt, kf, kd, alpha, mu, C, Tstart, TD);
sol = ode45(fh, [0, Tfin], xinit, options);
plot(sol.x,sol.y)
end
If these two functions are placed in the same file, venusflytrap
should come first, and the file should be named venusflytrap.m
.
Thanks so much! I don't have access to MATLAB at the minute to give this ago but as soon as I do I'll let you know how I get on!
– maria1991
Dec 1 at 17:04
add a comment |
If $X_I + X_O = 1$, then $frac{d}{dt}(X_I+X_O) = 0$, which implies $alpha = mu$. We can also simplify the problem by letting ode45
compute, say, $X_O(t)$ and then computing $X_I(t) = 1 - X_O(t)$.
MATLAB's ode45
and its siblings are passed a handle to a function of $t$ and $X_O$ that computes $frac{dX_O}{dt}$. That function may be built in two stages. First we write a function with all the parameters as inputs (besides $t$ and $X_O$). It may look like this:
function xdot = dxodt(t, x, kt, kf, kd, alpha, mu, C, Tstart, TD)
c = alpha * x^2 / (2*x^2 - 2*x + 1);
deltat = 1.0 * (0 <= t && t <= 0.3);
uh = kt*max(1-2*x,0)*deltat;
if C >= 14
ua = kf * (1-x) * deltat;
else
ua = 0;
end
deltac = 1.0 * (Tstart <= t && t <= Tstart + TD);
uc = kd * deltac;
xdot = c + uh + ua - uc - mu*x;
end
We then fix the values of the parameters and invoke the solver by some code like this:
function venusflytrap
%VENUSFLYTRAP simulate Venus Flytrap
kt = 10; % 1/s
kf = 2; % 1/s
kd = 0.000045; % 1/s
alpha = 3600; % 1/s
mu = alpha;
Tstart = 0;
TD = 15*3600;
% These values are uneducated guesses.
C = 1;
xinit = 0;
Tfin = 1;
options = odeset('stats','on');
fh = @(t,x) dxodt(t, x, kt, kf, kd, alpha, mu, C, Tstart, TD);
sol = ode45(fh, [0, Tfin], xinit, options);
plot(sol.x,sol.y)
end
If these two functions are placed in the same file, venusflytrap
should come first, and the file should be named venusflytrap.m
.
Thanks so much! I don't have access to MATLAB at the minute to give this ago but as soon as I do I'll let you know how I get on!
– maria1991
Dec 1 at 17:04
add a comment |
If $X_I + X_O = 1$, then $frac{d}{dt}(X_I+X_O) = 0$, which implies $alpha = mu$. We can also simplify the problem by letting ode45
compute, say, $X_O(t)$ and then computing $X_I(t) = 1 - X_O(t)$.
MATLAB's ode45
and its siblings are passed a handle to a function of $t$ and $X_O$ that computes $frac{dX_O}{dt}$. That function may be built in two stages. First we write a function with all the parameters as inputs (besides $t$ and $X_O$). It may look like this:
function xdot = dxodt(t, x, kt, kf, kd, alpha, mu, C, Tstart, TD)
c = alpha * x^2 / (2*x^2 - 2*x + 1);
deltat = 1.0 * (0 <= t && t <= 0.3);
uh = kt*max(1-2*x,0)*deltat;
if C >= 14
ua = kf * (1-x) * deltat;
else
ua = 0;
end
deltac = 1.0 * (Tstart <= t && t <= Tstart + TD);
uc = kd * deltac;
xdot = c + uh + ua - uc - mu*x;
end
We then fix the values of the parameters and invoke the solver by some code like this:
function venusflytrap
%VENUSFLYTRAP simulate Venus Flytrap
kt = 10; % 1/s
kf = 2; % 1/s
kd = 0.000045; % 1/s
alpha = 3600; % 1/s
mu = alpha;
Tstart = 0;
TD = 15*3600;
% These values are uneducated guesses.
C = 1;
xinit = 0;
Tfin = 1;
options = odeset('stats','on');
fh = @(t,x) dxodt(t, x, kt, kf, kd, alpha, mu, C, Tstart, TD);
sol = ode45(fh, [0, Tfin], xinit, options);
plot(sol.x,sol.y)
end
If these two functions are placed in the same file, venusflytrap
should come first, and the file should be named venusflytrap.m
.
If $X_I + X_O = 1$, then $frac{d}{dt}(X_I+X_O) = 0$, which implies $alpha = mu$. We can also simplify the problem by letting ode45
compute, say, $X_O(t)$ and then computing $X_I(t) = 1 - X_O(t)$.
MATLAB's ode45
and its siblings are passed a handle to a function of $t$ and $X_O$ that computes $frac{dX_O}{dt}$. That function may be built in two stages. First we write a function with all the parameters as inputs (besides $t$ and $X_O$). It may look like this:
function xdot = dxodt(t, x, kt, kf, kd, alpha, mu, C, Tstart, TD)
c = alpha * x^2 / (2*x^2 - 2*x + 1);
deltat = 1.0 * (0 <= t && t <= 0.3);
uh = kt*max(1-2*x,0)*deltat;
if C >= 14
ua = kf * (1-x) * deltat;
else
ua = 0;
end
deltac = 1.0 * (Tstart <= t && t <= Tstart + TD);
uc = kd * deltac;
xdot = c + uh + ua - uc - mu*x;
end
We then fix the values of the parameters and invoke the solver by some code like this:
function venusflytrap
%VENUSFLYTRAP simulate Venus Flytrap
kt = 10; % 1/s
kf = 2; % 1/s
kd = 0.000045; % 1/s
alpha = 3600; % 1/s
mu = alpha;
Tstart = 0;
TD = 15*3600;
% These values are uneducated guesses.
C = 1;
xinit = 0;
Tfin = 1;
options = odeset('stats','on');
fh = @(t,x) dxodt(t, x, kt, kf, kd, alpha, mu, C, Tstart, TD);
sol = ode45(fh, [0, Tfin], xinit, options);
plot(sol.x,sol.y)
end
If these two functions are placed in the same file, venusflytrap
should come first, and the file should be named venusflytrap.m
.
answered Nov 29 at 6:10
Fabio Somenzi
6,41221221
6,41221221
Thanks so much! I don't have access to MATLAB at the minute to give this ago but as soon as I do I'll let you know how I get on!
– maria1991
Dec 1 at 17:04
add a comment |
Thanks so much! I don't have access to MATLAB at the minute to give this ago but as soon as I do I'll let you know how I get on!
– maria1991
Dec 1 at 17:04
Thanks so much! I don't have access to MATLAB at the minute to give this ago but as soon as I do I'll let you know how I get on!
– maria1991
Dec 1 at 17:04
Thanks so much! I don't have access to MATLAB at the minute to give this ago but as soon as I do I'll let you know how I get on!
– maria1991
Dec 1 at 17:04
add a comment |
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What do you mean by $k_t(s^-1)$? Does it perhaps mean that the dimensions of $k_t$ are $s^{-1}$?
– Fabio Somenzi
Nov 28 at 18:18
It's some kind of rate so 'per second'
– maria1991
Dec 1 at 17:03