Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Express the 2nd order ODE

$$\begin{align}\mathrm d_t^2 u:=\frac{\mathrm d^2 u}{\mathrm dt^2}&=\sin(u)+\cos(\omega t)\qquad \omega \in \mathbb Z /\{0\} \\u(0)&=a\\\mathrm d_t u(0)&=b\end{align}$$

as a system of 1st order ODEs and verify there exists a global solution by invoking the global existence and uniqueness theorems.

I'm not sure how to express second order ODEs as first order ODEs, any tips?

Thanks.

share|improve this question
1  
Make a new variable, $v=\dfrac{\mathrm du}{\mathrm dt}$... –  J. M. Oct 9 '11 at 1:15
    
How do you show there exists a global solution byt invoking the theorem? Theorem states: An IVP has a unique solution if the function f is continuous with respect to the 1st variable and Lipshitz continuous with respect to the 2nd variable. –  Euden Sep 27 '12 at 12:14
add comment

2 Answers

Here's an example to get you started:

$$u^{(3)}(t)+t^3u''(t)+5u'(t)+\sin(t)u=e^{6t}$$

with initial values $u''(0)=1$, $u'(0)=2$, and $u(0)=3$

First, give new names to $u$ and its derivatives (stopping one short of the order of the ODE): $u=x_1$, $u'=x_2$, $u''=x_3$.

Substituting back into the DE (keeping in mind that $u^{(3)}(t)=x'_3(t)$) we get: $$x'_3(t)+t^3x_3(t)+5x_2(t)+\sin(t)x_1(t)=e^{6t}$$

Thus we have the equivalent system:

$$\begin{array}{ccrrrr} x'_1 & = & & x_2 & & \\ x'_2 & = & & & x_3 & \\ x'_3 & = & -\sin(t)x_1 & -5x_2 & -t^3x_3 & +e^{6t} \end{array}$$

Also, $x_1(0)=3$, $x_2(0)=2$, and $x_3(0)=1$.

share|improve this answer
add comment

To convert second-order ODE to a first-order system you have to introduce new variables:

$u_1=u$

$u_2=u'_t$

Now we can write following:

$(u_1)'_t=u_2$

$(u_2)'_t=u''_t=\sin(u_1)+\cos(\omega t)$

with the initial condition $u_1(0)=a$ , $u_2(0)=b$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.