Higher-Order Linear Equations

Technique: reduction to a first-order linear system plus the linear-systems existence–uniqueness theorem (Picard–Lindelöf for linear systems), Abel's formula for the Wronskian, and linear algebra. Step 1: Reduction to a system. Define $\mathbf{x}(t) = (x_1(t),\dots,x_n(t))^T$ by $x_i(t) = y^{(i-1)}(t)$ for $i=1,\dots,n$. Then $(\ast)$ is equivalent to the first-order linear system

with initial condition $\mathbf{x}(t_0) = (y_0, y_1, \dots, y_{n-1})^T$. This equivalence is immediate: differentiating the definitions of $x_1,\dots,x_{n-1}$ gives the first $n-1$ rows exactly, and the last row is $(\ast)$ itself rearranged. Every entry of $A$ and $\mathbf{b}$ is continuous on $I$ because each $p_i, f$ is. Step 2: Existence and uniqueness. For linear systems $\mathbf{x}' = A(t)\mathbf{x} + \mathbf{b}(t)$ with $A, \mathbf{b}$ continuous on an interval $I$ containing $t_0$, the Picard–Lindelöf theorem applies globally: the right-hand side $F(t,\mathbf{x}) = A(t)\mathbf{x}+\mathbf{b}(t)$ is continuous in $t$ and (being linear in $\mathbf{x}$ with continuous, hence locally bounded, coefficient matrix) satisfies a local Lipschitz condition in $\mathbf{x}$ uniformly on compact subintervals; linear growth in $\mathbf{x}$ then rules out finite-time blowup, so the unique local solution extends to all of $I$. This gives a unique $\mathbf{x} \in C^1(I;\mathbb{R}^n)$ solving the system with the prescribed initial vector. Setting $y = x_1$ recovers a unique $y \in C^n(I)$ solving $(\ast)$ with the prescribed initial data (uniqueness and existence transfer directly through the bijective correspondence of Step 1). This proves (a). Step 3: Superposition. If $L[y_1] = L[y_2] = 0$ then, since $L$ is built from linear combinations of derivatives with the same coefficient functions applied to both,

using linearity of differentiation term by term. So the homogeneous solution set $S$ is closed under linear combination, hence a vector subspace of $C^n(I)$. Step 4: $\dim S = n$. Define $\Phi: S \to \mathbb{R}^n$ by $\Phi(y) = (y(t_0), y'(t_0), \dots, y^{(n-1)}(t_0))$. This map is linear (differentiation and evaluation are linear operations). By part (a) applied with $f \equiv 0$: for every vector $(y_0,\dots,y_{n-1}) \in \mathbb{R}^n$ there is a unique $y \in S$ with $\Phi(y) = (y_0,\dots,y_{n-1})$. Existence of such $y$ for every target vector shows $\Phi$ is surjective; uniqueness of $y$ for each target vector shows $\Phi$ is injective (if $\Phi(y_1)=\Phi(y_2)$ then $y_1-y_2 \in S$ has zero initial data, so by uniqueness in (a), $y_1 - y_2 \equiv 0$). Hence $\Phi$ is a linear bijection, an isomorphism $S \cong \mathbb{R}^n$, so $\dim S = n$. This proves (b). Step 5: Wronskian criterion. Let $y_1,\dots,y_n \in S$ correspond under $\Phi$ to vectors $\mathbf{v}_1,\dots,\mathbf{v}_n \in \mathbb{R}^n$, where $\mathbf{v}_i$ is the $i$-th column of the matrix

so $\det M(t_0) = W(y_1,\dots,y_n)(t_0)$. Since $\Phi$ is a linear isomorphism (Step 4), $y_1,\dots,y_n$ are linearly independent in $S$ if and only if $\mathbf{v}_1,\dots,\mathbf{v}_n$ are linearly independent in $\mathbb{R}^n$, if and only if $\det M(t_0) \neq 0$, i.e. $W(t_0) \neq 0$. Because $\dim S = n$, $n$ linearly independent elements of $S$ automatically form a basis. This shows: $W(t_0)\neq 0$ for one $t_0$ $\iff$ $\{y_i\}$ independent $\iff$ fundamental set. To see that $W(t_0) \neq 0$ for one point forces $W(t) \neq 0$ for all $t \in I$ (Abel's formula), note that $W$ satisfies the first-order linear scalar ODE

a classical fact obtained by differentiating the determinant $W$ row by row: differentiating any row except the last produces a determinant with two equal rows (since row $k+1$ is the derivative of row $k$ for $k < n-1$), which vanishes; only differentiating the last row survives, and using $y_i^{(n)} = -p_{n-1}y_i^{(n-1)} - \cdots - p_0 y_i$ to eliminate $y_i^{(n)}$, all terms except the $-p_{n-1}(t) \times(\text{last row})$ term reduce to determinants with a repeated row and vanish. Solving this linear ODE explicitly gives

and since the exponential is never zero, $W(t_0) \neq 0 \implies W(t) \neq 0$ for all $t \in I$. This proves (c). Step 6: General solution formula. Suppose $y_p$ solves $(\ast)$ and $\{y_1,\dots,y_n\}$ is a fundamental set for $L[y]=0$. For any solution $y$ of $(\ast)$, linearity gives $L[y - y_p] = L[y] - L[y_p] = f - f = 0$, so $y - y_p \in S$. Since $\{y_1,\dots,y_n\}$ is a basis of $S$ (part (c) plus $\dim S = n$), there exist unique constants $c_1,\dots,c_n$ with $y - y_p = c_1y_1+\cdots+c_ny_n$, i.e. $y = y_p + \sum c_i y_i$. Conversely every expression of this form solves $(\ast)$ by superposition and linearity: $L[y_p + \sum c_iy_i] = L[y_p] + \sum c_i L[y_i] = f + 0 = f$. This proves (d), completing the proof. $\blacksquare$

Technique: operator factorization via $D = d/dt$, kernel computation for $(D-r)^k$, and Wronskian nondegeneracy via a Vandermonde-type argument. Step 1: Operator form. Write $(H)$ as $p(D)y = 0$ where $D = d/dt$ and $p(D) = a_nD^n + \cdots + a_0 I$ is a polynomial in $D$ with constant real coefficients. Because polynomials in $D$ with constant coefficients commute under composition (this is just commutativity and associativity of polynomial multiplication applied formally, valid since $D^iD^j = D^{i+j} = D^jD^i$ on smooth functions), we may factor

as an identity of operators on $C^\infty(\mathbb{R}, \mathbb{C})$, with the factors applied in any order. Step 2: Kernel of $(D-r)^k$. Claim: $\ker (D-r)^k = \operatorname{span}\{e^{rt}, te^{rt},\dots, t^{k-1}e^{rt}\}$, a $k$-dimensional space, for any $r \in \mathbb{C}$. Proof of claim, by induction on $k$. For $k=1$: $(D-r)y = 0 \iff y' = ry \iff (e^{-rt}y)' = 0 \iff y = ce^{rt}$, using the integrating factor $e^{-rt}$ and the fact that a function with zero derivative on an interval is constant. So $\ker(D-r) = \operatorname{span}\{e^{rt}\}$, dimension 1, matching the claim at $k=1$. Assume the claim for $k$; consider $(D-r)^{k+1}y = 0$, i.e. $(D-r)\big[(D-r)^ky\big] = 0$. Let $z = (D-r)^ky$. By the base case, $(D-r)z = 0 \implies z = c e^{rt}$ for some constant $c$. So $(D-r)^k y = ce^{rt}$, an inhomogeneous linear equation of order $k$ in $y$. By the induction hypothesis the homogeneous solutions $(D-r)^ky=0$ form a $k$-dimensional space spanned by $e^{rt},\dots,t^{k-1}e^{rt}$; we need one particular solution of $(D-r)^ky = ce^{rt}$. Substituting $y = w(t)e^{rt}$ and using the identity $(D - r)\big[w e^{rt}\big] = w' e^{rt}$ (direct computation: $(we^{rt})' - rwe^{rt} = w'e^{rt} + rwe^{rt} - rwe^{rt} = w'e^{rt}$) repeatedly gives $(D-r)^k[we^{rt}] = w^{(k)}e^{rt}$. So the equation becomes $w^{(k)}e^{rt} = ce^{rt}$, i.e. $w^{(k)} = c$, solved by $w(t) = \frac{c}{k!}t^k$. Hence $y_p = \frac{c}{k!}t^k e^{rt}$ is a particular solution, and by the general-solution structure (Fundamental Sets and General Solutions, part (d) above), every solution of $(D-r)^{k+1}y=0$ is

a $(k+1)$-dimensional space (dimension exactly $k+1$ because these $k+1$ monomials-times-exponential functions are linearly independent, verified directly: if $\sum_{s=0}^k c_st^se^{rt}\equiv 0$ then $\sum c_st^s \equiv 0$ as a polynomial in $t$, forcing all $c_s = 0$). This completes the induction. $\square$ (claim) Step 3: Assembling the kernel of $p(D)$. Since the factors $(D-r_j)^{k_j}$ commute, standard linear algebra for commuting operators gives

as a direct sum (this uses that the factors $(r - r_j)^{k_j}$ are pairwise coprime polynomials, so by Bézout there are polynomials $q_j(r)$ with $\sum_j q_j(r)\prod_{i\neq j}(r-r_i)^{k_i} = 1$; the corresponding operator identity provides the projections realizing the direct sum decomposition — this is the Chinese Remainder Theorem for the PID $\mathbb{C}[r]$ applied to the module $C^\infty(\mathbb{R};\mathbb{C})$ over $\mathbb{C}[D]$). By Step 2, $\dim \ker(D-r_j)^{k_j} = k_j$, so

matching the dimension $n$ guaranteed by the structure theorem above. A basis is the union over $j$ of $\{t^se^{r_jt}\}_{s=0}^{k_j-1}$, which is exactly $n$ functions (complex-valued, if some $r_j \notin \mathbb{R}$). Step 4: Realification. If the coefficients $a_i$ are real, the polynomial $p(r)$ has real coefficients, so nonreal roots occur in conjugate pairs $\alpha\pm i\beta$ with equal multiplicity $k$ (if $r_j = \alpha+i\beta$ is a root of multiplicity $k$, then $p(\bar r_j) = \overline{p(r_j)} = 0$, and the multiplicity of $\bar r_j$ equals that of $r_j$ by applying the same argument to all derivatives $p^{(l)}$ up to order $k-1$, all of which vanish at $r_j$ hence at $\bar r_j$ too). For such a pair, replace the complex basis elements $t^se^{(\alpha+i\beta)t}$ and $t^se^{(\alpha-i\beta)t}$ by the real linear combinations

This is an invertible linear change of basis within the 2-dimensional (for each fixed $s$) complex span, so the resulting set of $n$ real functions still spans $\ker p(D)$ and has the same cardinality $n$ as the dimension of the space, hence is linearly independent over $\mathbb{R}$ and forms a basis. Real roots need no modification. This produces precisely the basis stated in the theorem. Step 5: Fundamental set on every interval. The functions constructed are entire (real-analytic on all of $\mathbb{R}$) and their number equals $n = \dim S$ where $S$ is the (real) solution space of (H) by the previous theorem; since they are linearly independent (Step 3–4) and $\dim S = n$, they form a basis of $S$, hence by the Wronskian Criterion their Wronskian is nonzero at every point of every interval. $\blacksquare$