Q34E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Wronskian. For any two differentiable functions y₁ and y₂, the function (18) W[y₁,y₂](t) = y₁(t)_,y'₂(t) - y'₁(t),y₂(t) is called the Wronskian of y₁ and y₂. This function plays a crucial role in the proof of Theorem 2.

 

1. Show that W[y₁,y₂] can be conveniently expressed as the determinant $\mathit{W}\mathbf{[}{\mathbf{y}}_{\mathbf{1}}\mathbf{,}{\mathbf{y}}_{\mathbf{2}}\mathbf{]}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{\left[}\begin{array}{cc}{\mathbf{y}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}& {\mathbf{y}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\\ \mathbf{y}{\mathbf{\text{'}}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}& \mathbf{y}{\mathbf{\text{'}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\end{array}\mathbf{\right]}$.
2. Let y₁(t), y₂(t) be a pair of solutions to the homogeneous equation ay" + by' + cy = 0 (with $\mathit{a}\mathbf{\ne }\mathbf{0}$ on an open interval I. Prove that y₁(t) and y₂(t) are linearly independent on I if and only if their Wronskian is never zero on I. [Hint: This is just a reformulation of Lemma 1.]
3. Show that if y₁(t) and y₂(t) are any two differentiable functions that are linearly dependent on I, then their Wronskian is identically zero on I.

Step-by-Step Solution

Answer

1. Wronskian can be written as in determinant form $\mathit{W}\mathbf{[}{\mathbf{y}}_{\mathbf{1}}\mathbf{,}{\mathbf{y}}_{\mathbf{2}}\mathbf{]}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{\left[}\begin{array}{cc}{\mathbf{y}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}& {\mathbf{y}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\\ \mathbf{y}{\mathbf{\text{'}}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}& \mathbf{y}{\mathbf{\text{'}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\end{array}\mathbf{\right]}$.
2. y₁ and y₂ are linearly independent on I if and only if their Wronskian is never zero.
3. y₁ and y₂ are linearly dependent then Wronskian is identically zero.

Step 1: Wronskian can be written as in determinant.

We know that W[y₁,y₂](t) = y₁(t), y'₂(t) - y'₁(t), y₂(t)

And $\left|\begin{array}{cc}{y}_1\left(t\right)& y_2\left(t\right)\\ y{\text{'}}_1\left(t\right)& y{\text{'}}_2\left(t\right)\end{array}=\right|y_1\left(t\right),y{\text{'}}_2\left(t\right)-y{\text{'}}_1\left(t\right),y_2\left(t\right)$

Therefore, Wronskian can be written as;

$W[{\mathrm{y}}_1,{\mathrm{y}}_2]\left(t\right)=\left[\begin{array}{cc}{\mathrm{y}}_1\left(\mathrm{t}\right)& {\mathrm{y}}_2\left(\mathrm{t}\right)\\ \mathrm{y}{\text{'}}_1\left(\mathrm{t}\right)& \mathrm{y}{\text{'}}_2\left(\mathrm{t}\right)\end{array}\right]$

Step 2: Show both are linearly independent.

If and are linearly independent then $y_1\ne c{y}_2$ then  

$y_1\left(t\right),y{\text{'}}_2\left(t\right)-y{\text{'}}_1\left(t\right),y_2\left(t\right)\ne 0$

And

$\begin{array}{rcl}{y}_1\left(t\right),y{\text{'}}_2\left(t\right)-y{\text{'}}_1\left(t\right),y_2\left(t\right)& \ne & 0\\ y_1\left(t\right),y{\text{'}}_2\left(t\right)& \ne & y{\text{'}}_1\left(t\right),y_2\left(t\right)\\ y_1& \ne & c{y}_2\end{array}$

I.e. both are linearly independent.

 

Therefore y₁ and y₂ are linearly independent on I if and only if their Wronskian is never zero.

Step 3: Prove that Wronskian is identically zero

If and are linearly dependent then y₁ = cy₂ then y'₁ = cy'₂

$\begin{array}{rcl}W[y_1,y_2]\left(t\right)& =& y_1\left(t\right),y{\text{'}}_2\left(t\right)-y{\text{'}}_1\left(t\right),y_2\left(t\right)\\ & =& cy{\text{'}}_2{y}_2-cy{\text{'}}_2{y}_2\\ & =& 0\end{array}$ 

Therefore y₁ and y₂ are linearly dependent then Wronskian is identically zero.