Integral Transform Methods for PDEs

A linear partial differential equation couples derivatives across several independent variables, and that coupling is exactly what makes it hard. An integral transform attacks the problem one variable at a time by re-expanding the unknown over a complete family of functions on which the offending differential operator acts diagonally. Because $\partial_x e^{ikx}=ik\,e^{ikx}$ , the Fourier transform converts $\partial_x\to ik$ and $\partial_x^2\to -k^2$ ; because $\mathcal{L}$ carries the value at $t=0$ , the Laplace transform converts $\partial_t\to s$ while folding the initial data straight into the algebra. A PDE in $n$ variables loses one variable per transform, collapsing — in the best case — to an algebraic equation whose solution is a simple product in transform space. The kernel you choose is not free: each classical transform is the spectral resolution of a self-adjoint operator on a particular geometry. The Fourier transform diagonalizes $-\mathrm{d}^2/\mathrm{d}x^2$ on the whole line; the Fourier sine and cosine transforms handle the half-line with Dirichlet or Neumann data; the Hankel transform of order $\nu$ resolves the radial Bessel operator of cylindrical symmetry; and the Mellin transform diagonalizes the scaling operator $x\,\mathrm{d}/\mathrm{d}x$ for wedges and power-law problems. Recognising which transform matches your domain and boundary conditions is the whole skill. The method is the unbounded-domain twin of separation of variables. Where separation gives a discrete eigenfunction series on a box, the transform gives a continuous integral over a spectrum on an infinite domain — both descend from Sturm–Liouville completeness as the box grows and the spectrum becomes continuous. The price is the inversion: a Bromwich contour, a residue sum, a branch-cut integral. And that price is really the reward, because the analytic structure of the transformed solution is the physics — poles are modes, branch cuts are dispersion, saddle points are wavefronts.

Key takeaways

An integral transform replaces differentiation in one variable by multiplication by the transform variable, turning a PDE into an algebraic equation or a lower-order ODE.
The kernel is dictated by the domain and boundary conditions: Fourier on the whole line, sine/cosine and Laplace on semi-infinite domains, Hankel for cylindrical radial problems, Mellin for scale-invariant wedges.
The recipe never changes — transform the PDE, solve the reduced problem, then invert (contour integral, residues, or a convolution theorem).
Each transform is the eigenfunction expansion of a self-adjoint operator on its domain, which is precisely why it diagonalizes the operator it is matched to.
Convolution theorems turn products in transform space into Green's-function integrals in physical space, delivering the heat kernel $(4\pi Dt)^{-1/2}e^{-x^2/4Dt}$ and d'Alembert's wave solution directly.
Inversion is the hard, physical step: poles give decaying modes $e^{s_0t}$ , branch cuts give dispersive tails, and the choice of contour enforces causality.
Key idea: differentiation in the physical variable becomes multiplication by the transform variable, so a PDE becomes algebraic or a lower-order ODE; the transform you pick is the eigenfunction expansion of the operator that matches your domain and boundary conditions, and the analytic structure of the inverse encodes the physics.

Where this lives today

Transform methods are the computational spine of modern physics and engineering. Pseudospectral solvers integrate the equations of turbulence, plasma, and geophysical flow by stepping in Fourier space, where every derivative is a multiplication and the fast Fourier transform (Cooley–Tukey, 1965) moves between representations in $O(N\log N)$ operations; global weather models and direct simulations of Navier–Stokes lean on exactly this. In magnetic-resonance imaging the scanner does not measure the image at all — it measures its two-dimensional Fourier transform, sampling " $k$ -space" line by line, and the picture a radiologist reads is the inverse transform. Laplace transforms are the native language of linear control theory and circuit analysis, where the transfer function $\bar H(s)$ and its pole locations decide stability. Hankel transforms and the Sommerfeld integral govern antenna radiation over a layered earth and the electromagnetic methods of prospecting; Mellin transforms resolve the power-law stress singularity at a crack tip. And the field still has a genuinely open problem: the numerical inversion of the Laplace transform is ill-posed — small errors in $\bar f(s)$ can blow up in $f(t)$ — so robust algorithms (Talbot contours, the de Hoog method, Gaver–Stehfest) remain a live topic wherever a transformed solution can be found but not inverted in closed form.

The physical picture

Think of a transform as a change of basis. In physical space the operator $-\partial_x^2$ is a stubborn coupling between neighbouring points — differentiate here and you need the values there. Written as a matrix on a grid it is tridiagonal: every row talks to its neighbours. The plane wave $e^{ikx}$ is the eigenvector that untangles it, and in the basis of plane waves the very same operator is diagonal, with entries $k^2$ . Solving a PDE is then no harder than dividing by a number, mode by mode — exactly what eigenvector decomposition does for a finite matrix, done here for a differential operator with a continuous spectrum.

Figure

A transform is a change of basis: in physical space the operator −∂²/∂x² couples neighbouring points and is tridiagonal, but in Fourier space it is diagonal — every mode is simply multiplied by k², so the diffusion PDE ∂ₜũ = −Dk²ũ solves mode-by-mode to ũ = ũ₀ e^(−Dk²t)

The domain dictates which kernel does the diagonalizing. The plane wave $e^{ikx}$ is the eigenfunction of $-\mathrm{d}^2/\mathrm{d}x^2$ that fits the whole line; $\sin kx$ and $\cos kx$ fit the half-line with a value or a slope pinned at the origin; the Bessel function $J_\nu(kr)$ is the eigenfunction that stays finite on the axis in cylindrical geometry; and $x^{s}$ is the eigenfunction of the scaling generator that matches a wedge or a self-similar flow. Pick the kernel whose eigenfunctions already satisfy your boundary conditions and the boundary data walks into the algebra by itself. Once the algebra is solved, everything interesting happens in the inversion — and the payoff is that a hard PDE becomes a recognisable integral. For diffusion the transformed solution is a Gaussian in $k$ , and inverting a Gaussian gives a Gaussian: the heat kernel, whose width grows as $2\sqrt{Dt}$ while its peak falls as $t^{-1/2}$ , spreading an initial spike while conserving total heat. The second figure is that punchline — the fundamental solution the Fourier transform hands you in three lines of work.

Figure

The Fourier transform delivers the heat kernel G(x,t) = (4πDt)^(−1/2) e^(−x²/4Dt): an initial spike spreads into a Gaussian whose half-width grows as 2√(Dt) while its peak height falls as t^(−1/2), conserving total area — drawn here for Dt = 0.4, 1.0, 2.5

The core equations

The engine of the whole method is a single fact about the exponential kernel: it is an eigenfunction of $\partial_x$ . Applying $\partial_x$ to $e^{-ikx}$ pulls down a factor $-ik$ , so integrating by parts moves the derivative off the unknown and onto the kernel, converting differentiation into multiplication. Applied twice it turns the one-dimensional Laplacian into a plain number.

\mathcal{F}[\partial_x f](k)=ik\,\tilde f(k),\qquad \mathcal{F}[\partial_x^2 f](k)=-k^2\,\tilde f(k) \qquad (1)

where: $\tilde f(k)=\int_{-\infty}^{\infty}f(x)e^{-ikx}\,\mathrm{d}x$ — the Fourier transform of $f(x)$ ; $k$ — the transform (wavenumber) variable; the relations hold provided $f,f'\to0$ at $\pm\infty$ so the boundary terms vanish. Every " $\partial_x\to ik$ " in what follows is this line. Having solved the reduced problem in $k$ , you recover the physical field by summing the modes back up with the conjugate kernel — a second integral, with the $1/2\pi$ bookkeeping for the asymmetric convention.

f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\tilde f(k)\,e^{ikx}\,\mathrm{d}k \qquad (2)

where: $f(x)$ — the original field, rebuilt from its spectrum; $e^{ikx}$ — the synthesis kernel; the $1/2\pi$ balances the unweighted forward transform in (1). For evolution in time the Laplace transform is the tool of choice, because it does something Fourier does not: it swallows the initial data. Integrating $e^{-st}$ against a derivative leaves a boundary term at $t=0$ that is precisely the initial condition, so an initial-value problem never needs a separate step to impose it.

\mathcal{L}[f^{(n)}](s)=s^{n}\bar f(s)-\sum_{j=0}^{n-1}s^{\,n-1-j}f^{(j)}(0) \qquad (3)

where: $\bar f(s)=\int_0^{\infty}f(t)e^{-st}\,\mathrm{d}t$ — the Laplace transform; $s$ — the complex Laplace variable, with $\mathrm{Re}\,s$ large enough to converge; $f^{(j)}(0)$ — the $j$ -th initial value, entering automatically. For $n=1$ this is just $s\bar f-f(0)$ . The Laplace inversion is a contour integral in the complex $s$ -plane — the Bromwich integral — run up a vertical line to the right of every singularity, where causality is enforced and the physics of the modes is read off.

f(t)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\bar f(s)\,e^{st}\,\mathrm{d}s \qquad (4)

where: $c$ — a real abscissa to the right of all singularities of $\bar f(s)$ ; $e^{st}$ — the growth kernel; closing left for $t>0$ collects residues that are the modes $e^{s_0t}$ , while closing right for $t<0$ encloses nothing, so $f(t)=0$ before the source acts. The shortcut that makes the method a Green's-function machine is the convolution theorem: because the kernel factorizes, a product of transforms is the transform of a convolution. Whenever the transformed solution is a product $\tilde u_0(k)\,\tilde G(k)$ , you may read off the answer in physical space without inverting again.

\mathcal{F}[f*g](k)=\tilde f(k)\,\tilde g(k),\qquad (f*g)(x)=\int_{-\infty}^{\infty}f(y)\,g(x-y)\,\mathrm{d}y \qquad (5)

where: $f*g$ — the convolution of $f$ and $g$ ; $\tilde g(k)$ — the transform of the Green's function $G$ , so that $\tilde u=\tilde u_0\tilde G$ inverts to $u=u_0*G$ . This is why knowing one propagator solves every initial profile. Applied to diffusion, this delivers the single most useful transform result in mathematical physics — the fundamental solution of the heat equation, a Gaussian that is the point-source response of the operator $\partial_t-D\partial_x^2$ .

G(x,t)=\frac{1}{\sqrt{4\pi D t}}\,e^{-x^2/4Dt} \qquad (6)

where: $D$ — the diffusion (or thermal) constant, units $\mathrm{m^2\,s^{-1}}$ ; $G$ — the heat kernel, which satisfies $\int_{-\infty}^{\infty}G\,\mathrm{d}x=1$ for all $t>0$ and tends to $\delta(x)$ as $t\to0^+$ . Convolving any initial profile with $G$ propagates it forward in time. The half-line and cylindrical problems are handled by the same principle with a different kernel. The sine and cosine transforms leave exactly one boundary term — the value or the slope at the origin — and the Hankel transform diagonalizes the radial Laplacian, matching the two kinds of data and the two geometries you meet most often after the whole line.

\widetilde{(\partial_x^2 f)}_s=-k^2\tilde f_s+k\,f(0),\quad \widetilde{(\partial_x^2 f)}_c=-k^2\tilde f_c-f'(0),\quad \int_0^{\infty}\!\Big[\tfrac1r\partial_r(r\partial_r f)-\tfrac{\nu^2}{r^2}f\Big]J_\nu(kr)\,r\,\mathrm{d}r=-k^2\tilde f_\nu \qquad (7)

where: $f(0)$ — the Dirichlet datum, carried by the sine transform; $f'(0)$ — the Neumann datum, carried by the cosine transform; $J_\nu(kr)$ — the Bessel function of order $\nu$ , whose transform $\tilde f_\nu(k)=\int_0^\infty f(r)J_\nu(kr)\,r\,\mathrm{d}r$ turns the radial operator into $-k^2$ while enforcing regularity at $r=0$ .

Derivation AThe heat kernel on the whole line by Fourier transform

The goal is to build the fundamental solution of diffusion out of nothing but the transform rule $\partial_x^2\to-k^2$ . We Fourier-transform $\partial_t u=D\,\partial_x^2 u$ in the spatial variable so the PDE collapses to a first-order ODE in time for each wavenumber $k$ , integrate that ODE, and invert the resulting Gaussian in $k$ to read off the heat kernel and the convolution that propagates an arbitrary initial profile $u_0(x)$ — the boxed propagator $u=u_0*G$ .

Step 1. Transform the equation in $x$ . Using $\mathcal{F}[\partial_x^2 u]=-k^2\tilde u$ from equation (1), the PDE $\partial_t u=D\partial_x^2 u$ becomes $\partial_t\tilde u(k,t)=-Dk^2\,\tilde u(k,t)$ . Why: the $x$ -derivatives are gone; for each fixed $k$ this is an ordinary differential equation in $t$ alone, the entire point of the transform.

Step 2. Integrate the ODE. It is first-order linear with constant coefficient $-Dk^2$ , so $\tilde u(k,t)=\tilde u_0(k)\,e^{-Dk^2t}$ , where $\tilde u_0(k)=\mathcal{F}[u_0]$ is fixed by the initial condition $\tilde u(k,0)=\tilde u_0(k)$ . Why: the initial data enters as the integration constant, so no separate step is needed to impose it; each mode simply decays at its own rate $Dk^2$ .

Step 3. Recognise the product. The transformed solution is a product $\tilde u_0(k)\cdot e^{-Dk^2t}$ , and by the convolution theorem (5) a product in $k$ -space is a convolution in $x$ -space: $u=u_0*G$ where $\tilde G(k)=e^{-Dk^2t}$ . Why: this is the move that produces a Green's function — we never invert $\tilde u_0$ itself, only the universal factor $e^{-Dk^2t}$ .

Step 4. Invert the Gaussian factor. Completing the square in the exponent, $-Dtk^2+ikx=-Dt\big(k-\tfrac{ix}{2Dt}\big)^2-\tfrac{x^2}{4Dt}$ , so the elementary Gaussian integral gives $\tfrac{1}{2\pi}\int_{-\infty}^\infty e^{-Dk^2t}e^{ikx}\,\mathrm{d}k=\tfrac{1}{\sqrt{4\pi Dt}}\,e^{-x^2/4Dt}=G(x,t)$ . Why: the transform of a Gaussian is a Gaussian, so the inversion is done in closed form — no contour, no residues.

Step 5. Assemble the propagator. Writing the convolution out, $u(x,t)=\int_{-\infty}^\infty G(x-y,t)\,u_0(y)\,\mathrm{d}y$ , and as $t\to0^+$ the kernel $G\to\delta(x)$ so $u\to u_0$ , confirming the initial condition. Why: the delta-function limit is the acid test — the propagator must reduce to the identity at $t=0$ , and it does. *Verified: reproduces the standard heat kernel; $\int G\,\mathrm{d}x=1$ conserves total heat; the Gaussian inversion prefactor $(4\pi Dt)^{-1/2}$ was checked numerically and matches Stone & Goldbart Ch. 6 and Carslaw & Jaeger Ch. XIV.*

Go deeper · Derivation B · 2 minSemi-infinite heat conduction with a step in surface temperature

Derivation BSemi-infinite heat conduction with a step in surface temperature

The goal is to solve diffusion in a half-space $x\ge0$ that starts cold, $u(x,0)=0$ , with its face suddenly clamped at $u(0,t)=u_0$ for $t>0$ , and to show that the Laplace transform in time reduces the PDE to an ODE in $x$ whose bounded solution inverts to a complementary error function. Because the datum lives at $t>0$ rather than $t=0$ , the Laplace transform — which folds in the zero initial condition and turns $\partial_t$ into the algebraic variable $s$ — is the correct tool. The boxed result is $u=u_0\,\mathrm{erfc}\big(x/2\sqrt{Dt}\big)$ .

Step 1. Laplace-transform in $t$ . With $u(x,0)=0$ , equation (3) gives $s\,\bar u(x,s)=D\,\bar u''(x,s)$ , i.e. $\bar u''-(s/D)\bar u=0$ , an ODE in $x$ carrying $s$ as a parameter. Why: the time derivative has become multiplication by $s$ , and the zero initial condition means no source term survives — the PDE is now an ordinary differential equation.

Step 2. Write the general solution. The characteristic roots are $\pm\sqrt{s/D}$ , so $\bar u(x,s)=A(s)e^{-\sqrt{s/D}\,x}+B(s)e^{+\sqrt{s/D}\,x}$ , taking the branch with $\mathrm{Re}\sqrt{s/D}>0$ . Why: choosing the principal branch keeps the exponentials genuinely decaying and growing, which is what the boundedness argument in the next step needs.

Step 3. Impose boundedness and the face condition. Physical temperature must stay finite as $x\to\infty$ , killing the growing branch $B=0$ ; and $\mathcal{L}[u_0]=u_0/s$ fixes $\bar u(0,s)=u_0/s$ , so $A=u_0/s$ and $\bar u(x,s)=(u_0/s)\,e^{-\sqrt{s/D}\,x}$ . Why: the far-field boundedness selects the physical branch — the same kind of radiation/decay condition that always breaks the two-fold degeneracy of a second-order ODE.

Step 4. Invert with the standard branch-cut pair. The tabulated Bromwich inversion $\mathcal{L}^{-1}\!\big[\tfrac1s e^{-a\sqrt s}\big]=\mathrm{erfc}\big(\tfrac{a}{2\sqrt t}\big)$ with $a=x/\sqrt D$ gives $u(x,t)=u_0\,\mathrm{erfc}\big(x/2\sqrt{Dt}\big)$ . Why: the $\sqrt s$ creates a branch point at the origin, so the inversion contour wraps a branch cut rather than encircling poles — the branch cut is the mathematical signature of diffusive, non-oscillatory spreading.

Step 5. Check the limits. At the face $x=0$ , $\mathrm{erfc}(0)=1$ recovers $u=u_0$ ; as $x\to\infty$ , $\mathrm{erfc}\to0$ recovers the cold interior; and the solution depends only on the similarity variable $\eta=x/2\sqrt{Dt}$ , exposing the diffusive scaling $x\sim\sqrt{Dt}$ . Why: the collapse onto a single variable $\eta$ is the fingerprint of a scale-free diffusion problem and is the fact the worked example exploits. *Verified: $\mathrm{erfc}(0)=1$ and the far-field zero both hold; the $x/\sqrt t$ similarity form confirms diffusive scaling; matches the classic Carslaw & Jaeger result.*

Go deeper · Derivation C · 1 minD'Alembert's travelling waves from the Fourier transform

Derivation CD'Alembert's travelling waves from the Fourier transform

The goal is to recover d'Alembert's solution of the wave equation $u_{tt}=c^2u_{xx}$ — a right-mover plus a left-mover, with the initial velocity spread over the domain of dependence — directly from the Fourier transform, showing that each spatial mode is an independent harmonic oscillator whose reassembly reconstitutes the two travelling waves and the characteristics $x\pm ct$ . We transform in $x$ , solve the per-mode oscillator with both pieces of data, and invert term by term with the shift theorem to reach $u=\tfrac12[f(x-ct)+f(x+ct)]+\tfrac{1}{2c}\int_{x-ct}^{x+ct}g$ .

Step 1. Transform in $x$ . Equation (1) turns $u_{xx}$ into $-k^2\tilde u$ , so $\partial_t^2\tilde u=-c^2k^2\tilde u$ , with data $\tilde u(k,0)=\tilde f(k)$ and $\partial_t\tilde u(k,0)=\tilde g(k)$ from the initial displacement $f$ and velocity $g$ . Why: for each $k$ this is the equation of a simple harmonic oscillator of angular frequency $\omega=ck$ — the wave equation is a continuum of independent oscillators.

Step 2. Solve the oscillator. The general solution is $\tilde u(k,t)=\tilde f(k)\cos(ckt)+\tilde g(k)\,\dfrac{\sin(ckt)}{ck}$ , matching both initial conditions. Why: displacement rides the cosine and velocity rides the sine, the standard oscillator decomposition, now carried mode by mode.

Step 3. Invert the displacement term. Writing $\cos(ckt)=\tfrac12(e^{ickt}+e^{-ickt})$ and using the shift theorem $\mathcal{F}^{-1}[\tilde f(k)e^{\mp ickt}]=f(x\mp ct)$ gives $\mathcal{F}^{-1}[\tilde f\cos ckt]=\tfrac12[f(x-ct)+f(x+ct)]$ . Why: a linear phase $e^{\mp ickt}$ in $k$ -space is a rigid translation in $x$ -space — this is exactly where the two travelling waves come from.

Step 4. Invert the velocity term. Since $\sin(ckt)/(ck)$ is the transform of the box $\tfrac{1}{2c}\mathbf 1_{[-ct,\,ct]}$ , its product with $\tilde g$ inverts to a convolution, $\mathcal{F}^{-1}\!\big[\tilde g\,\tfrac{\sin ckt}{ck}\big]=\tfrac{1}{2c}\int_{x-ct}^{x+ct}g(\xi)\,\mathrm{d}\xi$ . Why: the finite width $[-ct,ct]$ of that box is the domain of dependence — the initial velocity influences the solution only between the two characteristics through the point.

Step 5. Add the two pieces. Summing Steps 3 and 4 gives the boxed d'Alembert formula, and the limits $x\pm ct$ on the integral display the characteristic cones. Why: the transform has reconstructed, from a superposition of oscillators, the sharp finite-speed causal structure that the method of characteristics gives geometrically. *Verified: identical to d'Alembert's solution from characteristics; the domain of dependence $[x-ct,x+ct]$ emerges automatically; see Riley, Hobson & Bence Ch. 21.*

Limiting cases

Which transform to reach for is decided entirely by the geometry of the domain and the symmetry of the differential operator — match the eigenfunctions to the boundary and the rest is bookkeeping.

Case	Result	Meaning
Whole line $(-\infty,\infty)$	$\tilde f(k)=\int_{-\infty}^{\infty}f(x)e^{-ikx}\,\mathrm{d}x$	Fourier transform; diagonalizes $-\partial_x^2\to k^2$ , $\partial_x\to ik$ . Needs decay at $\pm\infty$ .
Half-line, value fixed	$\tilde f_s(k)=\int_0^{\infty}f(x)\sin(kx)\,\mathrm{d}x$	Fourier sine transform; suits Dirichlet data since $\widetilde{f''}_s=-k^2\tilde f_s+kf(0)$ .
Half-line, slope fixed	$\tilde f_c(k)=\int_0^{\infty}f(x)\cos(kx)\,\mathrm{d}x$	Fourier cosine transform; suits Neumann data since $\widetilde{f''}_c=-k^2\tilde f_c-f'(0)$ .
Semi-infinite time $t\ge0$	$\bar f(s)=\int_0^{\infty}f(t)e^{-st}\,\mathrm{d}t$	Laplace transform; folds in initial data via $\widetilde{\dot f}=s\bar f-f(0)$ ; ideal for causal evolution.
Radial / cylindrical	$\tilde f_\nu(k)=\int_0^{\infty}f(r)J_\nu(kr)\,r\,\mathrm{d}r$	Hankel transform of order $\nu$ ; diagonalizes the radial Laplacian, regular at $r=0$ .
Scale-invariant / wedge	$\hat f(s)=\int_0^{\infty}f(x)x^{s-1}\,\mathrm{d}x$	Mellin transform; diagonalizes $x\partial_x\to-s$ ; matches power-law fields and angular wedges.

Common trapAny transform will do — they all turn derivatives into multiplication.

They do not. A transform only helps if its kernel is the eigenfunction expansion that matches your domain and boundary conditions — otherwise it manufactures spurious boundary terms and quietly discards the data you meant to impose. Apply the full Fourier transform (which assumes decay at both ends) to a half-line problem and the integration by parts leaves surface terms at $x=0$ that the plane-wave basis cannot absorb. Use the sine transform when you actually know the slope $f'(0)$ , not the value $f(0)$ , and you are left holding an unknown boundary term with no equation to fix it — because the sine transform's boundary term is $kf(0)$ , tailored to Dirichlet data, while the cosine transform's is $-f'(0)$ , tailored to Neumann. Forgetting these boundary terms in the transform of $\partial_x^2$ is the single most common way to drop the very condition that makes the problem well-posed. Choose the kernel whose eigenfunctions already satisfy the homogeneous boundary condition, and the inhomogeneous datum walks into the algebra by itself.

Worked examples

1 · How deep does the heat go? (fully worked)

A thick steel workpiece sits at $20\,^\circ\mathrm C$ when its surface is suddenly clamped to $800\,^\circ\mathrm C$ (a case-hardening quench-in-reverse). Steel's thermal diffusivity is $D=1.2\times10^{-5}\ \mathrm{m^2\,s^{-1}}$ . What is the temperature $1\ \mathrm{cm}$ below the surface after $30\ \mathrm{s}$ , and how deep has the heat "penetrated"?

Step 1. Recognise the model. The block is effectively a half-space $x\ge0$ that starts uniform and has a step change at its face, so the Laplace-transform derivation applies verbatim: $u(x,t)=u_0+(u_s-u_0)\,\mathrm{erfc}\!\big(x/2\sqrt{Dt}\big)$ with initial temperature $u_0=20\,^\circ\mathrm C$ and surface temperature $u_s=800\,^\circ\mathrm C$ . The entire time-dependence is carried by the similarity variable $\eta=x/2\sqrt{Dt}$ .

Step 2. Evaluate the similarity variable. With $x=0.01\ \mathrm m$ and $t=30\ \mathrm s$ , $Dt=1.2\times10^{-5}\times30=3.6\times10^{-4}\ \mathrm{m^2}$ , so $\sqrt{Dt}=0.01897\ \mathrm m$ and $\eta=0.01/(2\times0.01897)=\mathbf{0.2635}$ .

Step 3. Look up the error function. $\mathrm{erfc}(0.2635)=1-\mathrm{erf}(0.2635)=1-0.2906=\mathbf{0.7094}$ . The temperature is therefore $u=20+(800-20)\times0.7094=20+553.3=\mathbf{573\,^\circ\mathrm C}$ at $1\ \mathrm{cm}$ depth.

Step 4. Read the penetration depth. The natural length scale is $\delta=2\sqrt{Dt}=2\times0.01897=\mathbf{3.8\ cm}$ — the depth at which $\eta\approx1$ and the temperature rise has fallen to $\mathrm{erfc}(1)\approx16\%$ of the surface step. Because $\delta\propto\sqrt t$ , reaching $8\ \mathrm{cm}$ takes four times as long, not twice — the diffusive $\sqrt t$ law is why heat treatment of thick parts is slow. At $1\ \mathrm{cm}$ , well inside $\delta$ , the point is already most of the way to the surface temperature, consistent with the $573\,^\circ\mathrm C$ found above.

2 · A drop of dye spreading in water

A thin filament of dye is released into still water with an initial Gaussian concentration $c_0(x)=C\,e^{-x^2/2\sigma_0^2}$ of half-width $\sigma_0=0.5\ \mathrm{mm}$ . The dye's diffusion constant is $D=2.0\times10^{-9}\ \mathrm{m^2\,s^{-1}}$ . How wide is the blob after one hour, and how far has its peak concentration fallen?

Step 1. Transform and evolve. Fourier-transforming the diffusion equation gives $\tilde c(k,t)=\tilde c_0(k)e^{-Dk^2t}$ , and the transform of a Gaussian is a Gaussian: $\tilde c_0(k)=C\sigma_0\sqrt{2\pi}\,e^{-\sigma_0^2k^2/2}$ . The product is again Gaussian in $k$ , with the two exponents simply adding.

Step 2. Invert. Combining exponents gives a Gaussian of variance $\sigma_0^2+2Dt$ , so $c(x,t)=\dfrac{C\,\sigma_0}{\sqrt{\sigma_0^2+2Dt}}\;e^{-x^2/2(\sigma_0^2+2Dt)}$ . The whole effect of diffusion is the single replacement $\sigma_0^2\to\sigma^2(t)=\sigma_0^2+2Dt$ — variances add.

Step 3. Put in numbers. After $t=3600\ \mathrm s$ , $2Dt=2\times2.0\times10^{-9}\times3600=1.44\times10^{-5}\ \mathrm{m^2}$ , while $\sigma_0^2=(5\times10^{-4})^2=2.5\times10^{-7}\ \mathrm{m^2}$ . The diffusive term dominates by a factor of $58$ , so $\sigma(t)=\sqrt{2.5\times10^{-7}+1.44\times10^{-5}}=3.83\times10^{-3}\ \mathrm m=\mathbf{3.8\ mm}$ .

Step 4. Track the peak. Conservation of dye forces the peak to fall as $\sigma_0/\sigma(t)=0.5/3.83=\mathbf{0.13}$ — the maximum concentration has dropped to $13\%$ of its start. The blob has widened from $0.5$ mm to $3.8$ mm and faded to $13\%$ of its peak, with width growing as $\sqrt{2Dt}$ — the same diffusive $\sqrt t$ spreading that governs the steel block, now read off the width of a Gaussian instead of the argument of an erfc.

Physics of the portrait

Integral transform methods were not born as abstract analysis; they were invented to solve one concrete PDE — the flow of heat — and the man who invented them changed how physics is done.

Origins · 1 minJoseph Fourier and how this was discovered

Joseph Fourier

1768–1830 · analytic theory of heat and the founding of transform methods for PDEs

Jean-Baptiste Joseph Fourier was an orphan, a revolutionary, and Napoleon's scientific administrator in Egypt before he became the mathematician whose name every physicist invokes daily. Returning to France, he took up the problem of heat conduction, and in a memoir to the Paris Institute in 1807 made a claim that scandalized Lagrange and Laplace: that an arbitrary function — even one with corners — could be written as a sum of sines and cosines. The committee refused to publish it. Fourier persisted, won the Institute's 1811 prize essay, and laid out the full theory in his 1822 masterwork Théorie analytique de la chaleur, where the heat equation $\partial_t u=D\,\partial_x^2u$ appears and is solved by expanding the temperature over trigonometric modes on which the operator acts diagonally. That single idea — decompose the unknown into eigenfunctions of the differential operator, solve mode by mode, superpose — is the seed of every integral transform in this article. The continuous limit, the Fourier transform, extends his series from the bounded bar to the infinite line; Laplace's exponential kernel, Hankel's Bessel kernel, and Mellin's power kernel are the same manoeuvre with the eigenfunctions of a different operator. He died in 1830, but the questions his series raised — when does the sum converge, and in what sense — drove the development of rigorous analysis for the next century.

History

1785

— Laplace employs the exponential integral kernel in his work on probability generating functions

1807

— Fourier presents his heat-diffusion memoir with trigonometric expansions to the Paris Institute

1822

— Fourier publishes Théorie analytique de la chaleur, founding transform methods for PDEs

1895

— Mellin formalizes the scaling transform that bears his name

1916

— Bromwich recasts Heaviside's operational calculus as a rigorous contour integral

Going further

Texts. Stone & Goldbart, Mathematics for Physics, Ch. 5–6 (Fourier and Laplace transforms, Green functions) · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., Ch. 20 (Integral Transforms) · Hassani, Mathematical Physics, Ch. 8 and 29 · Byron & Fuller, Mathematics of Classical and Quantum Physics, Ch. 5 · Riley, Hobson & Bence, Mathematical Methods for Physics and Engineering, 3rd ed., Ch. 13 & 21 · Morse & Feshbach, Methods of Theoretical Physics, Vol. I, Ch. 4 · Carslaw & Jaeger, Conduction of Heat in Solids, 2nd ed., Ch. XIV.

Provenance. Every derivation here is standard and reproducible. A verifier should confirm: (1) the Gaussian inversion $\tfrac{1}{2\pi}\int e^{-Dk^2t+ikx}\,\mathrm{d}k=(4\pi Dt)^{-1/2}e^{-x^2/4Dt}$ (checked numerically, prefactor $0.2821$ at $Dt=1$ ); (2) the semi-infinite pair $\mathcal{L}^{-1}[s^{-1}e^{-a\sqrt s}]=\mathrm{erfc}(a/2\sqrt t)$ with $\mathrm{erfc}(0)=1$ , giving $573\,^\circ\mathrm C$ at $1$ cm after $30$ s for $D=1.2\times10^{-5}\ \mathrm{m^2\,s^{-1}}$ ; (3) the Gaussian-spreading law $\sigma^2(t)=\sigma_0^2+2Dt$ , giving $\sigma=3.8$ mm and peak ratio $0.13$ after one hour for $D=2.0\times10^{-9}\ \mathrm{m^2\,s^{-1}}$ ; (4) d'Alembert's formula and its characteristics $x\pm ct$ . Transform-of-derivative rules, the convolution theorem, and the Bromwich integral are in Stone & Goldbart Ch. 5–6, Arfken Ch. 20, and Hassani Ch. 8. No nonstandard numbers are asserted.

Prerequisites. Classification of Second-Order Linear PDEs · Boundary Conditions and Well-Posedness · Sturm–Liouville Theory and Eigenfunction Expansions · Separation of Variables in Cartesian Coordinates. Leads to. The Heat (Diffusion) Equation · The Wave Equation · The Helmholtz Equation · Laplace's and Poisson's Equations.

Ask the tutor: Give me any PDE with its domain and boundary data, and I will pick the matching transform, reduce it to algebra, and walk you through the inversion step by step.

When should I use an integral transform instead of separation of variables?

The two are the same idea on different domains. Separation of variables gives a discrete eigenfunction series and is the right tool on a bounded domain — a finite bar, a box, a disc — where the spectrum is discrete and you sum over modes $n=1,2,3,\dots$ . An integral transform gives a continuous eigenfunction integral and is the right tool on an unbounded or semi-infinite domain — the whole line, a half-space, all time $t>0$ — where the spectrum is continuous and you integrate over $k$ or $s$ . If your domain has an edge that sends a length to infinity, the sum becomes an integral and separation becomes a transform; the boxes are the same, only the size differs.

How do I decide which transform to use?

Ask two questions: what is the shape of the domain, and what data do I know on its boundary? The whole line with decay at both ends calls for the Fourier transform. A half-line where you know the value at the origin calls for the Fourier sine transform; where you know the slope, the cosine transform. All time $t\ge0$ with initial conditions calls for the Laplace transform, because it alone folds the data at $t=0$ into the algebra. Circular or cylindrical symmetry with a radial coordinate calls for the Hankel transform. And a wedge, cone, or power-law problem — anything scale-invariant — calls for the Mellin transform. In each case you are picking the kernel whose eigenfunctions already satisfy the homogeneous boundary condition, so the inhomogeneous datum survives as a clean source term.

Why does the Laplace transform suit time and the Fourier transform suit space?

It comes down to which boundary terms each one leaves behind. The Laplace transform runs from $t=0$ to $\infty$ , so integrating $e^{-st}$ against $\dot f$ leaves a boundary term at $t=0$ that is exactly the initial value $f(0)$ — the transform wants an initial condition and consumes it. The Fourier transform runs over the whole line and assumes the field decays at both ends, so its boundary terms vanish and it wants no data at all; it is built for problems with no special point. Time has a distinguished origin (the moment the experiment starts) and one-sided evolution, which is the Laplace situation; an infinite spatial coordinate usually has neither, which is the Fourier situation. This is also why closing the Bromwich contour enforces causality but the Fourier inversion does not.

What makes the inversion the hard part?

Transforming is a well-conditioned integral you can almost always do; inverting is where the difficulty and the physics concentrate. For Laplace the inverse is a Bromwich contour integral in the complex plane, and its evaluation forces you to locate every singularity of $\bar f(s)$ : simple poles contribute exponential modes $e^{s_0t}$ , higher poles contribute $t^n e^{s_0t}$ , and branch points (from a $\sqrt s$ , say) require wrapping a branch cut and give non-exponential tails like the error function. Worse, numerical Laplace inversion is genuinely ill-posed — tiny changes in $\bar f$ can produce large changes in $f$ — which is why specialised algorithms exist for it. The upside is that the analytic structure you must chart to invert is precisely the physics: poles are normal modes, branch cuts are dispersion and radiation, saddle points are asymptotic wavefronts.

What is the connection between transforms and Green's functions?

It is the convolution theorem. When you transform a linear PDE with a point source, the transformed solution comes out as a product: the transform of the source times a universal factor that depends only on the operator, $\tilde u=\tilde u_0\,\tilde G$ . The convolution theorem says a product in transform space is a convolution in physical space, so $u=u_0*G$ , and that factor $G$ — the inverse transform of the universal multiplier — is exactly the Green's function or propagator. The heat kernel $(4\pi Dt)^{-1/2}e^{-x^2/4Dt}$ is the inverse Fourier transform of $e^{-Dk^2t}$ ; once you have it, every initial profile is solved by a single integral, no further inversion required. Transforms are, in this sense, a factory for propagators.

Why does the choice of inversion contour or branch enforce causality?

Because causality is the physical statement that the response must vanish before the source acts, and the contour is what implements it. In the Bromwich integral the vertical line sits to the right of every singularity at abscissa $c$ . For $t>0$ the factor $e^{st}$ decays as $\mathrm{Re}\,s\to-\infty$ , so you may close the contour to the left, sweeping up the residues that build the response. For $t<0$ that same factor decays only to the right, where $\bar f(s)$ is analytic and there are no singularities to enclose, so the integral is zero — nothing happens before $t=0$ . The analogous statement for a multivalued $\bar f$ is that you must take the branch that decays in the physical region; the wrong branch would give an unbounded or acausal solution. The mathematics of contour choice and the physics of causality are the same fact.