advanced27 min studyLesson 234 in Physics

Resolve boundary structure into harmonic modes

Derive spatial decay of boundary patterns in a grounded strip.

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What you’ll understand

  • Separate Laplace’s equation in a Cartesian strip.
  • Determine Fourier coefficients from boundary data.
  • Explain why fine spatial structure decays quickly away from a boundary.

A boundary can have a complicated voltage pattern even when the interior contains no charge. Separation of variables replaces that pattern with simpler harmonic pieces, each satisfying the interior differential equation exactly. The boundary determines their amplitudes. For a strip, this method reveals a useful physical effect: short spatial wavelengths penetrate less deeply than long ones. This is spatial filtering by Laplace’s equation, not a time-dependent relaxation process.

Let the walls choose the basis

Consider the region 0<x<L and y>0, with zero potential on the side walls and a prescribed bottom potential f(x). Require decay as y tends to infinity. Assume a product of one function of x and another of y, substitute into Laplace’s equation, and divide by the product where it is nonzero. Since one resulting expression depends only on x and the other only on y, both must equal a separation constant.

Zero values at both side walls select sine functions with integer mode number n. Their second x derivatives are negative multiples of themselves, so the second y derivative must have the opposite sign. The vertical solutions are exponentials, not traveling sinusoids. Discard the growing exponential to meet the condition at infinity. The n=0 linear branch cannot satisfy both homogeneous side values unless it vanishes; it does not supply a missing constant mode here.

Vn(x,y)=bnsin⁡(nπxL)e−nπy/L,n=1,2,…V_n(x,y)=b_n\sin\left(\frac{n\pi x}{L}\right)e^{-n\pi y/L},\qquad n=1,2,\ldots
L, x, y are lengths in m; bn is a potential amplitude in V. Each term solves the charge-free Cartesian Laplace equation with grounded side walls and decay at infinity.↔ Scroll wide equations horizontally.

Project the boundary, then superpose

Linearity allows summing the separated modes. At the bottom boundary this becomes a Fourier sine expansion of f. Multiply by a chosen sine and integrate across the strip. Orthogonality removes every other mode and leaves its coefficient. This projection is why separation is practical: a coupled boundary condition becomes independent scalar integrals. The basis is chosen by geometry and boundary conditions, rather than by an arbitrary taste for trigonometric functions.

For a bottom electrode held at constant voltage between grounded side walls, the ideal corner values disagree. The sine series represents the constant on the open interval but is zero at the endpoints; the actual idealized corner is singular. That does not invalidate the interior harmonic solution. It does limit uniform convergence at the corner and can make field derivatives large there. A finite number of modes gives an approximation whose boundary error deserves attention, especially near discontinuities.

V=∑n=1∞bnsin⁡(nπxL)e−nπy/L,bn=2L∫0Lf(x)sin⁡(nπxL)dxV=\sum_{n=1}^{\infty}b_n\sin\left(\frac{n\pi x}{L}\right)e^{-n\pi y/L},\qquad b_n=\frac{2}{L}\int_0^Lf(x)\sin\left(\frac{n\pi x}{L}\right)dx
Fourier projection assumes an integrable boundary function with the usual convergence qualifications. The infinite series is the solution where it converges with the required interior regularity.↔ Scroll wide equations horizontally.

Penetration is set by spatial scale

The amplitude of mode n falls by a factor e after moving a distance L/(nπ) into the strip. A pattern with twice the mode number has half this penetration length. The total potential at a particular x can nevertheless rise or change sign with y because different modes may initially cancel; monotonic decay of every coefficient does not imply monotonic pointwise decay of the complete sum. Distinguish modal statements from statements about every observation point.

In the spectral lab, Reset and set mode count to 3 and coefficient decay to 0; inspect the initial sum. Reset and set mode count to 3 and coefficient decay to 2; predict suppression of higher harmonics. This illustrates how coefficients control spatial detail. The instrument uses sin(nx−t)/nᵖ, with a common temporal frequency, so playback does not implement the exponential y decay of a Laplace solution or a physical electrostatic evolution.

ℓn=Lnπ,An(y)An(0)=e−y/ℓn\ell_n=\frac{L}{n\pi},\qquad \frac{A_n(y)}{A_n(0)}=e^{-y/\ell_n}
ℓn is the e-folding penetration length in m and An is the modal potential amplitude. y is distance, not elapsed time.↔ Scroll wide equations horizontally.

Two prescribed boundary modes

A strip has L=π m and bottom potential 6 sin(x/(1 m))+2 sin(2x/(1 m)) volts. Find the potential at x=π/2 m, y=ln 2 m, and the decay factor of the second mode.

  1. The boundary is already a sine expansion. Its only nonzero coefficients are b₁=6 V and b₂=2 V. With L=π m, the associated vertical decay rates are 1 m⁻¹ and 2 m⁻¹.

  2. At y=ln 2 m the first mode is multiplied by 1/2 and the second by 1/4. Therefore the two modal amplitudes at that height are 3 V and 0.5 V, respectively; both have decayed without any time evolution.

  3. At x=π/2 m the first sine is one and the second is zero. The requested potential is consequently 3 V. The vanished second contribution is a node at that x, not evidence that its amplitude is zero everywhere.

The result

The potential at the specified point is 3 V. The second modal amplitude has fallen to one quarter of its bottom value.

Explore the model

Interactive lab

Interactive lab · spectral

Suppress higher spatial harmonics

Reset and set N=3, p=0; inspect the sine sum at time zero. Reset and set N=3, p=2; predict smaller high-mode contributions and less fine detail.

Suppress higher spatial harmonicsA finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Current readings: Sum at x = π/2: 0.867; Smallest wavelength: 2π / 5; Highest-mode coefficient: 0.2. All parameters are adjustable with the labeled controls below.01.573.144.716.28−2.56−1.2801.282.56x · radiansFinite sine sum
Finite sumHighest included mode
Sum at x = π/2
0.867
Smallest wavelength
2π / 5
Highest-mode coefficient
0.2
t = 0 s

A finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe lab sum is Σsin(nx−t)/nᵖ. It illustrates truncation and coefficients, not exponential penetration, spherical harmonics, or electrostatic time dynamics.

Check your understanding

Which penetration length belongs to mode n in this grounded strip?

Try it yourself

For L=π m, let the bottom potential be 8 sin(2x/(1 m)) volts. Find the height where its modal amplitude is 1 V, and the potential there at x=π/4 m.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Solve 8 exp(−2y/(1 m))=1 before substituting the spatial sine.

Compare with the solution

The required height is (ln 8)/2 m, approximately 1.040 m. At x=π/4 m the sine argument is π/2 and its value is one, so the potential is +1 V. At a node the potential would vanish at every height even though the mode retains a nonzero amplitude elsewhere.

Take it with you

  • Boundary geometry selects the separated eigenfunctions.
  • Orthogonal projection determines each mode independently.
  • Fine boundary structure has a short spatial penetration length.

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