Another PDE

  • #1
gtfitzpatrick
380
0

Homework Statement



w(r,[tex]\theta[/tex])= u(rcos [tex] \theta [/tex]),rsin([tex] \theta[/tex])) for some u(x,y)

express [tex]
\frac{ \partial w}{\partial r} [/tex] and [tex] \frac{ \partial w}{ \partial \theta} [/tex] in terms of [tex] \frac{ \partial u}{ \partial x} [/tex] and [tex] \frac{ \partial u}{ \partial y} [/tex]

Homework Equations



rewrite the following PDE and initial conditions in terms of w,r and [tex]\theta[/tex] and solve

y[tex] \frac{ \partial u}{ \partial x} [/tex] - x[tex] \frac{ \partial u}{ \partial y} [/tex] = 1 where u(x,0) = 0 for 0<x<[tex]\infty[/tex]

The Attempt at a Solution



[tex] \frac{ \partial w}{ \partial r} [/tex] = [tex] \frac{ \partial u}{ \partial x} [/tex] . [tex] \frac{ \partial x}{ \partial r} [/tex] + [tex] \frac{ \partial u}{ \partial y} [/tex] . [tex] \frac{ \partial y}{ \partial r} [/tex]

and


[tex] \frac{ \partial w}{ \partial \theta} [/tex] = [tex] \frac{ \partial u}{ \partial x} [/tex] . [tex] \frac{ \partial x}{ \partial \theta} [/tex] + [tex] \frac{ \partial u}{ \partial y} [/tex] . [tex] \frac{ \partial y}{ \partial \theta} [/tex]

but im not sure about part b...
 

Answers and Replies

  • #2
gtfitzpatrick
380
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any suggestions anyone?
 
  • #3
fzero
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The Attempt at a Solution



[tex] \frac{ \partial w}{ \partial r} [/tex] = [tex] \frac{ \partial u}{ \partial x} [/tex] . [tex] \frac{ \partial x}{ \partial r} [/tex] + [tex] \frac{ \partial u}{ \partial y} [/tex] . [tex] \frac{ \partial y}{ \partial r} [/tex]

and


[tex] \frac{ \partial w}{ \partial \theta} [/tex] = [tex] \frac{ \partial u}{ \partial x} [/tex] . [tex] \frac{ \partial x}{ \partial \theta} [/tex] + [tex] \frac{ \partial u}{ \partial y} [/tex] . [tex] \frac{ \partial y}{ \partial \theta} [/tex]

but im not sure about part b...

You should be able to simplify these by using the relationship between [tex]x,y[/tex] and [tex]r,\theta[/tex].

rewrite the following PDE and initial conditions in terms of w,r and [tex]\theta[/tex] and solve

y[tex] \frac{ \partial u}{ \partial x} [/tex] - x[tex] \frac{ \partial u}{ \partial y} [/tex] = 1 where u(x,0) = 0 for 0<x<[tex]\infty[/tex]

Did you try to write this in terms of [tex]r,\theta[/tex]? You'll find that the form of the equation becomes very simple.
 
  • #4
gtfitzpatrick
380
0
x= rcos [tex]\theta[/tex]
y = rsin [tex] \theta[/tex]
but how do i sub these in
 
  • #5
fzero
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As a first step you need to compute the derivatives [tex]\partial x/\partial r [/tex], etc. Then you need to determine the derivatives of u in terms of those of w and so on.
 
  • #6
gtfitzpatrick
380
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[tex]
\frac{ \partial x}{ \partial r}
[/tex] = 1
and [tex]
\frac{ \partial y}{ \partial r}
[/tex] = 1
so
[tex]
\frac{ \partial w}{ \partial r} = \frac{ \partial u}{ \partial x} + \frac{ \partial u}{ \partial y}
[/tex]
am i right in thinking this?
 
  • #7
fzero
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[tex]
\frac{ \partial x}{ \partial r}
[/tex] = 1
and [tex]
\frac{ \partial y}{ \partial r}
[/tex] = 1
so
[tex]
\frac{ \partial w}{ \partial r} = \frac{ \partial u}{ \partial x} + \frac{ \partial u}{ \partial y}
[/tex]
am i right in thinking this?

No. You just wrote that [tex]x=r \cos\theta[/tex], so how do you obtain [tex]\partial x/\partial r =1[/tex]?
 
  • #8
gtfitzpatrick
380
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sorrysorry i meant cos[tex]\theta[/tex]
 
  • #9
fzero
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OK, so you need to work out [tex]\partial w/\partial r[/tex] and [tex]\partial w/\partial \theta[/tex], then solve algebraically for [tex]\partial u/\partial x[/tex] and [tex]\partial u/\partial y[/tex].
 
  • #10
gtfitzpatrick
380
0
[tex]

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

[/tex]

and


[tex]

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

[/tex]

right?
 
  • #11
gtfitzpatrick
380
0
[tex]


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


[/tex]

and

[tex]


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


[/tex]

but im not sure where to go from here...
 
  • #12
fzero
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[tex]

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

[/tex]

and


[tex]

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

[/tex]

right?

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

[tex]y\frac{\partial u}{\partial x}-x\frac{\partial u}{\partial y}=1[/tex]

in terms of derivatives of [tex]w[/tex].
 
  • #13
gtfitzpatrick
380
0
from [tex]
\frac{ \partial w}{ \partial \mathbf{\theta}} = \frac{ \partial u}{ \partial x}rsin\theta - \frac{ \partial u}{ \partial y}rcos\theta
[/tex]

and

[tex]
y\frac{\partial u}{\partial x}-x\frac{\partial u}{\partial y}=1
[/tex]

[tex]
\frac{ \partial w}{ \partial \mathbf{\theta}} = 1
[/tex]

this is killing me. its probably staring me in the face...
 
  • #14
fzero
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[tex]
\frac{ \partial w}{ \partial \mathbf{\theta}} = 1
[/tex]

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?
 
  • #15
gtfitzpatrick
380
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[tex]
\frac{ \partial w}{ \partial \mathbf{\theta}} = 1
[/tex]

[tex] w(r,\theta) = \theta + f(r) [/tex]???

im digging a big hole here i feel :(
 
  • #16
fzero
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[tex]
\frac{ \partial w}{ \partial \mathbf{\theta}} = 1
[/tex]

[tex] w(r,\theta) = \theta + f(r) [/tex]???

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.
 
  • #17
gtfitzpatrick
380
0
thanks fzero for you patience!
ng code was used to generate this LaTeX image:


[tex]
\frac{ \partial w}{ \partial \theta} = \frac{ \partial u}{ \partial x}y - \frac{ \partial u}{ \partial y}x
[/tex]

[tex]
\frac{ \partial w}{ \partial r} = \frac{1}{r}( \frac{ \partial u}{ \partial x}x + \frac{ \partial u}{ \partial y}y)
[/tex]

so this the answer for part a?

and then i use

[tex]

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

[/tex]

[tex]
w(r,\theta) = \theta + f(r)
[/tex]

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.
 
  • #18
fzero
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[tex]
\frac{ \partial w}{ \partial \theta} = \frac{ \partial u}{ \partial x}y - \frac{ \partial u}{ \partial y}x
[/tex]

[tex]
\frac{ \partial w}{ \partial r} = \frac{1}{r}( \frac{ \partial u}{ \partial x}x + \frac{ \partial u}{ \partial y}y)
[/tex]

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.
 
  • #19
gtfitzpatrick
380
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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[tex]\theta[/tex] = 0 cant br just that can it?
 
  • #20
fzero
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when y=o gives rcos[tex]\theta[/tex] = 0 cant br just that can it?

Well [tex]y=r\sin\theta[/tex], so you want to find the solutions of [tex]r\sin\theta=0[/tex].
 
  • #21
gtfitzpatrick
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how do i find the solution when there are 2 variables though? [tex]\theta[/tex] and r ? could there not be a any solution?
 
  • #22
fzero
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how do i find the solution when there are 2 variables though? [tex]\theta[/tex] and r ? could there not be a any solution?

If the product

[tex]A(x) B(y)=0,[/tex]

then this is solved whenever either

[tex]A(x)=0[/tex]

or

[tex]B(y)=0.[/tex]
 
  • #23
gtfitzpatrick
380
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so either r=0
or [tex]\theta[/tex] = 0 or [tex]\pi[/tex] ?
 
  • #24
fzero
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so either r=0
or [tex]\theta[/tex] = 0 or [tex]\pi[/tex] ?

Right. Now you were also told that [tex]0< x < \infty[/tex], so are all of those compatible with this condition?
 
  • #25
gtfitzpatrick
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[tex]\theta[/tex] must = 0 as cos[tex]\pi[/tex] = -1?
 
  • #26
fzero
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[tex]\theta[/tex] must = 0 as cos[tex]\pi[/tex] = -1?

That's right. So what solution [tex]w(r,\theta)[/tex] satisfies the initial condition?
 
  • #27
gtfitzpatrick
380
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That's right. So what solution [tex]w(r,\theta)[/tex] satisfies the initial condition?

[tex]w(r,\theta)[/tex] = [tex]w(r,0)[/tex] ??
 
  • #28
fzero
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[tex]w(r,\theta)[/tex] = [tex]w(r,0)[/tex] ??

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

[tex]w(r,\theta) = \theta + f(r).[/tex]

What you want to do is use the initial condition to determine [tex]f(r)[/tex].
 
  • #29
gtfitzpatrick
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[tex]
w(r,\theta) = f(r).
[/tex]
 
  • #30
fzero
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[tex]
w(r,\theta) = f(r).
[/tex]

What value does that take when [tex]\theta=0[/tex]? Is it consistent with the information you were given?
 
  • #31
gtfitzpatrick
380
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cos[tex]\theta[/tex] = 1 when [tex]\theta[/tex] = 0

so that gives r = x, the initial condition?
 
  • #32
fzero
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cos[tex]\theta[/tex] = 1 when [tex]\theta[/tex] = 0

so that gives r = x, the initial condition?

The initial condition was [tex]u(x,0)=0[/tex], which we found was equivalent to [tex]w(r,\theta)=0[/tex]. The fact that [tex]r=x[/tex] when [tex]y=0[/tex] doesn't restrict the solution.
 
  • #33
gtfitzpatrick
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provided 0<r<[tex]\infty[/tex] ?
 
  • #34
fzero
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provided 0<r<[tex]\infty[/tex] ?

Yes, the initial condition doesn't pick out a value of r, though it does exclude [tex]r=0[/tex] for a reason that will probably become apparent when you have the complete solution in hand.
 
  • #35
gtfitzpatrick
380
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cool thank a mill
 

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