Firstly let us define what this Wronskian thing is.

Say you have 2 solutions to a second order homogenous equation, this is an important point, they have to be solutions to a second order homogeneous equation. Just a reminder, this is what a second order homogeneous equation looks like:

$$y”+py’+qy=0$$

Lets call the two solutions $y_1$ and $y_2$. The Wronskian of them will be given by:

So whenever you need to compute the Wronskian of something, you just need to find the determinant of the matrix above.

Now that is out of the way, let’s get to know our theorem, and as with any theorem you will see in MTH2032, there is a list of conditions you have to satisfy first, before you can use it. The conditions are:

- $p$ and $q$ are continuous functions on an interval $I$
- $y_1$ and $y_2$ are solutions to the homogeneous equation (we already talked about this)

Once these conditions are met, you can use the Wronskian Theorem which states:

- $y_1,y_2$ are linearly independent on $I$ if and only if $W(y_1,y_2)$ is never 0 in your interval $I$
- If the $W(y_1,y_2)=0$ at any point in $I$, then $W(y_1,y_2)=0$ for the entire interval

Before we jump into the proof, let’s take a look at a concrete example.

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*Example 1:*

$y_1=x^2$ *and *$y_2=x^4$* are two solutions to the homogeneous equation*

$$y”+3xy’-\frac{5}{x}y=0$$

*Are these two solutions linearly independent on the interval *$(0,\infty)$?

First thing we need to do is check our conditions, since $3x$ and $\frac{5}{x}$ are continuos on $(0,\infty)$ and they told us that $y_1$ and $y_2$ are solutions to it, we can use our theorem.

The Wronskian for these two solutions are:

On our interval $(0,\infty)$, $2x^5$ is never zero, so our two solutions are linearly independent. Note here that we don’t care that is zero at $x=0$, because that point is not in the interval we are interested in!

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Now that we know how to use our theorem let’s see if we can understand why it works with a proof.

proof:

Before we begin the proof just a reminder that the assumptions we can work with for this theorem are:

- $p$ and $q$ are continuous functions on an interval $I$
- $y_1$ and $y_2$ are solutions to the homogeneous equation

Keep these in mind as we will call upon these later!

Notice that the first conclusion of the Wronskian Theorem is an *iff *statement, so the proof that follows has to prove that the statement is true both ways. So we need to prove 2 things:

- If $W(y_1,y_2)$ is never 0 in your interval $I$ then $y_1,y_2$ are linearly independent on $I$.
- If $y_1,y_2$ are linearly independent on $I$ then $W(y_1,y_2)$ is never 0 in your interval $I$

To prove 1, we will use a proof by contrapositive! So we will prove that:

* if *$W(y_1,y_2)=0$

*$I$,*

**for some point in the interval***$x_0$,*

**lets call it***then*$y_1,y_2$

**are linearly dependent.**First note the following the system of equations:

Let’s think about this system. If we could find two solutions $c_1,c_2$ where at least one of which is non-zero, that would mean that they are linearly ** dependent**. Now, if you have found a solution to the system above, you have automatically found an

*infinite*number of solutions. You can see this by noting that since the right hand side of both equations are zero, any multiple of the $c_1,c_2$ pair you found is still going to give the desired solution of zero.

With that mind consider the co-efficient matrix of the above system:

And you should have realised that the determinant of this matrix is the Wronskian of $y_1$ and $y_2$. Furthermore with our discussion above (blue text) if there are infinitely many solutions to it, then the determinant has to be zero, which means the Wronskian has to be zero. This leads us to a remark that we will use to prove (1). (NOTE: we have not proved 1 but we are on our way to prove it. Furthermore the reasoning above is just to help us see why the following remark works and is not a full proof of it). The remark is that:

**The Wronskian is zero if and only if there exists** $c_1,c_2$ **where at least one of which is non-zero that solves the system**

To begin our proof of (1) we will first assume that $W(y_1,y_2)$ at our chosen point $x_0$ is zero. This means that there exists $c_1,c_2$ not both zero that solves our red and green system above. Now since $y_1$ and $y_2$ are solutions to a homogeneous equation $L(y)$ (check our assumptions up the top!) we can use the *superposition principle* to form another solution to $L(y)=0$, lets call this new solution, $z$, where: $$z=c_1y_1+c_2y_2$$ and $$L(z)=0.$$ Now I hope you noticed this, but if you didn’t, $z$ is just the red equation of our system, and no matter what you substitute into this equation, you will always get zero. This means that: $$z(x_0)=0.$$ Furthermore, the green equation is $z’$ and by similar reasoning: $$z'(x_0)=0.$$

Now this is where we need to get fancy, out of shear necessity, and not choice.

We note that our new solution $z$ is a solution to the following __Initial Value Problem:__ $$L(y)=0$$ with conditions: $$y(x_0)=0,y'(x_0)=0.$$

Furthermore if you tried $y=0$ you would find that it is also a solution to this IVP. Now the question is, does there exist a unique solution to this IVP or are there many solutions to it? If only there was a theorem to help us. Oh wait, there is! Enter the fundamental existence-uniqueness theorem for second order linear ordinary differential equations! I won’t go into too much detail here, but pretty much:

- $p$ and $q$ are continuous (check our assumptions!)
- $x_0$ is in that continuos interval
- our initial conditions $y(x_0)=0,y'(x_0)=0$ are constants (0 is a constant!)

means that there exists a unique solution to the IVP! So this means that in fact, $z=0$ which means that $$c_1y_1 +c_2y_2=0$$ for all $x$. This means that if the Wronskian is zero, then $y_1$ and $y_2$ are linearly dependent! Here we have just proven the contrapositive of (1) and hence have proven (1).

To prove (2) we will use the same contrapositive trick, we will prove the the following:

* if *$y_1,y_2$

**are linearly dependent on**$I$

**then**$W(y_1,y_2)=0$

**for every point in**$I$.

Here we will also begin by assuming our two solutions $y_1,y_2$ are ** linearly dependent** with the very definition of linear dependence which states that there exists $c_1,c_2$ not both zero (i.e non-trivial) such that:$$c_1y_1 +c_2y_2=0$$ for all $x \in I$. If we differentiate this, we get our green and red system of equations above, and since we assumed that $y_1,y_2$ are linearly dependent, $c_1,c_2$ are non-trivial solutions to that system. Again from the reasoning above, this means that the determinant of the co-efficient matrix is zero, and thus the wronskian is also zero. (check out the section with blue text if this is confusing!) This proves the contrapositive of (2) and hence proves (2).

So now we have proved the first part of our theorem! i.e we have proved that:

$y_1,y_2$ are linearly independent on $I$ if and only if $W(y_1,y_2)$ is never 0 in your interval $I$

Now we need to prove the second part of our theorem, namely:

If the $W(y_1,y_2)=0$ at any point in $I$, then $W(y_1,y_2)=0$ for the entire interval

This was already proved! If we look back at our proof of (1) we showed that if the Wronskian is 0, then $y_1,y_2$ are linearly dependent. Then in our proof of (2), we showed that if they are linearly dependent, the Wronskian is zero for all $x \in I$. Done! (if you couldn’t follow all those words take a look at the diagram below)

So we showed that if the Wronskian is zero at 1 point, it is zero at all points in your interval! DONE!

Okay so that was the Wronskian! And remember, the Wronskian is useful because it tells us if two solutions are linearly independent or not. Why would we want to know that? Enter the Basis theorem. The topic of the next post.

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