Quick Change of Variables Question

[thatboi]

Hey all,
I am currently struggling with a change of variables step in my calculations.
Suppose the solutions $f_{1}(x)$ and $f_{2}(x)$ of the following system of differential equations is known:

Now the system I wish to solve is:

Upon first glance, it seems that the association $f_{2}(-x) \leftrightarrow f_{1}(x)$ and $g_{2}(-x) \leftrightarrow g_{1}(x)$ will do the trick and we would be done.
However, I tried making the change of variables $y=-x$ in equation (117) and got the following:

which is now the same form as equation (118).
Suddenly, it seems like the solution is now $f_{2}(y) = f_{2}(-x) \leftrightarrow f_{1}(x)$ and $g_{2}(y) = g_{2}(-x) \leftrightarrow g_{1}(x)$
and I am left with a contradiction. Could someone point out to me where the mistake is? I feel like it is something simple that I am not seeing.
Thanks!

 

[thatboi]

Hey all,
I am currently struggling with a change of variables step in my calculations.
Suppose the solutions $f_{1}(x)$ and $f_{2}(x)$ of the following system of differential equations is known:
View attachment 313838
Now the system I wish to solve is:
View attachment 313839
Upon first glance, it seems that the association $f_{2}(-x) \leftrightarrow f_{1}(x)$ and $g_{2}(-x) \leftrightarrow g_{1}(x)$ will do the trick and we would be done.
However, I tried making the change of variables $y=-x$ in equation (117) and got the following:
View attachment 313840
which is now the same form as equation (118).
Suddenly, it seems like the solution is now $f_{2}(y) = f_{2}(-x) \leftrightarrow f_{1}(x)$ and $g_{2}(y) = g_{2}(-x) \leftrightarrow g_{1}(x)$
and I am left with a contradiction. Could someone point out to me where the mistake is? I feel like it is something simple that I am not seeing.
Thanks!

Sorry there was a typo here, the second sentence above should have said "Suppose the solutions $f_{1}(x)$ and $g_{1}(x)$ off the following system of differential equations is known..."

 

[PeroK]

Hey all,
I am currently struggling with a change of variables step in my calculations.
Suppose the solutions $f_{1}(x)$ and $f_{2}(x)$ of the following system of differential equations is known:
View attachment 313838
Now the system I wish to solve is:
View attachment 313839
Upon first glance, it seems that the association $f_{2}(-x) \leftrightarrow f_{1}(x)$ and $g_{2}(-x) \leftrightarrow g_{1}(x)$ will do the trick and we would be done.
However, I tried making the change of variables $y=-x$ in equation (117) and got the following:
View attachment 313840
which is now the same form as equation (118).
Suddenly, it seems like the solution is now $f_{2}(y) = f_{2}(-x) \leftrightarrow f_{1}(x)$ and $g_{2}(y) = g_{2}(-x) \leftrightarrow g_{1}(x)$
and I am left with a contradiction. Could someone point out to me where the mistake is? I feel like it is something simple that I am not seeing.
Thanks!

Where's the contradiction?

 

[pasmith]

These functions depend on the parameter $a$; suppressing that dependence may lead you astray.

If we define $f_3(x;a) = f_2(-x;a)$ then (117) becomes $$\begin{pmatrix} \frac{d}{dx}f_3(x;a) \\\frac{d}{dx}g_3(x;a)\end{pmatrix} = \begin{pmatrix} 1/x & -a \\ -a & 1/x\end{pmatrix} \begin{pmatrix} f_3(x;a) \\ g_3(x;a) \end{pmatrix}.$$ It is then obvious that this is the same as (118) with $a$ replaced by $-a$. So $$f_2(-x;a) = f_3(x;a) = f_1(x;-a).$$

 

[thatboi]

Where's the contradiction?

Sorry there was another typo in my post, it should say "Upon first glance, it seems the association $f_{2}(-x) \leftrightarrow f_{1}(x)$ and $g_{2}(-x) \leftrightarrow -g_{1}(x)$ will do the trick..."
So there is a missing negative sign for $g_{2}(-x)$ between the 2 associations.

 

[PeroK]

Sorry there was another typo in my post, it should say "Upon first glance, it seems the association $f_{2}(-x) \leftrightarrow f_{1}(x)$ and $g_{2}(-x) \leftrightarrow -g_{1}(-x)$ will do the trick..."
So there is a missing negative sign for $g_{2}(-x)$ between the 2 associations.

I'm not sure what you mean by these "associations". You're looking an equation, such as $f_2(x) = f_1(x)$ or $f_2(x) = -f_1(x)$ or something like that.

 

[thatboi]

I'm not sure what you mean by these "associations". You're looking an equation, such as $f_2(x) = f_1(x)$ or $f_2(x) = -f_1(x)$ or something like that.

Yes, by association I mean like $f_{2}(-x)$ would be the same as $f_{1}(x)$ in form but just with $x$ swapped for $-x$ and $g_{2}(-x) = -g_{1}(x)$

 

[PeroK]

Yes, by association I mean like $f_{2}(-x)$ would be the same as $f_{1}(x)$ in form but just with $x$ swapped for $-x$ and $g_{2}(-x) = -g_{1}(x)$

I don't know what that means. Note that $x$ and $y$ are dummy variables. They have no intrinsic meaning, other than as arguments for your functions.

 

[PeroK]

For example, if we have $f(-x) = -x$, then that is equivalent to $f(x) = x$. Draw the graphs if you need to. It's the same function.

 

[thatboi]

For example, if we have $f(-x) = -x$, then that is equivalent to $f(x) = x$. Draw the graphs if you need to. It's the same function.

I don't quite understand. So the situation was, if I knew the solutions $f_{1}(x)$ and $g_{1}(x)$ to the system in equation (118) and now I have a separate system equation (117) that I wish to find the solutions to. Then I can notice that because (118) and (117) have the same forms, the solution to equation (117) is just $f_{2}(-x) = f_{1}(x)$ and $g_{2}(-x) = -g_{1}(x)$, that is, I can write the solution to (117) in terms of solutions to (118). But once I made a change of variables to change equation (117) to (119) and I try to write the solution to equation (119) in terms of the solution to equation (118), I end up with a different set of solutions.

 

[PeroK]

I don't quite understand. So the situation was, if I knew the solutions $f_{1}(x)$ and $g_{1}(x)$ to the system in equation (118) and now I have a separate system equation (117) that I wish to find the solutions to. Then I can notice that because (118) and (117) have the same forms, the solution to equation (117) is just $f_{2}(-x) = f_{1}(x)$ and $g_{2}(-x) = -g_{1}(x)$, that is, I can write the solution to (117) in terms of solutions to (118). But once I made a change of variables to change equation (117) to (119) and I try to write the solution to equation (119) in terms of the solution to equation (118), I end up with a different set of solutions.

You have to do a correct change of variable. That means changing the derivative approriately. For example,$$\frac d {dx}f(-x) = -x$$is not the same as $$\frac d {dx}f(x) = x$$

 

[thatboi]

You have to do a correct change of variable. That means changing the derivative approriately. For example,$$\frac d {dx}f(-x) = -x$$is not the same as $$\frac d {dx}f(x) = x$$

Right, I thought there might be a mistake in my chain rule somewhere for (119), but if I write everything in terms of $y=-x$, is it not $\frac{d}{dx}f(-x) = \frac{dy}{dx}\frac{d}{dy}f(y) = -\frac{d}{dy}f(y)$, which is what I had in (119) already?

 

[PeroK]

Right, I thought there might be a mistake in my chain rule somewhere for (119), but if I write everything in terms of $y=-x$, is it not $\frac{d}{dx}f(-x) = \frac{dy}{dx}\frac{d}{dy}f(y) = -\frac{d}{dy}f(y)$, which is what I had in (119) already?

Yes.

Upon first glance, it seems that the association $f_{2}(-x) \leftrightarrow f_{1}(x)$ and $g_{2}(-x) \leftrightarrow g_{1}(x)$ will do the trick and we would be done.

If you mean $f_{2}(-x) = f_{1}(x)$, then that is not right.

However, I tried making the change of variables $y=-x$ in equation (117) and got the following:
View attachment 313840
which is now the same form as equation (118).

Yes. It's the same system of equations.

Suddenly, it seems like the solution is now $f_{2}(y) = f_{2}(-x) \leftrightarrow f_{1}(x)$ and $g_{2}(y) = g_{2}(-x) \leftrightarrow g_{1}(x)$

Now you are confusing yourself with trying to reconcile the dummy variables $x$ and $y$ and these "associations".

 

[thatboi]

Yes.

If you mean $f_{2}(-x) = f_{1}(x)$, then that is not right.

Yes. It's the same system of equations.

Now you are confusing yourself with trying to reconcile the dummy variables $x$ and $y$ and these "associations".

Sorry, I believe $f_{2}(-x) = f_{1}(x)$ and $g_{2}(-x) = -g_{1}(x)$ should be the correct solution.
That being said, if I were to work with the $y$ variables system, since they are the same form as (118), I could take the solution to (118), which we already know and swap out the $x$ variable for $y$ right, since these are just dummy variables anyway. Then in the final step could I not plug back my original transformation $y=-x$?

 

[PeroK]

Sorry, I believe $f_{2}(-x) = f_{1}(x)$ and $g_{2}(-x) = -g_{1}(x)$ should be the correct solution.

That's not right. Let's take an example that we can explicitly solve:
$$\frac d{dx} f_1(x) = af_1(x)$$Has a solution $f_1(x) = e^{ax}$.

Now, take:
$$\frac d{dx} f_2(-x) = -af_2(-x)$$This also has the same solution $f_2(x) = e^{ax}$

We can check this:
$$\frac d{dx} f_2(-x) = \frac d{dx}e^{-ax} = -ae^{-ax} = -af_2(-x)$$We can also see that your proposed solution: $f_2(-x) = f_1(x) = e^{ax}$ is not correct:
$$\frac d{dx} f_2(-x) = \frac d{dx}e^{ax} = ae^{ax} = af_2(-x)$$

 

[PeroK]

PS I suspect the root of your problem may be that you don't understand the concept of a dummy variable and you are unable to express an equation for $f_2(x)$. Instead, you are fixed on $-x$ as your dummy variable for $f_2$. For example. If we have:
$$f(-x) = -2x + 1$$then we can immediately write down:
$$f(x) = 2x + 1$$