Why does the Solution Set of a 2nd Order Homogeneous Equation have at most Two Dimensions?

Question

I believe that for a 2nd Order Homogeneous Differential Equation, the solution set has at most two dimensions. However, I want to be able to prove that rigorously.

I understand that for a 2nd Order Homogeneous Equation

$$ y'' + P(x)y' + Q(x)y = 0 $$

If y'', P(x) and Q(x) are differentiable along a closed interval [p, q]; and I pick an x₁ from that interval

$$x_1 \in [p,q] $$

And if I have two linearly independent solutions, y₁ and y₂, then for any third solution I come across, like y₃, I am able to prove that this is true

$$ y_3(x_1) = C_1y_1(x_1) + C_2y_2(x_1) $$

$$ y_3'(x_1) = C_1y_1'(x_1) + C_2y_2'(x_1) $$

This means that if there is a solution to our original 2nd Order Equation called y₁₂ such that

$$ y_{12}(x) = C_1y_1(x) + C_2y_2(x) $$

$$ y_3(x_1) = y_{12}(x_1) $$

$$ y_3'(x_1) = y_{12}'(x_1) $$

However, what I am stuck on is that, how do I know that y₃ equals y₁₂ for all x within [p, q]? And how can I rigorously prove that? I'm well aware that if we consider a small change in y₃, we get the following

$$ y_3(x_1 + dx) = y_{12}(x_1) + y_{12}'(x_1)dx $$

$$ y_3'(x_1 + dx) = y_{12}'(x_1) + [P(x_1)y'_{12} + Q(x_1)y_{12}(x_1)]dx $$

$$ y_3''(x_1 + dx) = P(x_1 + dx)y_3(x_1 + dx) + Q(x_1 + dx)y_3''(x_1 + dx) $$

and I can use recursion to figure out y₃(x₃ + 2dx) and other values after that. And we can see that all other values are based y₁₂. But that is only a hand-wavey approach to this proof using dx.

Can someone please explain how we can rigorously prove y₃ is a Linear Combination of the first two solutions, for all values of x?

Answer 1

You can use the Picard-Lindelöf theorem to prove this.

By setting $u_1 = y$ and $u_2 = y'$, you get $$ u' = \begin{pmatrix} u_1' \\ u_2' \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -Q(x) & -P(x) \end{pmatrix} \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} =A(x)\cdot u =: f(x, u) $$ The Picard-Lindelöf theorem now states that this initial value problem has a unique solution for an initial value $u(x_0)=u_0,$ if $f(x,u)$ is uniformly Lipschitz continuous in $u$, which means that there must be a constant $L$, which is independent of $x,$ such that $$ \left\| f(x,u)-f(x,v) \right\| = \left\| A(x) u - A(x) v \right\| \; \leq \; L\cdot \left\| u-v \right\| $$ This can obviously be achieved by setting $$ L = \max_{x\in[p,q]} \left\| A(x) \right\| $$ with a matrix norm that is consistent with the vector norm used above, because with this $L$, we get $$ \left\| A(x) u - A(x) v \right\| =\left\| A(x) (u-v) \right\| \leq\left\| A(x) \right\| \cdot \left\| u-v \right\| \leq\max_{x\in[p,q]} \left\| A(x) \right\| \cdot \left\| u-v \right\| = L\cdot \left\| u-v \right\| $$ As the space of initial values is two-dimensional, and there is a unique solution for each initial value, the space of solutions is two-dimensional, too.