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.