How many initial conditions are required?

Question

Consider the following system of ODE:

$$\begin{array}{ll}\ddot y + y + \ddot x + x = 0 \\ y+\dot x - x = 0 \end{array}$$

Question: How many initial conditions are required to determine a unique solution?

A naive reasoning leads to four: $y(0),\dot y(0), x(0)$ and $\dot x(0)$. However, if we write the system in a first-order form:

$$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} \dot x \\ \ddot x \\ \dot y \\ \ddot y\end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & 0 \\ -1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1\\ 1 & 0 & -1 & 0 \end{bmatrix}\begin{bmatrix} x \\ \dot x \\ y \\ \dot y\end{bmatrix}$$

the left matrix is not of full rank, which means the equations are not all independent. Indeed, by differentiating the second equation: $\dot y + \ddot x -\dot x=0$ which leads to $\ddot y + y + \dot x - \dot y + x =0$, or, in the first-order form: $$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} \dot x \\ \dot y \\ \ddot y \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 0 & 1 \\ -1 & -1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ \dot y \end{bmatrix}$$

With the above form, it shows only three initial conditions are required: $x(0),y(0),\dot y(0)$.

And, what if I had:

$$\begin{array}{ll}\ddot y + y + x^{(n)} + x = 0 \\ \dot x - x = 0 \end{array}$$

for some $n$. Then, I could solve the second equation for some $x(0)$, then differentiate $x$ $n$ times and inject in the first equation, so only $x(0), y(0), \dot y(0)$ are needed. Can this be seen directly, in a robust manner?

Answer 1

This problem depends on 3 arbitrary constants. Here is why.

Let $$\tag{1}\begin{cases}u&=&x+y\\v&=&x-y\end{cases}$$

The given differential system, written under the form:

$$\tag{2}\left\{\begin{array}{rclr}\ddot{(x+y)}& = & - (x+y) \ \ \ \ \ & (a) \\ \dot x& = & x-y \ \ \ \ \ & (b)\end{array}\right.$$

is equivalent to:

$$\tag{3}\begin{cases}\ddot u &=& -u \ \ \ \ \ &(a)\\\dot u +\dot v&=&2v \ \ \ \ \ &(b)\end{cases}$$

(equation (3b) comes from the addition of the two equations of (1), $2x=u+v$, then the differentiation of this relationship, and, at last, the use of (2b)).

The solution of equation (3a), the harmonic oscillator, depends on 2 arbitrary constants, for example under the form

$$u=A \cos(t)+B\sin(t)$$

This solution, plugged into equation (b) gives the first order linear differential equation:

$$-\dot v + 2v=-A \sin(t)+B \cos(t)$$

whose general solution depends on a supplementary constant $C$.

Knowing $u$ and $v$, it is immediate to deduce $x$ and $y$.

Answer 2

I guess that the first answer is $4$, even if the dynamical matrix is rank deficient. Consider this example: $$\left[\begin{array}{c}\dot{x}\\\dot{y} \end{array}\right] = \left[\begin{array}{cc}1 & 0\\ 1 & 0 \end{array}\right] \cdot \left[\begin{array}{c}x\\y \end{array}\right].$$ Following you thoughts, even in this case we need just $1$ initial condition instead of $2$. Anyway:

$$x(t) = A e^{-t}, y(t) = B + Ce^{-t}.$$

Then:

$$\dot{y}(t) = -Ce^{-t} = x(t) \Rightarrow A=-C,$$ and hence:

$$x(t) = Ae^{-t}, y(t) = B - Ae^{-t}.$$

This means that you still need $2$ initial conditions!!!

In this case, you have that $A = x(0)$ and $B = x(0)+y(0)$.

For the second case, you need $n+2$ initial conditions.

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Addition

If the dynamical matrix is rank deficient, then this means that it has at least one null eigenvalue. The presence of the null eigenvalue implies the presence of a "constant" term in the solution. This constant term still represents a degree of freedom of the system, even if the dynamical matrix lost a degree of freedom.