Solution for Expressing Derivatives in Terms of u

[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 I am not sure about part b...

 

[gtfitzpatrick]

any suggestions anyone?

 

[fzero]

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 I am 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

 

[fzero]

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?

 

[fzero]

$$\frac{ \partial x}{ \partial r}$$ = 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$$\theta$$

 

[fzero]

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 I am not sure where to go from here...

 

[fzero]

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

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...

 

[fzero]

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

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 :(

 

[fzero]

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

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

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.

 

[fzero]

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

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

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 can't br just that can it?

 

[fzero]

when y=o gives rcos$$\theta$$ = 0 can't 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?

 

[fzero]

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 $$\pi$$ ?

 

[fzero]

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?

 

[fzero]

$$\theta$$ 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?

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

 

[fzero]

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

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]

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

 

[fzero]

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

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?

 

[fzero]

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<$$\infty$$ ?

 

[fzero]

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.

 

[gtfitzpatrick]

cool thank a mill