# Another PDE

gtfitzpatrick

## 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...

gtfitzpatrick
any suggestions anyone?

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## 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.

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

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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.

gtfitzpatrick
$$\frac{ \partial x}{ \partial r}$$ = 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?

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$$\frac{ \partial x}{ \partial r}$$ = 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 $$\partial x/\partial r =1$$?

gtfitzpatrick
sorrysorry i meant cos$$\theta$$

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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 $$\partial u/\partial y$$.

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

and

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

right?

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

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...

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

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 $$w$$.

gtfitzpatrick
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...

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

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?

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

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

im digging a big hole here i feel :(

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

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

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.

gtfitzpatrick
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.

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

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

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.

gtfitzpatrick
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?

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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 $$r\sin\theta=0$$.

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

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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

$$B(y)=0.$$

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

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

Right. Now you were also told that $$0< x < \infty$$, so are all of those compatible with this condition?

gtfitzpatrick
$$\theta$$ must = 0 as cos$$\pi$$ = -1?

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$$\theta$$ must = 0 as cos$$\pi$$ = -1?

That's right. So what solution $$w(r,\theta)$$ satisfies the initial condition?

gtfitzpatrick
That's right. So what solution $$w(r,\theta)$$ satisfies the initial condition?

$$w(r,\theta)$$ = $$w(r,0)$$ ??

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$$w(r,\theta)$$ = $$w(r,0)$$ ??

You're told that $$u(x,y=0)=0$$ for all allowed x. You found that $$y=0$$ corresponds to $$\theta =0$$. So the initial condition is equivalent to $$w(r,\theta=0)=0$$. Your general solution is

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

What you want to do is use the initial condition to determine $$f(r)$$.

gtfitzpatrick
$$w(r,\theta) = f(r).$$

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

What value does that take when $$\theta=0$$? Is it consistent with the information you were given?

gtfitzpatrick
cos$$\theta$$ = 1 when $$\theta$$ = 0

so that gives r = x, the initial condition?

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cos$$\theta$$ = 1 when $$\theta$$ = 0

so that gives r = x, the initial condition?

The initial condition was $$u(x,0)=0$$, which we found was equivalent to $$w(r,\theta)=0$$. The fact that $$r=x$$ when $$y=0$$ doesn't restrict the solution.

gtfitzpatrick
provided 0<r<$$\infty$$ ?

provided 0<r<$$\infty$$ ?
Yes, the initial condition doesn't pick out a value of r, though it does exclude $$r=0$$ for a reason that will probably become apparent when you have the complete solution in hand.