# Another PDE

## Homework Statement

w(r,$$\theta$$)= u(rcos $$\theta$$),rsin($$\theta$$)) for some u(x,y)

express $$\frac{ \partial w}{\partial r}$$ and $$\frac{ \partial w}{ \partial \theta}$$ in terms of $$\frac{ \partial u}{ \partial x}$$ and $$\frac{ \partial u}{ \partial y}$$

## Homework Equations

rewrite the following PDE and initial conditions in terms of w,r and $$\theta$$ and solve

y$$\frac{ \partial u}{ \partial x}$$ - x$$\frac{ \partial u}{ \partial y}$$ = 1 where u(x,0) = 0 for 0<x<$$\infty$$

## The Attempt at a Solution

$$\frac{ \partial w}{ \partial r}$$ = $$\frac{ \partial u}{ \partial x}$$ . $$\frac{ \partial x}{ \partial r}$$ + $$\frac{ \partial u}{ \partial y}$$ . $$\frac{ \partial y}{ \partial r}$$

and

$$\frac{ \partial w}{ \partial \theta}$$ = $$\frac{ \partial u}{ \partial x}$$ . $$\frac{ \partial x}{ \partial \theta}$$ + $$\frac{ \partial u}{ \partial y}$$ . $$\frac{ \partial y}{ \partial \theta}$$

but im not sure about part b...

Related Calculus and Beyond Homework Help News on Phys.org
any suggestions anyone?

fzero
Homework Helper
Gold Member

## The Attempt at a Solution

$$\frac{ \partial w}{ \partial r}$$ = $$\frac{ \partial u}{ \partial x}$$ . $$\frac{ \partial x}{ \partial r}$$ + $$\frac{ \partial u}{ \partial y}$$ . $$\frac{ \partial y}{ \partial r}$$

and

$$\frac{ \partial w}{ \partial \theta}$$ = $$\frac{ \partial u}{ \partial x}$$ . $$\frac{ \partial x}{ \partial \theta}$$ + $$\frac{ \partial u}{ \partial y}$$ . $$\frac{ \partial y}{ \partial \theta}$$

but im not sure about part b...
You should be able to simplify these by using the relationship between $$x,y$$ and $$r,\theta$$.

rewrite the following PDE and initial conditions in terms of w,r and $$\theta$$ and solve

y$$\frac{ \partial u}{ \partial x}$$ - x$$\frac{ \partial u}{ \partial y}$$ = 1 where u(x,0) = 0 for 0<x<$$\infty$$
Did you try to write this in terms of $$r,\theta$$? You'll find that the form of the equation becomes very simple.

x= rcos $$\theta$$
y = rsin $$\theta$$
but how do i sub these in

fzero
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As a first step you need to compute the derivatives $$\partial x/\partial r$$, etc. Then you need to determine the derivatives of u in terms of those of w and so on.

$$\frac{ \partial x}{ \partial r}$$ = 1
and $$\frac{ \partial y}{ \partial r}$$ = 1
so
$$\frac{ \partial w}{ \partial r} = \frac{ \partial u}{ \partial x} + \frac{ \partial u}{ \partial y}$$
am i right in thinking this?

fzero
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Gold Member
$$\frac{ \partial x}{ \partial r}$$ = 1
and $$\frac{ \partial y}{ \partial r}$$ = 1
so
$$\frac{ \partial w}{ \partial r} = \frac{ \partial u}{ \partial x} + \frac{ \partial u}{ \partial y}$$
am i right in thinking this?
No. You just wrote that $$x=r \cos\theta$$, so how do you obtain $$\partial x/\partial r =1$$?

sorrysorry i meant cos$$\theta$$

fzero
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Gold Member
OK, so you need to work out $$\partial w/\partial r$$ and $$\partial w/\partial \theta$$, then solve algebraically for $$\partial u/\partial x$$ and $$\partial u/\partial y$$.

$$\frac{ \partial w}{ \partial r} = \frac{ \partial u}{ \partial x}cos\theta + \frac{ \partial u}{ \partial y}sin\theta$$

and

$$\frac{ \partial w}{ \partial r} = \frac{ \partial u}{ \partial x}rsin\theta - \frac{ \partial u}{ \partial y}rcos\theta$$

right?

$$\frac{ \partial w}{ \partial r} = \frac{1}{r}( \frac{ \partial u}{ \partial x}x + \frac{ \partial u}{ \partial y}y)$$

and

$$\frac{ \partial w}{ \partial \theta} = \frac{ \partial u}{ \partial x}x - \frac{ \partial u}{ \partial y}y$$

but im not sure where to go from here...

fzero
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Gold Member
$$\frac{ \partial w}{ \partial r} = \frac{ \partial u}{ \partial x}cos\theta + \frac{ \partial u}{ \partial y}sin\theta$$

and

$$\frac{ \partial w}{ \partial \mathbf{\theta}} = \frac{ \partial u}{ \partial x}rsin\theta - \frac{ \partial u}{ \partial y}rcos\theta$$

right?
Now, like I said before, you want to solve these algebraically for $$\partial u/\partial x$$ and $$\partial u/\partial y$$. You are then going to use those expressions to rewrite

$$y\frac{\partial u}{\partial x}-x\frac{\partial u}{\partial y}=1$$

in terms of derivatives of $$w$$.

from $$\frac{ \partial w}{ \partial \mathbf{\theta}} = \frac{ \partial u}{ \partial x}rsin\theta - \frac{ \partial u}{ \partial y}rcos\theta$$

and

$$y\frac{\partial u}{\partial x}-x\frac{\partial u}{\partial y}=1$$

$$\frac{ \partial w}{ \partial \mathbf{\theta}} = 1$$

this is killing me. its probably staring me in the face...

fzero
Homework Helper
Gold Member
$$\frac{ \partial w}{ \partial \mathbf{\theta}} = 1$$

this is killing me. its probably staring me in the face...
Well, what is the solution of that equation? Remember that this is a partial derivative, so the integration "constant" can depend on the other variable.

Next use the initial condition u(x,0) = 0 to fix the integration constant. What does the value y=0 correspond to in the polar coordinates?

$$\frac{ \partial w}{ \partial \mathbf{\theta}} = 1$$

$$w(r,\theta) = \theta + f(r)$$???

im digging a big hole here i feel :(

fzero
Homework Helper
Gold Member
$$\frac{ \partial w}{ \partial \mathbf{\theta}} = 1$$

$$w(r,\theta) = \theta + f(r)$$???

im digging a big hole here i feel :(
That's correct. Now you have to figure out what y=0 means in polar coordinates and then apply the condition u(x,0)=0 to this solution.

thanks fzero for you patience!
ng code was used to generate this LaTeX image:

$$\frac{ \partial w}{ \partial \theta} = \frac{ \partial u}{ \partial x}y - \frac{ \partial u}{ \partial y}x$$

$$\frac{ \partial w}{ \partial r} = \frac{1}{r}( \frac{ \partial u}{ \partial x}x + \frac{ \partial u}{ \partial y}y)$$

so this the answer for part a?

and then i use

$$\frac{ \partial w}{ \partial \mathbf{\theta}} = 1$$

$$w(r,\theta) = \theta + f(r)$$

and then as you say figure out what y=0 means in polar coordinates and then apply the condition u(x,0)=0 to this solution.

fzero
Homework Helper
Gold Member
$$\frac{ \partial w}{ \partial \theta} = \frac{ \partial u}{ \partial x}y - \frac{ \partial u}{ \partial y}x$$

$$\frac{ \partial w}{ \partial r} = \frac{1}{r}( \frac{ \partial u}{ \partial x}x + \frac{ \partial u}{ \partial y}y)$$

so this the answer for part a?
Yes, though I'm not sure whether your grader will care if you express it this way or in terms of the angle.

That's correct. Now you have to figure out what y=0 means in polar coordinates and then apply the condition u(x,0)=0 to this solution.

when y=o gives rcos$$\theta$$ = 0 cant br just that can it?

fzero
Homework Helper
Gold Member
when y=o gives rcos$$\theta$$ = 0 cant br just that can it?
Well $$y=r\sin\theta$$, so you want to find the solutions of $$r\sin\theta=0$$.

how do i find the solution when there are 2 variables though? $$\theta$$ and r ? could there not be a any solution?

fzero
Homework Helper
Gold Member
how do i find the solution when there are 2 variables though? $$\theta$$ and r ? could there not be a any solution?
If the product

$$A(x) B(y)=0,$$

then this is solved whenever either

$$A(x)=0$$

or

$$B(y)=0.$$

so either r=0
or $$\theta$$ = 0 or $$\pi$$ ?

fzero
or $$\theta$$ = 0 or $$\pi$$ ?
Right. Now you were also told that $$0< x < \infty$$, so are all of those compatible with this condition?
$$\theta$$ must = 0 as cos$$\pi$$ = -1?