Transform a non-linear second order ODE
Hi I'm stuck on the following problem:
Given a projectile with position $mathbf{r}(t)=[r_x(t),r_y(t)]$, velocity $mathbf{v}(t)=[v_x(t),v_y(t)]$ and acceleration $mathbf{a}(t)$, we can approximate the drag force $mathbf{F}_d$ acting in the opposite direction to the projectile's trajectory using
$$mathbf{F}_d = frac{1}{2} rho c_d A |mathbf{v}|mathbf{v},$$
where $rho$ is the air density, $c_d$ is the drag coefficiant, $A$ is the frontal area and $mathbf{v}$ is the velocity defined above.
Using Newton's second law, we can get
$$mathbf{F}=mmathbf{a} Rightarrow mathbf{a}(t) = -frac{1}{m}(-rho c_d A|mathbf{v}|mathbf{v} - mmathbf{g}),$$
Where m is the mass of the projectile and $mathbf{g} = (0,g)$ is the gravitational acceleration.
We can use the fact that
$$mathbf{a}(t)=frac{delta mathbf{v}}{delta t} = mathbf{dot{v}}(t)$$
and
$$mathbf{v}(t)=frac{delta mathbf{r}}{delta t} = mathbf{dot{r}}(t)$$
to give us the non-linear, second order Ordinary Differential Equation
$$mathbf{ddot{r}}(t)=frac{1}{m}(-frac{1}{2}rho c_d A |mathbf{dot{r}}|mathbf{dot{r}}-mmathbf{g}). ;;; (1)$$
We can also consider the initial conditions at time $t=0$. The projectile is initially fired from height $h$ at an angle of $alpha$ degrees with initial speed $v_0$, which gives:
$$mathbf{r}(0) = (0, h)$$ and $$mathbf{dot{r}}(0) = mathbf{v}(0)=[v_0 cos(frac{pialpha}{180}), v_0 sin(frac{pialpha}{180})].$$
At this point I need to transform equation (1) as a system of first-order equations for the two components of velocity and two components of position. Mathematically this will be expressed as
$$mathbf{dot{y}} = mathbf{F}(t, mathbf{y}(t)),$$
where $mathbf{y}(t) = [v_x(t), v_y(t), r_x(t), r_y(t)]$.
I'm familiar with a method to transfrom a second order ODE to a system of linear first-order ODEs, but the method I'm thinking of runs into a problem due to the fact we have $|mathbf{dot{r}}|mathbf{dot{r}}$ in the equation and I'm not sure how to handle this. Could anyone show me how to do this?
linear-algebra differential-equations physics
asked Nov 21 '18 at 7:03
Thomas M
185
add a comment |
Hi I'm stuck on the following problem:
Given a projectile with position $mathbf{r}(t)=[r_x(t),r_y(t)]$, velocity $mathbf{v}(t)=[v_x(t),v_y(t)]$ and acceleration $mathbf{a}(t)$, we can approximate the drag force $mathbf{F}_d$ acting in the opposite direction to the projectile's trajectory using
$$mathbf{F}_d = frac{1}{2} rho c_d A |mathbf{v}|mathbf{v},$$
where $rho$ is the air density, $c_d$ is the drag coefficiant, $A$ is the frontal area and $mathbf{v}$ is the velocity defined above.
Using Newton's second law, we can get
$$mathbf{F}=mmathbf{a} Rightarrow mathbf{a}(t) = -frac{1}{m}(-rho c_d A|mathbf{v}|mathbf{v} - mmathbf{g}),$$
Where m is the mass of the projectile and $mathbf{g} = (0,g)$ is the gravitational acceleration.
We can use the fact that
$$mathbf{a}(t)=frac{delta mathbf{v}}{delta t} = mathbf{dot{v}}(t)$$
and
$$mathbf{v}(t)=frac{delta mathbf{r}}{delta t} = mathbf{dot{r}}(t)$$
to give us the non-linear, second order Ordinary Differential Equation
$$mathbf{ddot{r}}(t)=frac{1}{m}(-frac{1}{2}rho c_d A |mathbf{dot{r}}|mathbf{dot{r}}-mmathbf{g}). ;;; (1)$$
We can also consider the initial conditions at time $t=0$. The projectile is initially fired from height $h$ at an angle of $alpha$ degrees with initial speed $v_0$, which gives:
$$mathbf{r}(0) = (0, h)$$ and $$mathbf{dot{r}}(0) = mathbf{v}(0)=[v_0 cos(frac{pialpha}{180}), v_0 sin(frac{pialpha}{180})].$$
At this point I need to transform equation (1) as a system of first-order equations for the two components of velocity and two components of position. Mathematically this will be expressed as
$$mathbf{dot{y}} = mathbf{F}(t, mathbf{y}(t)),$$
where $mathbf{y}(t) = [v_x(t), v_y(t), r_x(t), r_y(t)]$.
I'm familiar with a method to transfrom a second order ODE to a system of linear first-order ODEs, but the method I'm thinking of runs into a problem due to the fact we have $|mathbf{dot{r}}|mathbf{dot{r}}$ in the equation and I'm not sure how to handle this. Could anyone show me how to do this?
linear-algebra differential-equations physics
asked Nov 21 '18 at 7:03
Thomas M
185
1
Since your differential equation doesn't depend on $mathbf{r}$ directly, then just set $mathbf{v} = mathbf{dot r}$ to get a first order system.
– Jacky Chong
Nov 21 '18 at 7:07
@JackyChong So substituting $mathbf{v} = mathbf{dot{r}}$ into the equation tells gives use the first order equations for the velocity components, but what about the position components? We simply have that at time $t$, $mathbf{dot{r}}(t) = mathbf{v}$, yes?
– Thomas M
Nov 21 '18 at 7:37
Correct. Once you find the velocity, you just integrate to get the position.
– Andrei
Nov 21 '18 at 19:36
add a comment |
Hi I'm stuck on the following problem:
Given a projectile with position $mathbf{r}(t)=[r_x(t),r_y(t)]$, velocity $mathbf{v}(t)=[v_x(t),v_y(t)]$ and acceleration $mathbf{a}(t)$, we can approximate the drag force $mathbf{F}_d$ acting in the opposite direction to the projectile's trajectory using
$$mathbf{F}_d = frac{1}{2} rho c_d A |mathbf{v}|mathbf{v},$$
where $rho$ is the air density, $c_d$ is the drag coefficiant, $A$ is the frontal area and $mathbf{v}$ is the velocity defined above.
Using Newton's second law, we can get
$$mathbf{F}=mmathbf{a} Rightarrow mathbf{a}(t) = -frac{1}{m}(-rho c_d A|mathbf{v}|mathbf{v} - mmathbf{g}),$$
Where m is the mass of the projectile and $mathbf{g} = (0,g)$ is the gravitational acceleration.
We can use the fact that
$$mathbf{a}(t)=frac{delta mathbf{v}}{delta t} = mathbf{dot{v}}(t)$$
and
$$mathbf{v}(t)=frac{delta mathbf{r}}{delta t} = mathbf{dot{r}}(t)$$
to give us the non-linear, second order Ordinary Differential Equation
$$mathbf{ddot{r}}(t)=frac{1}{m}(-frac{1}{2}rho c_d A |mathbf{dot{r}}|mathbf{dot{r}}-mmathbf{g}). ;;; (1)$$
We can also consider the initial conditions at time $t=0$. The projectile is initially fired from height $h$ at an angle of $alpha$ degrees with initial speed $v_0$, which gives:
$$mathbf{r}(0) = (0, h)$$ and $$mathbf{dot{r}}(0) = mathbf{v}(0)=[v_0 cos(frac{pialpha}{180}), v_0 sin(frac{pialpha}{180})].$$
At this point I need to transform equation (1) as a system of first-order equations for the two components of velocity and two components of position. Mathematically this will be expressed as
$$mathbf{dot{y}} = mathbf{F}(t, mathbf{y}(t)),$$
where $mathbf{y}(t) = [v_x(t), v_y(t), r_x(t), r_y(t)]$.
I'm familiar with a method to transfrom a second order ODE to a system of linear first-order ODEs, but the method I'm thinking of runs into a problem due to the fact we have $|mathbf{dot{r}}|mathbf{dot{r}}$ in the equation and I'm not sure how to handle this. Could anyone show me how to do this?
linear-algebra differential-equations physics
asked Nov 21 '18 at 7:03
Thomas M
185
Hi I'm stuck on the following problem:
Given a projectile with position $mathbf{r}(t)=[r_x(t),r_y(t)]$, velocity $mathbf{v}(t)=[v_x(t),v_y(t)]$ and acceleration $mathbf{a}(t)$, we can approximate the drag force $mathbf{F}_d$ acting in the opposite direction to the projectile's trajectory using
$$mathbf{F}_d = frac{1}{2} rho c_d A |mathbf{v}|mathbf{v},$$
where $rho$ is the air density, $c_d$ is the drag coefficiant, $A$ is the frontal area and $mathbf{v}$ is the velocity defined above.
Using Newton's second law, we can get
$$mathbf{F}=mmathbf{a} Rightarrow mathbf{a}(t) = -frac{1}{m}(-rho c_d A|mathbf{v}|mathbf{v} - mmathbf{g}),$$
Where m is the mass of the projectile and $mathbf{g} = (0,g)$ is the gravitational acceleration.
We can use the fact that
$$mathbf{a}(t)=frac{delta mathbf{v}}{delta t} = mathbf{dot{v}}(t)$$
and
$$mathbf{v}(t)=frac{delta mathbf{r}}{delta t} = mathbf{dot{r}}(t)$$
to give us the non-linear, second order Ordinary Differential Equation
$$mathbf{ddot{r}}(t)=frac{1}{m}(-frac{1}{2}rho c_d A |mathbf{dot{r}}|mathbf{dot{r}}-mmathbf{g}). ;;; (1)$$
We can also consider the initial conditions at time $t=0$. The projectile is initially fired from height $h$ at an angle of $alpha$ degrees with initial speed $v_0$, which gives:
$$mathbf{r}(0) = (0, h)$$ and $$mathbf{dot{r}}(0) = mathbf{v}(0)=[v_0 cos(frac{pialpha}{180}), v_0 sin(frac{pialpha}{180})].$$
At this point I need to transform equation (1) as a system of first-order equations for the two components of velocity and two components of position. Mathematically this will be expressed as
$$mathbf{dot{y}} = mathbf{F}(t, mathbf{y}(t)),$$
where $mathbf{y}(t) = [v_x(t), v_y(t), r_x(t), r_y(t)]$.
I'm familiar with a method to transfrom a second order ODE to a system of linear first-order ODEs, but the method I'm thinking of runs into a problem due to the fact we have $|mathbf{dot{r}}|mathbf{dot{r}}$ in the equation and I'm not sure how to handle this. Could anyone show me how to do this?
linear-algebra differential-equations physics
linear-algebra differential-equations physics
asked Nov 21 '18 at 7:03
Thomas M
185
asked Nov 21 '18 at 7:03
Thomas M
185
asked Nov 21 '18 at 7:03
Thomas M
185
asked Nov 21 '18 at 7:03
Thomas M
185
asked Nov 21 '18 at 7:03
Thomas M
185
185
1
Since your differential equation doesn't depend on $mathbf{r}$ directly, then just set $mathbf{v} = mathbf{dot r}$ to get a first order system.
– Jacky Chong
Nov 21 '18 at 7:07
@JackyChong So substituting $mathbf{v} = mathbf{dot{r}}$ into the equation tells gives use the first order equations for the velocity components, but what about the position components? We simply have that at time $t$, $mathbf{dot{r}}(t) = mathbf{v}$, yes?
– Thomas M
Nov 21 '18 at 7:37
Correct. Once you find the velocity, you just integrate to get the position.
– Andrei
Nov 21 '18 at 19:36
add a comment |
1
Since your differential equation doesn't depend on $mathbf{r}$ directly, then just set $mathbf{v} = mathbf{dot r}$ to get a first order system.
– Jacky Chong
Nov 21 '18 at 7:07
@JackyChong So substituting $mathbf{v} = mathbf{dot{r}}$ into the equation tells gives use the first order equations for the velocity components, but what about the position components? We simply have that at time $t$, $mathbf{dot{r}}(t) = mathbf{v}$, yes?
– Thomas M
Nov 21 '18 at 7:37
Correct. Once you find the velocity, you just integrate to get the position.
– Andrei
Nov 21 '18 at 19:36
1
1
Since your differential equation doesn't depend on $mathbf{r}$ directly, then just set $mathbf{v} = mathbf{dot r}$ to get a first order system.
– Jacky Chong
Nov 21 '18 at 7:07
Since your differential equation doesn't depend on $mathbf{r}$ directly, then just set $mathbf{v} = mathbf{dot r}$ to get a first order system.
– Jacky Chong
Nov 21 '18 at 7:07
@JackyChong So substituting $mathbf{v} = mathbf{dot{r}}$ into the equation tells gives use the first order equations for the velocity components, but what about the position components? We simply have that at time $t$, $mathbf{dot{r}}(t) = mathbf{v}$, yes?
– Thomas M
Nov 21 '18 at 7:37
@JackyChong So substituting $mathbf{v} = mathbf{dot{r}}$ into the equation tells gives use the first order equations for the velocity components, but what about the position components? We simply have that at time $t$, $mathbf{dot{r}}(t) = mathbf{v}$, yes?
– Thomas M
Nov 21 '18 at 7:37
Correct. Once you find the velocity, you just integrate to get the position.
– Andrei
Nov 21 '18 at 19:36
Correct. Once you find the velocity, you just integrate to get the position.
– Andrei
Nov 21 '18 at 19:36
add a comment |
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1
Since your differential equation doesn't depend on $mathbf{r}$ directly, then just set $mathbf{v} = mathbf{dot r}$ to get a first order system.
– Jacky Chong
Nov 21 '18 at 7:07
@JackyChong So substituting $mathbf{v} = mathbf{dot{r}}$ into the equation tells gives use the first order equations for the velocity components, but what about the position components? We simply have that at time $t$, $mathbf{dot{r}}(t) = mathbf{v}$, yes?
– Thomas M
Nov 21 '18 at 7:37
Correct. Once you find the velocity, you just integrate to get the position.
– Andrei
Nov 21 '18 at 19:36