# Find a harmonic function in the first quadrant,

Find a harmonic function $\phi$(x,y) in the first quadrant with the boundary values $\phi$(x,0) = -1 for x>0, and $\phi$(0,y) = 1 for y>0. Is this function unique?

My attempt was this:

Consider the principal branch of log(z), cutting away the negative real axis. Then Log(z) is analytic away from the branch cut, and its real and imaginary parts are harmonic conjugates.

Look at the harmonic function Arg(x,y) in the first quadrant.

Consider the piecewise-defined function:

$$\phi = Arg(x,0) - 1$$
$$\phi = Arg(0,y)\frac{2}{\pi}$$

My gut feeling is that I haven't done it correctly, and that I may need to first construct some linear fractional transformation or other conformal mapping, and then compose the Arg function with it - and scaling Arg(f(z)) suitably to achieve -1, +1 on the boundary.

As for uniqueness, going back to my paragraph above, if indeed I have to construct some conformal first, then $\phi$ is not unique, since there can be many different compositions of conformal mappings that lead to the same outcome, e.g., there are many ways to compose a mapping that maps the UHP to the unit disk. Then, composing these different mappings with the harmonic function, Arg(z), produces a different harmonic function.

Any hints or solutions are welcome.

Thanks,

• Well, in what space are you looking for solutions? – bartgol Aug 19 '15 at 3:27
• Hi @bartgol, what do you mean by space? I think just the complex plane? Thanks, – User001 Aug 19 '15 at 3:29
• I mean, what kind of solution do you want? A $C^2$ function? An $H^1$ function? What's the set of functions that are candidate to be solutions (provided they are harmonic and satisfy the boundary conditions)? – bartgol Aug 19 '15 at 3:32
• Probably a function harmonic on the open quadrant $Q$ and continuous on $\overline Q \setminus \{0\}.$ – zhw. Aug 19 '15 at 3:34
• This might be beyond your scope, but if you know the Poisson integral representation of a harmonic function in the disk having specified boundary values, it's not too hard to do the same given any region conformally equivalent to the disk, i.e., the first quadrant. – Blake Aug 19 '15 at 3:41

## 3 Answers

Hint: Consider $\text {Arg}\ z - \pi/4.$

• Hi @zhw, I don't see how subtracting a $\frac{\pi}{4}$ helps? On the boundary, with y>0, Arg is $\frac{\pi}{2}$, so I should scale by $\frac{2}{\pi}$, and similarly, on the boundary with x>0, Arg = 0, so I should subtract -1 to achieve -1. Unless I am overlooking something really obvious... – User001 Aug 19 '15 at 3:41
• I offered up a harmonic function that equals $\pi/4, -\pi/4$ on the axes in question. No good? – zhw. Aug 19 '15 at 3:53
• Oh!! I had noticed that but wondered why it was useful, but I think I see it now!! :-) – User001 Aug 19 '15 at 3:54
• Thanks so much @zhw! must vacate the library, as it is closing. Have a great night! – User001 Aug 19 '15 at 3:55
• Hi @zhw., I posted a question last night on MSE regarding finding a harmonic function. I was wondering whether you could take a look at my solution, if you have some free time :-). – User001 Oct 11 '15 at 22:18

The function $f(x,y)=xy$ is harmonic and it vanishes on the coordinate lines $x=0$ and $y=0$. You can add this function to any solution you have and get a new solution. So you won't get uniqueness unless you restrict to some class of solutions, such as (perhaps) bounded functions.

You have the right idea for the solution you want: the arg function is constant on the positive x and y axes, and you can offset by a constant and then multiply by a constant to get what you want.

• Hi @TrialandError, really cool argument for showing that $\phi$ is not unique. Is my piecewise-defined function ok, or does it need to be modified? I'm trying to go for a single function that will do the job but don't see a way to do it... – User001 Aug 19 '15 at 3:51
• I need to scale to achieve +1 on y>0. But on the other hand, I need to subtract 1 on x>0. Hmmm...can I combine these two functions... – User001 Aug 19 '15 at 3:53
• @LebronJames : Yes, you can add constants because constants are solutions of the Laplace equation. Multiplying a solution of Laplace's equation by a scalar gives you another solution. So, if you have a function that is $A$ on the vertical axis and $B$ on the horizontal with $A \ne B$, then subtract $\frac{A+B}{2}$. to get a function that is $\frac{A-B}{2}$ on the vertical axis, and $-\frac{A-B}{2}$ on the horizontal. Then it's easy to scale to get $1$ in one place and $-1$ in the other. – DisintegratingByParts Aug 19 '15 at 4:44
• Ok, got it - thanks so much @TrialAndError :-) – User001 Aug 19 '15 at 5:37
• @LebronJames : You're welcome. – DisintegratingByParts Aug 19 '15 at 5:38

If we start with first principles, then we have the following PDE.

$$\nabla ^2 \phi(r,\theta) = 0 \,\,\text{for}\,\,0\le r, 0\le \theta\le \pi/2 \tag 1$$

with $\phi(r,\theta=0)=-1$ and $u(r,\theta =\pi/2)=1$. The problem is not well-posed since there is no condition imposed on the boundary $r=R\to \infty$.

We can write a general solution to $(1)$ as

$$\phi(r,\theta)=A+B\theta+\sum_{n=1}^{\infty}r^n(a_n\cos n\theta+b_n\sin n\theta)$$

Applying the boundary conditions reveals that

$$B+\sum_{n=1}^{\infty}a_nr^n=-1\implies a_n=0, B=-1$$

$$A\pi/2-1+\sum_{n=1}^{\infty}\sin(n\pi/2)b_nr^n=1\implies b_{2n-1}=0, A=\frac{4}{\pi}$$

Therefore, we can write

$$\bbox[5px,border:2px solid #C0A000]{\phi(r,\theta)=\frac{4}{\pi}\theta-1+\sum_{n=1}^{\infty}c_nr^{2n}\sin(2n\theta)}$$

for any suitable coefficients $c_n$.

• Hi @Dr.MV! Nice to hear from you -- and thanks so much for your answer :-). – User001 Aug 19 '15 at 5:40
• Please let me know how I can improve my answer. I really want to give you the best answer I can. – Mark Viola Aug 28 '15 at 19:38