Solving Differential Equations by Transform Methods

A differential equation is hard for one reason: it couples the value of a function to its slope, its curvature, and every higher derivative at every point at once. An integral transform breaks that coupling in a single stroke. Feed the equation through the Fourier kernel $e^{-ikx}$ or the Laplace kernel $e^{-st}$ and each derivative $\mathrm{d}/\mathrm{d}x$ turns into plain multiplication — by $ik$ for Fourier, by $s$ for Laplace — because the exponentials are the eigenfunctions of differentiation. What was a differential equation becomes an algebraic one. You solve the algebra, then transform back. That is the whole method, and it is one of the most powerful moves in mathematical physics. For an ordinary differential equation the transform gives a purely algebraic equation; solve, invert, done. For a partial differential equation you transform in some of the variables and are left with an ordinary differential equation — or even algebra — in the rest. The heat equation, the wave equation, and Laplace's equation are the three canonical trophies, each showing a different face: diffusion and the Gaussian kernel, causality and the light cone, boundary-value kernels on half-spaces. Two supporting theorems make it airtight. The differentiation rules say exactly how derivatives map, and for Laplace they carry the initial data $f(0), f'(0), \dots$ into the algebra automatically. The convolution theorem says dividing by the transformed operator is the same as convolving with its inverse transform — the Green's function — so transform methods and Green's-function methods are two views of one construction. The one genuine cost is inversion, which we treat honestly: residues, the Bromwich contour, and the distributional shortcuts that make $\hat\delta = 1$ .

Key takeaways

An integral transform trades every derivative for multiplication by the transform variable, so a linear constant-coefficient differential equation becomes an algebraic equation you can solve by hand.
Fourier suits the whole line $-\infty<x<\infty$ with decaying data; Laplace suits the half-line $t\ge0$ where initial conditions and causality matter — and it swallows $f(0),f'(0),\dots$ for free.
The recipe is always transform, solve the algebra (an ODE in transform space for a PDE), then invert — and inversion is usually the hardest step.
Dividing $\tilde u=\tilde g/P$ by the symbol polynomial $P$ is a product in transform space, so inverting it is a convolution: $u=g*G$ with $G$ the Green's function, the response to a point source.
The heat kernel $1/\sqrt{4\pi Dt}\,e^{-x^2/4Dt}$ , the damped-oscillator solution, and the retarded wave Green's function all drop out of the same three-line procedure.
Poles on the inversion contour and distributional inputs like $\delta(x)$ are handled rigorously by contour deformation and tempered distributions, not swept aside.
Key idea: differentiation is multiplication in transform space — $\widehat{f'}=ik\,\tilde f$ and $\mathcal{L}\{f'\}=s\tilde f-f(0)$ — so $Lu=g$ becomes $P\,\tilde u=\tilde g$ , and $u=\mathcal{F}^{-1}\{\tilde g/P\}=g*G$ reconstructs the answer as a convolution with the Green's function $G=\mathcal{F}^{-1}\{1/P\}$ .

The core insight

Differentiation is multiplication in transform space: $\widehat{f'}(k)=ik\,\tilde f(k)$ and $\mathcal{L}\{f'\}=s\,\tilde f-f(0)$ . A constant-coefficient linear differential equation $Lu=g$ therefore becomes an algebraic relation $P\,\tilde u=\tilde g$ with $P$ the symbol polynomial; solving gives $\tilde u=\tilde g/P$ , and inversion plus the convolution theorem reconstructs $u$ as $g$ convolved with the Green's function $\mathcal{F}^{-1}\{1/P\}$ .

Where this lives today

Transform methods became infrastructure. Every linear control system is designed in the Laplace domain: an aircraft autopilot, a disk-drive servo, a switching power supply are each specified by a transfer function $H(s)$ , and an engineer reads stability straight off the pole locations — any pole in the right half-plane means the loop rings itself apart. Signal processing runs on the discrete cousins of these transforms; the fast Fourier transform lets an MRI scanner reconstruct an image from data acquired literally in Fourier ( $k$ -space). Spectral methods integrate the equations of weather, climate, and turbulence by turning spatial derivatives into multiplication exactly as in equation (6). And the contour choice that made the retarded Green's function reappears at the foundations of quantum field theory: Feynman's $i\varepsilon$ prescription for a propagator is precisely a rule for which side of the poles the inversion contour passes. Choosing a contour, it turns out, is choosing a physics.

The physical picture

Picture the exponential $e^{ikx}$ as one coordinate axis in an infinite-dimensional space of functions. In this basis the derivative operator is diagonal — it multiplies the $k$ -th component by $ik$ and nothing more, so a differential equation that looks hopelessly entangled in $x$ becomes one short algebraic equation seen component by component. Solving a PDE this way is exactly like diagonalising a matrix: rotate into the basis where the operator is diagonal, divide, rotate back. The first figure draws that round trip.

Figure

The transform method as a round trip: a hard differential equation L u = g maps under the Fourier transform to the algebraic relation P(ik) û = ĝ, because the derivative ∂ₓ becomes multiplication by ik; you divide by the symbol polynomial P(ik) and then invert to land on the solution u = g convolved with the Green's function G

The Laplace transform is the same idea bent to obey causality. Restrict to $t\ge0$ and slip in the convergence factor $e^{-\sigma t}$ : now the transform copes with solutions that grow, and — the decisive bonus — the boundary term integration by parts leaves at $t=0$ delivers the initial conditions straight into the algebra. Laplace remembers where you started; Fourier only asks that the function fade at the two ends of the line. Division by the symbol $P$ is where the physics hides: zeros of $P$ are the normal modes and resonances, while the poles of $1/P$ , and which side of the contour they fall on, decide whether a disturbance runs forward in time (retarded) or backward (advanced). And because $\tilde u=\tilde g/P$ is a product, the inverse is a convolution with one universal object — the response $G$ to a unit point source. Compute $G$ once and every future forcing is one integral away. That is why the delta function is the natural probe, and why its spreading is worth drawing with real numbers.

Figure

The Gaussian heat kernel spreading from a point source with diffusivity D = 1.1×10⁻⁴ m²/s: the inverse Fourier transform of the damping factor e^(−Dk²t) is a Gaussian whose RMS width grows as σ = √(2Dt), so σ = 4.7 cm at t = 10 s, 14.8 cm at t = 100 s, and 33.2 cm at t = 500 s, while the peak height falls as 1 over √(4πDt)

The core equations

Everything begins with how a single derivative transforms. Integrate the Fourier integral of $f'$ by parts; on the whole line the boundary term dies because $f$ decays, and one factor of $ik$ is left behind per derivative.

\widehat{f^{(n)}}(k)=(ik)^n\,\tilde f(k) \qquad (1)

where: $\tilde f(k)=\int_{-\infty}^{\infty}f(x)e^{-ikx}\,\mathrm{d}x$ — the Fourier transform; $k$ — the wavenumber (transform) variable; $n$ — the order of the derivative. Reading: differentiating in $x$ is multiplying by $ik$ in $k$ , so an $n$ -th order operator becomes an $n$ -th degree polynomial. The Laplace transform lives on $[0,\infty)$ , so the same integration by parts keeps the boundary term at $t=0$ . That surviving term is the initial data, and it is why Laplace is the natural tool for initial-value problems.

\mathcal{L}\{f^{(n)}\}(s)=s^n\tilde f(s)-\sum_{j=0}^{n-1}s^{\,n-1-j}f^{(j)}(0) \qquad (2)

where: $\tilde f(s)=\int_0^{\infty}f(t)e^{-st}\,\mathrm{d}t$ — the Laplace transform; $s$ — the complex frequency, with $\mathrm{Re}\,s$ large enough for convergence; $f^{(j)}(0)$ — the $j$ -th initial value. Reading: each derivative contributes one power of $s$ and drags its Cauchy data along, so $\mathcal{L}\{f''\}=s^2\tilde f-sf(0)-f'(0)$ . Apply the differentiation rule to a constant-coefficient operator $L=\sum a_n\,\mathrm{d}^n/\mathrm{d}x^n$ and the whole equation collapses to arithmetic. The operator is replaced by its symbol polynomial and you simply divide.

Lu=g \;\xrightarrow{\;\mathcal{F}\;}\; P(ik)\,\tilde u(k)=\tilde g(k)\;\Rightarrow\;\tilde u(k)=\frac{\tilde g(k)}{P(ik)} \qquad (3)

where: $L$ — the differential operator; $P(ik)=\sum a_n(ik)^n$ — its symbol; $g$ — the source or forcing. Reading: solving the differential equation has become a single division in transform space — this line is the entire method in one arrow. Because equation (3) is a product $\tilde g\cdot(1/P)$ , the convolution theorem turns the inverse transform into a convolution with a universal kernel, the Green's function.

u(x)=(g*G)(x)=\int_{-\infty}^{\infty}g(x')\,G(x-x')\,\mathrm{d}x',\qquad G=\mathcal{F}^{-1}\!\left\{\frac{1}{P(ik)}\right\} \qquad (4)

where: $G$ — the Green's function, the response to a unit point source; $*$ — convolution. Reading: compute $G$ once and any forcing is one integral away — this is why transform methods and Green's-function methods are the same construction. The two inversion formulas close the loop: Fourier inversion is an integral over real $k$ ; Laplace inversion is the Bromwich integral up a vertical line to the right of every singularity.

f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\tilde f(k)\,e^{ikx}\,\mathrm{d}k,\qquad f(t)=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\tilde f(s)\,e^{st}\,\mathrm{d}s \qquad (5)

where: $\gamma$ — a real abscissa placed to the right of all singularities of $\tilde f(s)$ . Reading: where you put $\gamma$ (and, for Fourier, how you dodge poles on the real axis) is exactly where causality is decided. Finally, the impulse. The transform of a Dirac delta is the constant $1$ , which is why a point source is the cleanest possible probe and lands you directly on the Green's function.

\widehat{\delta}(k)=1,\qquad \mathcal{L}\{\delta(t)\}=1,\qquad \partial_t u=D\,\partial_x^2 u\;\xrightarrow{\;\mathcal{F}_x\;}\;\frac{\mathrm{d}\tilde u}{\mathrm{d}t}=-Dk^2\,\tilde u \qquad (6)

where: $\delta$ — the Dirac delta; $D$ — the diffusion constant; $\tilde u(k,t)$ — the spatial Fourier transform of $u$ . Reading: a spatial transform turns the heat PDE into the one-line ODE solved by $\tilde u(k,t)=\tilde u(k,0)\,e^{-Dk^2t}$ , with sharp high- $k$ modes decaying fastest — diffusion is a low-pass filter.

Derivation AFrom integration by parts to the differentiation rules

The goal is to prove the two facts every transform method rests on — that Fourier sends $f^{(n)}$ to $(ik)^n\tilde f$ with no leftover terms, while Laplace sends $f^{(n)}$ to $s^n\tilde f$ minus a tower of initial values — and to see that the difference between them is nothing but the location of the boundary in the integration by parts. Getting this straight is what lets you read off initial conditions for free and know exactly when a boundary term is allowed to vanish.

Step 1. Start from the Fourier definition applied to $f'$ : $\widehat{f'}(k)=\int_{-\infty}^{\infty}f'(x)e^{-ikx}\,\mathrm{d}x$ , and integrate by parts, differentiating the exponential. This gives $\big[f(x)e^{-ikx}\big]_{-\infty}^{\infty}+ik\int_{-\infty}^{\infty}f(x)e^{-ikx}\,\mathrm{d}x$ . Why: integration by parts is the only tool that moves the derivative off $f$ and onto the known kernel $e^{-ikx}$ , whose derivative is just $-ik$ times itself.

Step 2. For any $f$ in the Schwartz class the boundary term vanishes because $f(x)\to0$ as $|x|\to\infty$ , leaving $\widehat{f'}=ik\,\tilde f$ . Iterating $n$ times pulls down one factor of $ik$ each pass, so $\widehat{f^{(n)}}=(ik)^n\tilde f$ . Why: the clean multiplication rule is bought entirely by decay at infinity — this is precisely the condition Fourier requires and Laplace relaxes.

Step 3. Now redo it for Laplace on $[0,\infty)$ : $\mathcal{L}\{f'\}=\big[f(t)e^{-st}\big]_0^{\infty}+s\int_0^{\infty}f(t)e^{-st}\,\mathrm{d}t$ . The upper limit dies for $\mathrm{Re}\,s$ large, but the lower limit does not: it contributes $-f(0)$ , so $\mathcal{L}\{f'\}=s\tilde f-f(0)$ . Why: the half-line has a real boundary at $t=0$ , and the boundary term there is the initial value — causality enters not as an extra assumption but as an endpoint of the integral.

Step 4. Apply the rule twice for the second derivative: $\mathcal{L}\{f''\}=s\,\mathcal{L}\{f'\}-f'(0)=s^2\tilde f-sf(0)-f'(0)$ , and by induction $\mathcal{L}\{f^{(n)}\}=s^n\tilde f-\sum_{j=0}^{n-1}s^{n-1-j}f^{(j)}(0)$ . Why: every derivative you strip off leaves behind exactly one initial value, so an $n$ -th order initial-value problem needs precisely its $n$ Cauchy data — no more, no fewer — and they enter the algebra by themselves. *Note: term-by-term integration by parts reproduces equations (1) and (2); in particular $\mathcal{L}\{f''\}=s^2\tilde f-sf(0)-f'(0)$ is the standard second-derivative rule.*

Go deeper · Derivation B · 1 minThe heat kernel from a Fourier transform

Derivation BThe heat kernel from a Fourier transform

The aim is to solve the initial-value problem $\partial_t u=D\,\partial_x^2 u$ with $u(x,0)=u_0(x)$ on the whole line and recover the Gaussian heat kernel explicitly, so the abstract " $u=g*G$ " of equation (4) becomes a concrete formula. The route is canonical: transform in space to kill the $\partial_x^2$ , integrate the trivial ODE in time, and invert a Gaussian.

Step 1. Fourier transform in $x$ . The rule $\partial_x^2\to(ik)^2=-k^2$ turns the PDE into an ordinary differential equation in $t$ for each fixed $k$ : $\mathrm{d}\tilde u/\mathrm{d}t=-Dk^2\,\tilde u$ with $\tilde u(k,0)=\tilde u_0(k)$ . Why: the spatial transform diagonalises the only spatial derivative in the problem, so the different wavenumbers completely decouple and can be solved independently.

Step 2. Integrate the first-order linear ODE: $\tilde u(k,t)=\tilde u_0(k)\,e^{-Dk^2t}$ . Why: in transform space the evolution is just multiplication by a Gaussian damping factor — every spatial mode decays at rate $Dk^2$ , so the finest features (largest $k$ ) wash out first, which is diffusion.

Step 3. The solution is a product in $k$ , so inversion is a convolution: $u(x,t)=u_0*\mathcal{F}^{-1}\{e^{-Dk^2t}\}$ , and the inverse of a Gaussian is a Gaussian (complete the square): $\mathcal{F}^{-1}\{e^{-Dk^2t}\}=\dfrac{1}{\sqrt{4\pi Dt}}\,e^{-x^2/4Dt}$ . Why: the convolution theorem converts the trivial multiplicative evolution back into physical space, and the Gaussian is the fixed shape of the Fourier transform, so no special functions are needed.

Step 4. Assemble the answer, $u(x,t)=\displaystyle\int_{-\infty}^{\infty}\frac{e^{-(x-x')^2/4Dt}}{\sqrt{4\pi Dt}}\,u_0(x')\,\mathrm{d}x'$ . The kernel is the Green's function; as $t\to0^+$ it collapses to $\delta(x-x')$ and returns the initial data. Why: the point-source response — a Gaussian of width $\sigma=\sqrt{2Dt}$ and unit area — is all you ever need, because any initial profile is a superposition of point sources. *Note: substituting the kernel into $\partial_t-D\partial_x^2$ annihilates it for $t>0$ ; $\int K\,\mathrm{d}x'=1$ enforces $u(x,0^+)=u_0$ ; and $\sigma=\sqrt{2Dt}$ reproduces $4.7$ , $14.8$ , $33.2$ cm at $t=10,100,500$ s for $D=1.1\times10^{-4}$ m²/s (figure 2).*

Go deeper · Derivation C · 1 minThe retarded Green's function of the wave operator

Derivation CThe retarded Green's function of the wave operator

The goal is to show that causality is not an extra postulate bolted onto the wave equation but a choice of contour — that placing the Bromwich line to the right of the poles is what makes signals travel forward in time and confines them inside the light cone. We double-transform the point-impulse problem $(\partial_t^2-c^2\partial_x^2)G=\delta(x)\delta(t)$ , invert in $s$ with the causal abscissa, then invert in $k$ .

Step 1. Laplace transform in $t$ and Fourier transform in $x$ . The source $\delta(x)\delta(t)$ becomes $1$ , and the wave operator's symbol is $s^2+c^2k^2$ , so $(s^2+c^2k^2)\tilde G=1$ , giving $\tilde G(k,s)=1/(s^2+c^2k^2)$ . Why: two transforms reduce a second-order PDE in two variables all the way to a single algebraic reciprocal — there is nothing left to solve, only to invert.

Step 2. Invert in $s$ along the Bromwich contour with $\gamma>0$ . The poles sit at $s=\pm ic|k|$ , both to the left of the contour, so closing to the left for $t>0$ and finding nothing for $t<0$ gives $\mathcal{L}^{-1}\{(s^2+c^2k^2)^{-1}\}=\dfrac{\sin(c|k|t)}{c|k|}\,\theta(t)$ . Why: the step function $\theta(t)$ — the entire content of causality — is produced automatically by which side of the contour the poles lie on, not by any hand-imposed condition.

Step 3. Invert in $k$ . The Fourier integral of $\sin(c|k|t)/c|k|$ has support only where the phase can stay stationary, and the result is $G(x,t)=\dfrac{1}{2c}\,\theta(t)\,\theta(ct-|x|)$ . Why: the transform manufactures the light cone — the response is nonzero only inside $|x|<ct$ , so a disturbance spreads at exactly the speed $c$ and never outruns it. *Note: $G=\theta(t)\theta(ct-|x|)/2c$ satisfies $(\partial_t^2-c^2\partial_x^2)G=\delta(x)\delta(t)$ and vanishes for $t<0$ ; the advanced Green's function follows from choosing $\gamma<0$ .*

Limiting cases

Each transform is matched to a geometry and a class of data; picking the wrong one is the commonest way the method stalls.

Case	Result	Meaning
Whole line, decaying data	$u\to0$ as $\lvert x\rvert\to\infty$	Use Fourier; $\partial_x\to ik$ with no boundary terms because the surface terms vanish.
Half line, initial-value problem	$t\ge0$ , $u(0),u'(0)$ given	Use Laplace; the differentiation rule injects the Cauchy data by itself.
Finite interval or periodic	$u(x+L)=u(x)$	Use a Fourier series, discrete $k_n=2\pi n/L$ ; the sum replaces the integral and boundary conditions select the modes.
Singular or non-decaying forcing	$g=\delta(x)$ or $g=1$	Work in tempered distributions: $\hat\delta=1$ , all derivatives exist, and inversion gives the Green's function directly.
Symbol vanishes on the contour	$P(ik_0)=0$ for real $k_0$	Deform the contour or add an $i0^+$ prescription; otherwise the inverse integral is ill-defined.

Common trapDropping the initial-value terms in the Laplace derivative rule

The single most common error is writing $\mathcal{L}\{f''\}=s^2\tilde f$ and forgetting the $-sf(0)-f'(0)$ . Those terms are not a technicality — they are the initial conditions, and dropping them silently imposes homogeneous data $f(0)=f'(0)=0$ , handing you the wrong particular solution with no error message. The mirror-image trap lives in the inversion: when the symbol $P$ has a zero on the contour — a real $k_0$ with $P(ik_0)=0$ for Fourier, or a pole at the chosen abscissa for Laplace — the naive integral $\tilde g/P$ is undefined. You must deform the contour or add a causal prescription ( $k\to k+i0^+$ , $\gamma$ to the right of every pole), and the choice is not cosmetic: it decides retarded versus advanced, outgoing versus incoming. The algebra never warns you; the physics is entirely in the contour.

Worked examples

1 · Damped oscillator by Laplace transform (fully worked)

Solve $\ddot x+2\beta\dot x+\omega_0^2 x=0$ with $\omega_0=5\ \mathrm{s^{-1}}$ , $\beta=1\ \mathrm{s^{-1}}$ , and initial data $x(0)=0.1\ \mathrm{m}$ , $\dot x(0)=0$ . Find $x(t)$ and describe the motion.

Step 1. Transform the ODE with the Laplace derivative rule, letting the initial data ride along. Using $\mathcal{L}\{\ddot x\}=s^2\tilde x-sx_0-v_0$ and $\mathcal{L}\{\dot x\}=s\tilde x-x_0$ , the equation becomes $(s^2\tilde x-sx_0-v_0)+2\beta(s\tilde x-x_0)+\omega_0^2\tilde x=0$ . No boundary conditions have to be imposed by hand — they are already in the algebra.

Step 2. Solve for $\tilde x$ and complete the square in the denominator: $\tilde x(s)=\dfrac{(s+2\beta)x_0+v_0}{s^2+2\beta s+\omega_0^2}=\dfrac{(s+2\beta)x_0+v_0}{(s+\beta)^2+\omega_d^2}$ , with the damped frequency $\omega_d=\sqrt{\omega_0^2-\beta^2}=\sqrt{25-1}=\sqrt{24}=\mathbf{4.899\ s^{-1}}$ . Since $\beta<\omega_0$ the system is underdamped.

Step 3. Invert with the shift theorem $s\to s+\beta$ , which corresponds to the envelope $e^{-\beta t}$ . Splitting the numerator into $(s+\beta)$ and constant pieces gives $x(t)=e^{-\beta t}\!\left[x_0\cos\omega_d t+\dfrac{v_0+\beta x_0}{\omega_d}\sin\omega_d t\right]$ .

Step 4. Put the numbers in: $\dfrac{v_0+\beta x_0}{\omega_d}=\dfrac{(1)(0.1)}{4.899}=0.02041\ \mathrm{m}$ , so $x(t)=e^{-t}\!\left[0.1\cos(4.899\,t)+0.02041\sin(4.899\,t)\right]$ m. The bracket is a single sinusoid of amplitude $\sqrt{0.1^2+0.02041^2}=\mathbf{0.1021\ m}$ , so the mass oscillates at $4.899$ rad/s under an envelope $e^{-t}$ that drops to $1/e$ in exactly $1$ s; setting $\beta=0$ recovers simple harmonic motion at $\omega_0=5$ s⁻¹. Answer: $x(t)=e^{-t}[0.1\cos 4.899t+0.0204\sin 4.899t]$ m, underdamped with $\omega_d=4.90\ \mathrm{s^{-1}}$ .

2 · How fast a hot spot spreads (fully worked)

A thin metal bar of thermal diffusivity $D=1.1\times10^{-4}\ \mathrm{m^2/s}$ is given a sharp pulse of heat at the origin. Using the heat kernel, find the RMS width of the temperature profile at $t=100$ s, and estimate when the spot has spread to $\sigma=0.33$ m.

Step 1. For a point-source initial condition $u_0(x)=Q\,\delta(x)$ the heat-kernel formula from the derivation gives directly $u(x,t)=\dfrac{Q}{\sqrt{4\pi Dt}}\,e^{-x^2/4Dt}$ — a Gaussian of variance $\sigma^2=2Dt$ .

Step 2. At $t=100$ s, $\sigma=\sqrt{2Dt}=\sqrt{2(1.1\times10^{-4})(100)}=\sqrt{0.022}=\mathbf{0.148\ m}$ (about $15$ cm), while the peak has fallen to $1/\sqrt{4\pi Dt}=2.69$ times $Q$ — matching the amber curve in figure 2.

Step 3. Diffusion spreads as $\sqrt{t}$ , so reaching $\sigma=0.33$ m needs $t=\sigma^2/2D=0.33^2/(2\times1.1\times10^{-4})=\mathbf{495\ s}$ , essentially the $500$ s blue curve — doubling the width takes four times as long. Swap $D\to i\hbar/2m$ and the identical kernel becomes the free-particle Schrödinger propagator, describing a spreading quantum wave packet. Answer: $\sigma=15$ cm at $100$ s; $\sigma=33$ cm at $\approx495$ s; width grows as $\sqrt{t}$ .

Physics of the portrait

Transform methods are a two-century collaboration between three very different temperaments: a heat theorist who needed to sum sines, a probabilist who needed to tame integrals, and a self-taught telegraph engineer who used the method brilliantly before anyone could say why it worked.

Origins · 1 mintransform methods and how this was discovered

transform methods

turning calculus into algebra since 1807 · integral transforms and operational calculus

The story starts with heat. In his 1807 memoir and the 1822 Théorie analytique de la chaleur, Joseph Fourier claimed any temperature profile could be written as a sum of sines and cosines, and that under this decomposition the heat equation solved itself mode by mode — the first transform method, met with deep suspicion by Lagrange and Laplace for its cavalier convergence. Pierre-Simon Laplace had, around 1785, introduced the integral $\int f(t)e^{-st}\,\mathrm{d}t$ as a tool in probability, though he never turned it on differential equations. The person who did that, spectacularly, was Oliver Heaviside: in the 1890s the reclusive English engineer invented an "operational calculus" that treated $\mathrm{d}/\mathrm{d}t$ as an algebraic symbol $p$ and solved telegraph and circuit transients by literally dividing by polynomials in $p$ — methods that horrified the Royal Society, to which he shot back that "mathematics is an experimental science." Heaviside was right but unrigorous, and it fell to Thomas Bromwich in 1916 to show that his operational fractions were exactly inverse Laplace transforms along a contour in the complex plane. That fusion — Fourier's modes, Laplace's kernel, Heaviside's audacity, Bromwich's contour — is the transform method as physicists use it today.

History

1785

— Laplace introduces the integral transform in his work on probability

1807

— Fourier presents his sine-series solution of heat conduction to the Paris Academy

1822

— Fourier publishes Théorie analytique de la chaleur, founding harmonic analysis

1893

— Heaviside develops the operational calculus for circuit and telegraph transients

1916

— Bromwich recasts operational calculus rigorously as contour integrals of the Laplace transform

Going further

Texts. Stone & Goldbart, Mathematics for Physics, chapters on Fourier and Laplace methods and Green's functions · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., Ch. 20 (integral transforms) · Hassani, Mathematical Physics, chapters on transforms and Green's functions · Byron & Fuller, Mathematics of Classical and Quantum Physics, Ch. 5 · Riley, Hobson & Bence, Mathematical Methods for Physics and Engineering, 3rd ed., Ch. 13 and 15 · Morse & Feshbach, Methods of Theoretical Physics (Green's functions and contour prescriptions) · Whitham, Linear and Nonlinear Waves (transform solution of wave problems). Key results to keep in mind: the heat kernel $1/\sqrt{4\pi Dt}\,e^{-x^2/4Dt}$ is derived by completing the square, with $\int K\,\mathrm{d}x'=1$ and $\sigma=\sqrt{2Dt}$ giving $4.7$ , $14.8$ , $33.2$ cm at $t=10,100,500$ s for $D=1.1\times10^{-4}$ m²/s; the damped-oscillator solution satisfies $x(0)=0.1$ , $\dot x(0)=0$ with $\omega_d=\sqrt{24}=4.899$ s⁻¹; the RC step response $V_0(1-e^{-t/\tau})$ reaches $0.632\,V_0$ at $\tau=1$ ms with cutoff $f_c=159$ Hz; and the 1D retarded Green's function $G=\theta(t)\theta(ct-|x|)/2c$ solves the wave equation with a point impulse and vanishes for $t<0$ .

Prerequisites. The Fourier Transform · Properties of the Fourier Transform · The Laplace Transform · The Convolution Theorem · The Dirac Delta Function. Leads to. Green's Functions via Integral Transforms · Laplace Transform Inversion · Tempered Distributions and Fourier Transforms · Parseval's and Plancherel's Theorems.

Ask the tutor: Want to see any of these worked live — pick your own ODE or PDE and I will transform it, solve the algebra, and walk you through the inversion contour step by step.

Why do exponentials make differentiation into multiplication?

Because $e^{ikx}$ and $e^{-st}$ are eigenfunctions of the derivative operator: $\mathrm{d}/\mathrm{d}x\,e^{ikx}=ik\,e^{ikx}$ . An integral transform expands a function in this eigenbasis, and in the basis where an operator is diagonal it acts by simple scaling. So writing $f$ as a superposition of exponentials automatically diagonalises every constant-coefficient derivative — each mode is multiplied by its eigenvalue $ik$ or $s$ . It is the same reason diagonalising a matrix turns matrix equations into scalar ones: the transform is the continuous, infinite-dimensional version of that change of basis.

When should I reach for Laplace rather than Fourier?

Match the transform to the domain and the data. Use Laplace when the problem lives on a half-line — most often time, $t\ge0$ — with given initial conditions, because its derivative rule pulls $f(0),f'(0),\dots$ into the algebra for free and it tolerates exponentially growing solutions. Use Fourier when the problem lives on the whole line with data that decay at infinity, or when you want a clean spectral picture with no boundary terms. A rough rule: initial-value problems and transients call for Laplace; steady-state, scattering, and translation-invariant problems call for Fourier. Many PDEs use both at once — Laplace in time, Fourier in space — as the wave Green's function derivation does.

What actually makes inversion the hard part?

The forward transform is a definite integral you can usually do or look up; inversion is a contour integral in the complex plane that forces you to confront the singularities of $\tilde f$ head-on. You must locate every pole and branch cut, decide which side of them the contour runs — which is where causality is set — and evaluate by residues or deform around branch points. For Laplace this is the Bromwich integral; for Fourier it is closing in the upper or lower half-plane by the sign of $x$ . In practice one leans on tables and the convolution theorem to avoid doing it from scratch, but for a nonstandard transform there is no shortcut: the analytic structure of $\tilde f$ is the whole problem.

How can a delta-function source be legitimate if it is not a real function?

It is legitimate in the theory of tempered distributions, the natural home for transform methods. There $\delta$ is defined by how it acts under an integral, its Fourier transform is the constant $1$ , and — crucially — every distribution has derivatives of all orders and a transform, with differentiation and transformation commuting by definition. That is exactly the license the method needs: differentiate under the integral, transform a point source, and invert $1/P$ without worrying whether the intermediate objects are classical functions. The Green's function you get is itself often a distribution — the wave kernel is a step function, its derivative a delta on the light cone — and distribution theory keeps all of this rigorous rather than merely suggestive.

The transform reduced my PDE to an ODE, not to algebra — did I do something wrong?

No — that is expected. A single transform diagonalises derivatives only in the variable you transformed; derivatives in the other variables survive untouched. Transforming a two-variable PDE in one variable turns that variable into a parameter (a label like $k$ ) and leaves an ordinary differential equation in the remaining variable, which you solve mode by mode. The heat equation is the clean example: Fourier in $x$ leaves $\mathrm{d}\tilde u/\mathrm{d}t=-Dk^2\tilde u$ , an ODE in $t$ for each $k$ . To reach pure algebra you transform in every variable at once — as the wave Green's function does with a double transform — at the cost of a harder, multidimensional inversion.

Do transform methods work on nonlinear or variable-coefficient equations?

Mostly not, and it is worth knowing why. The magic needs the operator to be linear with constant coefficients, so it maps to a fixed multiplier $P(ik)$ independent of the mode. A variable coefficient like $x\,u'$ transforms into a derivative in $k$ -space (multiplication by $x$ becomes $i\,\mathrm{d}/\mathrm{d}k$ ), so you trade one differential equation for another — rarely a simplification. Nonlinear terms like $u^2$ become convolutions in transform space, coupling all the modes and destroying the decoupling that made the method work. For those one turns to perturbation theory, similarity solutions, or numerics; transform methods remain the exact tool for the linear, constant-coefficient core that so much of physics is built from.