Question

Re f=u = -y sinh y cos x + x sin x cosh y,
find f ?
(Re: real part of f)

Answer 1

if i remember correctly you have
\(u_x=v_y\) so you could try differentiating and then integrate, but the derivative is such a big mess i think maybe there is some other method

Answer 2

but f = u(x,y) + i v(x,y)

Answer 3

i don't know the imaginary part of f

Answer 4

ux=(ysinh(y)+cosh(y))sin(x)+cosh(y)xcos(x)

Answer 5

v=integral of vy= sin(x)(sinh(y)+(y−1)ey−(−y−1)e−y2)+xcos(x)sinh(y)+C

Answer 6

and then f= u+iv is that correct ?

Answer 7

@oldrin.bataku

Answer 8

so we want the harmonic conjugate of \(u\)
$$u(x,y) = -y\sinh y\cos x + x\sin x\cosh y$$
if \(f(x+iy)=u(x,y)+iv(x,y)\) is holomorphic we satisfy the Cauchy-Riemann equations: $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\\\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$ so we want a \(v\) satisfying $$\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}=y\sinh y\sin x+\sin x\cosh y+x\cos x\cosh y\\\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}=\sinh y\cos x+y\cosh y\cos x-x\sin x\sinh y$$

Answer 9

so: $$v(x,y)=\int(y\sinh y\sin x+\sin x\cosh y+x\cos x\cosh y)\, dy\\\qquad=\sin x\int y\sinh y\, dy+(\sin x+x\cos x)\sinh y\\\qquad=(y\cosh y-\sinh y)\sin x+(\sin x+x\cos x)\sinh y+f(x)\\\qquad=y\cosh y\sin x+x\sinh y\cos x+f(x)\\v(x,y)=\int(\sinh y \cos x+y\cosh y\cos x-x\sin x\sinh y)dx\\\qquad=\sinh y\sin x+y\cosh y\sin x-\sinh y\int x\sin x\,dx\\\qquad=(\sinh y+y\cosh y)\sin x+(x\cos x-\sin x)\sinh y+g(y)\\\qquad=y\cosh y\sin x+x\cos x\sinh y+g(y)$$

so valid \(v\) are of the form $$v(x,y)=y\cosh y\sin x+x\sinh y\cos x+C$$

Answer 10

thank you so much