Harmonic functions and harmonic conjugate

At a Glance

• Frequency: 7 sub-parts across 7 of 13 years (2015, 2016, 2017, 2018, 2022, 2023, 2024)
• Priority tier: T2
• Marks (count): 10 (5), 15 (2)
• Average solve time: ~9 min
• Difficulty mix: easy 4, medium 2, hard 1
• Section: A | Dominant type: computation

Why This Chapter Matters

Harmonic conjugates appear in 7 of the last 13 years — always in Section A — making this the most reliable Complex Analysis atom in the paper. Five of the seven questions ask you to: (i) verify a given function is harmonic, (ii) find its harmonic conjugate by integrating the Cauchy–Riemann equations, and (iii) express the analytic function $f(z)=u+iv$ in closed form. The Milne–Thomson method is the fastest route for step (iii) and avoids direct integration entirely for the conjugate. The two harder questions (2017, 2024) ask for the theoretical proof that real/imaginary parts of an analytic function are harmonic.

Minimum Theory

Harmonic function. A twice-differentiable real function $\phi(x,y)$ is harmonic iff $\nabla^2\phi=\phi_{xx}+\phi_{yy}=0$ everywhere on its domain.

Harmonic conjugate. If $f=u+iv$ is analytic, then $u$ and $v$ are harmonic and satisfy the Cauchy–Riemann equations: $u_x=v_y,\qquad u_y=-v_x.$ Given $u$ (harmonic), the function $v$ satisfying these equations is the harmonic conjugate of $u$, and $f=u+iv$ is analytic. The conjugate is unique up to an additive real constant.

Recovering the conjugate by CR integration. Given $u$:

1. Integrate $v_y=u_x$ w.r.t. $y$: $v=\int u_x\,dy+g(x)$.
2. Differentiate and match $v_x=-u_y$: this determines $g'(x)$, hence $g(x)$ up to a constant.

Milne–Thomson method. For an analytic $f=u+iv$: $f'(z)=u_x+iv_x=u_x-iu_y.$ Substitute $y=0$, $x=z$ in $u_x-iu_y$ (valid because $f'$ is analytic, and the formula on the real axis extends by analyticity). Then integrate $f'(z)$ to recover $f(z)$.

Equivalently, if $v$ is given: $f'(z)=v_y+iv_x.$ Substitute $y=0$, $x=z$ and integrate.

Why $u,v$ are harmonic. Differentiate CR: $\partial_x(u_x=v_y)\Rightarrow u_{xx}=v_{yx}$ and $\partial_y(u_y=-v_x)\Rightarrow u_{yy}=-v_{xy}$. Adding and using $v_{xy}=v_{yx}$ (analyticity guarantees $C^\infty$): $u_{xx}+u_{yy}=0$. Similarly $v_{xx}+v_{yy}=0$.

Useful identities. $\ln(x^2+y^2)=2\operatorname{Re}\ln z$, $\arctan(y/x)=\operatorname{Im}\ln z$, so $\ln z=\tfrac{1}{2}\ln(x^2+y^2)+i\arctan(y/x)$.

Question Archetypes

Two patterns cover every harmonic-function question in the corpus.

harmonic-conjugate (5 question(s); 2015, 2016, 2018, 2022, 2023)

Recognition Cues

• “Show/prove that $u$ (or $v$) is harmonic.” Then “find its harmonic conjugate $v$ (or $u$).”
• “Express the analytic function $f(z)=u+iv$ in terms of $z$.”
• Sometimes: given $u-v$, find $f(z)$ subject to an initial condition.

Solution Template

Standard form (given $u$ or $v$ explicitly):

1. Harmonicity: compute $\phi_{xx}+\phi_{yy}$ and show it is zero. Show each second partial, then add.
2. Conjugate (CR route): integrate $v_y=u_x$ w.r.t. $y$; differentiate and use $v_x=-u_y$ to find the integration function $g(x)$; set the constant to $0$.
3. Analytic function: assemble $f=u+iv$; recognise the closed form. Common patterns:
   • $u+iv = z^n$ (polynomial): spot $\text{Re}(z^n)$ in $u$.
   • $u+iv = ze^z$, $e^z$, $\sin z$, $\cos z$: use $e^x(\cos y+i\sin y)=e^z$, $x\pm iy=z$.
   • $u+iv$ involves $\arctan$/$\log$: use $\ln z=\tfrac{1}{2}\ln(x^2+y^2)+i\arctan(y/x)$.
   • Milne–Thomson (fastest for complicated $u$): write $f'(z)=u_x-iu_y$, set $y=0,x=z$, integrate.

Given $u-v$ (2022 type):

1. Note $\operatorname{Re}[(1+i)f]=u-v$ (since $(1+i)(u+iv)=(u-v)+i(u+v)$).
2. Set $g(z)=(1+i)f(z)$; apply Milne–Thomson to $\operatorname{Re}(g)=u-v$.
3. Simplify $g(z)$ on the real axis ($y=0$), analytically continue to get $g(z)$, then $f(z)=g(z)/(1+i)$.
4. Fix any additive constant via the initial condition.

Worked Example(s)

2016 Paper 2, 2016-P2-Q1d (10 marks, compulsory)

Is $v=x^3-3xy^2+2y$ harmonic? If yes, find its conjugate $u$ and the analytic function.

Harmonicity. $v_{xx}=6x$, $v_{yy}=-6x$; $v_{xx}+v_{yy}=0$ ✓.

Conjugate $u$ (CR integration). CR equations with $v$ as imaginary part: $u_x=v_y=-6xy+2$, $u_y=-v_x=-(3x^2-3y^2)$.

Integrate $u_x$ w.r.t. $x$: $u=-3x^2y+2x+\varphi(y).$ Differentiate: $u_y=-3x^2+\varphi'(y)$. Match $u_y=-3x^2+3y^2$: $\varphi'(y)=3y^2$, $\varphi(y)=y^3+C$. $\boxed{u=-3x^2y+y^3+2x.}$

Analytic function. $f=u+iv=(-3x^2y+y^3)+i(x^3-3xy^2)+(2x+2iy)$. Recognise $iz^3=-(3x^2y-y^3)+i(x^3-3xy^2)$ and $2z=2x+2iy$: $\boxed{f(z)=iz^3+2z.}$

2018 Paper 2, 2018-P2-Q1c (10 marks, compulsory)

Prove $u=(x-1)^3-3xy^2+3y^2$ is harmonic. Find the conjugate $v$ and $f(z)$.

Harmonicity. $u_{xx}=6(x-1)$, $u_{yy}=-6(x-1)$; sum $=0$ ✓.

Conjugate (Milne–Thomson). $f'(z)=u_x-iu_y=\bigl[3(x-1)^2-3y^2\bigr]-i\bigl[-6xy+6y\bigr]$.

Set $y=0$, $x=z$: $f'(z)=3(z-1)^2$. Integrate: $f(z)=(z-1)^3+C.$ Take $C=0$: $\boxed{f(z)=(z-1)^3.}$

Conjugate. $\operatorname{Im}(z-1)^3=3(x-1)^2y-y^3=3x^2y-6xy-y^3+3y$: $\boxed{v=3x^2y-6xy-y^3+3y.}$

2023 Paper 2, 2023-P2-Q3b (15 marks)

Prove $u=e^x(x\cos y-y\sin y)$ is harmonic. Find its conjugate $v$ and $f(z)$.

Harmonicity. $u_x=e^x[(x+1)\cos y-y\sin y]$, $u_{xx}=e^x[(x+2)\cos y-y\sin y]$. $u_y=-e^x[(x+1)\sin y+y\cos y]$, $u_{yy}=-e^x[(x+2)\cos y-y\sin y]$. Sum: $u_{xx}+u_{yy}=0$ ✓.

Conjugate (CR integration). Integrate $v_y=u_x=e^x[(x+1)\cos y-y\sin y]$ w.r.t. $y$: $v=e^x\bigl[(x+1)\sin y+y\cos y-\sin y\bigr]+g(x)=e^x\bigl[x\sin y+y\cos y\bigr]+g(x).$ Differentiate: $v_x=e^x[x\sin y+y\cos y+\sin y]+g'(x)=e^x[(x+1)\sin y+y\cos y]+g'(x)$. Match $v_x=-u_y=e^x[(x+1)\sin y+y\cos y]$: $g'(x)=0$, so $g(x)=C$. $\boxed{v=e^x(x\sin y+y\cos y).}$

Analytic function. $f=e^x[(x\cos y-y\sin y)+i(x\sin y+y\cos y)]=e^x e^{iy}(x+iy)=(x+iy)e^{x+iy}$: $\boxed{f(z)=ze^z.}$

Common Traps

• The integration “constant” when integrating $v_y$ w.r.t. $y$ is a function of $x$, not a number. It must be determined using the second CR equation.
• The Milne–Thomson substitution ($y=0$, $x=z$) must be done on $u_x-iu_y$ (where $u$ is the real part), or $v_y+iv_x$ (where $v$ is the imaginary part). Mixing up $u$ and $v$ gives the wrong sign.
• When assembling $f(z)$: the pattern $e^z=e^x(\cos y+i\sin y)$, $z=x+iy$, $iz^n,z^n$ account for essentially all standard cases. If you cannot spot the pattern, Milne–Thomson is more reliable than direct identification.
• For the $u-v$ type (2022): $(1+i)(u+iv)=(u-v)+i(u+v)$, so $\operatorname{Re}[(1+i)f]=u-v$, not $u+v$. Getting this factor wrong leads to $f$ being off by a factor of $i$.

harmonic-proof (2 question(s); 2017, 2024)

Recognition Cues

• “If $f=u+iv$ is analytic, show that $\nabla^2 u=\nabla^2 v=0$.”
• “If $w=f(z)$ is analytic, show $\nabla^2\log|f'(z)|=0$.”

Solution Template

Both parts harmonic (2017 type):

1. State the CR equations $u_x=v_y$, $u_y=-v_x$ (follow from analyticity).
2. Differentiate CR-1 in $x$ and CR-2 in $y$; add: $u_{xx}+u_{yy}=v_{xy}-v_{yx}=0$ (using $v_{xy}=v_{yx}$ since $f\in C^\infty$).
3. Similarly differentiate the other way to show $v_{xx}+v_{yy}=0$.

Log of derivative (2024 type):

1. Since $f$ analytic and $f'\ne0$, the function $\log f'(z)$ is locally analytic.
2. $\operatorname{Re}(\log f'(z))=\log|f'(z)|$.
3. Real part of an analytic function is harmonic: done.

Worked Example(s)

2017 Paper 2, 2017-P2-Q3b (15 marks)

$f=u+iv$ analytic on the unit disc $D$. Show $\nabla^2u=0=\nabla^2v$.

Since $f$ is analytic on $D$, at every point $u,v$ satisfy the Cauchy–Riemann equations: $u_x=v_y\quad(\text{CR-1}),\qquad u_y=-v_x\quad(\text{CR-2}).$ Analyticity implies $f\in C^\infty(D)$, so all mixed partials are equal ($u_{xy}=u_{yx}$, etc.).

$u$ is harmonic. Differentiate CR-1 in $x$: $u_{xx}=v_{yx}$. Differentiate CR-2 in $y$: $u_{yy}=-v_{xy}$. Add: $u_{xx}+u_{yy}=v_{yx}-v_{xy}=0$ (mixed partials equal). $\blacksquare$

$v$ is harmonic. Differentiate CR-1 in $y$: $u_{yx}=v_{yy}$. Differentiate CR-2 in $x$: $u_{xy}=-v_{xx}$. Subtract: $v_{yy}+v_{xx}=u_{yx}-u_{xy}=0$. $\blacksquare$

$\boxed{\;\nabla^2u=\nabla^2v=0\text{ on }D.\;}$

2024 Paper 2, 2024-P2-Q1b (10 marks, compulsory)

If $w=f(z)$ is analytic, show $\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)\log|f'(z)|=0$.

Suppose $f'\ne0$ in the region of interest. Choose a local branch of $\log$: $\log f'(z)=\log|f'(z)|+i\arg f'(z).$ Since $f$ is analytic, $f'$ is analytic; since $f'\ne0$, $\log f'(z)$ is locally analytic. Its real part is $\log|f'(z)|$. A real part of an analytic function is harmonic (by the theorem proved in 2017-Q3b above): $\nabla^2\log|f'(z)|=0.\qquad\blacksquare$

Common Traps

• Justify mixed-partial equality. The cancellation $v_{xy}=v_{yx}$ (which kills $u_{xx}+u_{yy}$) requires $v\in C^2$. For analytic functions $f\in C^\infty$, so this is automatic — but state it.
• For the log proof, note the assumption $f'\ne0$: at zeros of $f'$, $\log|f'|=-\infty$ and the equation makes no sense. State this at the outset.
• Show both $\nabla^2u=0$ and $\nabla^2v=0$ for the 2017 question — both are required.

Marks-Aware Writing

10-mark questions (2015, 2016, 2018, 2022, 2024): Three clear steps: verify Laplace, find the conjugate (show the CR integration or Milne–Thomson step explicitly), write $f(z)$ in closed form. For the log proof: 3–4 lines are sufficient.

15-mark questions (2017, 2023): For 2023, show all second-partial computations in full — each one is worth marks. For 2017, label the four differentiation steps and the mixed-partial equality clearly; the examiner is looking for the logical chain, not just the final equation.

Practice Set