Laplace equation with piece-wise constant boundary conditions, using change of variables

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I have a problem trying to find a solution of this problem:

Find a harmonic function $u(x,y)$ in the region $operatornameIm(z)>0$ with:
$$ u(x,0) = begincases 1 & textrmif xge 0\ 0 & textrmif x<0 \ endcases $$

then I'd like to use that idea for solving this other problem:

Find a harmonic function $u(x,y)$ in the region $operatornameIm(z)>0$ with:
$$ u(x,0) = begincases T_1 & textrmif x < a\ T_2 & textrmif ale xle b \ T_2 & textrmif x > b endcases $$

I did this for the first problem:

I defined a $g(z)= z = x+iy$

then I said, for an $h$ in polar coordinates:

$h(g,0)=0$

$h(g,pi)=1$

so $h(g,theta)=Atheta+B$

where $A=frac1pi$ and $B=0$

then I said:

$$ u(x,y)=h(operatornameRe(g),operatornameIm(g))=frac1piarctanleft(fracyxright) $$

but when I compute $u(x,0)$ it looks like this:

$$u(x,0)=frac1piarctan(0)=0, forall x$$

so it must not be a solution.

What should I do in this situation? What's wrong in the method that I proposed?

pde boundary-value-problem

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edited Jul 19 at 8:31

Dylan

11.4k31026

asked Jul 18 at 23:56

Verónica

235

add a comment | 

up vote
-1
down vote

favorite

I have a problem trying to find a solution of this problem:

Find a harmonic function $u(x,y)$ in the region $operatornameIm(z)>0$ with:
$$ u(x,0) = begincases 1 & textrmif xge 0\ 0 & textrmif x<0 \ endcases $$

then I'd like to use that idea for solving this other problem:

Find a harmonic function $u(x,y)$ in the region $operatornameIm(z)>0$ with:
$$ u(x,0) = begincases T_1 & textrmif x < a\ T_2 & textrmif ale xle b \ T_2 & textrmif x > b endcases $$

I did this for the first problem:

I defined a $g(z)= z = x+iy$

then I said, for an $h$ in polar coordinates:

$h(g,0)=0$

$h(g,pi)=1$

so $h(g,theta)=Atheta+B$

where $A=frac1pi$ and $B=0$

then I said:

$$ u(x,y)=h(operatornameRe(g),operatornameIm(g))=frac1piarctanleft(fracyxright) $$

but when I compute $u(x,0)$ it looks like this:

$$u(x,0)=frac1piarctan(0)=0, forall x$$

so it must not be a solution.

What should I do in this situation? What's wrong in the method that I proposed?

pde boundary-value-problem

share|cite|improve this question

edited Jul 19 at 8:31

Dylan

11.4k31026

asked Jul 18 at 23:56

Verónica

235

add a comment | 

up vote
-1
down vote

favorite

up vote
-1
down vote

favorite

I have a problem trying to find a solution of this problem:

Find a harmonic function $u(x,y)$ in the region $operatornameIm(z)>0$ with:
$$ u(x,0) = begincases 1 & textrmif xge 0\ 0 & textrmif x<0 \ endcases $$

then I'd like to use that idea for solving this other problem:

Find a harmonic function $u(x,y)$ in the region $operatornameIm(z)>0$ with:
$$ u(x,0) = begincases T_1 & textrmif x < a\ T_2 & textrmif ale xle b \ T_2 & textrmif x > b endcases $$

I did this for the first problem:

I defined a $g(z)= z = x+iy$

then I said, for an $h$ in polar coordinates:

$h(g,0)=0$

$h(g,pi)=1$

so $h(g,theta)=Atheta+B$

where $A=frac1pi$ and $B=0$

then I said:

$$ u(x,y)=h(operatornameRe(g),operatornameIm(g))=frac1piarctanleft(fracyxright) $$

but when I compute $u(x,0)$ it looks like this:

$$u(x,0)=frac1piarctan(0)=0, forall x$$

so it must not be a solution.

What should I do in this situation? What's wrong in the method that I proposed?

pde boundary-value-problem

share|cite|improve this question

edited Jul 19 at 8:31

Dylan

11.4k31026

asked Jul 18 at 23:56

Verónica

235

I have a problem trying to find a solution of this problem:

Find a harmonic function $u(x,y)$ in the region $operatornameIm(z)>0$ with:
$$ u(x,0) = begincases 1 & textrmif xge 0\ 0 & textrmif x<0 \ endcases $$

then I'd like to use that idea for solving this other problem:

Find a harmonic function $u(x,y)$ in the region $operatornameIm(z)>0$ with:
$$ u(x,0) = begincases T_1 & textrmif x < a\ T_2 & textrmif ale xle b \ T_2 & textrmif x > b endcases $$

I did this for the first problem:

I defined a $g(z)= z = x+iy$

then I said, for an $h$ in polar coordinates:

$h(g,0)=0$

$h(g,pi)=1$

so $h(g,theta)=Atheta+B$

where $A=frac1pi$ and $B=0$

then I said:

$$ u(x,y)=h(operatornameRe(g),operatornameIm(g))=frac1piarctanleft(fracyxright) $$

but when I compute $u(x,0)$ it looks like this:

$$u(x,0)=frac1piarctan(0)=0, forall x$$

so it must not be a solution.

What should I do in this situation? What's wrong in the method that I proposed?

pde boundary-value-problem

share|cite|improve this question

edited Jul 19 at 8:31

Dylan

11.4k31026

asked Jul 18 at 23:56

Verónica

235

share|cite|improve this question

share|cite|improve this question

edited Jul 19 at 8:31

Dylan

11.4k31026

edited Jul 19 at 8:31

Dylan

11.4k31026

edited Jul 19 at 8:31

Dylan

11.4k31026

11.4k31026

asked Jul 18 at 23:56

Verónica

235

asked Jul 18 at 23:56

Verónica

235

asked Jul 18 at 23:56

Verónica

235

235

add a comment | 

add a comment | 

1 Answer
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oldest

votes

up vote
0
down vote



accepted

There are two problems. First, your boundary conditions are wrong.

$$u(x > 0,0) = h(r,0) = 1$$
$$u(x<0,0) = h(r,pi) = 0$$

therefore $h(r,theta) = 1- dfracthetapi$

Second, converting back to Cartesian isn't as straightforward as taking the arctangent, since the function $arctanleft(fracyxright)$ only gives values between $(-pi/2,pi/2]$, and it isn't the case here. You have to split up the domain like this

$$ theta = begincases arctanleft(fracyxright), & 0 < theta < fracpi2 \ arctanleft(fracyxright) + pi, & fracpi2 < theta < pi endcases $$

because when $x < 0$, $y > 0$, $arctanleft(fracyxright)$ gives an angle in the range $(-fracpi2,0]$, which is not what we want.

Finally, we have

$$ u(x,y) = -frac1piarctanleft(fracyxright) + begincases 1, & x> 0 \ 0, & x < 0 endcases $$

Alternatively, you can also write

$$ u(x,y) = frac12 + frac1piarctanleft(fracxyright) $$

by noting that $theta = fracpi2 - phi$ where $fracxy = cottheta = tanphi$

share|cite|improve this answer

edited Jul 21 at 7:28

answered Jul 19 at 7:39

Dylan

11.4k31026

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you! Shouldn't it be: $$ u(x,y) = -frac1piarctanleft(fracxyright)+ begincases 1, & x> 0 \ 0, & x < 0 endcases $$ ?
  – Verónica
  Jul 21 at 1:47
  
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

There are two problems. First, your boundary conditions are wrong.

$$u(x > 0,0) = h(r,0) = 1$$
$$u(x<0,0) = h(r,pi) = 0$$

therefore $h(r,theta) = 1- dfracthetapi$

Second, converting back to Cartesian isn't as straightforward as taking the arctangent, since the function $arctanleft(fracyxright)$ only gives values between $(-pi/2,pi/2]$, and it isn't the case here. You have to split up the domain like this

$$ theta = begincases arctanleft(fracyxright), & 0 < theta < fracpi2 \ arctanleft(fracyxright) + pi, & fracpi2 < theta < pi endcases $$

because when $x < 0$, $y > 0$, $arctanleft(fracyxright)$ gives an angle in the range $(-fracpi2,0]$, which is not what we want.

Finally, we have

$$ u(x,y) = -frac1piarctanleft(fracyxright) + begincases 1, & x> 0 \ 0, & x < 0 endcases $$

Alternatively, you can also write

$$ u(x,y) = frac12 + frac1piarctanleft(fracxyright) $$

by noting that $theta = fracpi2 - phi$ where $fracxy = cottheta = tanphi$

share|cite|improve this answer

edited Jul 21 at 7:28

answered Jul 19 at 7:39

Dylan

11.4k31026

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you! Shouldn't it be: $$ u(x,y) = -frac1piarctanleft(fracxyright)+ begincases 1, & x> 0 \ 0, & x < 0 endcases $$ ?
  – Verónica
  Jul 21 at 1:47
  
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

There are two problems. First, your boundary conditions are wrong.

$$u(x > 0,0) = h(r,0) = 1$$
$$u(x<0,0) = h(r,pi) = 0$$

therefore $h(r,theta) = 1- dfracthetapi$

Second, converting back to Cartesian isn't as straightforward as taking the arctangent, since the function $arctanleft(fracyxright)$ only gives values between $(-pi/2,pi/2]$, and it isn't the case here. You have to split up the domain like this

$$ theta = begincases arctanleft(fracyxright), & 0 < theta < fracpi2 \ arctanleft(fracyxright) + pi, & fracpi2 < theta < pi endcases $$

because when $x < 0$, $y > 0$, $arctanleft(fracyxright)$ gives an angle in the range $(-fracpi2,0]$, which is not what we want.

Finally, we have

$$ u(x,y) = -frac1piarctanleft(fracyxright) + begincases 1, & x> 0 \ 0, & x < 0 endcases $$

Alternatively, you can also write

$$ u(x,y) = frac12 + frac1piarctanleft(fracxyright) $$

by noting that $theta = fracpi2 - phi$ where $fracxy = cottheta = tanphi$

share|cite|improve this answer

edited Jul 21 at 7:28

answered Jul 19 at 7:39

Dylan

11.4k31026

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you! Shouldn't it be: $$ u(x,y) = -frac1piarctanleft(fracxyright)+ begincases 1, & x> 0 \ 0, & x < 0 endcases $$ ?
  – Verónica
  Jul 21 at 1:47
  
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

There are two problems. First, your boundary conditions are wrong.

$$u(x > 0,0) = h(r,0) = 1$$
$$u(x<0,0) = h(r,pi) = 0$$

therefore $h(r,theta) = 1- dfracthetapi$

Second, converting back to Cartesian isn't as straightforward as taking the arctangent, since the function $arctanleft(fracyxright)$ only gives values between $(-pi/2,pi/2]$, and it isn't the case here. You have to split up the domain like this

$$ theta = begincases arctanleft(fracyxright), & 0 < theta < fracpi2 \ arctanleft(fracyxright) + pi, & fracpi2 < theta < pi endcases $$

because when $x < 0$, $y > 0$, $arctanleft(fracyxright)$ gives an angle in the range $(-fracpi2,0]$, which is not what we want.

Finally, we have

$$ u(x,y) = -frac1piarctanleft(fracyxright) + begincases 1, & x> 0 \ 0, & x < 0 endcases $$

Alternatively, you can also write

$$ u(x,y) = frac12 + frac1piarctanleft(fracxyright) $$

by noting that $theta = fracpi2 - phi$ where $fracxy = cottheta = tanphi$

share|cite|improve this answer

edited Jul 21 at 7:28

answered Jul 19 at 7:39

Dylan

11.4k31026

There are two problems. First, your boundary conditions are wrong.

$$u(x > 0,0) = h(r,0) = 1$$
$$u(x<0,0) = h(r,pi) = 0$$

therefore $h(r,theta) = 1- dfracthetapi$

Second, converting back to Cartesian isn't as straightforward as taking the arctangent, since the function $arctanleft(fracyxright)$ only gives values between $(-pi/2,pi/2]$, and it isn't the case here. You have to split up the domain like this

$$ theta = begincases arctanleft(fracyxright), & 0 < theta < fracpi2 \ arctanleft(fracyxright) + pi, & fracpi2 < theta < pi endcases $$

because when $x < 0$, $y > 0$, $arctanleft(fracyxright)$ gives an angle in the range $(-fracpi2,0]$, which is not what we want.

Finally, we have

$$ u(x,y) = -frac1piarctanleft(fracyxright) + begincases 1, & x> 0 \ 0, & x < 0 endcases $$

Alternatively, you can also write

$$ u(x,y) = frac12 + frac1piarctanleft(fracxyright) $$

by noting that $theta = fracpi2 - phi$ where $fracxy = cottheta = tanphi$

share|cite|improve this answer

edited Jul 21 at 7:28

answered Jul 19 at 7:39

Dylan

11.4k31026

share|cite|improve this answer

share|cite|improve this answer

edited Jul 21 at 7:28

edited Jul 21 at 7:28

edited Jul 21 at 7:28

answered Jul 19 at 7:39

Dylan

11.4k31026

answered Jul 19 at 7:39

Dylan

11.4k31026

answered Jul 19 at 7:39

Dylan

11.4k31026

11.4k31026

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you! Shouldn't it be: $$ u(x,y) = -frac1piarctanleft(fracxyright)+ begincases 1, & x> 0 \ 0, & x < 0 endcases $$ ?
  – Verónica
  Jul 21 at 1:47
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you! Shouldn't it be: $$ u(x,y) = -frac1piarctanleft(fracxyright)+ begincases 1, & x> 0 \ 0, & x < 0 endcases $$ ?
  – Verónica
  Jul 21 at 1:47
  
  

  
  
  
  

Thank you! Shouldn't it be: $$ u(x,y) = -frac1piarctanleft(fracxyright)+ begincases 1, & x> 0 \ 0, & x < 0 endcases $$ ?
– Verónica
Jul 21 at 1:47

Thank you! Shouldn't it be: $$ u(x,y) = -frac1piarctanleft(fracxyright)+ begincases 1, & x> 0 \ 0, & x < 0 endcases $$ ?
– Verónica
Jul 21 at 1:47

add a comment | 

 

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