3.3 參考解答

  1. ϕ(x,y)=ax2+bxy+cy2\phi (x,y)=ax^2+bxy+cy^2

    ϕxx+ϕyy=0\phi_{xx}+\phi_{yy}=0

    ϕx=2ax+by\phi_x=2ax+by, ϕy=bx+2cy\phi_y=bx+2cy

    ϕxx=2a\phi_{xx}=2a, ϕyy=2c\phi_{yy}=2c

    2a+2c=02a+2c=0, i.e. a=−ca=-c

    ϕ(x,y)=a(x2−y2)+bxy\phi(x,y)=a(x^2-y^2)+bxy

5. (a) u(x,y)=y3−3x2yu(x,y)=y^3-3x^2y

Since f(z)f(z) is analytic function satisfy C.R.E.

{ux=vyuy=−vx\begin{cases} u_x=v_y \\ u_y=-v_x \end{cases} {vy=−6xyvx=−(3y2−3x2)\begin{cases} v_y=-6xy \\ v_x=-(3y^2-3x^2) \end{cases}

v(x,y)=∫−6xy dy=−3xy2+g(x) v(x,y)=\int -6xy \space dy=-3xy^2+g(x)

vx=−3y2+g’(x)=−3y2+3x2v_x=-3y^2+g’(x)=-3y^2+3x^2 

∴g’(x)=3x2\therefore g’(x)=3x^2 , g(x)=x3+Cg(x)=x^3+C (CC is constant)

∴v(x,y)=x3−3xy2+C\therefore v(x,y)=x^3-3xy^2+C.

Hence, f(z)=y3−3x2y+i(3−3xy2+C)f(z)=y^3-3x^2y+i(^3-3xy^2+C).

(b) u(x,y)=sin⁡ysinh⁡xu(x,y)=\sin{y}\sinh{x}

Since f(z)f(z) is analytic function satisfy C.R.E.

{ux=vyuy=−vx\begin{cases} u_x=v_y \\ u_y=-v_x \end{cases} {vy=sin⁡ycosh⁡xvx=−(cos⁡ysinh⁡x)\begin{cases} v_y=\sin{y}\cosh{x} \\ v_x=-(\cos{y}\sinh{x}) \end{cases}

v(x,y)=∫sin⁡ycosh⁡x dy=cosh⁡x⋅(−cos⁡y)+g(x)v(x,y)=\int \sin{y}\cosh{x} \space dy=\cosh{x}\cdot (-\cos{y})+g(x)

vx=sinh⁡x(−cos⁡y)+g’(x)=−(cos⁡ysinh⁡x)v_x= \sinh{x}(-\cos{y})+g’(x)=-(\cos{y}\sinh{x})

∴g’(x)=0\therefore g’(x)=0 , g(x)=Cg(x)=C (CC is constant)

∴v(x,y)=−cos⁡ycosh⁡x+C\therefore v(x,y)=-\cos{y}\cosh{x}+C

Hence, f(z)=sin⁡ysinh⁡x+i(−cos⁡ycosh⁡x+C)f(z)=\sin{y}\sinh{x}+i(-\cos{y}\cosh{x}+C)

(c) v(x,y)=eysin⁡xv(x,y)=e^y \sin{x}

Since f(z)f(z) is analytic function satisfy C.R.E.

{ux=vyuy=−vx\begin{cases} u_x=v_y \\ u_y=-v_x \end{cases} {ux=eysin⁡xuy=−eycos⁡x\begin{cases} u_x=e^y \sin{x} \\ u_y=-e^y \cos{x} \end{cases}

u(x,y)=∫e6sin⁡x dx=−eycos⁡x+g(y)u(x,y)=\int e^6 \sin{x} \space dx=-e^y \cos{x}+g(y)

uy=−eycos⁡x+g’(y)=−eycos⁡xu_y= -e^y \cos{x}+g’(y)=-e^y \cos{x}

∴g’(y)=0\therefore g’(y)=0 , g(y)=Cg(y)=C (CC is constant)

∴u(x,y)=−eycos⁡x+C\therefore u(x,y)=-e^y \cos{x}+C.

Hence, f(z)=−eycos⁡x+C+i(eysin⁡x)f(z)=-e^y \cos{x}+C+i(e^y \sin{x}).

(d) v(x,y)=sin⁡xcosh⁡yv(x,y)=\sin{x}\cosh{y}

Since f(z)f(z) is analytic function satisfy C.R.E.

{ux=vyuy=−vx\begin{cases} u_x=v_y \\ u_y=-v_x \end{cases} {ux=sin⁡xsinh⁡yuy=−cos⁡xcosh⁡y\begin{cases} u_x=\sin{x}\sinh{y} \\ u_y=-\cos{x}\cosh{y} \end{cases}

u(x,y)=∫sin⁡xsinh⁡y dx=−cos⁡xsinh⁡y+g(y)u(x,y)=\int \sin{x}\sinh{y} \space dx=-\cos{x}\sinh{y}+g(y)

uy=−cos⁡xcosh⁡y+g’(y)=−cos⁡xcosh⁡yu_y=-\cos{x}\cosh{y}+g’(y)=-\cos{x}\cosh{y}

∴g’(y)=0\therefore g’(y)=0 , g(y)=Cg(y)=C (CC is constant)

∴u(x,y)=−cos⁡xsinh⁡y+C\therefore u(x,y)=-\cos{x}\sinh{y}+C

Hence, f(z)=cos⁡xsinh⁡y+C+i(sin⁡xcosh⁡y)f(z)=\cos{x}\sinh{y}+C+i(\sin{x}\cosh{y}).

  1. Since vv is harmonic conjugate of uu, then f(z)=u(x,y)+iv(x,y)f(z)=u(x,y)+iv(x,y) is analytic

    & satisfy {ux=vyuy=−vx\begin{cases} u_x=v_y \\ u_y=-v_x \end{cases}

    ⇒{vy=uxvx=−uy⇔i.e.f∗(z)=(−i)(u+iv)=v+i(−u)\Rightarrow \begin{cases} v_y=u_x \\ v_x=-u_y \end{cases} \xLeftrightarrow{\text{i.e.}} f^*(z)=(-i)(u+iv)=v+i(-u)

    ∴−u\therefore -u is a harmonic conjugate of vv.

  1. Since vv be a harmonic conjugate of uu.

    Let f(z)=u(x,y)+i(x,y)f(z)=u(x,y)+i(x,y) is analytic

    & f2(z)=u2−v2+i(2uv)f^2(z)=u^2-v^2+i(2uv) is analytic

    & by Theorem 3.8, let h=u2−v2h=u^2-v^2 is a harmonic function.

  1. Since vv is harmonic conjugate of u⟶u \longrightarrow satisfy {ux=vyuy=−vx\begin{cases} u_x=v_y \\ u_y=-v_x \end{cases}

    & uu is harmonic conjugate of v⟶v \longrightarrow satisfy {vx=uyvy=−ux\begin{cases} v_x=u_y \\ v_y=-u_x \end{cases}.

    Then f(z)=u+ivf(z)=u+iv is analytic, ux=uy=vx=vy=0u_x=u_y=v_x=v_y=0.

    ⇒u\Rightarrow u, vv constant functions