• Support PF! Buy your school textbooks, materials and every day products Here!

Another PDE

  • #1

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
any suggestions anyone?
 
  • #3
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289

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
x= rcos [tex]\theta[/tex]
y = rsin [tex] \theta[/tex]
but how do i sub these in
 
  • #5
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289
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
[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
Science Advisor
Homework Helper
Gold Member
3,119
289
[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
sorrysorry i meant cos[tex]\theta[/tex]
 
  • #9
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289
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
[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
[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
Science Advisor
Homework Helper
Gold Member
3,119
289
[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
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
Science Advisor
Homework Helper
Gold Member
3,119
289
[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
[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
Science Advisor
Homework Helper
Gold Member
3,119
289
[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
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
Science Advisor
Homework Helper
Gold Member
3,119
289
[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
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
Science Advisor
Homework Helper
Gold Member
3,119
289
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
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
Science Advisor
Homework Helper
Gold Member
3,119
289
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
so either r=0
or [tex]\theta[/tex] = 0 or [tex]\pi[/tex] ?
 
  • #24
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289
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
[tex]\theta[/tex] must = 0 as cos[tex]\pi[/tex] = -1?
 

Related Threads on Another PDE

  • Last Post
Replies
2
Views
506
  • Last Post
Replies
1
Views
536
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
1
Views
742
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
814
  • Last Post
2
Replies
34
Views
2K
  • Last Post
Replies
14
Views
1K
  • Last Post
Replies
1
Views
1K
Top