Question

Show that u(x, y) is harmonic in some domain and find a harmonic conjugate v(x, y)
when
(a) u(x, y) = 2x(1 − y); (b) u(x, y) = 2x − x3 + 3xy2;
(c) u(x, y) = sinh x sin y; (d) u(x, y) = y/(x2 + y2)

Answer 1

Answer:

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Answer 2

Concept:

The harmonic conjugate to a actual valued feature is an imaginary feature, such that the entire feature is differentiable

Given:

We are given that

(a)$u(x,y)=2x(1-y)$

(b)$u(x,y)=2x-x^3+3xy^2$

(c)$u(x,y)=\sinh x\sin y$

(d)$u(x,y)=\frac{y}{x^2+y^2}$

Find:

We have to expose that $u(x,y)$  iis harmonic in a few area and discover a harmonic conjugate  $v(x,y)$ for every of the given.

Solution:

(a) Given that $u(x,y)=2x(1-y)$

Firstly, we can discover $u_{xx}$ and $u_{yy}$ and display that the sum of $u_{xx}+u_{yy}=0$.

$u_{xx}=0$ and $u_{yy}=0$ and their sum is also $0$.

So, it's far demonstrated that  $u(x,y)$  is harmonic in a few area.

Now, we can discover the harmonic conjugate of $v(x,y)$.

Let $v(x, y)$ be a harmonic conjugate. Then $u(x, y)$ and $v(x, y)$ fulfill the Cauchy-Riemann conditions.

$\begin{aligned}v(x,y)&=\int v_{y}dy\\ &=\int u_{x}dy\\ &=\int (2-2y)dy\\ &=2y-y^2+\phi(x)\end$

where $\phi(x)$ is an undetermined feature of $x$. Since$v_{x}=-u_{y}$, we have

$\phi'(x)=2x$

Further, we can discover the value of $\phi (x)$ and replacement in $v(x,y),$ we get

$\phi(x)=x^2+c$ and $v(x,y)=x^2-y^2+2y+c$

Hence, $v(x,y)=x^2-y^2+2y+c$ when $u(x,y)=2x(1-y).$

(b) Given that $u(x,y)=2x-x^3+3xy^2$.

Firstly, we can discover $u_{xx}$ and $u_{yy}$ and display that the sum of $u_{xx}+u_{yy}=0$.

$u_{xx}=-6x$ and $u_{yy}=6x$ and their sum is also $0$.

So, it's far demonstrated that  $u(x,y)$  is harmonic in a few area.

Now, we can discover the harmonic conjugate of $v(x,y)$.

Let $v(x, y)$ be a harmonic conjugate. Then $u(x, y)$ and $v(x, y)$ fulfill the Cauchy-Riemann conditions.

$\begin{aligned}v(x,y)&=\int v_{y}dy\\ &=\int u_{x}dy\\ &=\int (2-3x^2+3y^2)dy\\ &=2y-3x^2y+y^3+\phi(x)\end$

where $\phi(x)$ is an undetermined function of $x$. Since$v_{x}=-u_{y}$, we have

$-6xy+\phi'(x)=-6xy\\\phi'(x)=0$

Further, we can discover the value of $\phi (x)$ and replacement in $v(x,y),$ we get

$\phi(x)=c$ and $v(x,y)=2y-3x^2y+y^3+c$

Hence, $v(x,y)=2y-3x^2y+y^3+c$ when $u(x,y)=2x-x^3+3xy^2.$

(c) Given that $u(x,y)=\sinh (x)\sin y$

Firstly, we can discover $u_{xx}$ and $u_{yy}$ and display that the sum of $u_{xx}+u_{yy}=0$.

$u_{xx}=\sinh x\sin y$ and $u_{yy}=-\sinh x\sin y$ and their sum is also $0$.

So, it's far demonstrated that  $u(x,y)$  is harmonic in a few area.

Now, we can discover the harmonic conjugate of $v(x,y)$.

Let $v(x, y)$ be a harmonic conjugate. Then $u(x, y)$ and $v(x, y)$ fulfill the Cauchy-Riemann conditions.

$\begin{aligned}v(x,y)&=\int v_{y}dy\\ &=\int u_{x}dy\\ &=\int (\cosh x \sin y)dy\\ &=-\cosh x\cos y+\phi(x)\end$

where $\phi(x)$ is an undetermined function of $x$. Since$v_{x}=-u_{y}$, we have

$-\sinh x\cos y+\phi'(x)=-\sinh x\cos y\\\phi'(x)=0$

Further, we can discover the value of $\phi (x)$ and replacement in $v(x,y),$ we get

$\phi(x)=c$ and $v(x,y)=-\cosh x\cos y+c$

Hence, $v(x,y)=-\cosh x\cos y+c$ when $u(x,y)=\sinh x\sin y$.

(d) given that $u(x,y)=\frac{y}{x^2+y^2}$

Firstly, we can discover $u_{xx}$ and $u_{yy}$ and display that the sum of $u_{xx}+u_{yy}=0$.

$u_{xx}=\frac{6x^2y-2y^3}{(x^2+y^2)^{3}}$ and $u_{yy}=-\frac{6x^2y-2y^3}{(x^2+y^2)^{3}}$ and their sum is also $0$.

So, it's far demonstrated that $u(x,y)$  is harmonic in a few area.

Now, we can discover the harmonic conjugate of $v(x,y)$.

Let $v(x, y)$ be a harmonic conjugate. Then $u(x, y)$ and $v(x, y)$ fulfill the Cauchy-Riemann conditions.

$\begin{aligned}v(x,y)&=\int v_{y}dy\\ &=\int u_{x}dy\\ &=\int -\frac{2xy}{(x^2+y^2)^2}dy\\ &=\frac{x}{x^2+y^2}+\phi(x)\end$

where $\phi(x)$ is an undetermined function of $x$. Since$v_{x}=-u_{y}$, we have

$-\frac{x^2-y^2}{x^2+y^2}+\phi'(x)=-\frac{x^2-y^2}{x^2+y^2}\\\phi'(x)=0$

Further, we can discover the value of $\phi (x)$ and replacement in $v(x,y),$ we get

$\phi(x)=c$ and $v(x,y)=\frac{x}{x^2+y^2}+c$

Hence, $v(x,y)=\frac{x}{x^2+y^2}+c$when $u(x,y)=\frac{y}{x^2+y^2}$.

#SPJ3