Transform a non-linear second order ODE












0














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?










share|cite|improve this question


















  • 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
















0














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?










share|cite|improve this question


















  • 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














0












0








0







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?










share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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















active

oldest

votes











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3007363%2ftransform-a-non-linear-second-order-ode%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown






























active

oldest

votes













active

oldest

votes









active

oldest

votes






active

oldest

votes
















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3007363%2ftransform-a-non-linear-second-order-ode%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

How to change which sound is reproduced for terminal bell?

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents