Mixing
I have a problem with $v(x, y)=2xy+ λy$
$begingroup$
$(x, y) in mathbb{R}^2$ and $lambda in mathbb{R}$ and $v(x, y)=2 x y+ lambda y$. Determine for which value of $lambda$ a function $u(x, y)$ exists such that $$f(x+iy) =u(x, y) +iv(x, y) $$ is holomorphic. If this is possible find $u(x, y)$ such that $f(0)=0$ and $f(i) =1.$
Solution
$v(x, y)=2xy+ lambda y$. I use the rule $u_x=v_y$ and $u_y=-v_x$, so we have $v_x=2y$ and $v_y=2x+ lambda$, and so we can write $u_x=2x+ lambda$ and $u_y=-2y$ and later $u_{xx} +u_{yy} =2-2=0 $. Hence it is holomorphic, this is possible, but this means that $lambda$ is any number $in mathbb R$??
And later, when I calculate, I got $v(x, y) =(2x+1)y+c(x)$ and $v(x, y) =-2yx+c(y)$.
But how to show that $f(0)=0$ and $f(i) =1$?
complex-analysis
edited Jan 19 at 20:32
jordan_glen
1
asked Jan 19 at 19:37
KamiloKamilo
43
$endgroup$
add a comment |
$begingroup$
$(x, y) in mathbb{R}^2$ and $lambda in mathbb{R}$ and $v(x, y)=2 x y+ lambda y$. Determine for which value of $lambda$ a function $u(x, y)$ exists such that $$f(x+iy) =u(x, y) +iv(x, y) $$ is holomorphic. If this is possible find $u(x, y)$ such that $f(0)=0$ and $f(i) =1.$
Solution
$v(x, y)=2xy+ lambda y$. I use the rule $u_x=v_y$ and $u_y=-v_x$, so we have $v_x=2y$ and $v_y=2x+ lambda$, and so we can write $u_x=2x+ lambda$ and $u_y=-2y$ and later $u_{xx} +u_{yy} =2-2=0 $. Hence it is holomorphic, this is possible, but this means that $lambda$ is any number $in mathbb R$??
And later, when I calculate, I got $v(x, y) =(2x+1)y+c(x)$ and $v(x, y) =-2yx+c(y)$.
But how to show that $f(0)=0$ and $f(i) =1$?
complex-analysis
edited Jan 19 at 20:32
jordan_glen
1
asked Jan 19 at 19:37
KamiloKamilo
43
$endgroup$
$begingroup$
Do you mean that $lambda in mathbb R$?
$endgroup$
– jordan_glen
Jan 19 at 19:43
$begingroup$
This is my question
$endgroup$
– Kamilo
Jan 19 at 19:45
$begingroup$
Because calculating derivtives i cant find vaule λ
$endgroup$
– Kamilo
Jan 19 at 19:47
add a comment |
$begingroup$
$(x, y) in mathbb{R}^2$ and $lambda in mathbb{R}$ and $v(x, y)=2 x y+ lambda y$. Determine for which value of $lambda$ a function $u(x, y)$ exists such that $$f(x+iy) =u(x, y) +iv(x, y) $$ is holomorphic. If this is possible find $u(x, y)$ such that $f(0)=0$ and $f(i) =1.$
Solution
$v(x, y)=2xy+ lambda y$. I use the rule $u_x=v_y$ and $u_y=-v_x$, so we have $v_x=2y$ and $v_y=2x+ lambda$, and so we can write $u_x=2x+ lambda$ and $u_y=-2y$ and later $u_{xx} +u_{yy} =2-2=0 $. Hence it is holomorphic, this is possible, but this means that $lambda$ is any number $in mathbb R$??
And later, when I calculate, I got $v(x, y) =(2x+1)y+c(x)$ and $v(x, y) =-2yx+c(y)$.
But how to show that $f(0)=0$ and $f(i) =1$?
complex-analysis
edited Jan 19 at 20:32
jordan_glen
1
asked Jan 19 at 19:37
KamiloKamilo
43
$endgroup$
$(x, y) in mathbb{R}^2$ and $lambda in mathbb{R}$ and $v(x, y)=2 x y+ lambda y$. Determine for which value of $lambda$ a function $u(x, y)$ exists such that $$f(x+iy) =u(x, y) +iv(x, y) $$ is holomorphic. If this is possible find $u(x, y)$ such that $f(0)=0$ and $f(i) =1.$
Solution
$v(x, y)=2xy+ lambda y$. I use the rule $u_x=v_y$ and $u_y=-v_x$, so we have $v_x=2y$ and $v_y=2x+ lambda$, and so we can write $u_x=2x+ lambda$ and $u_y=-2y$ and later $u_{xx} +u_{yy} =2-2=0 $. Hence it is holomorphic, this is possible, but this means that $lambda$ is any number $in mathbb R$??
And later, when I calculate, I got $v(x, y) =(2x+1)y+c(x)$ and $v(x, y) =-2yx+c(y)$.
But how to show that $f(0)=0$ and $f(i) =1$?
complex-analysis
complex-analysis
edited Jan 19 at 20:32
jordan_glen
1
asked Jan 19 at 19:37
KamiloKamilo
43
edited Jan 19 at 20:32
jordan_glen
1
asked Jan 19 at 19:37
KamiloKamilo
43
edited Jan 19 at 20:32
jordan_glen
1
edited Jan 19 at 20:32
jordan_glen
1
edited Jan 19 at 20:32
jordan_glen
1
1
asked Jan 19 at 19:37
KamiloKamilo
43
asked Jan 19 at 19:37
KamiloKamilo
43
asked Jan 19 at 19:37
KamiloKamilo
43
43
$begingroup$
Do you mean that $lambda in mathbb R$?
$endgroup$
– jordan_glen
Jan 19 at 19:43
$begingroup$
This is my question
$endgroup$
– Kamilo
Jan 19 at 19:45
$begingroup$
Because calculating derivtives i cant find vaule λ
$endgroup$
– Kamilo
Jan 19 at 19:47
add a comment |
$begingroup$
Do you mean that $lambda in mathbb R$?
$endgroup$
– jordan_glen
Jan 19 at 19:43
$begingroup$
This is my question
$endgroup$
– Kamilo
Jan 19 at 19:45
$begingroup$
Because calculating derivtives i cant find vaule λ
$endgroup$
– Kamilo
Jan 19 at 19:47
$begingroup$
Do you mean that $lambda in mathbb R$?
$endgroup$
– jordan_glen
Jan 19 at 19:43
$begingroup$
Do you mean that $lambda in mathbb R$?
$endgroup$
– jordan_glen
Jan 19 at 19:43
$begingroup$
This is my question
$endgroup$
– Kamilo
Jan 19 at 19:45
$begingroup$
This is my question
$endgroup$
– Kamilo
Jan 19 at 19:45
$begingroup$
Because calculating derivtives i cant find vaule λ
$endgroup$
– Kamilo
Jan 19 at 19:47
$begingroup$
Because calculating derivtives i cant find vaule λ
$endgroup$
– Kamilo
Jan 19 at 19:47
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Use Cauchy-Riemann equations.
Using
The results are:
Integrate to find u(x,y)
Now we have
Recall the two values f(0)=0 and f(i)=1
For the f(0)=0 case this implies that x=0 and y=0.
In this case it is true, but there is no lambda to be recovered.
For the f(i)=1 case this implies that x=0 and y=1.
Lambda can be recovered here. Consider the following:
Solving for lambda we find that
answered Jan 19 at 21:07
Erock BroxErock Brox
23628
$endgroup$
$begingroup$
Thanks! Your solution was very useful.
$endgroup$
– Kamilo
Jan 19 at 22:10
add a comment |
$begingroup$
Integrating the Cauchy-Riemann equations gives
$$u(0,y)=c -y^2$$
and then
$$u(x,y)=u(0,y)+x^2+lambda x=c-y^2+x^2+lambda x$$
Finally, for all $x,yinmathbb{R}$,
$$u(x,y)+iv(x,y)=x^2-y^2+lambda x+ 2ixy +ilambda y +c$$
or, equivalently, for all $zinmathbb{C}$, $f(z)=z^2+lambda z +c$.
So, $f(0)=0Longleftrightarrow c=0$ and $f(i)=1Longleftrightarrow -1+lambda i= 1 Longleftrightarrow lambda=frac{2}{i}=-2i$.
answered Jan 19 at 21:14
AyoubAyoub
33118
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Use Cauchy-Riemann equations.
Using
The results are:
Integrate to find u(x,y)
Now we have
Recall the two values f(0)=0 and f(i)=1
For the f(0)=0 case this implies that x=0 and y=0.
In this case it is true, but there is no lambda to be recovered.
For the f(i)=1 case this implies that x=0 and y=1.
Lambda can be recovered here. Consider the following:
Solving for lambda we find that
answered Jan 19 at 21:07
Erock BroxErock Brox
23628
$endgroup$
$begingroup$
Thanks! Your solution was very useful.
$endgroup$
– Kamilo
Jan 19 at 22:10
add a comment |
$begingroup$
Use Cauchy-Riemann equations.
Using
The results are:
Integrate to find u(x,y)
Now we have
Recall the two values f(0)=0 and f(i)=1
For the f(0)=0 case this implies that x=0 and y=0.
In this case it is true, but there is no lambda to be recovered.
For the f(i)=1 case this implies that x=0 and y=1.
Lambda can be recovered here. Consider the following:
Solving for lambda we find that
answered Jan 19 at 21:07
Erock BroxErock Brox
23628
$endgroup$
$begingroup$
Thanks! Your solution was very useful.
$endgroup$
– Kamilo
Jan 19 at 22:10
add a comment |
$begingroup$
Use Cauchy-Riemann equations.
Using
The results are:
Integrate to find u(x,y)
Now we have
Recall the two values f(0)=0 and f(i)=1
For the f(0)=0 case this implies that x=0 and y=0.
In this case it is true, but there is no lambda to be recovered.
For the f(i)=1 case this implies that x=0 and y=1.
Lambda can be recovered here. Consider the following:
Solving for lambda we find that
answered Jan 19 at 21:07
Erock BroxErock Brox
23628
$endgroup$
Use Cauchy-Riemann equations.
Using
The results are:
Integrate to find u(x,y)
Now we have
Recall the two values f(0)=0 and f(i)=1
For the f(0)=0 case this implies that x=0 and y=0.
In this case it is true, but there is no lambda to be recovered.
For the f(i)=1 case this implies that x=0 and y=1.
Lambda can be recovered here. Consider the following:
Solving for lambda we find that
answered Jan 19 at 21:07
Erock BroxErock Brox
23628
answered Jan 19 at 21:07
Erock BroxErock Brox
23628
answered Jan 19 at 21:07
Erock BroxErock Brox
23628
answered Jan 19 at 21:07
Erock BroxErock Brox
23628
23628
$begingroup$
Thanks! Your solution was very useful.
$endgroup$
– Kamilo
Jan 19 at 22:10
add a comment |
$begingroup$
Thanks! Your solution was very useful.
$endgroup$
– Kamilo
Jan 19 at 22:10
$begingroup$
Thanks! Your solution was very useful.
$endgroup$
– Kamilo
Jan 19 at 22:10
$begingroup$
Thanks! Your solution was very useful.
$endgroup$
– Kamilo
Jan 19 at 22:10
add a comment |
$begingroup$
Integrating the Cauchy-Riemann equations gives
$$u(0,y)=c -y^2$$
and then
$$u(x,y)=u(0,y)+x^2+lambda x=c-y^2+x^2+lambda x$$
Finally, for all $x,yinmathbb{R}$,
$$u(x,y)+iv(x,y)=x^2-y^2+lambda x+ 2ixy +ilambda y +c$$
or, equivalently, for all $zinmathbb{C}$, $f(z)=z^2+lambda z +c$.
So, $f(0)=0Longleftrightarrow c=0$ and $f(i)=1Longleftrightarrow -1+lambda i= 1 Longleftrightarrow lambda=frac{2}{i}=-2i$.
answered Jan 19 at 21:14
AyoubAyoub
33118
$endgroup$
add a comment |
$begingroup$
Integrating the Cauchy-Riemann equations gives
$$u(0,y)=c -y^2$$
and then
$$u(x,y)=u(0,y)+x^2+lambda x=c-y^2+x^2+lambda x$$
Finally, for all $x,yinmathbb{R}$,
$$u(x,y)+iv(x,y)=x^2-y^2+lambda x+ 2ixy +ilambda y +c$$
or, equivalently, for all $zinmathbb{C}$, $f(z)=z^2+lambda z +c$.
So, $f(0)=0Longleftrightarrow c=0$ and $f(i)=1Longleftrightarrow -1+lambda i= 1 Longleftrightarrow lambda=frac{2}{i}=-2i$.
answered Jan 19 at 21:14
AyoubAyoub
33118
$endgroup$
add a comment |
$begingroup$
Integrating the Cauchy-Riemann equations gives
$$u(0,y)=c -y^2$$
and then
$$u(x,y)=u(0,y)+x^2+lambda x=c-y^2+x^2+lambda x$$
Finally, for all $x,yinmathbb{R}$,
$$u(x,y)+iv(x,y)=x^2-y^2+lambda x+ 2ixy +ilambda y +c$$
or, equivalently, for all $zinmathbb{C}$, $f(z)=z^2+lambda z +c$.
So, $f(0)=0Longleftrightarrow c=0$ and $f(i)=1Longleftrightarrow -1+lambda i= 1 Longleftrightarrow lambda=frac{2}{i}=-2i$.
answered Jan 19 at 21:14
AyoubAyoub
33118
$endgroup$
Integrating the Cauchy-Riemann equations gives
$$u(0,y)=c -y^2$$
and then
$$u(x,y)=u(0,y)+x^2+lambda x=c-y^2+x^2+lambda x$$
Finally, for all $x,yinmathbb{R}$,
$$u(x,y)+iv(x,y)=x^2-y^2+lambda x+ 2ixy +ilambda y +c$$
or, equivalently, for all $zinmathbb{C}$, $f(z)=z^2+lambda z +c$.
So, $f(0)=0Longleftrightarrow c=0$ and $f(i)=1Longleftrightarrow -1+lambda i= 1 Longleftrightarrow lambda=frac{2}{i}=-2i$.
answered Jan 19 at 21:14
AyoubAyoub
33118
answered Jan 19 at 21:14
AyoubAyoub
33118
answered Jan 19 at 21:14
AyoubAyoub
33118
answered Jan 19 at 21:14
AyoubAyoub
33118
33118
add a comment |
add a comment |
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$begingroup$
Do you mean that $lambda in mathbb R$?
$endgroup$
– jordan_glen
Jan 19 at 19:43
$begingroup$
This is my question
$endgroup$
– Kamilo
Jan 19 at 19:45
$begingroup$
Because calculating derivtives i cant find vaule λ
$endgroup$
– Kamilo
Jan 19 at 19:47