Partial differentiation

**vuze88** · April 4th 2010, 02:42 AM

Suppose that f is a differentiable function of a single variable and $F(x,y)$ is defined by $F(x,y)=f(x^2-y)$ .

Show that $F$ satisfies the partial differential equation $\frac{\partial F}{\partial x}+2x\frac{\partial F}{\partial y}=0$ .

Given that $F(0,y)=\sin y$ for all $y$ , find a formula for $F(x,y)$ .

I'm pretty sure im supposed to use the chain rule for this but i dont know how to apply it. Can someone care to explain?

**HallsofIvy** · April 4th 2010, 04:03 AM

Originally Posted by vuze88

Suppose that f is a differentiable function of a single variable and $F(x,y)$ is defined by $F(x,y)=f(x^2-y)$ .

Show that $F$ satisfies the partial differential equation $\frac{\partial F}{\partial x}+2x\frac{\partial F}{\partial y}=0$ .

Given that $F(0,y)=\sin y$ for all $y$ , find a formula for $F(x,y)$ .

I'm pretty sure im supposed to use the chain rule for this but i dont know how to apply it. Can someone care to explain?

If we let $u= x^2- y$ then $\frac{\partial F}{\partial x}= \frac{dF}{du}\frac{\partial u}{\partial x}$ . What is that? $\frac{\partial F}{\partial y}= \frac{dF}{du}\frac{\partial u}{\partial y}$ . What is that? What is the sum?

If $u= x^2- y$ , then when x= 0, u= -y, y= -u. Saying that F(0,y)= sin y means that F(u)= sin(-u).

**vuze88** · April 4th 2010, 04:34 AM

Thanks for the useful reply. I'm just a little confused with when to use the normal d's and the curly d's. Say for example, we have $u(x,t)=g(x+\lambda t)$ where $\lambda$ is a constant and we have to find $u_{tt}$ . Is the following working correct?

let $w=x+\lambda t$

$u_{t}=\frac{\mathrm{d} u}{\mathrm{d} w}\frac{\partial w}{\partial t}=\lambda\frac{\mathrm{d} u}{\mathrm{d} w}$

$u_{tt}=\frac{\partial }{\partial t}(\lambda\frac{\mathrm{du} }{\mathrm{d} w})=\lambda\frac{\partial }{\partial w}(\frac{\partial u}{\partial t})=\lambda\frac{\partial }{\partial w}(\lambda\frac{\partial u}{\partial w})=\lambda^2g''(x+\lambda t)$

Also, how come the chain rule works with both normal d's and curley d's. Is there some sort of proof for the result dealing with more than one variable. Sorry, im new to this topic so any help would be appreciated.

**drumist** · April 4th 2010, 06:11 AM

Basically, anytime you have a function $f(x)$ that is defined by a single variable ( $x$ in this case), you would take what is called the "total" derivative which is

$\frac{d}{dx}f(x)=\frac{df}{dx}=f'$

But, when dealing with a function of more than one variable, i.e., $g(y,z)$ , you must use the partial derivatives which are

$\frac{\partial}{\partial y}g(y,z)=\frac{\partial g}{\partial y}=g_y$

and

$\frac{\partial}{\partial z}g(y,z)=\frac{\partial g}{\partial z }=g_z$

You probably understand this, but it may seem odd to use both notations in the same context. For example, the problem you gave was

$u(x,t)=g(x+\lambda t)$

with $w=x+\lambda t$

Notice that when you make the substitution, you can write the function as $g(w)$ which is entirely defined by a single variable $w$ . So:

$u(w) = g(w)$

Therefore, $u$ is now a single variable function, so when you take a derivative of $u$ with respect to $w$ , you would write it as $\frac{du}{dw}$ .

But when you take a derivative of $u$ with respect to $x$ , you would write it as $\frac{\partial u}{\partial x}$ .

**vuze88** · April 4th 2010, 03:01 PM

Oh ok, so when i was working out $u_{tt}$ , i shouldve taken the "total" differentiation for everything, since $u_t$ was defined in terms on one variable?

**drumist** · April 4th 2010, 07:54 PM

Originally Posted by vuze88

Oh ok, so when i was working out $u_{tt}$ , i shouldve taken the "total" differentiation for everything, since $u_t$ was defined in terms on one variable?

Notice that $u_t = \lambda \frac{du}{dw} = \lambda g'(w) = \lambda g'(x+\lambda t)$

So even though it may have been disguised, it still is a function of both x and t.

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