PDE Types, Conditions, and Well-Posedness, differential equations lesson

The temperature along a rod depends on position and on time, so the unknown is a function $u(x, t)$ and its rates of change are partial derivatives. Writing down the right equation is only half of the problem. The same equation can come with initial data, boundary data or both, and only some choices give exactly one solution that changes little when the data change a little. This lesson classifies second-order equations in two variables by their highest derivatives, checks a model solution of each type, and separates a problem that merely has solutions from one that is well posed.

Key terms

Principal part

Definition Let $u = u(x, y)$ be an unknown function of two independent variables, with partial derivatives written $u_x = \partial u/\partial x$ , $u_{xx} = \partial^2 u/\partial x^2$ , $u_{xy} = \partial^2 u/\partial x\,\partial y$ and so on. Consider a second-order equation

\begin{aligned} &Au_{xx} + 2Bu_{xy} + Cu_{yy} \\[0.4em] &\quad = F(x, y, u, u_x, u_y), \end{aligned}

where the coefficients $A$ , $B$ and $C$ are functions of $(x, y)$ , not all zero at the point under study, and the right side contains no second derivatives. Its principal part is $Au_{xx} + 2Bu_{xy} + Cu_{yy}$ , the part made of the highest derivatives, and its discriminant is $B^2 - AC$ . The factor $2$ in front of $B$ is part of this convention. A book that writes the mixed term as $Bu_{xy}$ uses $B^2 - 4AC$ instead, with the same signs. The letters $x$ and $y$ stand for whichever two variables the problem has, such as a position $x$ and a time $t$ , where $u_t$ is the rate of change in time at a fixed position. The order of a partial differential equation is the order of its highest derivative, which has nothing to do with the number of independent variables.

In plain words Only the highest derivatives enter the classification. First derivatives, the unknown itself and any source term are left out, however much they matter for the behavior of solutions.

Example The heat equation $u_t = u_{xx}$ in the variables $(x, t)$ reads $u_{xx} - u_t = 0$ . Its principal part is $u_{xx}$ alone, with $A = 1$ and $B = C = 0$ , because no second derivative in $t$ appears. The term $u_t$ is lower order, even though it carries the evolution in time. For $u_{xx} + 4u_{xy} + u_{yy} = 0$ the mixed coefficient is $2B = 4$ , so $B = 2$ and the discriminant is $B^2 - AC = 4 - 1 = 3$ .

Elliptic, parabolic and hyperbolic equations

Definition At a point where $A$ , $B$ and $C$ are not all zero, the equation above is elliptic if $B^2 - AC < 0$ , parabolic if $B^2 - AC = 0$ and hyperbolic if $B^2 - AC > 0$ . It has a type on a region when it has that type at every point of the region, and the type can change from point to point when the coefficients vary. A smooth change of the independent variables with nonzero Jacobian $J$ multiplies the discriminant by $J^2$ , so it never changes the type. The model equations are

\begin{aligned} u_{xx} + u_{yy} &= 0 \quad \text{(Laplace)}, \\[0.4em] u_t &= k\,u_{xx} \quad \text{(heat)}, \\[0.4em] u_{tt} &= c^2 u_{xx} \quad \text{(wave)}, \end{aligned}

with constants $k > 0$ and $c > 0$ . Their discriminants are $-1$ , $0$ and $c^2$ , so they are elliptic, parabolic and hyperbolic. For the wave equation in the variables $(x, t)$ , the coefficient of $u_{xx}$ is $A = -c^2$ and that of $u_{tt}$ is $C = 1$ . This sign test is a local classification of a single second-order equation in two variables. Systems, higher-order equations and equations in more variables need broader definitions.

In plain words At any one point, a linear change of variables brings the principal part to two second derivatives of the same sign (elliptic), a single second derivative (parabolic), or two of opposite signs (hyperbolic). The model elliptic equation describes steady states fixed by data on the whole boundary, the model parabolic equation describes diffusion running forward from an initial profile, and the model hyperbolic equation describes waves, which need an initial position and an initial velocity. The type suggests which data fit, but by itself it proves nothing about a particular problem.

Example Three model solutions on $0 \le x \le \pi$ show the three behaviors, each with its own data. The heat profile $u = e^{-t}\sin x$ has $u_t = -e^{-t}\sin x = u_{xx}$ , so it solves $u_t = u_{xx}$ for $t \ge 0$ , with initial profile $\sin x$ and value $0$ at both ends. The wave $u = \cos t\sin x$ has $u_{tt} = -\cos t\sin x = u_{xx}$ , so it solves $u_{tt} = u_{xx}$ , with initial displacement $\sin x$ , initial velocity $u_t(x, 0) = 0$ and fixed ends. The function $u = e^{-y}\sin x$ has $u_{xx} = -e^{-y}\sin x$ and $u_{yy} = e^{-y}\sin x$ , so it solves Laplace’s equation on the rectangle $[0, \pi] \times [0, 1]$ , with value $0$ on the sides $x = 0$ and $x = \pi$ , value $\sin x$ on the bottom and value $e^{-1}\sin x$ on the top. The first decays, the second oscillates, and the third involves no time at all. A variable coefficient can change the type. For $u_{xx} + y\,u_{yy} = 0$ the discriminant is $-y$ , so the equation is elliptic for $y > 0$ , hyperbolic for $y < 0$ and parabolic on the line $y = 0$ , where the $u_{yy}$ term drops out.

Compatibility of data

Definition In an initial-boundary value problem the initial data and the boundary data meet at the corners of the domain. Consider the heat equation $u_t = k\,u_{xx}$ on $0 < x < L$ , $t > 0$ , with $u(0, t) = g_0(t)$ , $u(L, t) = g_1(t)$ and $u(x, 0) = \varphi(x)$ . The data are compatible at the corners if

\begin{gathered} \varphi(0) = g_0(0), \\[0.4em] \varphi(L) = g_1(0). \end{gathered}

A solution that is continuous on the closed set $0 \le x \le L$ , $t \ge 0$ , corners included, can exist only if these conditions hold. A solution whose derivatives $u_t$ and $u_{xx}$ are also continuous up to the corners needs the first-order conditions $g_0'(0) = k\,\varphi''(0)$ and $g_1'(0) = k\,\varphi''(L)$ as well, because at a corner the equation ties the time derivative of the boundary data to the second derivative of the initial data. Data that fail these conditions can still lead to a useful solution that is smooth for $t > 0$ and approaches the data in a weaker sense, but that solution does not have the stated continuity at the corners.

In plain words At a corner the starting profile and the boundary rule both claim to give the value of $u$ . If they disagree, no continuous function can satisfy both.

Example Take $k = 1$ and $L = \pi$ . The initial profile $\varphi(x) = \sin x$ with zero boundary values is compatible, since $\sin 0 = \sin \pi = 0$ , and it also meets the first-order conditions, since $\varphi''(x) = -\sin x$ vanishes at both ends. The constant profile $\varphi(x) = 1$ with zero boundary values is not compatible, since $\varphi(0) = 1 \ne 0$ . Its half-range sine series still produces the solution

u(x, t) = \frac{4}{\pi}\sum_{n\ \text{odd}}\frac{e^{-n^2 t}\sin nx}{n},

which satisfies $u_t = u_{xx}$ for $t > 0$ and $u(0, t) = u(\pi, t) = 0$ , and $u(x, t) \to 1$ as $t \to 0^+$ for each $x$ in $(0, \pi)$ . Arbitrarily close to the corner $(0, 0)$ it takes values near $0$ and values near $1$ , so it cannot be continuous there.

Well-posedness

Definition A problem, made of an equation, a domain and side conditions, is well posed in the sense of Hadamard, for a stated class of data and a stated class of solutions with stated norms, if a solution exists for every admissible datum, the solution is unique, and the solution depends continuously on the data. A problem that fails any of the three conditions is ill posed. The classes and the norms are part of the statement, so the same equation can be well posed with one set of side conditions and ill posed with another. Finding one explicit solution settles existence for that single datum and says nothing about uniqueness or continuous dependence.

In plain words A good model problem has an answer, has only one, and gives nearly the same answer when the measured data are slightly off. Each of these can fail on its own.

Example The heat equation $u_t = u_{xx}$ on $0 < x < \pi$ with zero boundary values and a continuous initial profile $\varphi$ with $\varphi(0) = \varphi(\pi) = 0$ is well posed in the maximum norm. A solution exists, the maximum principle gives uniqueness, and the next section gives the continuous dependence. Missing data destroy uniqueness. For the wave equation $u_{tt} = u_{xx}$ with fixed ends at $x = 0$ and $x = \pi$ and only the displacement $u(x, 0) = \sin x$ prescribed, both $\cos t\sin x$ and $(\cos t + \sin t)\sin x$ are solutions, and so is $\cos t\sin x + a\sin t\sin x$ for every constant $a$ . Prescribing the initial velocity $u_t(x, 0) = 0$ as well removes this freedom.

Continuous dependence on data

Definition Fix a norm for the data and a norm for the solutions on a time interval $[0, T]$ or on a region. The solution depends continuously on the data if for every $\varepsilon > 0$ there is a $\delta > 0$ such that any two admissible data closer than $\delta$ give solutions closer than $\varepsilon$ . For a linear problem with unique solutions, this is the same as a bound $\lVert u \rVert \le C\,\lVert \text{data} \rVert$ with one constant $C$ for all data. For the heat equation $u_t = k\,u_{xx}$ on $0 < x < L$ with zero boundary values, the maximum principle gives such a bound with $C = 1$ . Write $\lVert w \rVert_\infty$ for the largest value of $\lvert w(x) \rvert$ on $[0, L]$ . If $u$ and $v$ are solutions continuous on $0 \le x \le L$ , $t \ge 0$ , with initial profiles $\varphi$ and $\psi$ , then

\begin{aligned} &\lVert u(\cdot, t) - v(\cdot, t) \rVert_\infty \\[0.4em] &\quad \le \lVert \varphi - \psi \rVert_\infty \end{aligned}

for every $t \ge 0$ .

In plain words Small errors in the data cause small errors in the solution, measured in the chosen norm and over the chosen time interval. For forward diffusion with the same boundary values, the largest error never grows, because the flow of heat smooths out every bump.

Example Perturb the heat profile $\sin x$ on $[0, \pi]$ by $\varepsilon\sin x$ . The solution changes by $\delta u(x, t) = \varepsilon e^{-t}\sin x$ , so

\begin{aligned} \lVert \delta u(\cdot, t) \rVert_\infty &= \lvert \varepsilon \rvert e^{-t} \\[0.4em] &\le \lvert \varepsilon \rvert. \end{aligned}

A perturbation $\varepsilon\sin nx$ of higher frequency decays even faster, as $\varepsilon e^{-n^2 t}\sin nx$ . One mode only illustrates the bound, and the maximum principle proves it for all continuous data with zero boundary values.

Backward heat equation

Definition The backward heat equation is $u_t = -k\,u_{xx}$ with $k > 0$ . It is the heat equation with time reversed. If $v$ solves $v_t = k\,v_{xx}$ , then $u(x, t) = v(x, T - t)$ solves $u_t = -k\,u_{xx}$ . Its principal part $-k\,u_{xx}$ has discriminant $0$ , so it is parabolic, just like the forward equation. The initial-boundary value problem for it on $0 < x < \pi$ with zero boundary values is ill posed. Solutions do not depend continuously on the data, in the maximum norm or in the mean-square norm, and for many ordinary profiles no solution exists for any positive time. The profile $x(\pi - x)$ is one, since a solution up to time $T > 0$ would make it the value at time $T$ of a forward heat solution, which forces its sine coefficients to be at most a constant times $e^{-n^2 kT}$ , while its coefficients are $8/(\pi n^3)$ for odd $n$ .

In plain words Forward diffusion damps the mode $\sin nx$ by the factor $e^{-n^2 kt}$ , so fine detail fades fast. Running time backward multiplies it by $e^{n^2 kt}$ instead, so the tiniest fine-scale error in the data grows enormously. Recovering an earlier temperature profile from a measured later one is this backward problem, and solving it stably takes extra information or regularization.

Example With $k = 1$ , take $u_n(x, t) = e^{-n + n^2 t}\sin nx$ for a positive integer $n$ . Then $u_{n,t} = n^2 u_n$ and $u_{n,xx} = -n^2 u_n$ , so $u_{n,t} = -u_{n,xx}$ , and $u_n$ vanishes at $x = 0$ and $x = \pi$ . The sizes at the start and at time $1$ are

\begin{gathered} \lVert u_n(\cdot, 0) \rVert_\infty = e^{-n} \to 0, \\[0.4em] \lVert u_n(\cdot, 1) \rVert_\infty = e^{n^2 - n} \to \infty. \end{gathered}

For $n = 5$ the data have size $e^{-5} \approx 0.0067$ , and the solution at $t = 1$ has size $e^{20} \approx 4.9 \times 10^8$ . Data that tend to zero produce solutions that blow up at a fixed time, so no bound $\lVert u(\cdot, 1) \rVert_\infty \le C\,\lVert u(\cdot, 0) \rVert_\infty$ can hold. Every $u_n$ is an exact solution, which shows that solvability for particular data is not well-posedness.

Check yourself

Classify the equations $u_t = 4u_{xx} - u_x$ and $u_{tt} + u_t = 9u_{xx}$ , both for $u(x, t)$ , and $u_{xx} + u_{yy} = xy$ for $u(x, y)$ .

Show the answer

In the first equation the principal part is $4u_{xx}$ , so $A = 4$ and $B = C = 0$ , the discriminant is $0$ and the equation is parabolic. The terms $u_t$ and $u_x$ are lower order. In the second the principal part is $u_{tt} - 9u_{xx}$ , with $A = -9$ for $u_{xx}$ , $C = 1$ for $u_{tt}$ and $B = 0$ , so the discriminant is $9 > 0$ and the equation is hyperbolic. The damping term $u_t$ does not affect the type. The third is Poisson’s equation, with $A = C = 1$ and $B = 0$ , so the discriminant is $-1$ and it is elliptic. The source $xy$ does not affect the type.

Classify $u_{xx} + 2u_{xy} + 2u_{yy} = 0$ . A classmate reads $B = 2$ from the mixed term and concludes that the equation is hyperbolic. Where is the mistake?

Show the answer

The mixed coefficient is $2B = 2$ , so $B = 1$ , and the discriminant is $B^2 - AC = 1 - 2 = -1 < 0$ . The equation is elliptic at every point. Reading $B = 2$ gives $4 - 2 = 2$ , which has the wrong sign, because the convention puts the factor $2$ in front of $B$ . A change of variables confirms the answer. With $\xi = x$ , $\eta = y - x$ and $u(x, y) = w(\xi, \eta)$ , the chain rule gives

\begin{aligned} &u_{xx} + 2u_{xy} + 2u_{yy} \\[0.4em] &\quad = w_{\xi\xi} + w_{\eta\eta}, \end{aligned}

so the equation is Laplace’s equation in the new variables.

Consider $u_t = u_{xx}$ on $0 < x < \pi$ , $t > 0$ , with $u(0, t) = u(\pi, t) = 0$ and $u(x, 0) = x(\pi - x)$ . Are the data compatible at the corners? What does the maximum principle say about the solution?

Show the answer

The zero-order conditions hold, since $\varphi(x) = x(\pi - x)$ is continuous and vanishes at $x = 0$ and $x = \pi$ . So the well-posed problem described above applies, and its solution is continuous up to the corners and smooth for $t > 0$ . Its sine coefficients are $8/(\pi n^3)$ for odd $n$ and $0$ for even $n$ , so

u(x, t) = \frac{8}{\pi}\sum_{n\ \text{odd}}\frac{e^{-n^2 t}\sin nx}{n^3}.

The first-order conditions fail. The boundary data are constant, so $g_0'(0) = 0$ , while $\varphi''(x) = -2$ at both ends. Along $x = 0$ the time derivative $u_t$ is $0$ , while along $t = 0$ the second derivative $u_{xx}$ is $-2$ . If both were continuous at the corner, the equation $u_t = u_{xx}$ would fail there, so they cannot both be continuous up to the corners. The maximum principle gives $\lvert u(x, t) \rvert \le \max \varphi = \pi^2/4$ for every $t \ge 0$ . Since the problem is linear, it also shows that an error of at most $\delta$ in the initial profile, with the same zero boundary values, changes the solution by at most $\delta$ at every later time.

PDE Types, Conditions, and Well-Posedness

Key terms

Principal part

Elliptic, parabolic and hyperbolic equations

Compatibility of data

Well-posedness

Continuous dependence on data

Backward heat equation

Check yourself

How this lesson connects

Builds on

Used later in