1. ## 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!

2. 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!

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

Now if you plug back into the original PDE you get

$\displaystyle \displaystyle x\left[ \frac{1}{y}g'\left( \frac{x}{y}\right)\right]+y\left[ -\frac{x}{y^2}g'\left( \frac{x}{y}\right)\right]=0$

3. 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!

4. 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 $\displaystyle g(u)$ is any $\displaystyle C^1$ function.

5. 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
$\displaystyle x \frac{\partial f}{\partial x}+ y\frac{\partial f}{\partial y}= 0$

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,
$\displaystyle \frac{\partial f}{\partial x}= \frac{dg(u)}{du}\frac{\partial u}{\partial x}= \frac{1}{y}\frac{df}{du}$
and
$\displaystyle \frac{\partial f}{\partial y}= \frac{dg(u)}{du}\frac{\partial u}{\partial y}= -\frac{x}{y^2}\frac{df}{du}$
where u= x/y. Putting those into the equation gives
$\displaystyle x\frac{\partial f}{\partial x}+ y\frac{\partial f}{\partial y}= x\frac{1}{y}\frac{df}{du}+ y\left(-\frac{x}{y^2}\right)\frac{df}{du}= \frac{x}{y}\frac{df}{du}- \frac{x}{y}\frac{df}{du}= 0$
so your solution is, in fact, correct.

6. 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?

7. 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. $\displaystyle C^1$ 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.

8. Thanks again for everything

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