Initial, boundary and terminal conditions

Why a PDE needs extra conditions

Lesson 6.3.1 found many solutions of each equation. The heat equation $u_t = u_{xx}$ is solved by $e^{-t}\sin x$, by $e^{-4t}\sin 2x$, by $x^2 + 2t$ and by any sum of these. The transport equation $u_t + 2u_x = 0$ is solved by $f(x - 2t)$ for every function $f$.

An ODE has the same problem on a smaller scale. In lesson 6.1.1, $\frac{dy}{dt} = -2y$ has the solutions $Ce^{-2t}$, and one number, the starting value $y(0)$, picks out one of them. A second-order ODE needs two numbers (lesson 6.2.2).

A PDE has a whole function's worth of freedom, like the $f$ in $f(x - 2t)$. So its extra conditions are functions too, each given along an edge of the region. The region is the set of points $(x, t)$ where the equation must hold.

The initial condition

An initial condition gives the whole profile at the start: $u(x, 0) = f(x)$ for every $x$ in the region. It is the PDE version of $y(0)$.

On the whole line, an initial condition is enough for both of lesson 6.3.1's time-dependent equations. For the transport equation, $u(x, 0) = f(x)$ fixes the shape, so the solution is $f(x - ct)$. For the heat equation it fixes the solution too, among solutions that stay bounded (stated here without proof).

Example. Start the heat equation from $u(x, 0) = \sin 2x$. Lesson 6.3.1 found that $e^{-4t}\sin 2x$ is a solution, and at $t = 0$ it equals $\sin 2x$. So it is the solution: at $(\frac{\pi}{4}, 0.25)$ it is $e^{-1} = 0.3679$.

Boundary conditions

On a finite stretch $0 \le x \le L$, such as a rod of length $L$, heat can flow in or out through the ends. So the problem must also say what happens there, at every time. These are boundary conditions: $u(0, t) = a(t)$ and $u(L, t) = b(t)$, two given functions of time.

A boundary condition can give a slope instead of a value. An insulated end, through which no heat flows, has $u_x = 0$ there.

Example 1. On $0 \le x \le \pi$, start from $u(x, 0) = \sin x$ and hold both ends at 0. Then $u = e^{-t}\sin x$ fits all three conditions: it is 0 at $x = 0$ and $x = \pi$ for every $t$, and $\sin x$ at $t = 0$.

Example 2. Now hold the ends of a rod $0 \le x \le 1$ at 20 and 80. Start it at $u(x, 0) = 20 + 60x + 10\sin \pi x$: a straight line with a bump. The solution is

The line solves the heat equation, since its $u_t$ and $u_{xx}$ are both 0. The bump is a solution of lesson 6.3.1's kind, $e^{-k^2t}\sin kx$ with $k = \pi$, and it is 0 at both ends. The sum solves the equation (lesson 6.3.1) and fits every condition.

The bump dies away and the rod settles on the straight line from 20 to 80, its steady state. A steady state does not change, so $u_t = 0$ and the equation becomes $u_{xx} = 0$. That ODE's solutions are straight lines, fixed by the two end values as in lesson 6.2.2.

Which ends need a condition

The heat equation is second order in $x$, like the ODE $u_{xx} = 0$, so it needs a condition at each end. The transport equation is first order in $x$, and it needs a condition at one end only: the end where the shape comes in.

Take $u_t + u_x = 0$ on $x \ge 0$, a shape moving right at speed 1. Start with an empty stream, $u(x, 0) = 0$, and feed in $u(0, t) = t$ at the left end. Whatever enters at time $s$ reaches $x$ at time $s + x$, so

At $t = 2$: $u(0.5, 2) = 1.5$, $u(1.5, 2) = 0.5$ and $u(3, 2) = 0$. Feeding in $1 - e^{-t}$ instead gives 0.7769, 0.3935 and 0: a different solution with the same initial condition. So the condition at the inflow end is needed.

At the far end of a stretch $0 \le x \le L$, the values arriving are already fixed by what came in. A second condition there would usually contradict them, and then there is no solution at all.

Conditions that disagree at a corner

The initial and boundary conditions meet at the corners of the region, such as $(0, 0)$. If they disagree there, the solution inherits the clash.

Feed $u(0, t) = 1$ into the empty stream above, where $u(x, 0) = 0$. The transport equation carries a jump from 0 to 1 along the line $x = t$. The heat equation instead smooths such a clash at once (module 6.4).

Terminal conditions: solving backwards

Some problems fix the end instead of the start. A terminal condition gives the profile at a final time $T$: $u(x, T) = g(x)$. The task is then to find $u$ at earlier times, down to $t = 0$, by solving backwards.

The natural equation for this is the heat equation with the sign of $u_t$ flipped: $u_t + u_{xx} = 0$, that is, $u_t = -u_{xx}$. Run backwards from $T$, it spreads and smooths just as the heat equation does forwards.

Example. On $0 \le x \le \pi$ with both ends at 0, take $T = 1$ and $u(x, 1) = \sin x$. The solution is $u = e^{-(1 - t)}\sin x$: it equals $\sin x$ at $t = 1$, and $u_t = e^{-(1 - t)}\sin x = -u_{xx}$.

Read the table from left to right. The profile shrinks as you go back from the final time, as the heat equation's does going forward.

Time to go

The link is exact. Measure time backwards from the end, with $\tau = T - t$, the time to go. By the chain rule (lesson 2.3.3), $u_t = -u_\tau$, since $\tau$ falls by 1 when $t$ rises by 1.

So $u_t + u_{xx} = 0$ becomes $u_\tau = u_{xx}$: the heat equation in $\tau$. The terminal condition at $t = T$ is its initial condition at $\tau = 0$. In the example, $u = e^{-\tau}\sin x$, the solution of Example 1 with $\tau$ for $t$.

Case study 6.3.4 meets an equation of this kind: an option's value, fixed by its payoff at expiry.

Well-posed problems

A problem made of an equation, a region and its conditions is well-posed when three things hold. A solution exists; there is only one; and a small change in the conditions changes the solution only a little. The last matters because real data are measured, and measurements are never exact.

The wrong direction

The heat equation run backwards fails the third test. Suppose $u_t = u_{xx}$ and the profile at $t = 1$ is $\sin x$, or $\sin x + 0.001\sin 10x$: two profiles that differ by at most 0.001.

Going back, the $\sin x$ part grows by $e^{1 - t}$, only 1.1052 by $t = 0.9$. But the ripple $0.001\sin 10x$ came from $0.001e^{100(1 - t)}\sin 10x$, since sharp wiggles die fast going forward. So going back, it explodes:

An invisible change in the data changes the answer by 22 a tenth of a unit earlier. The same ripple in a terminal condition for $u_t + u_{xx} = 0$ shrinks instead: that problem is well-posed.

Common mistakes

1. The equation alone does not fix the solution. Without conditions, the heat equation has $e^{-k^2t}\sin kx$ for every $k$, and more.
2. Give the condition at the end the equation runs from. Initial for $u_t = u_{xx}$, terminal for $u_t = -u_{xx}$. The wrong way round is not well-posed.
3. Count the ends. The heat equation needs a condition at both ends, the transport equation only at the inflow end. A condition at the outflow end too usually leaves no solution.
4. Keep the sign in $u_t = -u_\tau$. Time to go falls as time rises.
5. A condition holds along a whole edge. $u(0, t) = 0$ means for every $t$, not only at $t = 0$.
6. Check the corners. If the initial and boundary values disagree there, the solution has a jump or a corner.