Finding Solutions for ar''(t) + br'(t) + cr(t) = 0 with Vector r

[Jhenrique]

Exist solution for ar''(t) + br'(t) + cr(t) = 0 ? If yes how is?

PS: r is a vector.

 

[Matterwave]

That form of differential equation is basically for a damped harmonic oscillator: http://en.wikipedia.org/wiki/Damped_harmonic_oscillator#Damped_harmonic_oscillator.

The fact that [bf]r[/bf] is a vector is not really important to this problem since the problem decouples to 3 identical equations for x y and z (there's no term in that equation that couples the different directions to each other).

Solving just for x (the solutions for y and z are exactly the same): Try the solution $x=e^{\lambda t}$, then this equation becomes:

$$a\lambda^2 e^{\lambda t}+b\lambda e^{\lambda t}+c e^{\lambda t}=0$$

Canceling the common factor $e^{\lambda t}$ we get:

$$a\lambda^2+b\lambda+c=0$$

This is a quadratic equation for $\lambda$ with the usual solution:

$$\lambda=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

The differential equation is linear so we can add two different solutions together and still get a solution. The general solution is then:

$$x=A \exp\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)+B\exp\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)$$

More detail can be found here: http://en.wikipedia.org/wiki/Damping

 

[Jhenrique]

The fact of the differential equation be vectorial and not scalar changes everything!

The solution for $y'(x)=ky(x)$ is $y=C \exp(kx)$ and graphically is an exponential curve, but the solution for $\vec{r}'(t) = K \vec{r}(t)$ is $\vec{r} = C_1 \exp(\lambda_1 t) \hat{v}_1 + C_2 \exp(\lambda_2 t) \hat{v}_2$ and graphically is or a saddle, or a node, or a proper node, etc...

Realize that my doubt, in matrix terms, will be so:
\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{bmatrix} \begin{bmatrix} x''\\ y''\\ \end{bmatrix} + \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \\ \end{bmatrix} \begin{bmatrix} x'\\ y'\\ \end{bmatrix} + \begin{bmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \\ \end{bmatrix} \begin{bmatrix} x\\ y\\ \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ \end{bmatrix}

 

[HallsofIvy]

Well, you do what Matterwave suggested just taking those coefficients to be matrices instead of numbers. All of those operations, exponential, square root, etc. can be defined for matrices.

 

[Chestermiller]

Another way a problem like this is sometimes done is to define a new vector x by:

x_2n-1=r_n
x_2n=r'_n

such that dx_2n-1/dt=x_2n

Then you can rewrite the differential equation in the form dx/dt = Ax, where A is a 2n x 2n matrix. Then, when you make the substitution x = ce^λt, it leads to an ordinary eigenvalue value problem (A-λI)x = 0.

Chet