Introduction to PDEs

Every equation so far has had one independent variable. But heat spreads through space and time, a wave depends on position and time, and an electric potential varies across a whole region. Their unknown is a function of several variables, so the equation involves partial derivatives — it is a partial differential equation (PDE).

Three linear second-order PDEs underpin an astonishing share of physics. Writing u_t = \partial u/\partial t, u_{xx} = \partial^2 u/\partial x^2, they are:

\underbrace{u_t = \alpha\, u_{xx}}_{\text{heat}}, \qquad \underbrace{u_{tt} = c^2\, u_{xx}}_{\text{wave}}, \qquad \underbrace{u_{xx} + u_{yy} = 0}_{\text{Laplace}}.

Heat diffuses (smooths out), waves propagate (travel undamped), and Laplace describes the steady state when nothing changes in time. One technique cracks all three: separation of variables.

Separating the heat equation, line by line

Solve u_t = \alpha\, u_{xx} by hunting for solutions that split into a function of x times a function of t.

Step 1 — assume a product.

u(x, t) = X(x)\,T(t).

Step 2 — take the derivatives the equation needs. Only T depends on t, only X on x:

u_t = X\,T', \qquad u_{xx} = X''\,T.

Step 3 — substitute into u_t = \alpha\, u_{xx}.

X\,T' = \alpha\,X''\,T.

Step 4 — separate the variables. Divide by \alpha\,X\,T so each side depends on one variable only:

\frac{T'}{\alpha\,T} = \frac{X''}{X}.

Step 5 — the separation constant. The left side is a function of t alone, the right of x alone, yet they are equal for all x, t. A function of t can equal a function of x only if both equal the same constant — call it -\lambda (the sign chosen for convenience):

\frac{T'}{\alpha\,T} = \frac{X''}{X} = -\lambda.

Step 6 — two ordinary equations fall out.

X'' = -\lambda X, \qquad T' = -\alpha\lambda\,T.

Step 7 — solve each. The space equation is our old friend with oscillating solutions; the time equation decays exponentially:

X(x) = \sin\!\big(\sqrt{\lambda}\,x\big),\ \cos\!\big(\sqrt{\lambda}\,x\big), \qquad T(t) = e^{-\alpha\lambda t}.

Step 8 — a product solution (a "mode"). With boundary conditions u(0,t) = u(L,t) = 0 forcing \lambda_n = (n\pi/L)^2,

u_n(x, t) = e^{-\alpha\lambda_n t}\,\sin\!\Big(\frac{n\pi x}{L}\Big).

Step 9 — superpose into a Fourier series. The equation is linear, so sums of modes still solve it. The general solution matching an initial temperature u(x, 0) = f(x) is

u(x, t) = \sum_{n=1}^{\infty} b_n\, e^{-\alpha\lambda_n t}\,\sin\!\Big(\frac{n\pi x}{L}\Big),

with the b_n chosen so the t = 0 sum equals f(x) — the Fourier coefficients. Each mode decays at its own rate; higher modes (larger n) vanish fastest, which is why sharp features smooth out first.

Separation only ever produces the building blocks \sin(n\pi x/L) and \cos(n\pi x/L). The miracle — Fourier's 1822 claim, doubted at the time — is that any reasonable initial profile f(x) is a sum of these:

f(x) = \sum_{n=1}^{\infty} b_n \sin\!\Big(\frac{n\pi x}{L}\Big), \qquad b_n = \frac{2}{L}\int_0^L f(x)\sin\!\Big(\frac{n\pi x}{L}\Big)\,dx.

The sines are orthogonal — like perpendicular axes — so the coefficient b_n is just the projection of f onto the nth mode, computed by that integral. This is what makes separation a complete method and not a lucky guess: the boundary conditions pick which modes \lambda_n are allowed, and the initial condition sets their weights b_n. Specify both, and the solution is determined.

Worked example: check a candidate solution by substitution

You don't always need the full separation-of-variables machinery. If someone simply hands you a candidate u(x, t), checking whether it solves a PDE is just calculus: compute the partial derivatives the equation asks for, on both sides, and see if they agree. Take u(x, t) = \sin(x)\, e^{-kt} and the heat equation u_t = k\, u_{xx} (the diffusion constant is called k here — the same role \alpha played above, just a different letter, as different books use different symbols).

Step 1 — the time derivative. Freeze x and differentiate with respect to t only:

u_t = \frac{\partial}{\partial t}\big[\sin(x)\, e^{-kt}\big] = -k\,\sin(x)\, e^{-kt}.

Step 2 — the first space derivative. Now freeze t and differentiate with respect to x:

u_x = \frac{\partial}{\partial x}\big[\sin(x)\, e^{-kt}\big] = \cos(x)\, e^{-kt}.

Step 3 — the second space derivative. Differentiate again with respect to x, still with t frozen:

u_{xx} = \frac{\partial}{\partial x}\big[\cos(x)\, e^{-kt}\big] = -\sin(x)\, e^{-kt}.

Step 4 — compare both sides of the equation.

k\, u_{xx} = k\big(-\sin(x)\, e^{-kt}\big) = -k\sin(x)\, e^{-kt} = u_t. \ \checkmark

The two sides agree for every x and every t, so u(x,t) = \sin(x)e^{-kt} genuinely solves the heat equation — it is exactly the n=1, L=\pi mode from the derivation above, caught here by a quick direct check instead of a full derivation.

But notice this check alone doesn't hand you the answer to any specific physical problem. u = 0 also solves the heat equation. So does u = \sin(2x)e^{-4kt}, and so does any sum of such modes with any weights at all. Compare this to an ODE: y' = y needed exactly one extra fact, the number y(0), to pick a single curve out of the family y = a_0 e^x. A PDE needs much more: a whole function's worth of information along the edges of the region for all time (the boundary conditions) together with a whole function's worth of information at the starting instant (the initial condition) before "the" solution becomes unique.

Two mix-ups trip up almost everyone meeting PDEs for the first time.

Run the heat equation forward on a sharp temperature spike — one hot point in an otherwise cold rod — and watch it. Within an instant the spike is already a smooth little bump; a moment later it is a gentle rounded hill, spreading ever wider and flatter. Every corner, every sharp edge, gets sanded away almost immediately. That smoothing only runs one way: play the film backwards and it looks impossible, because it is — you cannot run heat diffusion in reverse and recover the spike. This is one reason physicists call heat flow irreversible, and it is baked directly into the mathematics of u_t = \alpha u_{xx}.

The wave equation could not behave more differently. Pluck a guitar string into a sharp triangular kink and let go: the kink travels cleanly down the string, reflects off the far end, and comes back — still recognisably a kink, not smeared into a blur. Because u_{tt} = c^2 u_{xx} involves an even number of time derivatives, running the film backwards looks exactly as valid as running it forwards: the wave equation is time-reversible. Two equations that look almost identical on paper describe two utterly different kinds of physics — one destroys information about the past, the other keeps every bit of it.

Watch a heat mode decay

A single mode of the heated rod, u(x, t) = e^{-\alpha\lambda t}\sin(kx) with k = n\pi/L. Step time forward and the sine wave keeps its shape but its amplitude melts toward zero; raise the mode number n and the same wave decays faster (the rate \alpha\lambda \propto n^2), so fine wiggles disappear first.

See it explained