# Partial Diff (2)

• Feb 16th 2008, 12:57 AM
MilK
Partial Diff (2)
If V =x f(u) and u = y/x,

determine the value of n such that

x^2(∂^2V / ∂x^2) + nxy( ∂^2V / ∂x∂y) + y^2 ( ∂^2V / ∂y^2) = 0

Need help to check my working..

∂V/∂x = xf'(u) + f(u)

∂^2V/∂x^2 = xf"(u) + f'(u) +f'(u)

For u = y/x,
∂u/∂x = -y/x^2
∂^2u/∂x^2 = 2y/x^3

Sub into ∂V/∂x and ∂^2V/∂x^2,

I get ∂V/∂x= -y/x + y/x =0
∂^2V/∂x^2= x(2y/x^3) - (2y/x^2) =0

∂V/∂y = xf'(u)
∂^2V/∂y^2 = xf"(u)

∂u/∂y = 1/x
∂^2u/∂x^2 = 0

Sub into ∂V/∂y and ∂^2V/∂y^2,

∂V/∂y = xf'(u) = 1
∂^2V/∂y^2 = xf"(u) = 0

And,
∂^2V / ∂x∂y= (∂/ ∂x)(∂V/∂y)<----------(Do i need to find ∂^2u / ∂x∂y?)
= (∂/ ∂x)xf'(u)<-----------( i am using ∂u/∂y for this function)
= x(0) + 1/x = 1/x

Sub into given eqn,

x^2(0) + nxy(1/x) + y^2(0) = 0
ny = 0
n=0

Is this correct?
• Feb 16th 2008, 02:01 AM
TwistedOne151
Errors
Right away I can see a glaring error:

Quote:

Need help to check my working..

∂V/∂x = xf'(u) + f(u)
You forgot that u is a function of x and y. This should be
$\displaystyle \frac{\partial V}{\partial x}=xf'(u)\frac{\partial u}{\partial x}+f(u)$
$\displaystyle \frac{\partial V}{\partial x}=xf'(u)\cdot\left(-\frac{y}{x^2}\right)+ f(u)$
$\displaystyle \frac{\partial V}{\partial x}=f(u)-\frac{yf'(u)}{x}$

Shortly thereafter, you substitute $\displaystyle \frac{\partial u}{\partial x}$ in as f'(u). This is simply wrong. f(u) is some unknown (arbitrary) function, and thus f'(u) is simply the derivative of that function, and cannot be substituted with anything else. (Try picturing the sample case $\displaystyle f(u)=\sin{u}$. Then $\displaystyle f'(u)=\cos{u}$, which you can see is most definitely not $\displaystyle \frac{\partial u}{\partial x}=-\frac{y}{x^2}$

--Kevin C.
• Feb 16th 2008, 02:40 AM
MilK
Ok i understand what you are saying

but that means the final answer will be in terms of derivatives of f(u)?

or how should i go about solving this question?

Thks
• Feb 16th 2008, 03:49 AM
TwistedOne151
You will be able to combine like terms with regards to f(u), f'(u), and f''(u); and, in fact, some of them will cancel, but mostly, you will get an equation involving one or more of those, but there will be one (and only one) value of n for which that equation will be true for all f(u).

--Kevin C.
• Feb 16th 2008, 06:42 AM
MilK
∂V/∂x = f(u) - yf'(u)/x

∂^2V/∂x^2 = f"(u)y/x^3

∂V/∂y = f'(u)
∂^2V/∂y^2 = f"(u)/x

∂^2V / ∂x∂y= (∂/ ∂x)(∂V/∂y) = f'(u)(-y/x^2)

Sub into given eqn and factorise,

f"(u)(y/x) (1 - ny+y) = 0

therefore

1-ny+y = 0
n=(y+1)/y

• Feb 16th 2008, 02:47 PM
TwistedOne151
No
No.

You were correct on $\displaystyle \frac{\partial V}{\partial x}=f(u)-\frac{y}{x}f'(u)$, but not $\displaystyle \frac{\partial^2 V}{\partial x^2}$.
$\displaystyle \frac{\partial^2 V}{\partial x^2}=\frac{\partial}{\partial x}\frac{\partial V}{\partial x}$
$\displaystyle \,\,=\frac{\partial}{\partial x}\left(f(u)-\frac{y}{x}f'(u)\right)$
$\displaystyle \,\,=f'(u)\frac{\partial u}{\partial x}+\frac{y}{x^2}f'(u)-\frac{y}{x}f''(u)\frac{\partial u}{\partial x}$
$\displaystyle \,\,=-\frac{y}{x^2}f'(u)+\frac{y}{x^2}f'(u)-\left(\frac{y}{x}\right)\left(-\frac{y}{x^2}\right)f''(u)$
$\displaystyle \,\,=\frac{y^2}{x^3}f''(u)$.

Next,
$\displaystyle \frac{\partial^2 V}{\partial x\partial y}=\frac{\partial}{\partial x}\frac{\partial V}{\partial y}$
$\displaystyle \,\,=\frac{\partial}{\partial x}f'(u)$
$\displaystyle \,\,=f''(u)\frac{\partial u}{\partial x}$
$\displaystyle \,\,=-\frac{y}{x^2}f''(u)$

Thus:
$\displaystyle x^2\frac{\partial^2 V}{\partial x^2}+nxy\frac{\partial^2 V}{\partial x\partial y}+y^2\frac{\partial^2 V}{\partial y^2}=0$
becomes:
$\displaystyle x^2\cdot\frac{y^2}{x^3}f''(u)+nxy\cdot\left(-\frac{y}{x^2}f''(u)\right)+y^2\cdot\frac{1}{x}f''( u)=0$
$\displaystyle \frac{y^2}{x}f''(u)-n\frac{y^2}{x}f''(u)+\frac{y^2}{x}f''(u)=0$
$\displaystyle (2-n)\frac{y^2}{x}f''(u)=0$
And the answer should be obvious from that.

--Kevin C.
• Feb 16th 2008, 04:10 PM
MilK
now i get it, thanks