You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
Proving that a homogeneous expression satisfies a partial differential equation
- Thread starter aruwin
- Start date
SlipEternal
Junior Member
- Joined
- Jan 4, 2012
- Messages
- 114
My inclination would be to consider the binomial expansion and see what you get.
My inclination would be to consider the binomial expansion and see what you get.
How to do that exactly??Give me hints,please.
SlipEternal
Junior Member
- Joined
- Jan 4, 2012
- Messages
- 114
Sure, I would google Binomial Expansion. When I did, the first link gave me everything I would need to know about it: Binomial Theorem
SlipEternal
Junior Member
- Joined
- Jan 4, 2012
- Messages
- 114
I just realized an induction argument would likely work better. Consider the case where \(\displaystyle p=1\). Then you would have:
\(\displaystyle x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y} = nf\)
And, you also have \(\displaystyle f(tx, ty) = t^n f(x,y)\)
Try taking partials of both sides and see what you get.
\(\displaystyle x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y} = nf\)
And, you also have \(\displaystyle f(tx, ty) = t^n f(x,y)\)
Try taking partials of both sides and see what you get.
I just realized an induction argument would likely work better. Consider the case where \(\displaystyle p=1\). Then you would have:
\(\displaystyle x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y} = nf\)
And, you also have \(\displaystyle f(tx, ty) = t^n f(x,y)\)
Try taking partials of both sides and see what you get.
I am not sure how to partial differentiate that kind of expression.
This is my attempt,though:
∂f(t)/∂x + ∂f(t)/∂y = t^n(∂f(1)/∂x + ∂f(1)/∂y)
SlipEternal
Junior Member
- Joined
- Jan 4, 2012
- Messages
- 114
I am not sure how to partial differentiate that kind of expression.
This is my attempt,though:
∂f(t)/∂x + ∂f(t)/∂y = t^n(∂f(1)/∂x + ∂f(1)/∂y)
Apply the chain rule. \(\displaystyle tx, ty\) are both functions of \(\displaystyle x\) and \(\displaystyle y\).
SlipEternal
Junior Member
- Joined
- Jan 4, 2012
- Messages
- 114
In case you forget, the multivariable chain rule looks like this:
\(\displaystyle \frac{\partial f(u(x,y),v(x,y))}{\partial{x}} = \frac{\partial{f}}{\partial{u}}\frac{\partial{u}}{\partial{x}} + \frac{\partial{f}}{\partial{v}}\frac{\partial{v}}{\partial{x}}\)
Since \(\displaystyle t\) may be a function of both \(\displaystyle x\text{ and }y\), I add in all possible factors. You know more about your problem than I do. If \(\displaystyle t\) is considered to be independent from both \(\displaystyle x\text{ and }y\), then ignore that part.
\(\displaystyle \frac{\partial f(u(x,y),v(x,y))}{\partial{x}} = \frac{\partial{f}}{\partial{u}}\frac{\partial{u}}{\partial{x}} + \frac{\partial{f}}{\partial{v}}\frac{\partial{v}}{\partial{x}}\)
Since \(\displaystyle t\) may be a function of both \(\displaystyle x\text{ and }y\), I add in all possible factors. You know more about your problem than I do. If \(\displaystyle t\) is considered to be independent from both \(\displaystyle x\text{ and }y\), then ignore that part.
In case you forget, the multivariable chain rule looks like this:
\(\displaystyle \frac{\partial f(u(x,y),v(x,y))}{\partial{x}} = \frac{\partial{f}}{\partial{u}}\frac{\partial{u}}{\partial{x}} + \frac{\partial{f}}{\partial{v}}\frac{\partial{v}}{\partial{x}}\)
Since \(\displaystyle t\) may be a function of both \(\displaystyle x\text{ and }y\), I add in all possible factors. You know more about your problem than I do. If \(\displaystyle t\) is considered to be independent from both \(\displaystyle x\text{ and }y\), then ignore that part.
How to determine whether t is independent or not?
SlipEternal
Junior Member
- Joined
- Jan 4, 2012
- Messages
- 114
Honestly, I have only studied undergrad partial differential equations, and the course did not go this far in depth. I would recommend playing with the chain rule and seeing what you can come up with. I don't have the time to figure it out, but I think that is the right track.
Edit:
To solve this, I would assume \(\displaystyle t = t(x,y)\). That way, you can have \(\displaystyle u(x,t) = tx, v(y,t) = ty\) and the multivariable chain rule can be extended to that:
\(\displaystyle \frac{\partial{f}}{\partial{x}} = \frac{\partial{f}}{\partial{u}}\frac{\partial{u}}{\partial{x}} + \frac{\partial{f}}{\partial{u}}\frac{\partial{u}}{\partial{t}}\frac{\partial{t}}{\partial{x}} + \frac{\partial{f}}{\partial{v}}\frac{\partial{v}}{\partial{t}}\frac{\partial{t}}{\partial{x}}\)
Moreover, you know that \(\displaystyle \frac{\partial{u}}{\partial{x}} = t, \frac{\partial{u}}{\partial{t}} = x, \frac{\partial{v}}{\partial{t}} = y\).
So, taking partials with respect to \(\displaystyle x\) on both sides:
\(\displaystyle \frac{\partial{f}}{\partial{u}}\left(t+x\frac{\partial{t}}{\partial{x}}\right)+y\frac{\partial{f}}{\partial{v}}\frac{\partial{t}}{\partial{x}} = f\frac{\partial{t^n}}{\partial{x}} + t^n\frac{\partial{f}}{\partial{x}}\) (I think)
Edit:
To solve this, I would assume \(\displaystyle t = t(x,y)\). That way, you can have \(\displaystyle u(x,t) = tx, v(y,t) = ty\) and the multivariable chain rule can be extended to that:
\(\displaystyle \frac{\partial{f}}{\partial{x}} = \frac{\partial{f}}{\partial{u}}\frac{\partial{u}}{\partial{x}} + \frac{\partial{f}}{\partial{u}}\frac{\partial{u}}{\partial{t}}\frac{\partial{t}}{\partial{x}} + \frac{\partial{f}}{\partial{v}}\frac{\partial{v}}{\partial{t}}\frac{\partial{t}}{\partial{x}}\)
Moreover, you know that \(\displaystyle \frac{\partial{u}}{\partial{x}} = t, \frac{\partial{u}}{\partial{t}} = x, \frac{\partial{v}}{\partial{t}} = y\).
So, taking partials with respect to \(\displaystyle x\) on both sides:
\(\displaystyle \frac{\partial{f}}{\partial{u}}\left(t+x\frac{\partial{t}}{\partial{x}}\right)+y\frac{\partial{f}}{\partial{v}}\frac{\partial{t}}{\partial{x}} = f\frac{\partial{t^n}}{\partial{x}} + t^n\frac{\partial{f}}{\partial{x}}\) (I think)
Last edited: