Partial Differential Equation

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February 25th 2011, 08:50 AM
Darkprince

Partial Differential Equation

For this exercise I will use the symbol of normal derivative for the partial derivatives:

Assume that f(x,y) is a differentiable function that obeys the partial differential equation

x*(df/dx)+y*(df/dy) = 0 (1)
in the region x>0, y>0. Suppose that the variables u and v are related to x and y by u=x/y and v=x*y

Use the chain rule to express the partial derivatives df/dx and df/dy in terms of derivatives with respect to u and v.

Hence write down the equation (1) in terms of u and v, and obtain the general solution of (1), expressing your final answer in terms of the original variables x and y.

My solution:

chain rule for f=f(u,v)

df/dx=df/du*du/dx + df/dv * dv/dx=df/du*(1/y) + df/dv * y

df/dy=df/du * du/dy + df/dv * dv/dy = df/du * (-x/y^2) + df/dv * x

Then (1) becomes (after some cancelling):

2*x*y*df/dv = 0, but x*y=v so we have 2v*df/dv=0, implies df/dv=0

=> f=g(u) for any function g(u)
=> f(x,y)=g(x/y)

I would appreciate if anyone can check my solution and let me know if it misses something! Thanks in advance for any help!
February 25th 2011, 09:22 AM
TheEmptySet

Quote:

Originally Posted by Darkprince

For this exercise I will use the symbol of normal derivative for the partial derivatives:

Assume that f(x,y) is a differentiable function that obeys the partial differential equation

x*(df/dx)+y*(df/dy) = 0 (1)
in the region x>0, y>0. Suppose that the variables u and v are related to x and y by u=x/y and v=x*y

Use the chain rule to express the partial derivatives df/dx and df/dy in terms of derivatives with respect to u and v.

Hence write down the equation (1) in terms of u and v, and obtain the general solution of (1), expressing your final answer in terms of the original variables x and y.

My solution:

chain rule for f=f(u,v)

df/dx=df/du*du/dx + df/dv * dv/dx=df/du*(1/y) + df/dv * y

df/dy=df/du * du/dy + df/dv * dv/dy = df/du * (-x/y^2) + df/dv * x

Then (1) becomes (after some cancelling):

2*x*y*df/dv = 0, but x*y=v so we have 2v*df/dv=0, implies df/dv=0

=> f=g(u) for any function g(u)
=> f(x,y)=g(x/y)

I would appreciate if anyone can check my solution and let me know if it misses something! Thanks in advance for any help!

Your answer is correct. As always If can be directly verified.

$\displaystyle \frac{\partial f}{\partial x}=\frac{\partial }{\partial x}g\left( \frac{x}{y}\right)=\frac{1}{y}g'\left( \frac{x}{y}\right)$
and
$\displaystyle \frac{\partial f}{\partial y}=\frac{\partial }{\partial y}g\left( \frac{x}{y}\right)=-\frac{x}{y^2}g'\left( \frac{x}{y}\right)$

Now if you plug back into the original PDE you get

$\displaystyle x\left[ \frac{1}{y}g'\left( \frac{x}{y}\right)\right]+y\left[ -\frac{x}{y^2}g'\left( \frac{x}{y}\right)\right]=0$
February 25th 2011, 09:25 AM
Darkprince

Thank you very much! But should I verify that in my solution, for a complete solution or I am ok with exactly what I wrote? Thanks again!
February 25th 2011, 09:29 AM
TheEmptySet

Quote:

Originally Posted by Darkprince

Thank you very much! But should I verify that in my solution, for a complete solution or I am ok with exactly what I wrote? Thanks again!

I am not 100% sure by what you mean complete solution. You have given the general form for the soluiton. Unless you were given a boundary condition or some other information this is the best solution. It may be good to mention that $g(u)$ is any $C^1$ function.
February 25th 2011, 09:31 AM
HallsofIvy

You can use LaTeX to make formulas look really good. On this board, start with [ math ] and end with [ /math ] (without the spaces). For example,
[ math ]x \frac{\partial f}{\partial x}+ y\frac{\partial f}{\partial y}= 0 [ /math ]
without the spaces is
$x \frac{\partial f}{\partial x}+ y\frac{\partial f}{\partial y}= 0$

And it's easy to check if f(x,y)= g(x/y), for g any differentiable function of a single variable.

By the chain rule,
$\frac{\partial f}{\partial x}= \frac{dg(u)}{du}\frac{\partial u}{\partial x}= \frac{1}{y}\frac{df}{du}$
and
$\frac{\partial f}{\partial y}= \frac{dg(u)}{du}\frac{\partial u}{\partial y}= -\frac{x}{y^2}\frac{df}{du}$
where u= x/y. Putting those into the equation gives
$x\frac{\partial f}{\partial x}+ y\frac{\partial f}{\partial y}= x\frac{1}{y}\frac{df}{du}+ y\left(-\frac{x}{y^2}\right)\frac{df}{du}= \frac{x}{y}\frac{df}{du}- \frac{x}{y}\frac{df}{du}= 0$
so your solution is, in fact, correct.
February 25th 2011, 09:39 AM
Darkprince

Thanks for your replies! I mean for a "complete" solution do I have to verify my solution by substituting in the given equation the solution I deduced? Also what do you mean that g(u) is any C^1?
February 25th 2011, 09:47 AM
TheEmptySet

Quote:

Originally Posted by Darkprince

Thanks for your replies! I mean for a "complete" solution do I have to verify my solution by substituting in the given equation the solution I deduced? Also what do you mean that g(u) is any C^1?

It is always a good idea to check your solution. $C^1$ just means that the function has one continous derivative. As I mentioned, without boundary data we don't really know anything about g. I used the fact that I could take its derivative to verify the solution.
February 25th 2011, 09:54 AM
Darkprince

Thanks again for everything :)