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Is the following reasoning correct?

Determine a harmonic conjugate to the function \begin{equation} f(x,y)=2y^{3}-6x^{2}y+4x^{2}-7xy-4y^{2}+3x+4y-4 \end{equation}

We first of all check if $f(x,y)$ is indeed a harmonic function. This amounts to show $f(x,y)$ satisfy the two-dimensional Laplace equation \begin{equation} \frac{\partial^{2 }f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}=0 \tag{1} \end{equation} We have $\frac{\partial^{2}f}{\partial x^{2}}=8-12y$ and $\frac{\partial^{2} f}{\partial y^{2}}=12y-8$. Thus, (1) is fulfilled, and so $f(x,y)$ is harmonic.

Next, we seek to determine a harmonic conjugate to the given function. Let $u(x,y)=2y^{3}-6x^{2}y+4x^{2}-7xy-4y^{2}+3x+4y-4$. \begin{equation*} u_{x}=v_{y} \iff -12xy+8x-7y+3=v_{y} \end{equation*} Integrate with respect to $y$ \begin{equation} v=-6xy^{2}+8xy-\frac{7}{2}y^{2}+3y+h(x) \tag{2} \end{equation} where $h(x)$ is a function of $x$ alone. To determine this, we use the second Cauchy-Riemann equation $v_{x}=-u_{y}$ \begin{align*} -u_{y}=v_{x} &\iff 6x^{2}+7x-6y^{2}+8y-4=h'(x)-6y^{2}+8y \\ &\iff h'(x)=6x^{2}+7x-4 \end{align*} Integrating with respect to $x$ we have \begin{equation} h(x)=2x^{3}+\frac{7}{2}x^{2}-4x+C \end{equation} where $C$ is an arbitrary constant. Therefore, if we let $C=0$, then one harmonic conjugate of $u$ is given as: \begin{equation} v=2x^{3}+\frac{7}{2}x^{2}-6xy^{2}+8xy-4x-\frac{7}{2}y^{2}+3y \end{equation}

Zero Pancakes
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harmOnic17
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  • The reasoning is correct. Unless I overlooked something, you have also made no computation errors. – Daniel Fischer Sep 13 '14 at 18:46
  • I would try to find a polynomial in $z=x+iy$ such that your given function is the imaginary part; for you, degree 3 suffices. Then the real part is a conjugate – Will Jagy Sep 13 '14 at 18:48
  • It's always something, anyway this is correct:$$ p(z) = -2 z^3 + \left(4i - \frac{7}{2} \right) z^2 + (3i+4)z - 4i. $$ The polynomial you were given is the imaginary part – Will Jagy Sep 13 '14 at 19:10
  • Thanks for taking the time... – harmOnic17 Sep 13 '14 at 19:53

2 Answers2

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Yet another shortcut. Since $u$ is harmonic (on the simply connected domain $\mathbb{C}$), there has to be a harmonic conjugate $v$. Let $F = u+iv$ be the corresponding holomorphic function. It follows from (the derivation of) Cauchy-Riemann's equations that: $$ F' = u'_x - i\,u'_y = -12xy + 8x -7y + 3 + i(6x^2+7x-6y^2+8y-4). $$ Let $G(z) = 3 + 8z + i(6z^2+7z-4)$. Then $G(z) = F'(z)$ if $z$ is real, so by the identity theorem, $G = F'$ for all $z$. Hence $$ F(z) = 3z + 4z^2 - 4 + i(2z^3+\frac72z^2-4z+C) $$ for some real constant $C$ (the real part of the constant of integration has to be $4$ to match $u$). Finally $$ v = \operatorname{Im}(F(z)). $$

mrf
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  • Could you please elaborate on how the first statement (the one following from the C-R equations' derivation) came around? A few hints would be appreciated! – William R. Ebenezer Nov 04 '19 at 08:55
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Your reasoning is correct. I'll give an alternative solution, which works only for polynomials, but can be carried out mechanically: substitute $x=(z+\bar z)/2$ and $y=(z-\bar z)/(2i)$. Result: $$ f(z) = iz^3+ (2+7i/4)z^2+(3/2-2i)z -i\bar z^3 + (2-7i/4) \bar z^2 + (3/2+2i)\bar z - 4$$ The reason this polynomial is harmonic is that no monomial has $z$ and $\bar z$ together. The fact that we started with something real lends it certain symmetry: namely, the terms $$-i\bar z^3 + (2-7i/4) \bar z^2 + (3/2+2i)\bar z$$ are just the complex conjugates of $iz^3+ (2+7i/4)z^2+(3/2-2i)z$. Therefore, $$ f(z) = \operatorname{Re}(2iz^3+ 2(2+7i/4)z^2+2(3/2-2i)z -4)$$ and the imaginary part of same complex polynomial gives the harmonic conjugate of $f$.