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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.

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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
up vote 2 down vote accepted

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

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


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

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!!!

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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)

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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 $$ u(x,y)=c_0+c_1x\tag{7} $$

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