Understanding Wronskian of solutions to homoeneous 2nd order linear DE

[zenterix]

Homework Statement

Below are a few calculations involving the Wronskian of two solutions $y_1$ and $y_2$ of the homogeneous equation

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

on some interval $[a,b]$.

Relevant Equations

I would like to understand the result that the Wronskian can only either be identically zero or else identically non-zero.

The Wronskian of these two solutions is also a function of $x$.

$$W=y_1y_2'-y_1'y_2$$

$$W'=y_1y_2''+y_1'y_2'-y_1'y_2'-y_1''y_2$$

$$=y_1y_2''-y_1''y_2$$

The two solutions satisfy

$$y_1''+Py_1'+Qy_1=0$$

$$y_2''+Py_2'+Qy_2=0$$

Multiply the first by $y_2$ and the second by $y_1$ and subtract the first from the second to obtain

$$(y_1y_2''-y_2y_1''')+P(y_1y_2'-y_2y_1')=0$$

$$W'+PW=0$$

This is a differential equation in the Wronskian.

The solution is

$$W(x)=ce^{-\int Pdx}\tag{1}$$

At this point we can conclude that the Wronskian is either identically zero or else identically non-zero.

What determines the two cases is the constant $c$.

My question is about this constant.

Let's solve (2) in more steps

$$\frac{1}{W}W'=-P$$

$$\ln{\left (\frac{W(x)}{W(x_i)}\right )}=-\int_{x_i}^x Pdx$$

$$\frac{W(x)}{W(x_i)}=e^{-\int_{x_i}^x Pdx}$$

$$W(x)=W(x_i)e^{-\int_{x_i}^x Pdx}$$

$$=W(x_i)e^{-\int Pdx + C}$$

$$=W(x_i)e^Ce^{-\int Pdx}$$

If $W(x)=0$ then it must be that $W(x_i)=0$. Since we can choose any $x_i$ in the interval then it must be that $W(x)=0$ on the interval.

If $W(x)\neq 0$ then it must be that $W(x_i)\neq 0$ and so $W(x)\neq 0$ for all $x$.

Therefore, if the Wronskian is zero at any point in the interval, it is zero at all points in the interval, and if it is nonzero at any point then it is non-zero at all points.

Another important result is that if the two solutions on an interval are linearly dependent, then their Wronskian is identically zero on this interval.

I think I have lost track of what my question is.

The result we derived for the Wronskian is actually a general result, right.

If we have any equation $y'+ay=0$ then the solution is $y(x)=y(x_i)e^{ax}$ and so if $y(x)=0$ at any $x$ then it must be the case that $y=0$.

 

[fresh_42]

I think I have lost track of what my question is.

So did I.

The result we derived for the Wronskian is actually a general result, right.

Not sure what you meant by result.

All I can see was that you analyzed an equation $a=b\cdot e^c$ and concluded that $a=0$ if and only if $b=0.$ I guess, that is a general result.

If we have any equation $y'+ay=0$ then the solution is $y(x)=y(x_i)e^{ax}$ and so if $y(x)=0$ at any $x$ then it must be the case that $y=0$.

Isn't $y(x)=0$ at any $x$ the definition of $y=0$ without using a differential equation?

Maybe your question was what happens if $\int Pdx \to \infty \;$?

 

[zenterix]

Not sure what you meant by result.

By result I mean that all we did in the end was solve a very simple differential equation $W'+PW=0$ and conclude that the solution can either be identically zero or identically non-zero.

Since I hadn't seen this interpretation of a solution to this differential equation I thought it was some specific characteristic of a Wronskian, but in fact it is a characteristic of any function that satisfies this very simple and common differential equation.

So yes, indeed all I did is as you describe. But note that the context for arriving at these simple calculations was a slightly more complicated result: that the Wronskian of any two solutions to a homogeneous 2nd order linear DE is identically zero or identically non-zero.