Intro to finite difference methods for PDEs

WHAT is a finite difference method?

Notation convention (memorise this): the subscript is space, the superscript is time. $u_i^n \approx u(x_i,\,t_n), \qquad x_i = a + i\,h,\qquad t_n = n\,k.$

HOW do we turn a derivative into a difference? (Derivation from scratch)

Everything comes from Taylor's theorem. Expand $u$ about $x_i$ , stepping by $\pm h$ :

$u(x_i+h) = u_i + h\,u_i' + \tfrac{h^2}{2}u_i'' + \tfrac{h^3}{6}u_i''' + O(h^4)\tag{A}$ $u(x_i-h) = u_i - h\,u_i' + \tfrac{h^2}{2}u_i'' - \tfrac{h^3}{6}u_i''' + O(h^4)\tag{B}$

First derivative — three flavours

Forward difference. From (A), solve for $u_i'$ : $u_i' = \frac{u_{i+1}-u_i}{h} - \frac{h}{2}u_i'' - \dots \;\Rightarrow\; \boxed{u_i' \approx \frac{u_{i+1}-u_i}{h}}$ Why this step? We isolate $u_i'$ and drop everything multiplied by $h$ or higher; the first dropped term is $O(h)$ , so this is first-order accurate, error $\sim h$ .

Backward difference. From (B): $\;u_i' \approx \dfrac{u_i - u_{i-1}}{h}$ , also $O(h)$ .

Central difference. Subtract (B) from (A). The even-power ( $u_i''$ ) terms cancel: $u(x_i+h)-u(x_i-h) = 2h\,u_i' + \tfrac{h^3}{3}u_i''' + \dots$ $\Rightarrow \boxed{u_i' \approx \frac{u_{i+1}-u_{i-1}}{2h}},\qquad \text{error } O(h^2).$ Why this step? Cancelling the $u''$ term is why central is one order more accurate for the same grid — a free upgrade.

Second derivative

Add (A) and (B). The odd-power ( $u'$ , $u'''$ ) terms cancel: $u(x_i+h)+u(x_i-h) = 2u_i + h^2 u_i'' + \tfrac{h^4}{12}u_i'''' + \dots$ Solve for $u_i''$ : $\boxed{u_i'' \approx \frac{u_{i+1} - 2u_i + u_{i-1}}{h^2}},\qquad \text{error } O(h^2).$

Formula The three you must own

quantity	stencil	order
$u_x$ forward	$\dfrac{u_{i+1}-u_i}{h}$	$O(h)$
$u_x$ central	$\dfrac{u_{i+1}-u_{i-1}}{2h}$	$O(h^2)$
$u_{xx}$ central	$\dfrac{u_{i+1}-2u_i+u_{i-1}}{h^2}$	$O(h^2)$

Figure — Intro to finite difference methods for PDEs

WHY does this work? (The numbers behind accuracy)

Worked Example 1 — Heat equation (explicit FTCS scheme)

Solve the heat equation $u_t = \alpha\,u_{xx}$ .

Step 1 — Forward in time at node $(i,n)$ : $u_t \approx \frac{u_i^{n+1}-u_i^n}{k}.$ Why? Forward in time means the new value $u_i^{n+1}$ appears alone — we can solve for it explicitly without inverting a matrix.

Step 2 — Central in space: $u_{xx}\approx \frac{u_{i+1}^n - 2u_i^n + u_{i-1}^n}{h^2}.$ Why? Best accuracy ( $O(h^2)$ ) at no extra cost.

Step 3 — Substitute into the PDE and rearrange: $\frac{u_i^{n+1}-u_i^n}{k} = \alpha\,\frac{u_{i+1}^n - 2u_i^n + u_{i-1}^n}{h^2}$ $\boxed{u_i^{n+1} = u_i^n + r\big(u_{i+1}^n - 2u_i^n + u_{i-1}^n\big)},\qquad r = \frac{\alpha k}{h^2}.$ Why this step? Everything on the right is at the known time level $n$ , so we march forward in time, one row at a time. This is FTCS (Forward-Time Central-Space) and it is explicit.

Worked Example 2 — One Poisson step (boundary value problem)

Solve $u'' = f(x)$ on $[0,1]$ with $u(0)=u(1)=0$ , using $h=0.25$ (so interior nodes $x_1,x_2,x_3$ ), $f\equiv -2$ .

Step 1 — Discretise: $\dfrac{u_{i+1}-2u_i+u_{i-1}}{h^2} = f_i$ . Why? Replace the only derivative present by the central second-difference.

Step 2 — Write equations for each interior node ( $u_0=u_4=0$ , $h^2=0.0625$ , $f_i h^2 = -0.125$ ): $-2u_1 + u_2 = -0.125,\quad u_1 - 2u_2 + u_3 = -0.125,\quad u_2 - 2u_3 = -0.125.$ Why this step? The boundary values are known constants, so they move to the right-hand side. We get a small linear system $A\mathbf{u}=\mathbf{b}$ with the tridiagonal matrix $A=\begin{pmatrix}-2&1&0\\1&-2&1\\0&1&-2\end{pmatrix}.$

Step 3 — Solve. By symmetry $u_1=u_3$ . From eq.1: $u_2 = 2u_1 - 0.125$ . Sub into eq.2: $u_1 - 2(2u_1-0.125)+u_1 = -0.125 \Rightarrow -2u_1 +0.25 = -0.125 \Rightarrow u_1 = 0.1875$ . Then $u_2 = 0.25$ . Check: exact solution of $u''=-2,\ u(0)=u(1)=0$ is $u=x(1-x)$ ; at $x=0.25,0.5$ it gives $0.1875, 0.25$ — exact, because the true solution is a quadratic and the central scheme is exact for cubics.

Recall Feynman: explain it to a 12-year-old

Imagine a curve drawn on graph paper. You only mark dots where the grid lines cross. To guess how steep the curve is at a dot, you look at the dot just left and just right and ask "how much did the height change between them?" — that's a slope-from-neighbours. To guess how much it bends, you compare the middle dot to the average of its two neighbours. The whole "finite difference" idea is just: don't measure slopes and bends exactly — estimate them from nearby dots. The closer the dots (smaller $h$ ), the better the guess. The danger: if you step forward in time too greedily, your guesses pile up errors and the picture explodes.

Flashcards

What replaces a derivative in a finite difference method?

A difference quotient of neighbouring grid values, derived from Taylor expansion.

u_i^n

, which index is space and which is time?

Subscript

i

= space (

x_i

), superscript

n

= time (

t_n

Central difference for

u_x

and its order?

\dfrac{u_{i+1}-u_{i-1}}{2h}

, order

O(h^2)

Forward difference for

u_x

and its order?

\dfrac{u_{i+1}-u_i}{h}

, order

O(h)

Central second-difference for

u_{xx}

\dfrac{u_{i+1}-2u_i+u_{i-1}}{h^2}

, order

O(h^2)

Why is central first-difference

O(h^2)

but forward only

O(h)

Subtracting symmetric expansions cancels the

u''

term, removing the leading

O(h)

error.

FTCS update for

u_t=\alpha u_{xx}

u_i^{n+1}=u_i^n+r(u_{i+1}^n-2u_i^n+u_{i-1}^n)

r=\alpha k/h^2

Stability condition for explicit FTCS heat scheme?

r=\alpha k/h^2 \le \tfrac12

If you halve

h

, how does an

O(h^2)

error change?

Drops by a factor of about 4.

What kind of system does discretising

u''=f

(BVP) produce?

A tridiagonal linear system

A\mathbf{u}=\mathbf{b}

with boundary values moved to the RHS.

Connections

Taylor's Theorem — the engine that generates every difference formula and its error order.
Heat Equation — primary parabolic test case (FTCS, implicit schemes).
Wave Equation — hyperbolic case; uses central-in-time, leads to CFL condition.
Laplace and Poisson Equations — elliptic BVPs giving tridiagonal/banded systems.
Von Neumann Stability Analysis — why the $r\le\tfrac12$ rule exists.
Tridiagonal Matrix Algorithm (Thomas) — fast solver for the linear systems FDM produces.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, computer ke paas infinite memory nahi hai, toh woh poore continuous function $u(x,t)$ ko store nahi kar sakta. Isliye hum domain ko ek grid mein chop kar dete hain — sirf grid points pe values rakhte hain: $u_i^n$ , jahan subscript $i$ space hai aur superscript $n$ time hai. Ab PDE mein jo derivatives hain, unko hum Taylor expansion se banaye gaye "difference quotients" se replace kar dete hain. Bas yahi pura khel hai — derivative ko nearby points ke difference se approximate karna.

Taylor se teen cheezein nikalti hain jo ratta maar lo: forward difference $\frac{u_{i+1}-u_i}{h}$ (accuracy $O(h)$ ), central difference $\frac{u_{i+1}-u_{i-1}}{2h}$ (accuracy $O(h^2)$ , kyunki symmetric subtract karne se $u''$ term cancel ho jaata hai — yeh free upgrade hai), aur second derivative $\frac{u_{i+1}-2u_i+u_{i-1}}{h^2}$ . $O(h^2)$ ka matlab: $h$ aadha karo toh error ek-chauthai ho jaata hai — bahut powerful.

Heat equation $u_t=\alpha u_{xx}$ ke liye FTCS scheme banega: $u_i^{n+1}=u_i^n+r(u_{i+1}^n-2u_i^n+u_{i-1}^n)$ , jahan $r=\alpha k/h^2$ . Isme right side sab known time level pe hai, toh aage march karte jao. Lekin ek trap hai: yeh tabhi stable hai jab $r\le \tfrac12$ . Agar $h$ chhota karoge toh $k$ ko $h^2$ ki tarah aur chhota karna padega, warna solution oscillate karke explode ho jaayega.

Aur jab BVP type problem ho jaise $u''=f$ , toh discretise karne par ek tridiagonal linear system $A\mathbf{u}=\mathbf{b}$ ban jaata hai — boundary values known hain isliye unhe right side bhej do, aur solve kar lo. Yaad rakho: FDM = grid banao, derivatives ko differences banao, algebra solve karo, aur stability check karna mat bhoolo.

Test yourself — Partial Differential Equations

Connections