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I know $\frac{\partial^2 u}{\partial x^2}=6xy$ and $\frac{\partial ^2 u}{\partial y^2} =-6xy$ and adding these together I get 0 which tells me they are harmonic functions.

To determine the harmonic conjugate, I know that $\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}=3x^2y^2-y^3$ Integrating this with respect to y gives me $v=\frac{3}{2}x^2y-\frac{y^4}{4} + f(x)$

and I know that $\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}=-x^3+3xy^2$. Integrating this with respect to x gives $v=\frac{-x^4}{4}+\frac{3x^2 y^2}{2}+f(y)$

Where do I go from here?

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  • $\begingroup$ Your integration wrt $\;y\;$ is wrong: it should be $$v=\frac32x^2\color{red}{y^2}-\frac14y^4+f(x)$$ $\endgroup$
    – Timbuc
    Commented May 21, 2015 at 19:55
  • $\begingroup$ @Aljabra Please let me know how I can improve my answer. I just want to give you the best answer I can. $\endgroup$
    – Mark Viola
    Commented May 21, 2015 at 20:18

3 Answers 3

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I'd rather go with the second Cauchy-Riemann equation after you did the first integration (corrected down here):

$$u_y=x^3-3xy^2=-v_x=-3xy^2-f'(x)\implies f'(x)=-x^3\implies f(x)=-\frac14x^4+K$$

and etc.

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Start with $u=x^3y-xy^3$. Then, $u_x=3x^2y-y^3$ and $u_y=x^3-3xy^2$.

Thus, from the Cauchy-Riemann equation $u_x=v_y$ we find that

$$u_x=v_y \implies v=\frac32 x^2y^2-\frac14 y^4 +C_1(x)$$

Next, we use this expression for $v$ in the Cauchy-Riemann equation $u_y=-v_x$ to find that

$$v_x=-u_y \implies C_1'(x) =-x^3$$

whereupon we see that $C_1(x) = -\frac14 x^4+C_2$.

Putting this together we have $v=\frac32 x^2y^2-\frac14 y^4 -\frac14 x^4+C_2$.

Therefore, the complex conjugate of $f$ is

$$\bar f=(x^3y-xy^3)-i\left(\frac32 x^2y^2-\frac14 y^4 +-\frac14 x^4\right)-iC_2$$

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  • $\begingroup$ how did you get $C'_1(x)=-x^3$? $\endgroup$
    – Al jabra
    Commented May 21, 2015 at 21:24
  • $\begingroup$ @Aljabra Apology for taking so long to revert. My oversight. To answer your question, I took the partial of $v=\frac32 x^2y^2-\frac14 y^4+C_1(x)$ with respect to $x$, which is $3xy^2+C_1'(x)$ and equated it to the negative of the partial of $u=x^3y-xy^3$ with respect to $y$, which is $-x^3+3xy^2$. Thus, $C_1'=-x^3$. Please let me know how I can improve my answer. I really just want to give you the best answer I can! $\endgroup$
    – Mark Viola
    Commented May 27, 2015 at 20:56
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Since it is a homogeneous polynomial, meaning exponent sums $3+1=1+3=4,$ this is guaranteed to be the imaginary part of some constant times $z^4,$ where $z = x + i y.$ I wrote it out, $$z^4 = x^4 - 6 x^2 y^2 + y^4 + 4(x^3y - x y^3)i$$ Your initial polynomial is $1/4$ the imaginary part of this, so the conjugate can be $$ \left(x^4 - 6 x^2 y^2 + y^4 \right) / 4 $$

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