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Given a harmonic function on a region with no holes, the Cauchy-Riemann equations can be integrated to build the missing imaginary part, unique up to a constant — so on such a region harmonic and real part of an analytic function are the same condition.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to construct a harmonic conjugate by integrating the first Cauchy-Riemann equation and fixing the unknown function of one variable with the second, assemble the analytic function from its two parts, say why the conjugate is unique only up to a constant, and say what goes wrong on a region with a hole in it.
That both parts of an analytic function are harmonic. This lesson runs that sentence backwards: given one harmonic function, build the other part, and so build the analytic function neither half was handed to you as.
$v$ is a harmonic conjugate of $u$ on a region when $u + iv$ is analytic there — so the order matters, and conjugate here has nothing to do with the complex conjugate $\bar z$. A region is simply connected when it has no holes, so that every loop in it can be shrunk to a point inside it; a disc is simply connected and a punctured disc is not.
The theorem. If $u$ is harmonic on a simply connected region, then a harmonic conjugate $v$ exists there, and it is unique up to an additive constant. So on such a region, harmonic and the real part of an analytic function say the same thing.
The construction is the two Cauchy-Riemann equations, used one after the other.
$$v(x, y) = \int u_x\,dy + g(x).$$
The constant of integration is a function of $x$, because anything independent of $y$ integrates to zero in $y$.
Why simple connectivity is a hypothesis and not decoration. On the punctured plane, $u = \ln|z|$ is harmonic. Its conjugate would have to be the argument, and the argument cannot be defined continuously round a loop enclosing the origin — it grows by $2\pi$ each time round. So $\ln|z|$ has no single-valued harmonic conjugate there, and the analytic function it belongs to, $\log z$, is exactly the multivalued object of lesson 5.
What this is for. A boundary value problem for Laplace's equation — the steady temperature of a plate, the potential around a conductor — becomes a question about an analytic function the moment the conjugate is built, and the whole apparatus of unit 3 becomes available to it. That is the route unit 5 takes at the end of the course.
Another way: picture
Draw the level curves of $u$ and the level curves of its conjugate $v$ on the same picture: they cross at right angles, everywhere. If $u$ is the temperature of a plate, the curves of constant $u$ are the isotherms and the curves of constant $v$ are the lines along which heat flows. One construction produces both families at once, and the right angles are the Cauchy-Riemann equations seen geometrically.
Another way: steps
Three small points that cause most of the lost marks here.
The constant is real freedom. A conjugate is determined up to an additive real constant, which shifts $f$ by a purely imaginary constant. A question that wants one particular conjugate has to say so, usually by prescribing a value at a point, and this lesson's questions do.
The order is not symmetric. If $v$ is a conjugate of $u$, then $u + iv$ is analytic — but $v + iu$ generally is not. What is true is that $-u$ is a conjugate of $v$, because $i(u + iv) = -v + iu$ is analytic. Swapping the two halves without changing a sign is the commonest error in this lesson.
The unknown is a function, not a number. Integrating in $y$ leaves $g(x)$, and writing a constant $C$ there instead loses every conjugate whose $x$ behaviour is non-trivial. The whole second half of the method exists to determine $g$, and a solution that writes $C$ at step one has nothing left to do at step two and gets the wrong answer confidently.
There is also a shortcut worth knowing once the method is secure. If $u$ is harmonic, then $f(z) = 2u\left(\tfrac{z}{2}, \tfrac{z}{2i}\right) + C$ is analytic with real part $u$, which recovers $f$ in one line from the formula for $u$. It is a check rather than a method: it hides the two equations that the construction makes visible.
Integrating $v_y = u_x$ with respect to $y$ adds an arbitrary quantity that does not depend on $y$ — and does not depend on $y$ is not the same as is a constant. It may be any function of $x$ at all, and it usually is one.
Writing a bare constant there is the error that quietly ruins the method. The second Cauchy-Riemann equation is then over-determined: it demands something of $g'(x)$ that a constant cannot supply, and the usual response is to notice a contradiction and conclude the conjugate does not exist. It does; the wrong integration constant hid it.
Two smaller confusions are worth naming. The harmonic conjugate has no relationship to the complex conjugate: $v$ is not $\bar u$ and not $-u$. And the existence theorem is local — it says a conjugate exists on a simply connected region. On a region with a hole in it, a harmonic function may genuinely have no conjugate at all, and $\ln|z|$ on the punctured plane is the example that shows the hypothesis is doing work.
$u = x^{2} - y^{2} + x$ is harmonic, since $2 - 2 = 0$. First equation: $v_y = u_x = 2x + 1$.
Check harmonic, then start with the first equation.
Integrating in $y$: $v = 2xy + y + g(x)$.
The constant of integration depends on $x$.
Second equation: $v_x = 2y + g'(x)$ must equal $-u_y = 2y$, so $g' = 0$ and $v = 2xy + y$. Then $f = z^{2} + z$.
The second equation fixes the unknown function.
$u = \ln|z| = \tfrac12\ln(x^{2} + y^{2})$ is harmonic on the plane with the origin removed.
Harmonic where it is defined.
Its conjugate would have to be an argument of $z$, and going once round the origin increases that by $2\pi$: no single-valued choice exists.
The region has a hole.
On any disc missing the origin there is no problem, and the conjugate is a branch of the argument. Simple connectivity is exactly the difference.
Local existence, global failure.
$u_x = 6xy$, so $v_y = 6xy$ and $v = 3xy^{2} + g(x)$.
Integrate the first equation in $y$.
$v_x = 3y^{2} + g'$ must equal $-u_y = 3y^{2} - 3x^{2}$, so $g' = -3x^{2}$ and $v = 3xy^{2} - x^{3}$. Here $f = -iz^{3}$.
Let $u = x^{2} - y^{2} + 2x$, which is harmonic. Build its harmonic conjugate $v$ with $v(0, 0) = 0$, and fill in the table at the point $x = 1$, $y = 4$.
| Value | |
|---|---|
| The partial of u in x at the point | |
| The partial of u in y at the point | |
| The derivative of the unknown function of x | |
| The value of v at the point |
$u = 2xy$ is harmonic. Its harmonic conjugate $v$ with $v(0, 0) = 0$ takes what value at $x = 4$, $y = 7$?
Answer:
Put the steps of finding a harmonic conjugate of a given function in order.
Number the steps in order (write the number in the box):
Match each harmonic function to its harmonic conjugate, taking the constant to be zero.
| $2xy$ | $y^{2} - x^{2}$ | $9y$ | $e^{x}\sin y$ | |
|---|---|---|---|---|
| $x^{2} - y^{2}$ | ||||
| $2xy$ | ||||
| $9x$ | ||||
| $e^{x}\cos y$ |
$u$ is harmonic on a disc. Counting two conjugates that differ by a constant as the same one, how many harmonic conjugates does $u$ have there?
Answer:
Build the proof that two harmonic conjugates of the same function on a disc differ by a constant.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Let $u = x^{2} - y^{2} + x$, which is harmonic. Build its harmonic conjugate $v$ with $v(0, 0) = 0$, and fill in the table at the point $x = 4$, $y = 3$.
| Value | |
|---|---|
| The partial of u in x at the point | |
| The partial of u in y at the point | |
| The derivative of the unknown function of x | |
| The value of v at the point |
You can build a harmonic conjugate and say what is left undetermined. Say in your own words why the constant of integration is a function rather than a number. Next: contours, and the integrals that will prove everything this unit has been asserting.
9. Your turn: a harmonic conjugate of $u = 3x^{2}y - y^{3}$, step 2