Wronskian, Existence and Uniqueness of Solutions

TL;DR

For a second-order homogeneous linear differential equation with arbitrary continuous variable coefficients $p$ and $q$ on an interval $I$:

\[y^{\prime\prime} + p(x)y^{\prime} + q(x)y = 0\]

and initial conditions

\[y(x_0)=K_0, \qquad y^{\prime}(x_0)=K_1\]

the following four theorems hold:

1. Existence and Uniqueness Theorem for Initial Value Problems: The initial value problem consisting of the given equation and initial conditions has a unique solution $y(x)$ on interval $I$.
2. Determining Linear Dependence/Independence Using the Wronskian: For two solutions $y_1$ and $y_2$ of the equation, if there exists an $x_0$ in interval $I$ where the Wronskian $W(y_1, y_2) = y_1y_2^{\prime} - y_2y_1^{\prime}$ equals zero, then the two solutions are linearly dependent. Furthermore, if there exists an $x_1$ in interval $I$ where $W\neq 0$, then the two solutions are linearly independent.
3. Existence of General Solution: The given equation has a general solution on interval $I$.
4. Non-existence of Singular Solutions: This general solution includes all solutions of the equation (i.e., singular solutions do not exist).

Prerequisites

• Solution of First-Order Linear ODE
• Homogeneous Linear ODEs of Second Order
• Homogeneous Linear ODEs with Constant Coefficients
• Euler-Cauchy Equation
• Inverse matrices and singular matrices, determinants

Homogeneous Linear Differential Equations with Arbitrary Continuous Variable Coefficients

Previously, we explored the general solutions of homogeneous linear ODEs with constant coefficients and Euler-Cauchy equations. In this post, we extend our discussion to the more general case of second-order homogeneous linear differential equations with arbitrary continuous variable coefficients $p$ and $q$:

\[y^{\prime\prime} + p(x)y^{\prime} + q(x)y = 0 \label{eqn:homogeneous_linear_ode_with_var_coefficients}\tag{1}\]

We will examine the existence and form of general solutions to this equation. Additionally, we will investigate the uniqueness of the initial value problem consisting of differential equation ($\ref{eqn:homogeneous_linear_ode_with_var_coefficients}$) and the following two initial conditions:

\[y(x_0)=K_0, \qquad y^{\prime}(x_0)=K_1 \label{eqn:initial_conditions}\tag{2}\]

To preview our conclusion, the key insight is that linear differential equations with continuous coefficients do not possess singular solutions (solutions that cannot be derived from the general solution).

Existence and Uniqueness Theorem for Initial Value Problems

Existence and Uniqueness Theorem for Initial Value Problems
If $p(x)$ and $q(x)$ are continuous functions on some open interval $I$, and $x_0$ is a point in interval $I$, then the initial value problem consisting of equation ($\ref{eqn:homogeneous_linear_ode_with_var_coefficients}$) and conditions ($\ref{eqn:initial_conditions}$) has a unique solution $y(x)$ on interval $I$.

We will only address the proof of uniqueness here, as proving existence is typically more complex.
If you’re not interested in the proof, you can skip to the Linear Dependence and Linear Independence of Solutions section.

Proof of Uniqueness

Assume that the initial value problem consisting of differential equation ($\ref{eqn:homogeneous_linear_ode_with_var_coefficients}$) and initial conditions ($\ref{eqn:initial_conditions}$) has two solutions $y_1(x)$ and $y_2(x)$ on interval $I$. We need to show that their difference

\[y(x) = y_1(x) - y_2(x)\]

is identically zero on interval $I$, which would imply that $y_1 \equiv y_2$ on interval $I$, proving uniqueness.

Since equation ($\ref{eqn:homogeneous_linear_ode_with_var_coefficients}$) is a homogeneous linear differential equation, the linear combination $y$ of $y_1$ and $y_2$ is also a solution on interval $I$. Since $y_1$ and $y_2$ satisfy the same initial conditions ($\ref{eqn:initial_conditions}$), $y$ satisfies the conditions

\[\begin{align*} & y(x_0) = y_1(x_0) - y_2(x_0) = 0, \\ & y^{\prime}(x_0) = y_1^{\prime}(x_0) - y_2^{\prime}(x_0) = 0 \end{align*} \label{eqn:initial_conditions_*}\tag{3}\]

Now consider the function

\[z(x) = y(x)^2 + y^{\prime}(x)^2\]

and its derivative

\[z^{\prime} = 2yy^{\prime} + 2y^{\prime}y^{\prime\prime}\]

From the differential equation, we have

\[y^{\prime\prime} = -py^{\prime} - qy\]

Substituting this into the expression for $z^{\prime}$, we get

\[z^{\prime} = 2yy^{\prime} - 2p{y^{\prime}}^2 - 2qyy^{\prime} \label{eqn:z_prime}\tag{4}\]

Since $y$ and $y^{\prime}$ are real numbers, we have

\[(y\pm y^{\prime})^2 = y^2 \pm 2yy^{\prime} + {y^{\prime}}^2 \geq 0\]

From this and the definition of $z$, we obtain two inequalities:

\[(a)\ 2yy^{\prime} \leq y^2 + {y^{\prime}}^2 = z, \qquad (b)\ 2yy^{\prime} \geq -(y^2 + {y^{\prime}}^2) = -z \label{eqn:inequalities}\tag{5}\]

These inequalities imply that $|2yy^{\prime}|\leq z$, and therefore for the last term in equation ($\ref{eqn:z_prime}$), we have:

\[\pm2qyy^{\prime} \leq |\pm 2qyy^{\prime}| = |q||2yy^{\prime}| \leq |q|z.\]

Using this result along with $-p \leq |p|$ and applying inequality ($\ref{eqn:inequalities}$a) to the term $2yy^{\prime}$ in equation ($\ref{eqn:z_prime}$), we get:

\[z^{\prime} \leq z + 2|p|{y^{\prime}}^2 + |q|z\]

Since ${y^{\prime}}^2 \leq y^2 + {y^{\prime}}^2 = z$, we obtain:

\[z^{\prime} \leq (1 + 2|p| + |q|)z\]

If we denote the function in parentheses as $h = 1 + 2|p| + |q|$, we have:

\[z^{\prime} \leq hz \quad \forall x \in I \label{eqn:inequality_6a}\tag{6a}\]

Similarly, from equations ($\ref{eqn:z_prime}$) and ($\ref{eqn:inequalities}$), we get:

\[\begin{align*} -z^{\prime} &= -2yy^{\prime} + 2p{y^{\prime}}^2 + 2qyy^{\prime} \\ &\leq z + 2|p|z + |q|z = hz \end{align*} \label{eqn:inequality_6b}\tag{6b}\]

These two inequalities ($\ref{eqn:inequality_6a}$) and ($\ref{eqn:inequality_6b}$) are equivalent to:

\[z^{\prime} - hz \leq 0, \qquad z^{\prime} + hz \geq 0 \label{eqn:inequalities_7}\tag{7}\]

The integrating factors for the left sides of these inequalities are:

\[F_1 = e^{-\int h(x)\ dx} \qquad \text{and} \qquad F_2 = e^{\int h(x)\ dx}\]

Since $h$ is continuous, the indefinite integral $\int h(x)\ dx$ exists, and since $F_1$ and $F_2$ are positive, from inequalities ($\ref{eqn:inequalities_7}$) we get:

\[F_1(z^{\prime} - hz) = (F_1 z)^{\prime} \leq 0, \qquad F_2(z^{\prime} + hz) = (F_2 z)^{\prime} \geq 0\]

This means that $F_1 z$ is non-increasing and $F_2 z$ is non-decreasing on interval $I$. From equation ($\ref{eqn:initial_conditions_*}$), we know that $z(x_0) = 0$, so:

\[\begin{cases} \left(F_1 z \geq (F_1 z)_{x_0} = 0\right)\ \& \ \left(F_2 z \leq (F_2 z)_{x_0} = 0\right) & (x \leq x_0) \\ \left(F_1 z \leq (F_1 z)_{x_0} = 0\right)\ \& \ \left(F_2 z \geq (F_2 z)_{x_0} = 0\right) & (x \geq x_0) \end{cases}\]

Finally, dividing these inequalities by the positive values $F_1$ and $F_2$, we can prove uniqueness:

\[(z \leq 0) \ \& \ (z \geq 0) \quad \forall x \in I\] \[z = y^2 + {y^{\prime}}^2 = 0 \quad \forall x \in I\] \[\therefore y \equiv y_1 - y_2 \equiv 0 \quad \forall x \in I. \ \blacksquare\]

Linear Dependence and Linear Independence of Solutions

Let’s recall what we discussed in Homogeneous Linear ODEs of Second Order. The general solution on an open interval $I$ is constructed from a basis $y_1$, $y_2$ - a pair of linearly independent solutions on $I$. Two functions $y_1$ and $y_2$ are linearly independent on interval $I$ if for all $x$ in the interval:

\[k_1y_1(x) + k_2y_2(x) = 0 \Leftrightarrow k_1=0\text{ and }k_2=0 \label{eqn:linearly_independent}\tag{8}\]

If this condition is not satisfied, and there exist non-zero constants $k_1$, $k_2$ such that $k_1y_1(x) + k_2y_2(x) = 0$, then $y_1$ and $y_2$ are linearly dependent on interval $I$. In this case, for all $x$ in interval $I$:

\[\text{(a) } y_1 = ky_2 \quad \text{or} \quad \text{(b) } y_2 = ly_1 \label{eqn:linearly_dependent}\tag{9}\]

meaning that $y_1$ and $y_2$ are proportional.

Now let’s examine the method for determining linear dependence/independence of solutions:

Determining Linear Dependence/Independence Using the Wronskian
i. If differential equation ($\ref{eqn:homogeneous_linear_ode_with_var_coefficients}$) has continuous coefficients $p(x)$ and $q(x)$ on an open interval $I$, then a necessary and sufficient condition for two solutions $y_1$ and $y_2$ of equation ($\ref{eqn:homogeneous_linear_ode_with_var_coefficients}$) to be linearly dependent is that their Wronski determinant, or Wronskian

\[W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1^{\prime} & y_2^{\prime} \\ \end{vmatrix} = y_1y_2^{\prime} - y_2y_1^{\prime} \label{eqn:wronskian}\tag{10}\]

equals zero at some point $x_0$ in interval $I$.

\[\exists x_0 \in I: W(x_0)=0 \iff y_1 \text{ and } y_2 \text{ are linearly dependent}\]

ii. If the Wronskian $W=0$ at one point $x=x_0$ in interval $I$, then $W=0$ at all points in interval $I$.

\[\exists x_0 \in I: W(x_0)=0 \implies \forall x \in I: W(x)=0\]

In other words, if there exists a point $x_1$ in interval $I$ where $W\neq 0$, then $y_1$ and $y_2$ are linearly independent on interval $I$.

\[\begin{align*} \exists x_1 \in I: W(x_1)\neq 0 &\implies \forall x \in I: W(x)\neq 0 \\ &\implies y_1 \text{ and } y_2 \text{ are linearly independent} \end{align*}\]

The Wronskian was first introduced by the Polish mathematician Józef Maria Hoene-Wroński, and was named after him by the Scottish mathematician Sir Thomas Muir in 11882 HE, after Wroński’s death.

Proof

i. (a)

Suppose $y_1$ and $y_2$ are linearly dependent on interval $I$. Then either equation ($\ref{eqn:linearly_dependent}$a) or ($\ref{eqn:linearly_dependent}$b) holds on interval $I$. If equation ($\ref{eqn:linearly_dependent}$a) holds, then:

\[W(y_1, y_2) = y_1y_2^{\prime} - y_2y_1^{\prime} = ky_2ky_2^{\prime} - y_2ky_2^{\prime} = 0\]

Similarly, if equation ($\ref{eqn:linearly_dependent}$b) holds:

\[W(y_1, y_2) = y_1y_2^{\prime} - y_2y_1^{\prime} = y_1ly_1^{\prime} - ly_1y_1^{\prime} = 0\]

Thus, we’ve shown that the Wronskian $W(y_1, y_2)=0$ for all $x$ in interval $I$.

i. (b)

Conversely, suppose that $W(y_1, y_2)=0$ at some point $x = x_0$. We’ll show that $y_1$ and $y_2$ are linearly dependent on interval $I$. Consider the system of linear equations in unknowns $k_1$, $k_2$:

\[\begin{gather*} k_1y_1(x_0) + k_2y_2(x_0) = 0 \\ k_1y_1^{\prime}(x_0) + k_2y_2^{\prime}(x_0) = 0 \end{gather*} \label{eqn:linear_system}\tag{11}\]

This can be expressed as the vector equation:

\[\left[\begin{matrix} y_1(x_0) & y_2(x_0) \\ y_1^{\prime}(x_0) & y_2^{\prime}(x_0) \end{matrix}\right] \left[\begin{matrix} k_1 \\ k_2 \end{matrix}\right] = 0 \label{eqn:vector_equation}\tag{12}\]

The coefficient matrix is:

\[A = \left[\begin{matrix} y_1(x_0) & y_2(x_0) \\ y_1^{\prime}(x_0) & y_2^{\prime}(x_0) \end{matrix}\right]\]

and its determinant is $W(y_1(x_0), y_2(x_0))$. Since $\det(A) = W=0$, $A$ is a singular matrix without an inverse matrix, so the system of equations ($\ref{eqn:linear_system}$) has a non-trivial solution $(c_1, c_2)$ where at least one of $c_1$ or $c_2$ is non-zero. Now consider the function:

\[y(x) = c_1y_1(x) + c_2y_2(x)\]

Since equation ($\ref{eqn:homogeneous_linear_ode_with_var_coefficients}$) is homogeneous and linear, by the superposition principle, this function is a solution to equation ($\ref{eqn:homogeneous_linear_ode_with_var_coefficients}$) on interval $I$. From equation ($\ref{eqn:linear_system}$), this solution satisfies the initial conditions $y(x_0)=0$, $y^{\prime}(x_0)=0$.

However, the trivial solution $y^* \equiv 0$ also satisfies the same initial conditions $y^*(x_0)=0$, ${y^*}^{\prime}(x_0)=0$. Since the coefficients $p$ and $q$ in equation ($\ref{eqn:homogeneous_linear_ode_with_var_coefficients}$) are continuous, the Existence and Uniqueness Theorem for Initial Value Problems guarantees that the solution is unique, so $y \equiv y^*$. This means that on interval $I$:

\[c_1y_1 + c_2y_2 \equiv 0\]

Since at least one of $c_1$ or $c_2$ is non-zero, condition ($\ref{eqn:linearly_independent}$) is not satisfied, which means that $y_1$ and $y_2$ are linearly dependent on interval $I$.

ii.

If the Wronskian $W(x_0)=0$ at some point $x_0$ in interval $I$, then by i.(b), $y_1$ and $y_2$ are linearly dependent on interval $I$, and by i.(a), $W\equiv 0$ on interval $I$. Therefore, if there exists a point $x_1$ in interval $I$ where $W(x_1)\neq 0$, then $y_1$ and $y_2$ must be linearly independent. $\blacksquare$

The General Solution Includes All Solutions

Existence of General Solution

If $p(x)$ and $q(x)$ are continuous on an open interval $I$, then differential equation ($\ref{eqn:homogeneous_linear_ode_with_var_coefficients}$) has a general solution on interval $I$.

Proof

By the Existence and Uniqueness Theorem for Initial Value Problems, differential equation ($\ref{eqn:homogeneous_linear_ode_with_var_coefficients}$) has a solution $y_1(x)$ on interval $I$ satisfying the initial conditions:

\[y_1(x_0) = 1, \qquad y_1^{\prime}(x_0) = 0\]

and a solution $y_2(x)$ on interval $I$ satisfying the initial conditions:

\[y_2(x_0) = 0, \qquad y_2^{\prime}(x_0) = 1\]

The Wronskian of these two solutions at $x=x_0$ is non-zero:

\[W(y_1(x_0), y_2(x_0)) = y_1(x_0)y_2^{\prime}(x_0) - y_2(x_0)y_1^{\prime}(x_0) = 1\cdot 1 - 0\cdot 0 = 1\]

By the Wronskian method for determining linear dependence/independence, $y_1$ and $y_2$ are linearly independent on interval $I$. Therefore, these two solutions form a basis for the solutions of equation ($\ref{eqn:homogeneous_linear_ode_with_var_coefficients}$) on interval $I$, and the general solution $y = c_1y_1 + c_2y_2$ with arbitrary constants $c_1$, $c_2$ exists on interval $I$. $\blacksquare$

Non-existence of Singular Solutions

If differential equation ($\ref{eqn:homogeneous_linear_ode_with_var_coefficients}$) has continuous coefficients $p(x)$ and $q(x)$ on an open interval $I$, then every solution $y=Y(x)$ of equation ($\ref{eqn:homogeneous_linear_ode_with_var_coefficients}$) on interval $I$ has the form:

\[Y(x) = C_1y_1(x) + C_2y_2(x) \label{eqn:particular_solution}\tag{13}\]

where $y_1$, $y_2$ form a basis for the solutions of equation ($\ref{eqn:homogeneous_linear_ode_with_var_coefficients}$) on interval $I$, and $C_1$, $C_2$ are appropriate constants.
In other words, equation ($\ref{eqn:homogeneous_linear_ode_with_var_coefficients}$) does not have singular solutions (solutions that cannot be obtained from the general solution).

Proof

Let $y=Y(x)$ be any solution of equation ($\ref{eqn:homogeneous_linear_ode_with_var_coefficients}$) on interval $I$. By the Existence of General Solution theorem, equation ($\ref{eqn:homogeneous_linear_ode_with_var_coefficients}$) has a general solution on interval $I$:

\[y(x) = c_1y_1(x) + c_2y_2(x) \label{eqn:general_solution}\tag{14}\]

We need to show that for any $Y(x)$, there exist constants $c_1$, $c_2$ such that $y(x)=Y(x)$ on interval $I$. First, let’s show that we can find values of $c_1$, $c_2$ such that $y(x_0)=Y(x_0)$ and $y^{\prime}(x_0)=Y^{\prime}(x_0)$ for any chosen $x_0$ in interval $I$. From equation ($\ref{eqn:general_solution}$), we have:

\[\begin{gather*} \left[\begin{matrix} y_1(x_0) & y_2(x_0) \\ y_1^{\prime}(x_0) & y_2^{\prime}(x_0) \end{matrix}\right] \left[\begin{matrix} c_1 \\ c_2 \end{matrix}\right] = \left[\begin{matrix} Y(x_0) \\ Y^{\prime}(x_0) \end{matrix}\right] \end{gather*} \label{eqn:vector_equation_2}\tag{15}\]

Since $y_1$ and $y_2$ form a basis, the determinant of the coefficient matrix, $W(y_1(x_0), y_2(x_0))\neq 0$, so equation ($\ref{eqn:vector_equation_2}$) can be solved for $c_1$ and $c_2$. Let’s call the solution $(c_1, c_2) = (C_1, C_2)$. Substituting these values into equation ($\ref{eqn:general_solution}$), we get the particular solution:

\[y^*(x) = C_1y_1(x) + C_2y_2(x).\]

Since $C_1$ and $C_2$ satisfy equation ($\ref{eqn:vector_equation_2}$):

\[y^*(x_0) = Y(x_0), \qquad {y^*}^{\prime}(x_0) = Y^{\prime}(x_0)\]

By the uniqueness part of the Existence and Uniqueness Theorem, $y^* \equiv Y$ on interval $I$. $\blacksquare$