Laplace on rectangle — separation of variables

WHAT is the problem?

WHY only one nonzero side at a time? Laplace's equation is linear and homogeneous, so sums of solutions are solutions. If all four sides are nonzero, split into four sub-problems each with three zero edges, solve each, and add. So mastering the one-side case masters everything.

HOW: derive it from scratch

Step 1 — Separate

Assume $u(x,y)=X(x)Y(y)$ . Plug in: $X''(x)Y(y)+X(x)Y''(y)=0.$ Divide by $X(x)Y(y)$ (legal where neither is zero): $\frac{X''}{X}+\frac{Y''}{Y}=0 \quad\Longrightarrow\quad \frac{X''}{X}=-\frac{Y''}{Y}.$

$\frac{X''}{X}=-\lambda,\qquad \frac{Y''}{Y}=\lambda.$

Step 2 — Choose the sign using the homogeneous direction

We have $u(0,y)=u(a,y)=0$ , i.e. $X(0)=X(a)=0$ . Two zero boundary conditions in $x$ ⇒ we need an oscillating $X$ (a sine), which forces $\lambda>0$ . Write $\lambda=k^2$ .

Step 3 — Solve the $X$ eigenvalue problem

$X''+k^2X=0 \Rightarrow X=A\cos kx + B\sin kx.$ $X(0)=0 \Rightarrow A=0$ . So $X=B\sin kx$ . $X(a)=0 \Rightarrow \sin ka=0 \Rightarrow ka=n\pi$ , $n=1,2,3,\dots$

$\boxed{k_n=\frac{n\pi}{a},\qquad X_n(x)=\sin\frac{n\pi x}{a}}$

These are the eigenfunctions; $\lambda_n=(n\pi/a)^2$ the eigenvalues.

Step 4 — Solve $Y$ with the matching $\lambda$

$Y''-k_n^2 Y=0 \Rightarrow Y=C\cosh\frac{n\pi y}{a}+D\sinh\frac{n\pi y}{a}.$ Apply $u(x,0)=0 \Rightarrow Y(0)=0 \Rightarrow C=0$ . So $Y_n(y)=\sinh\frac{n\pi y}{a}.$

Step 5 — Superpose and match the last edge

Each $u_n=\sin\frac{n\pi x}{a}\sinh\frac{n\pi y}{a}$ solves Laplace with three zero edges. Sum them: $u(x,y)=\sum_{n=1}^\infty B_n\,\sin\frac{n\pi x}{a}\,\sinh\frac{n\pi y}{a}.$ Impose the final condition $u(x,b)=f(x)$ : $f(x)=\sum_{n=1}^\infty \Big(B_n\sinh\frac{n\pi b}{a}\Big)\sin\frac{n\pi x}{a}.$ This is just the Fourier sine series of $f$ . The bracket is the coefficient: $B_n\sinh\frac{n\pi b}{a}=\frac{2}{a}\int_0^a f(x)\sin\frac{n\pi x}{a}\,dx.$

Figure — Laplace on rectangle — separation of variables

Worked examples

Worked example Example 1 — constant top edge

$f(x)=u_0$ Square plate $a=b=1$ , $u(x,1)=u_0$ , other three sides $0$ . Coefficient integral. $\int_0^1 u_0\sin(n\pi x)\,dx = u_0\cdot\frac{1-\cos n\pi}{n\pi}=\frac{u_0(1-(-1)^n)}{n\pi}$ . Why this step? Matching the top edge is a Fourier sine expansion of the constant $u_0$ . This is $0$ for even $n$ , and $\frac{2u_0}{n\pi}$ for odd $n$ . Divide by $\sinh n\pi$ (the $Y$ normalization at $y=b=1$ ): $B_n=\frac{2}{\sinh n\pi}\cdot\frac{u_0(1-(-1)^n)}{n\pi}.$ Why? Because $u(x,b)$ carries the factor $\sinh\frac{n\pi b}{a}$ that must be peeled off. Answer: $u(x,y)=\frac{4u_0}{\pi}\sum_{n\ \text{odd}}\frac{1}{n}\,\frac{\sinh(n\pi y)}{\sinh(n\pi)}\,\sin(n\pi x).$ Sanity check: at $y=1$ the sinh ratio $=1$ , giving the classic square-wave Fourier series for $u_0$ . ✓ At $y=0$ , $\sinh 0=0\Rightarrow u=0$ . ✓

Worked example Example 3 — nonzero LEFT edge instead

$u(0,y)=g(y)$ , other three sides zero, on $0<x<a,0<y<b$ . Why this step? Now the homogeneous direction is $y$ (two zero edges at $y=0,b$ ), so swap roles: eigenfunctions $\sin\frac{n\pi y}{b}$ , and the $x$ -solution is hyperbolic. Because the nonzero edge is at $x=0$ (not $x=a$ ), use $\sinh\frac{n\pi (a-x)}{b}$ so it vanishes at $x=a$ : $u(x,y)=\sum_n B_n\,\frac{\sinh\frac{n\pi(a-x)}{b}}{\sinh\frac{n\pi a}{b}}\,\sin\frac{n\pi y}{b},\quad B_n=\frac{2}{b}\int_0^b g(y)\sin\frac{n\pi y}{b}\,dy.$ Why the shifted argument? We need the hyperbolic factor $=0$ on the opposite (zero) edge $x=a$ and $=1$ ·matching on $x=0$ . Choosing $\sinh\frac{n\pi(a-x)}{b}$ does exactly that.

Recall Feynman: explain to a 12-year-old

Imagine a metal plate that's cold (0°) on three sides and hot on the top. After a while the heat spreads out and settles. We want to know how warm each inside point is. The smart trick: pretend the temperature is "a left-right pattern" times "an up-down pattern." The left-right pattern must be $0$ at both side-walls, so it's a wave that fits exactly (like a guitar string: 1 hump, 2 humps, 3 humps...). The up-down pattern must die away as you go down toward the cold bottom — that's the $\sinh$ , which is $0$ at the bottom and grows toward the hot top. Add up all the guitar-string waves with the right loudness to match the top edge — done!

Recall — Active flashcards

#flashcards/maths

Why does the separation constant have to be a constant?

One side depends only on

x

, the other only on

y

; equal for all

x,y

⇒ both equal a common constant.

Which direction gets sine eigenfunctions and why?

The direction with TWO homogeneous (zero) boundaries; two zero endpoints force oscillation, only sines vanish at both.

What are the eigenvalues/eigenfunctions for

X(0)=X(a)=0

\lambda_n=(n\pi/a)^2

X_n=\sin(n\pi x/a)

n=1,2,\dots

Why

\sinh

(not

\cosh

or sine) in the

y

-direction here?

Sign of

\lambda

is fixed positive by

x

, giving

Y''=k^2Y

⇒ hyperbolic; need

Y(0)=0

so pick

\sinh

since

\sinh 0=0

Final form of

u(x,y)

for top edge

f(x)

\sum_n B_n\sin\frac{n\pi x}{a}\sinh\frac{n\pi y}{a}

Formula for

B_n

B_n=\dfrac{2}{a\sinh(n\pi b/a)}\int_0^a f(x)\sin\frac{n\pi x}{a}\,dx

How do you handle all four edges nonzero?

Split into four problems (each one nonzero edge), solve separately, superpose (linearity).

f(x)=5\sin(3\pi x/a)

, what is

u

5\,\dfrac{\sinh(3\pi y/a)}{\sinh(3\pi b/a)}\sin\frac{3\pi x}{a}

(single mode, no integral).

What goes wrong if you use

\cosh

for

Y

\cosh 0=1\ne0

, violates

u(x,0)=0

Physical meaning of Laplace's equation?

Steady state; each point's value is the average of its neighbours (no sources, no time change).

Connections

Fourier sine series — the engine that computes $B_n$ .
Separation of variables — the general method (also used for Heat equation 1D, Wave equation 1D).
Sturm–Liouville eigenvalue problems — why $\sin(n\pi x/a)$ are orthogonal eigenfunctions.
Hyperbolic functions sinh cosh — the decaying/growing $y$ -profiles.
Superposition principle — splitting the four-edge problem.
Harmonic functions and mean value property — the "average of neighbours" picture.
Heat equation 1D — same eigenfunctions, but $e^{-\lambda t}$ instead of $\sinh$ .

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Laplace equation $u_{xx}+u_{yy}=0$ ka matlab hai ek steady state — jaise ek metal plate jiska temperature ab settle ho gaya, ya electric potential. Har andar ke point ki value uske aas-paas ke points ka average hoti hai. Humein chaaron edges pe values di jaati hain (boundary conditions) aur andar ka pattern nikaalna hota hai.

Trick simple hai: maan lo $u(x,y)=X(x)\cdot Y(y)$ — yaani ek "left-right pattern" times ek "up-down pattern". Isse ek mushkil PDE do aasaan ODE ban jaati hain. Jis direction mein do zero edges hain, wahan solution ek sine wave hota hai (jaise guitar ki string — 1 hump, 2 hump...), kyunki sirf sine hi dono ends pe zero ho sakta hai. Doosri direction mein, jahan ek hi zero edge hai, solution sinh hota hai — kyunki $\sinh 0=0$ aur woh exponentially grow/decay karta hai, oscillate nahi. Yaad rakho: "do zero sing, ek zero swing".

Last step: nonzero edge $f(x)$ ko match karne ke liye sabhi sine modes ko jod do — yeh bilkul Fourier sine series hai. Coefficient $B_n$ nikaalte waqt yaad rakho ki usmein $\sinh(n\pi b/a)$ ka factor divide karna hai, warna galat answer aayega. Agar $f(x)$ already ek single sine ho (jaise $5\sin 3\pi x$ ), toh integral karne ki zaroorat hi nahi — seedha ek hi mode! Yeh 80/20 wala point exam mein bahut kaam aata hai. Agar chaaron edges nonzero hain, toh linearity se chaar alag problems mein tod do, har ek solve karo, phir add kar do.

Test yourself — Partial Differential Equations

Connections