# Partial differentiation

• Apr 4th 2010, 02:42 AM
vuze88
Partial differentiation
Suppose that f is a differentiable function of a single variable and $\displaystyle F(x,y)$ is defined by $\displaystyle F(x,y)=f(x^2-y)$.

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

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

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?
• Apr 4th 2010, 04:03 AM
HallsofIvy
Quote:

Originally Posted by vuze88
Suppose that f is a differentiable function of a single variable and $\displaystyle F(x,y)$ is defined by $\displaystyle F(x,y)=f(x^2-y)$.

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

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

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

If $\displaystyle u= x^2- y$, then when x= 0, u= -y, y= -u. Saying that F(0,y)= sin y means that F(u)= sin(-u).
• Apr 4th 2010, 04:34 AM
vuze88
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 $\displaystyle u(x,t)=g(x+\lambda t)$ where $\displaystyle \lambda$ is a constant and we have to find $\displaystyle u_{tt}$. Is the following working correct?

let $\displaystyle w=x+\lambda t$

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

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

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.
• Apr 4th 2010, 06:11 AM
drumist
Basically, anytime you have a function $\displaystyle f(x)$ that is defined by a single variable ($\displaystyle x$ in this case), you would take what is called the "total" derivative which is

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

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

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

and

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

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

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

with $\displaystyle w=x+\lambda t$

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

$\displaystyle u(w) = g(w)$

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

But when you take a derivative of $\displaystyle u$ with respect to $\displaystyle x$, you would write it as $\displaystyle \frac{\partial u}{\partial x}$.
• Apr 4th 2010, 03:01 PM
vuze88
Oh ok, so when i was working out $\displaystyle u_{tt}$, i shouldve taken the "total" differentiation for everything, since $\displaystyle u_t$ was defined in terms on one variable?
• Apr 4th 2010, 07:54 PM
drumist
Quote:

Originally Posted by vuze88
Oh ok, so when i was working out $\displaystyle u_{tt}$, i shouldve taken the "total" differentiation for everything, since $\displaystyle u_t$ was defined in terms on one variable?

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

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