Separation of Variables in Cartesian Coordinates

The uniqueness theorems handed us an extraordinary license: any function that satisfies Laplace's equation and matches the boundary data — found by any means whatsoever — is the potential. What they did not hand us is merchandise: stare at a grounded slot with a charged strip across its mouth, and no amount of theorem-quoting produces $V(x,y)$ . Separation of variables is the factory that finally delivers — the most systematic solution technique in this part of physics, and, because the same move solves the heat equation, the vibrating membrane, and the Schrödinger equation in a box, arguably the most reused calculation in a physicist's education. The method opens with what looks like an unforgivable cheat: we assume the solution is a product of one-variable functions, $V=X(x)Y(y)$ . The PDE promptly shatters into ordinary differential equations linked by separation constants; the boundary conditions select the allowed eigenfunctions and eigenvalues, producing oscillatory dependence in the bounded directions and exponential or hyperbolic dependence in the remaining direction. Superposition of these separated solutions then forms a Fourier series capable of representing the relevant boundary data. The cheat is redeemed at the end: completeness guarantees the sum can match the boundary, and uniqueness guarantees that once it does, we are finished. Cartesian coordinates are where the machinery runs cleanest — slots, channels, and boxes bounded by planes — and where the sines and exponentials wear no disguise. Master the logic here and the spherical and cylindrical versions are the same play with a different cast.

Key takeaways

To solve $\nabla^2V=0$ in rectangular geometry, assume a product $V=X(x)Y(y)Z(z)$ ; the PDE splits into ODEs linked by separation constants that must sum to zero.
The zero-sum forces a split personality: some factors oscillate, at least one decays or grows. Grounded wall-pairs select the sinusoids and quantize $k=n\pi/a$ ; the leftover direction gets the exponentials.
Superposition plus completeness of the sine family lets the mode-sum match any boundary potential; Fourier's trick (sine orthogonality) extracts each coefficient by an integral, and the uniqueness theorems (their own article) certify the result as the answer.
The classic slot — grounded plates at $y=0,a$ , a strip at $V_0$ across $x=0$ — yields $V=\frac{4V_0}{\pi}\sum_{n\,\text{odd}}\frac1n e^{-n\pi x/a}\sin(n\pi y/a)$ , which resums to $\frac{2V_0}{\pi}\arctan[\sin(\pi y/a)/\sinh(\pi x/a)]$ .
Mode $n$ decays as $e^{-n\pi x/a}$ : one slot-width deep the potential is a single pure sine at 4.3% strength — fine boundary detail is a near-field phenomenon.
Key idea: assume $V=X(x)Y(y)Z(z)$ ; the PDE splits into ODEs whose solutions are sinusoids on coordinates with homogeneous boundaries and exponentials on the remaining coordinate; superpose the modes into a Fourier series and fix the coefficients by orthogonality on the inhomogeneous boundary.

The core insight

Assume $V=X(x)Y(y)Z(z)$ ; the PDE splits into ordinary differential equations, with the boundary conditions selecting sinusoidal eigenfunctions in the bounded directions and exponential or hyperbolic dependence in the remaining direction. Superpose the separated modes into a Fourier series, and determine their coefficients by orthogonality from the inhomogeneous boundary condition.

Where this lives today

Separation of variables graduated from hand technique to industrial infrastructure. Spectral methods — the workhorses of weather models, turbulence simulation, and gravitational-wave source modelling — are this article's move at scale: expand the unknown in eigenfunctions adapted to the geometry, convert the PDE to algebra on coefficients, and let the fast Fourier transform apply Fourier's trick to millions of modes in milliseconds. Every FFT-based Poisson solver inside a plasma or galaxy-formation code is equation (5) executed on a grid, and the exponential convergence that made three terms suffice in worked example 1 is why spectral codes crush finite differences whenever the geometry cooperates. Where it does not, finite-element solvers take over — and this article still supervises: closed forms like the slot's arctangent are the benchmarks against which commercial field solvers are validated, an exact harmonic function leaving error nowhere to hide. The slot's eigenfunctions lead a second life in microwave engineering: the $\sin(n\pi y/a)$ patterns are the transverse mode profiles of rectangular waveguides, where oscillate-versus-decay returns as propagating modes versus evanescent ones below cutoff. Griffiths' chalkboard slot is, mode for mode, the mathematics inside every radar feed.

The physical picture

Here is the geometry that carries the whole article — Griffiths' slot. Two semi-infinite conducting plates lie parallel at $y=0$ and $y=a$ , both grounded. The open end at $x=0$ is closed by a strip, insulated from the plates by hairline gaps and held at potential $V_0$ . The interior is charge-free, so $V$ obeys Laplace's equation there, and far down the channel the potential must fade to nothing.

Figure

The slot — two grounded plates, a hot strip, and a potential that must fade down the channel

Between the grounded plates the potential behaves like a plucked string: pinned to zero at $y=0$ and $y=a$ , whatever it does in between is a standing wave with nodes at the walls — $\sin(n\pi y/a)$ , a whole number of half-wavelengths fitting the gap. Perpendicular to the plates there is no second wall to support another oscillatory mode; instead, the condition $V\to0$ as $x\to\infty$ , together with Laplace's equation, selects exponential decay. As Laplace's Equation established, the direction-curvatures must cancel. A sine curves toward zero, so if $V$ bends like a sine across the slot it must bend the other way along it — exponential behavior. Oscillation in one direction is paid for by decay in the other. The payment is made at a precise exchange rate — the most physical fact in the article. The mode with $n$ half-wavelengths across the slot decays along it as $e^{-n\pi x/a}$ : the wigglier the pattern, the faster it dies. One slot-width in, the fundamental keeps $e^{-\pi}\approx4.3\%$ of its strength; the $n=3$ harmonic keeps $e^{-3\pi}\approx0.008\%$ — effectively gone. Whatever jagged potential the strip imposes at the mouth, only the smoothest component quickly comes to dominate a short distance in. This is why sharp electrode detail blurs with distance, why fringing fields heal within about one plate-gap — the mean-value smoothness of harmonic functions made quantitative. The strategy is now visible before any algebra: build the pinned standing waves the walls allow, attach the decaying exponential Laplace's equation demands, and blend them to reproduce whatever the strip does at the mouth. The resulting equipotential map — the payoff picture — shows exactly the predicted healing: contours crowd against the strip and relax into the smooth single-hump fundamental within a width.

Figure

Equipotentials in the slot, from the closed-form solution — sharp near the strip, smooth within a width

From one PDE to ordinary equations

Inside the slot — and inside any charge-free region bounded by coordinate planes — the governing equation is the one inherited from Laplace's Equation, written in the coordinates that match the walls:

\nabla^2 V=\frac{\partial^2 V}{\partial x^2}+\frac{\partial^2 V}{\partial y^2}+\frac{\partial^2 V}{\partial z^2}=0 \qquad (1)

where: $V$ — the electrostatic potential in the charge-free region; $x,y,z$ — Cartesian coordinates aligned with the bounding planes. The alignment is the point: each boundary condition then constrains one coordinate at a time (" $V=0$ at $y=0$ "), exactly what a one-variable factor can absorb. Geometry chooses the coordinates; the coordinates choose the special functions. The opening move is the product ansatz — the educated gamble that solutions exist of the form

V(x,y,z)=X(x)\,Y(y)\,Z(z) \qquad (2)

where: $X,Y,Z$ — functions of a single variable each, to be determined. Honesty about its status: the general solution is not a product, and the slot's final answer will visibly fail to be one — we are really hunting for a complete set of product-shaped building blocks (the warning box settles the accounts). The slot needs only the two-dimensional version $V=X(x)Y(y)$ , since nothing depends on the coordinate running along the strip.

Derivation AWhy the product ansatz shatters the PDE into ordinary equations

The goal is the engine of the whole method: substituting $V=X(x)Y(y)$ into Laplace's equation forces each factor to satisfy an ordinary differential equation, with a constant coupling them — and the sign of that constant makes one factor oscillate while the other decays. The strategy: substitute, then isolate the $x$ -dependence on one side of an equals sign and the $y$ -dependence on the other, and ask what kind of function of $x$ can equal a function of $y$ . The step to watch is the leap from "these are equal" to "both are constant" — it looks like sleight of hand and is actually the theorem.

Step 1. Substitute the product into $\partial_x^2V+\partial_y^2V=0$ — each derivative sees only its own factor, giving $YX''+XY''=0$ — then divide through by $V=XY$ and shuffle:

\frac{1}{X}\frac{\mathrm{d}^2X}{\mathrm{d}x^2}=-\frac{1}{Y}\frac{\mathrm{d}^2Y}{\mathrm{d}y^2}

Step 2. Stare at what has happened: the left side is a function of $x$ alone, the right of $y$ alone — and they are equal for every $(x,y)$ in the region.

Step 3. Freeze $y$ and let $x$ roam: the right side does not move, so the left side cannot vary at all. It is a constant, and by the mirror argument so is the right side, the same constant. Call it $k^2$ :

\frac{\mathrm{d}^2X}{\mathrm{d}x^2}=k^2X,\qquad \frac{\mathrm{d}^2Y}{\mathrm{d}y^2}=-k^2Y

Why this is the crux: a function of $x$ that equals a function of $y$ is a function of neither — independent variables can only agree on a number. That number, the separation constant, is the sole thread still tying the two ODEs together.

Step 4. Read the sign structure. The two constants are equal and opposite, so the factors cannot both oscillate and cannot both decay. With $k^2>0$ as written, $X=Ae^{kx}+Be^{-kx}$ while $Y=C\sin ky+D\cos ky$ ; $k^2<0$ swaps the roles. Which direction gets which is not ours to decide: the slot's $Y$ must vanish at two walls, and no real exponential combination vanishes twice — the oscillatory factor always belongs to the direction pinned at both ends. In three dimensions the same argument yields three constants obeying $k_x^2+k_y^2+k_z^2=0$ , so at least one factor is always exponential — three sines cannot conspire to cancel curvature.

The derivation compresses into a template every coordinate system fills with its own cast. In three dimensions the separated system reads

\frac{X''}{X}=-k_x^2=\lambda_x,\qquad \frac{Y''}{Y}=-k_y^2=\lambda_y,\qquad \frac{Z''}{Z}=-k_z^2=\lambda_z,\qquad \boxed{\lambda_x+\lambda_y+\lambda_z=0} \qquad (3)

where: $\lambda_x,\lambda_y,\lambda_z$ — the separation constants, each with dimensions of inverse length squared. When a separation constant is negative, it is convenient to write it as $-k^2$ , with $k>0$ having dimensions of inverse length; the corresponding factor is then sinusoidal. A positive separation constant produces exponential or hyperbolic dependence. The zero-sum constraint expresses the curvature trade-off imposed by Laplace's equation. The signs of the separation constants determine whether each factor is oscillatory or exponential, while the boundary conditions determine which sign is physically admissible. The most useful cases are summarized below:

the boundary situation on a coordinate	the factor you need	why
grounded walls at both ends	$\sin(n\pi y/a)$ , $k=n\pi/a$ quantized	only sinusoids vanish twice; wall spacing fixes the wavelengths
region runs to infinity, $V\to0$	$e^{-kx}$ only	the growing branch is evicted at infinity
finite run, live electrode at one end	$\sinh kx$ , $\cosh kx$ combinations	both exponentials kept, recombined to fit each end
no oscillation demanded ( $k=0$ sector)	$A+Bx$	the ideal parallel-plate potential hiding inside the method

A single product mode is far too rigid to match a real boundary function, and here the linearity of (1) earns its keep: any sum of solutions is a solution, so the honest general solution in the slot superposes every allowed mode,

V(x,y)=\sum_{n=1}^{\infty} C_n\,e^{-n\pi x/a}\,\sin\!\Big(\frac{n\pi y}{a}\Big) \qquad (4)

where: $C_n$ — one free coefficient per standing-wave mode. The two hard questions about (4) have named answers. Can such a sum match an arbitrary strip potential? Yes — the completeness of the sine eigenfunctions. Is a match the answer? Yes — the first uniqueness theorem, recalled by name from Uniqueness Theorems for the Potential. All that remains is finding the $C_n$ , and orthogonality makes that mechanical:

C_n=\frac{2}{a}\int_0^a V_0(y)\,\sin\!\Big(\frac{n\pi y}{a}\Big)\,\mathrm{d}y \qquad (5)

where: $V_0(y)$ — the prescribed potential on the strip at $x=0$ . Equation (5) is Fourier's trick: distinct sines integrate against each other to zero, so multiplying the boundary condition by a chosen sine and integrating filters out one coefficient at a time — the dot-product move of vector algebra, executed in a function space. The derivation below runs the whole slot problem, trick included.

Go deeper · Derivation B · 2 minThe slot problem end to end, from boundary conditions to Fourier coefficients

Derivation BThe slot problem end to end, from boundary conditions to Fourier coefficients

The goal is the complete solution of the hero geometry: grounded plates at $y=0$ and $y=a$ , the strip across $x=0$ at constant $V_0$ , and $V\to0$ down the channel — the problem Griffiths §3.3.1 made canonical. The strategy is the method's standard rhythm: homogeneous boundaries select and quantize the eigenfunctions, infinity prunes the exponentials, superpose, then spend the one inhomogeneous condition on the coefficients. Watch each of the four boundary conditions get consumed exactly once — a well-posed problem leaves none over and comes up none short.

Step 1. Assign the signs. The conditions are: $V=0$ at $y=0$ ; $V=0$ at $y=a$ ; $V=V_0$ at $x=0$ for $0<y<a$ ; $V\to0$ as $x\to\infty$ . The $y$ -direction is pinned at both ends, so $Y$ oscillates: $Y''=-k^2Y$ , forcing $X''=+k^2X$ with $X=Ae^{kx}+Be^{-kx}$ .

Step 2. Spend the grounded walls. $Y(0)=0$ kills the cosine; $Y(a)=0$ then demands $\sin ka=0$ , so $k=n\pi/a$ with $n=1,2,3,\dots$ (negative $n$ duplicates, $n=0$ is trivial). Why quantization appears: a whole number of half-wavelengths must fit between the walls, like a clamped string. Boundary conditions on a finite interval always discretize the separation constant — the mechanism that quantizes energies in the quantum square well.

Step 3. Spend the condition at infinity: the branch $e^{+n\pi x/a}$ blows up down the channel, so its coefficient vanishes. Each surviving mode is $e^{-n\pi x/a}\sin(n\pi y/a)$ — the decay rate slaved to the transverse wavenumber, higher $n$ dying faster.

Step 4. Superpose and face the last wall. Linearity assembles $V(x,y)=\sum_n C_n e^{-n\pi x/a}\sin(n\pi y/a)$ ; setting $x=0$ turns the final condition into a pure expansion problem, $V(0,y)=V_0(y)=\sum_n C_n\sin(n\pi y/a)$ for $0<y<a$ . The expansion is required on the open interval $0<y<a$ ; the idealized endpoint values are not imposed because the strip is separated from the grounded plates by the hairline insulating gaps.

Step 5. Fourier's trick. Multiply by $\sin(m\pi y/a)$ and integrate across the slot; the orthogonality relation

\int_0^a \sin\!\Big(\frac{n\pi y}{a}\Big)\sin\!\Big(\frac{m\pi y}{a}\Big)\,\mathrm{d}y=\frac{a}{2}\,\delta_{nm}

annihilates every term except $n=m$ , leaving $C_m=\frac{2}{a}\int_0^a V_0(y)\sin(m\pi y/a)\,\mathrm{d}y$ — equation (5).

Step 6. Evaluate: $\int_0^a\sin(n\pi y/a)\,\mathrm{d}y=\frac{a}{n\pi}(1-\cos n\pi)$ is zero for even $n$ , $2a/n\pi$ for odd. Hence $C_n=4V_0/n\pi$ for odd $n$ only, and

V(x,y)=\frac{4V_0}{\pi}\sum_{n=1,3,5,\dots}\frac{1}{n}\,e^{-n\pi x/a}\,\sin\!\Big(\frac{n\pi y}{a}\Big)

Why only odd harmonics: the boundary data is symmetric about the midplane $y=a/2$ , and even- $n$ sines are antisymmetric about it — the geometry's symmetry pre-filters the spectrum before any integral is computed. *Numerically, at $x=y=a/2$ with $V_0=100$ V the partial sums through $n=1,3,5$ give $26.468$ , $26.087$ , $26.097$ V, converging on the closed form's $26.096$ V — worked example 1.*

For the constant strip the series hides a small miracle: it can be summed exactly. The closed form is worth owning — a boundary-value problem you can hold in one hand, and a benchmark for every numerical solver you will ever write.

Go deeper · Derivation C · 1 minResumming the slot series into an arctangent

Derivation CResumming the slot series into an arctangent

The goal is to collapse the infinite series into $V=\frac{2V_0}{\pi}\arctan[\sin(\pi y/a)/\sinh(\pi x/a)]$ . The strategy is a complex-variable classic: each term is the imaginary part of $z^n$ for one complex number $z$ , so the odd- $n$ sum of $z^n/n$ becomes a logarithm, whose imaginary part is an arctangent. The watch-point is the polar bookkeeping — everything hinges on $|z|<1$ , which the decay factor guarantees for all $x>0$ .

Step 1. Set $z=e^{-\pi x/a}e^{i\pi y/a}$ , so $z^n=e^{-n\pi x/a}e^{in\pi y/a}$ and the series is an imaginary part:

V=\frac{4V_0}{\pi}\,\mathrm{Im}\sum_{n\,\text{odd}}\frac{z^n}{n}

Step 2. Sum the odd-power log series. From $-\ln(1-z)=\sum_{n\ge1}z^n/n$ , subtracting the same series at $-z$ cancels the even powers:

\sum_{n\,\text{odd}}\frac{z^n}{n}=\frac{1}{2}\ln\!\Big(\frac{1+z}{1-z}\Big)

Step 3. Take the imaginary part — the phase of the log's argument. With $\mathrm{Im}\,z=e^{-\pi x/a}\sin(\pi y/a)$ and $|z|^2=e^{-2\pi x/a}$ , this gives $V=\frac{2V_0}{\pi}\arctan\big[2e^{-\pi x/a}\sin(\pi y/a)\big/\big(1-e^{-2\pi x/a}\big)\big]$ . Why the phase formula holds: $\frac{1+z}{1-z}=\frac{(1+z)(1-\bar z)}{|1-z|^2}$ has a real positive denominator, so the phase is the arctangent of $2\,\mathrm{Im}\,z$ over $1-|z|^2$ — both positive in the slot, keeping the arctangent on its principal branch.

Step 4. Recognize the hyperbolic sine. Divide top and bottom by $2e^{-\pi x/a}$ : the denominator becomes $(e^{\pi x/a}-e^{-\pi x/a})/2=\sinh(\pi x/a)$ , giving Griffiths' Eq. 3.36:

V(x,y)=\frac{2V_0}{\pi}\,\arctan\!\Big(\frac{\sin(\pi y/a)}{\sinh(\pi x/a)}\Big)

Step 5. Interrogate the limits. As $x\to0^+$ the arctangent tends to $\pi/2$ , so $V\to V_0$ across the strip; as $x\to\infty$ the argument dies as $e^{-\pi x/a}$ and $V\to0$ ; on the plates $\sin(\pi y/a)=0$ gives $V=0$ identically. All four boundary conditions, read off by eye. As a numerical spot-check, at $r=e^{-0.37\pi}$ , $\theta=0.61\pi$ both sides of the resummation identity equal $0.288994$ ; the closed form at $x=y=a/2$ , $V_0=100$ V gives $\frac{200}{\pi}\arctan(1/2.3013)=26.096$ V, matching the series.

One dimension remains to be restored. In a genuinely three-dimensional problem — the box of worked example 2 — two coordinates carry grounded wall-pairs, so two factors oscillate with independently quantized wavenumbers, and the zero-sum rule (3) fixes the third constant to $\gamma_{nm}^2=k_x^2+k_y^2>0$ : hyperbolic. For a box $0<x,y,z<a$ with five faces grounded and the lid $z=a$ held at $V_0(x,y)$ ,

V=\sum_{n,m=1}^{\infty} C_{nm}\,\sin\!\Big(\frac{n\pi x}{a}\Big)\sin\!\Big(\frac{m\pi y}{a}\Big)\,\frac{\sinh(\gamma_{nm}z)}{\sinh(\gamma_{nm}a)},\qquad \gamma_{nm}=\frac{\pi}{a}\sqrt{n^2+m^2} \qquad (6)

where: $\gamma_{nm}$ — the longitudinal inverse-length scale set by the $(n,m)$ transverse mode; the $\sinh$ ratio vanishes on the grounded floor and equals one on the lid, so the live condition falls entirely on the coefficients. Even the gentlest pattern $(1,1)$ penetrates less far than the slot's fundamental ( $\gamma_{11}=\sqrt2\,\pi/a$ ). Fourier's trick, applied once in $x$ and once in $y$ , extracts

C_{nm}=\frac{4}{a^2}\int_0^a\!\!\int_0^a V_0(x,y)\,\sin\!\Big(\frac{n\pi x}{a}\Big)\sin\!\Big(\frac{m\pi y}{a}\Big)\,\mathrm{d}x\,\mathrm{d}y \qquad (7)

where: the prefactor $4/a^2$ is two copies of the slot's $2/a$ , one per orthogonality integral. For a uniform lid both integrals repeat Step 6 of the slot derivation, giving $C_{nm}=16V_0/\pi^2nm$ for $n,m$ both odd, zero otherwise. The same logic runs in curved geometries, where walls trade sines for new special functions — the business of Separation of Variables in Spherical Coordinates and Separation of Variables in Cylindrical Coordinates, the next two articles, with Legendre Polynomials supplying the spherical cast.

Limiting cases

case	result	meaning
deep in the slot, $x\gtrsim a$	$V\approx\frac{4V_0}{\pi}e^{-\pi x/a}\sin(\pi y/a)$	single-mode dominance: one width in, the fundamental mode is about $1600$ times larger than the $n=3$ mode in amplitude
at the mouth, $x\to0^+$	series $\to V_0$ for $0<y<a$	completeness at work: the sine series reconstructs the constant boundary value on the open interval, with the endpoint discontinuities set by the grounded plates
on the plates, $y=0,a$	$V=0$ exactly, term by term	eigenfunctions carry the homogeneous conditions identically
slot width $a\to\infty$	quantized sum $\to$ integral over $k$	Fourier series matures into the Fourier transform; discreteness was the walls' doing
geometry uniform in $z$	2D problem: one sine, one exponential	the slot itself; adding a $z$ -wall pair upgrades sums to double sums as in (6)

Common trapA sum of products is not a product — why the ansatz looks illegal and is not

The slot's answer is manifestly not of the form $X(x)Y(y)$ — no factoring tidies an infinite sum of distinct products into one. So the opening assumption is not a statement about the form of the final solution, and students are right to smell a swindle. The method's actual claim is subtler: separable solutions form a complete set of building blocks, the true solution can be assembled from them, and uniqueness certifies the assembly. Treat "assume $V=XY$ " as shorthand for "hunt for basis functions of this shape." Two corollaries of the same confusion: never impose the inhomogeneous condition on a single mode (only the full superposition can satisfy it — spend it last), and never let the exponential direction steal the sine (a factor required to vanish at two walls cannot be built from $e^{\pm kx}$ ; if your assignment demands that, your sign is backwards).

Worked examples

1 · The slot, evaluated: series versus closed form (confidence builder)

A slot with plate separation $a=2.0$ cm has its strip held at $V_0=100$ V. Find the potential at $(x,y)=(1.0\ \mathrm{cm},\,1.0\ \mathrm{cm})$ from the series and from the closed form, then the centreline potential a full width in.

Step 1. In natural units $x/a=y/a=1/2$ , so each term needs $e^{-n\pi/2}$ and $\sin(n\pi/2)$ . The coefficients are $C_n=4V_0/n\pi$ : $C_1=127.32$ V, $C_3=42.44$ V, $C_5=25.46$ V — a slowly decaying spectrum at the mouth.

Step 2. Sum term by term: $n=1$ gives $127.32\times e^{-\pi/2}=127.32\times0.20788=+26.468$ V; $n=3$ gives $42.44\times e^{-3\pi/2}\times(-1)=-0.381$ V; $n=5$ gives $25.46\times3.882\times10^{-4}=+0.010$ V. Running totals: $26.468\to26.087\to26.097$ V — three terms and the answer is set to better than a millivolt.

Step 3. Check against the closed form. With $\sin(\pi y/a)=1$ and $\sinh(\pi/2)=2.3013$ ,

V=\frac{2(100)}{\pi}\arctan\!\Big(\frac{1}{2.3013}\Big)=63.662\times0.40992=\boxed{26.10\ \mathrm{V}}

The series' $26.097$ V and the arctangent's $26.096$ V agree — the same function in two costumes.

Step 4. Go a full width deep: $(x,y)=(2.0,\,1.0)$ cm. The closed form gives $\frac{200}{\pi}\arctan(1/\sinh\pi)=63.662\times\arctan(0.08659)=\boxed{5.50\ \mathrm{V}}$ ; the $n=1$ term alone gives $127.32\,e^{-\pi}=5.5022$ V — high by 0.06%. One width in, the potential has forgotten everything about the strip except its fundamental: a pure sine arch at 5.5% of $V_0$ .

2 · The cube's centre: symmetry versus the double series (exam level)

A cubical box of side $a$ has five faces grounded and the sixth held at $V_0=600$ V. Find the potential at the centre $(a/2,a/2,a/2)$ — first without solving anything, then by summing the double series of equation (6) as a check.

Step 1. The symmetry argument. Take six copies of the cube, each with a different single face at $V_0$ and the rest grounded. Superposing all six gives a cube with every face at $V_0$ — whose solution is simply $V=V_0$ throughout, since the constant is harmonic and matches the boundary (the first uniqueness theorem, recalled by name, forbids any rival).

Step 2. The centre is the one point every cube rotation fixes, and the rotations map the six single-face problems into one another, so each contributes equally there. Six equal shares summing to $V_0$ leave

V(\mathrm{centre})=\frac{V_0}{6}=\boxed{100\ \mathrm{V}}

exact, with no series in sight — superposition and uniqueness wielded as weapons.

Step 3. Now make the machine confirm it. For the uniform lid, equation (7) gives $C_{nm}=16V_0/\pi^2nm$ for $n,m$ odd. At the centre, $\sin(n\pi/2)\sin(m\pi/2)=\pm1$ and the height factor is $\sinh(\gamma_{nm}a/2)/\sinh(\gamma_{nm}a)$ . The leading term has $\gamma_{11}a=\sqrt2\,\pi=4.4429$ , giving $\sinh(2.2214)/\sinh(4.4429)=0.10719$ , so

V_{11}=\frac{16(600)}{\pi^2}\times0.10719=1.6211\times600\times0.10719=104.26\ \mathrm{V}

— within 4.3% of the truth one term into a double series, because $\gamma_{11}$ suppresses everything else brutally.

Step 4. Add shells of odd modes: through $n,m\le3$ the sum is $99.887$ V; through $5$ , $100.0035$ V; through $7$ , $99.99988$ V; through $11$ , $100.0000$ V. The double series converges, alternating gently, to $\boxed{V_0/6=100\ \mathrm{V}}$ — vindicating the symmetry argument to as many digits as we care to take. When an exact shortcut and the general machine both exist, run both: agreement is the cheapest error-detection in physics.

Origins · 1 minJoseph Fourier and how this was discovered

Joseph Fourier

1768–1830 · Auxerre, Grenoble, Paris — heat, Egypt, and a scandalous series

Fourier was a tailor's son, orphaned at nine, who nearly became a priest before the Revolution swept him into mathematics, into prison twice, and then into Napoleon's Egyptian expedition as scientific secretary. Appointed prefect of Isère, he drained swamps by day and studied heat by night — and in 1807 delivered to the Paris Institute a memoir claiming that any function could be written as a sum of sines and cosines. The committee — Lagrange and Laplace among them — refused it; Lagrange considered the claim unproven and unbelievable. Fourier reworked it, won the Institute's 1811 prize (with reservations attached), and published the full theory in 1822 as the Théorie analytique de la chaleur, the book in which separation of variables and the coefficient-extraction trick bearing his name became standard equipment. He was elected the Academy's permanent secretary, and in 1824 identified what we now call the greenhouse effect. The rigor the committee demanded arrived a generation later, with Dirichlet — but the physics never had to wait.

History

1807

— Fourier's memoir on heat claims arbitrary functions expand in trigonometric series; the Institute committee of Lagrange and Laplace declines to publish it.

1811

— A reworked version wins the Institute's grand prize, with an official note deploring its lack of rigor.

1822

— The Théorie analytique de la chaleur appears: separation of variables and coefficient extraction become the standard method for linear PDEs.

1829

— Dirichlet publishes the first rigorous convergence conditions for Fourier series, supplying the proof the 1807 committee had demanded.

1899

— Gibbs explains the 9% overshoot of truncated Fourier series at jumps — the ringing in every partial sum at the slot's mouth.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §3.3.1 · Jackson, Classical Electrodynamics, 3rd ed., §2.9 · Purcell & Morin, Electricity and Magnetism, 3rd ed., §3.7 · Zangwill, Modern Electrodynamics, §7.5 Key self-consistency checks: the separation constraint $k_x^2+k_y^2+k_z^2=0$ follows directly from the product ansatz; the Fourier coefficient formula $C_n=\tfrac{2}{a}\int_0^a V_0\sin(n\pi x/a)\mathrm{d}x$ rests on $\int_0^a\sin^2(n\pi x/a)\,\mathrm{d}x=a/2$ ; the resummation identity $\sum_{n\,\mathrm{odd}}n^{-1}r^n\sin n\theta=\tfrac12\arctan[2r\sin\theta/(1-r^2)]$ underlies the arctangent closed form; and the box solution satisfies $V\to0$ on the grounded faces and $V\to V_0$ on $z=c$ .

Builds on: Laplace's Equation · Boundary Conditions for the Potential · Uniqueness Theorems for the Potential Leads to: Separation of Variables in Spherical Coordinates · Separation of Variables in Cylindrical Coordinates · Legendre Polynomials

How can we assume V is a product when the final answer clearly is not one?

Because the assumption is a recruiting poster, not a description. The method hunts for product-shaped building blocks; the theorems do the real work: completeness guarantees the blocks span everything the boundary can demand, so some superposition matches the data, and the first uniqueness theorem certifies that match as the only solution. The bricks are products; the house is not; the house is still built entirely of bricks. Compare vectors: "the answer points along an axis" would be false, but expanding it in axis-aligned unit vectors is exact.

Why does a function of x equalling a function of y force both to be constant?

Because $x$ and $y$ are independent — you can change one while nailing the other down. If $f(x)=g(y)$ everywhere, pick any fixed $y_0$ : then $f(x)=g(y_0)$ for every $x$ , so $f$ is constant, and $g$ equals the same constant. The step fails only where the variables are not independent (on a boundary curve relating them), which is why the argument runs in the region's interior. This little move, not the algebra around it, is the entire licence for replacing a PDE by ODEs.

How do I decide which coordinate gets the sines and which gets the exponentials?

You do not decide — you read it off the boundary conditions. A factor that must vanish at two walls has to pass through zero twice, and no real combination $Ae^{kx}+Be^{-kx}$ does that (at most one zero). So the doubly-pinned direction takes the sinusoids, quantized by the wall spacing, and the zero-sum rule forces the remaining direction into exponentials: the decaying branch alone if the region runs to infinity, both branches as $\sinh$ / $\cosh$ if it is finite. Guessing backwards is self-punishing — you will be demanding two zeros of an exponential, and no constants will comply.

What happens at the corners of the slot, where the strip at V0 meets the grounded plates?

The boundary data itself is discontinuous there — $V_0$ jammed against $0$ — so no continuous function can honor both values at the corner points. The series does the best mathematics allows: it converges to the true potential at every interior point, and on the mouth line to $V_0$ everywhere except the corners, where it splits the difference. Truncated sums also show the Gibbs phenomenon — a roughly 9% overshoot that crowds toward the corners but never shrinks. None of this contaminates the interior: one step in, the exponentials smooth everything.

How many terms of the series do I actually need in practice?

Depth decides. At depth $x$ , mode $n$ arrives pre-multiplied by $e^{-n\pi x/a}$ , so successive odd terms are suppressed by $e^{-2\pi x/a}$ — a factor of about 535 at $x=a$ , about 23 at $x=a/2$ . Half a width in, three terms gave millivolt accuracy in worked example 1; a full width in, one term is good to 0.06%. Only near the mouth is convergence slow, needing roughly $a/x$ terms to rebuild the discontinuous edge. Fine boundary detail is a near-field phenomenon; the far field is always cheap.

Does separation of variables work when the region contains charge?

Not directly — the method solves Laplace's equation, and the superposition step leans on homogeneity. With charge present you solve Poisson's equation instead: find any particular solution producing the source term, subtract it, and the remainder satisfies Laplace's equation with adjusted boundary data — which this article's machinery handles. The heavier artillery for sources, Green functions, comes later in the chapter, and eigenfunction expansions reappear there: the sine family that diagonalized the slot also diagonalizes the point-source problem.

Is this the same separation of variables used in quantum mechanics?

The very same, so the investment pays twice. The time-independent Schrödinger equation in a rectangular box separates identically: doubly-pinned directions produce quantized sinusoids — the particle-in-a-box states, with the slot's own $n\pi/a$ wavenumbers — and orthogonality extracts expansion coefficients exactly as Fourier's trick extracted the $C_n$ here. The differences are bookkeeping: quantum problems separate time from space, and the separation constant becomes the energy. When quantum courses reach spherical potentials, the Legendre cast introduced two articles from now walks on stage again, unchanged.