Change of Variables for Elliptic Integral

[McCoy13]

Homework Statement

Given the differential equation

$$u_{xx}+3u_{yy}-2u_{x}+24u_{y}+5u=0$$

use the substitution of dependent variable

$$u=ve^{ \alpha x + \beta y}$$

and a scaling change of variables

$$y'= \gamma y$$

to reduce the differential equation to

$$v_{xx}+v_{yy}+cv=0$$

Homework Equations

I have no idea

The Attempt at a Solution

I tried a direct substitution of both variables:

$$u_{x}=\alpha ve^{\alpha x+\beta y}$$
$$u_{xx}=\alpha^{2}ve^{\alpha x+\beta y}$$
$$u_{y}=\beta ve^{\alpha x+\beta y}$$
$$u_{yy}=\beta^{2} ve^{\alpha x+\beta y}$$

Plugging in this gives

$$\alpha^{2} ve^{\alpha x+\beta y}+3\beta^{2}ve^{\alpha x+\beta y}-2\alpha ve^{\alpha x+\beta y}+24\beta ve^{\alpha x+\beta y}+5ve^{\alpha x+\beta y}$$

You can obviously factor out $ve^{\alpha x+\beta y}$, but that doesn't really do much for you. I also tried doing this with the y' substitution. It also occurred to me that since v is probably supposed to be understood as v(x,y), I tried this set of substitutions:

$$u_{x}=\alpha ve^{\alpha x+\beta y}+v_{x}e^{\alpha x+\beta y}$$
$$u_{xx}=\alpha^{2} ve^{\alpha x+\beta y}+\alpha v_{x}e^{\alpha x+\beta y}+v_{xx}e^{\alpha x+\beta y}$$
$$u_{x}=\beta ve^{\alpha x+\beta y}+v_{y}e^{\alpha x+\beta y}$$
$$u_{xx}=\beta^{2} ve^{\alpha x+\beta y}+\beta v_{y}e^{\alpha x+\beta y}+v_{yy}e^{\alpha x+\beta y}$$

None of these attempts gave me any insight into the problem.

 

[gabbagabbahey]

You need to treat $v$ as a function of $x$ and $y$ and use the product rule to differentiate $u(x,y)=v(x,y)e^{\alpha x+\beta y}$.

Also, your $\LaTeX$ isn't displaying properly because you aren't puuting spaces between \alpha and x (or \beta and y )

 

[McCoy13]

You need to treat $v$ as a function of $x$ and $y$ and use the product rule to differentiate $u(x,y)=v(x,y)e^{\alpha x+\beta y}$.

Also, your $\LaTeX$ isn't displaying properly because you aren't puuting spaces between \alpha and x (or \beta and y )

I fixed the LaTeX and (hopefully now that it's displaying correctly), you'll see that I used the product rule at the bottom of my attempted solution. However, it is non-obvious to me how making this correction by using the product rule helps me. I will have lots of first order partial derivatives floating around that are not in the desired equation.

 

[gabbagabbahey]

However, it is non-obvious to me how making this correction by using the product rule helps me. I will have lots of first order partial derivatives floating around that are not in the desired equation.

You'll end up with a bunch of terms involving $v$ and its partial derivatives, all multyiplied by $e^{\alpha x+\beta y}$, which you can factor out of your DE (since it is never zero, getting rid of this factor doesn't exclude any solutions)... giving you a different DE for $v$...one that you can simplify by choosing nice values for $\alpha$ and $\beta$.

Give it a shot and post your attempt.

 

[McCoy13]

You'll end up with a bunch of terms involving $v$ and its partial derivatives, all multyiplied by $e^{\alpha x+\beta y}$, which you can factor out of your DE (since it is never zero, getting rid of this factor doesn't exclude any solutions)... giving you a different DE for $v$...one that you can simplify by choosing nice values for $\alpha$ and $\beta$.

Give it a shot and post your attempt.

Haha, it did not occur to me to simply pick alpha and beta, even though they are arbitrary parameters introduced in the change of variables.

I ended up with $v_{xx}+3v_{yy}-31v$, and I'm assuming you can take care of the factor of 3 in front of $3v_{yy}$ by simply correctly setting $\gamma$ when you substitute in y'.

 

[gabbagabbahey]

Haha, it did not occur to me to simply pick alpha and beta, even though they are arbitrary parameters introduced in the change of variables.

I ended up with $v_{xx}+3v_{yy}-31v$, and I'm assuming you can take care of the factor of 3 in front of $3v_{yy}$ by simply correctly setting $\gamma$ when you substitute in y'.

I think the factor of -31 might be a little off (my back of envelope calc gave me a factor of -49), so you'll probably want to double check that. But otherwise, yes...can you see what value of $\gamma$ will get rid of the factor of 3?

 

[McCoy13]

I used $\gamma = \sqrt{3}$. This is correct? If not, perhaps $\frac{1}{\sqrt{3}}$.

 

[gabbagabbahey]

I used $\gamma = \sqrt{3}$. This is correct? If not, perhaps $\frac{1}{\sqrt{3}}$.

You tell me...the chain rule tells you that $v_y(x,y'(y))=\frac{dy'}{dy}v_{y'}$, so what value of $\gamma$ makes $v_{y'y'}=3v_{yy}$?

 

[McCoy13]

You tell me...the chain rule tells you that $v_y(x,y'(y))=\frac{dy'}{dy}v_{y'}$, so what value of $\gamma$ makes $v_{y'y'}=3v_{yy}$?

$$\frac{d^{2}y'}{dy^{2}}v_{y'}+\frac{dy'}{dy}v_{y'y'}=v_{yy}$$
$$\frac{d^{2}y'}{dy^{2}}=0 \Rightarrow \gamma v_{y'y'}=v_{yy}$$
$$\gamma = 1/3$$

 

[gabbagabbahey]

$$\frac{d^{2}y'}{dy^{2}}v_{y'}+\frac{dy'}{dy}v_{y'y'}=v_{yy}$$
$$\frac{d^{2}y'}{dy^{2}}=0 \Rightarrow \gamma v_{y'y'}=v_{yy}$$
$$\gamma = 1/3$$

Ermm... shouldn't you have

$$v_{yy}=\left(\frac{dy'}{dy}\right)^2v_{y'y'}$$

??

 

[McCoy13]

Yes, I should. Forgot to apply chain rule while using product rule. Bah! Thanks for all the help.

Also, I rechecked the 31, and I got it right unless I made a mistake in my substitution or if I missed a term while I was gathering terms. I'll double check it from the start before I hand it in.

EDIT: Upon reviewing the work, 49 is the correct number, not 31.