# Proof verification for the following differential equation!

The question is as follows

Let $f_1$ & $f_2$ be two solutions to the following second order homogeneous linear differential equation

$$a_0(x)\frac{d^2y}{dx^2}+a_1(x)\frac{dy}{dx}+a_2(x)y=0$$

a)Show that $f_1$ & $f_2$ has a common zero at a point $x_0$ of the interval $a \leq x \leq b$ then $f_1$ & $f_2$ are linearly dependent on $a \leq x \leq b$.

My work is as follows

We know that they can be written as a linear combination of solution!

$$c_1f_1(x)+c_2f_2(x)=0$$

$$c_1f_1'(x)+c_2f_2'(x)=0$$

where $c_1$ & $c_2$ are not both zeroes

Since

$f_1(x_0)=0, f_2(x_0)=0$ on interval $a \leq x \leq b$

Then $f(x)=0$

$f_1'(x_0)=m_1,f_2'(x_0)=m_2$

For point $x_0$ on interval $a \leq x \leq b$

We have the theorem where if a system of two homogeneous linear algebraic equations has a non-trivial solution if the determinant of the system is zero.

Non-trivial means to mean me that $c_1$ & $c_2$ are not both zeroes.

Checking the determinant of the homogeneity of the linear equation!

$$c_1f_1(x_0)+c_2f_2(x_0)=f(x)$$

$$c_1f_1'(x_0)+c_2f_2'(x_0)=f'(x)$$

$$W[f_1(x),f_2(x)]=\begin{bmatrix}f_1(x_0) & f_2(x_0) \\f_1'(x_0) & f_2'(x_0) \end{bmatrix}$$

$$W[f_1(x),f_2(x)]=0$$ on interval $a \leq x \leq b$

If the system is linearly independent then it will be a contradiction.

2) Show that $f_1$ & $f_2$ have relative maxima at common point $x_0$ of interval $a \leq x \leq b$ then $f_1$ & $f_2$ are linearly dependent!

Relative maxima occurs at $x_0$ when $f_1'(x_0)$ is undefined or zeroes.

$$c_1f_1(x_0)+c_2f_2(x_0)=0$$

$$c_1f_1'(x_0)+c_2f_2'(x_0)=0$$

Suppose that $f_1(x_0)=k_1$ and $f_2(x_0)=k_2$

$f_1'(x_0)=0$ & $f_2'(x_0)=0$

$$W[f_1(x),f_2(x)]=f_1(x)f_2'(x)-f_2(x)f_1'(x)$$

One condition for the Wronskian definition to be zeroes that is when $f_1'(x_0)=0$ & $f_2'(x_0)=0$ if and only if $x_0$ is a relative extremum for the two function! Therefore, it is linearly dependent since wronskian is again 0.

Can someone please provide a better way of doing this?

## 1 Answer

In both cases, you have $W[f_1,f_2](x_0)=0$, so let's assume this and show that $W[f_1,f_2](x)=0$ everywhere. Let us also suppose $a_0(x)$ is nonzero on $[a,b]$. Your two solutions $f_1$ and $f_2$ satisfy \begin{align} a_0 f_2''+a_1 f_2'+a_2 f_2&=0,\\ a_0 f_1''+a_1 f_1'+a_2 f_1&=0. \end{align} Multiply the first equation by $f_1$, the second equation by $f_2$, and subtract. If $W:=f_1f_2'-f_2f_1'$, then the difference of the equations can be rewritten as a first order ODE for $W$: $$a_0 W'+a_1W=0.$$ Observe that $W(x)=0$ is a solution to this equation on $[a,b]$ which satisfies $W(x_0)=0$. Since $a_0\neq 0$, this solution is unique, and we conclude $W(x)=0$ on $[a,b]$, implying $f_1$ and $f_2$ are linearly dependent.

• How about the the second part of the question? – Crazy Jul 10 '17 at 1:48
• @Crazy If $f_1$ and $f_2$ have relative maximas at $x_0$, then $f_1'(x_0)=f_2'(x_0)=0$, so $W[f_1,f_2](x_0)=0$ as in the first part. – user254433 Jul 10 '17 at 1:49
• Same logic to tackle the question? – Crazy Jul 10 '17 at 1:50
• Yes, the same in both parts. – user254433 Jul 10 '17 at 1:53