# find all real valued harmonic functions on the plane that are constant on all vertical lines

find all real valued harmonic functions on the plane that are constant on all vertical lines.

Harmonic function is a twice continuously differentiable function that satisfies Laplace's equation.

vertical: $f(x)=y$

harmonic: $\frac{\partial^2{f}}{\partial{x}^2}=0$

implies that $f(x,y)=ax+b$ is a line in the plane.

-
Well, what does harmonic mean formally, and what does constant on vertical lines mean formally? – Hagen von Eitzen Jul 7 '14 at 5:21
Correct; constant on vertical lines means $\partial f/\partial y=0$, after which laplace's equation becomes $\partial^2 f/\partial x^2=0$ whose solutions are precisely $f(x,y)=ax+b$ for constants $a$ and $b$ (don't put the letter $y$ on the left, though - we're talking about $f$, not $y$). – blue Jul 7 '14 at 5:48
Your presentation of the reasoning is not very intelligible though. You want to start by writing the equation for being harmonic before anything - the full equation is $\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=0$. After that, you write down the equation for constancy on vertical lines - that is $\frac{\partial f}{\partial y}=0$. (Writing $f(x)=y$ is nonsense.) Then you simplify Laplace's down to $\frac{\partial^2 f}{\partial x^2}=0$. Finally you write $f(x,y)=ax+b$ (don't lose the $y$ there; it's still a function on the plane, so there is a $y$ argument). – blue Jul 7 '14 at 6:03
@blue: I just visually saw that $f(x)=y$ if all vertical line is constant since each vertical line hits only one point on the curve ( the harmonic function ). you are right. formally, it should be $\frac{\partial{f}}{\partial{y}}=0$. – claire Jul 7 '14 at 6:33

I take the plane to be $\Bbb R^2$, and, using "plain" old fashioned $x$-$y$ coordinates, I then take the vertical lines to be parallel to the $y$-axis, that is, the lines are the sets of the form $\{(x_0, y)\}$ where $x_0 \in \Bbb R$ is fixed and $y \in \Bbb R$ may take any value.

Proceeding somewhat formally:

A differentialble function $u(x, y)$ on the plane $\Bbb R^2$ is then constant on the vertical lines if $u(x_0, y) = c, \; \text{a constant}$ for all fixed $x_0$. Thus we must have

$\dfrac{\partial u(x, y)}{\partial y} = 0 \tag{1}$

for all $x$; and thus

$\dfrac{\partial^2 u(x, y)}{\partial y^2} = 0 \tag{2}$

as well. Since $u(x, y)$ is harmonic, we must have

$\dfrac{\partial^2 u(x, y)}{\partial x^2} + \dfrac{\partial^2 u(x, y)}{\partial y^2} = 0, \tag{3}$

and so by (2),

$\dfrac{\partial^2 u(x, y)}{\partial x^2} = 0. \tag{4}$

(4) implies that

$\dfrac{\partial u(x, y)}{\partial x} = f(y), \tag{5}$

for some function $f(y)$ of $y$ alone, and thus integrating (5) with respect to $x$ we see that

$u(x, y) - u(x_0, y) = \int_{x_0}^x \dfrac{\partial u(s, y)}{\partial s}ds = \int_{x_0}^x f(y) ds = f(y)(x - x_0); \tag{6}$

or

$u(x, y) = u(x_0, y) + f(y)(x - x_0). \tag{7}$

If we differentiate (7) with respect to $y$ and use (1) we find that

$f'(y)(x - x_0) = 0 \tag{8}$

for all $x$, whence $f'(y) = 0$ and $f(y) = f_0$, also a constant. Then

$u(x, y) = u(x_0, y) + f_0(x - x_0). \tag{9}$

Using (1) again and a similar argument to (6) but applied to $y$ we have

$u(x_0, y) = u(x_0, y_0), \tag{10}$

(though indeed we were given that $u(x, y)$ is constant on vertical lines) and so

$u(x, y) = u(x_0, y_0) + f_0(x - x_0), \tag{11}$

which holds for any $(x_0, y_0) \in \Bbb R$.

That's how I see it, in any event. (11) describes a line in the $x$-$u$ plane, so I would say our OP claire is basically correct.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

-
Wow! That was fast! – Robert Lewis Jul 7 '14 at 6:09

A function of the plane is harmonic if [equation one].

A function is constant on all vertical lines if [equation two].

If a function is both, [equation one] simplifies to [equation three].

And then [equation three] can be solved by ... (proceed from there)

-

This is similar in idea to Robert Lewis's answer, but I tried to distill the main ideas.

A harmonic function satisfies $$\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}=0\tag{1}$$ at all points.

If $u$ is constant on vertical lines, then $$u(x,y)=u(x,0)=v(x)\tag{2}$$ Therefore, at all points, we have $$\frac{\partial u}{\partial y}=0\qquad\text{and thus}\qquad\frac{\partial^2u}{\partial y^2}=0\tag{3}$$ Combining $(1)$ and $(3)$, we get that $$\frac{\partial^2u}{\partial x^2}=0\tag{4}$$ at all points.

Applying $(4)$ to $(2)$ yields $$\frac{\mathrm{d}^2}{\mathrm{d}x^2}v(x)=0\tag{5}$$ which implies that for some constants $c_0$ and $c_1$ $$v(x)=c_0+c_1x\tag{6}$$ Applying $(6)$ to $(2)$ shows that $$u(x,y)=c_0+c_1x\tag{7}$$

-