1.2 Classification of the PDEs
1.2.1 Standard classification of linear PDEs 9
To start in our task of classification, assume for simplicity a 2-dimensional domain d = 2, and a constant coefficient second order PDE:
auxx+ buxy + cuyy+ e = 0. (1.10)
We look for a curve γ : R2 7→ R that is sufficiently regular and such that when we write the PDE along this curve it turns into and Ordinary Differential Equation (ODE). We write this curve in parametric form as γ(σ) (fig. 1.1) as follows:
γ = x = x(σ) y = y(σ)
Writing the above equations on a local reference system, we obtain:
dux
Writing uxx from the previous system and substituting it in eq. (1.10), we have:
uxy
This equation is a re-definition of the PDE on the curve γ(σ), or, in other words, the equation is satisfied on γ. Now we can choose γ so that the first term in square brackets is zero, obtaining an equation for uxand uy where only ordinary derivatives appear:
a dy dx
2
− bdy
dx+ c = 0.
We note that dy/dx is the slope of γ, which can then be obtained by solving the ODE:
dy
dx = b±√
b2− 4ac
2a . (1.11)
The solution of this ODE yields families of curves, that are called characteristic curves. Different families arise depending on the sign of the discriminant ∆ = b2 − 4ac. We then call the equations depending on this sign, obtaining the following classification:
• b2− 4ac < 0: two complex solutions: the equation is “elliptic”;
• b2− 4ac = 0: one real solution: the equation is “parabolic”;
• b2− 4ac > 0: two real solutions: the equation is “hyperbolic”.
Examples
Laplace equation
∆u = ∂2u
∂x2 +∂2u
∂y2 = 0
a = c = 1 b = 07→ b2 − 4ac < 0 is an elliptic equation;
Wave equation
∂2u
∂t2 −∂2u
∂x2 = 0 (1.12)
a = 1 b = 0 c =−1 7→ b2− 4ac > 0 is a hyperbolic equation;
diffusion or heat equation
∂u
∂t −∂2u
∂x2 = 0
a =−1 b = c = 0 7→ b2− 4ac = 0 is a parabolic equation.
1.2.2 Simple examples and solutions
We show in this paragraph some simple but clarifying examples of PDEs and their exact analytical solution. From these solutions we will extrapolate some typical characteristics of the solutions of PDEs.
Example 1.3. Find u : [0, 1]7→ R such that:
−u00= 0 x∈ [0, 1]
u(0) = 1;
u(1) = 0.
This is a elliptic equation. In this simple case the solution is obtained directly by integration between x = 0 and x = 1. We have:
u(x) = 1− x.
Example 1.4. Find u : [0, 1]7→ R such that:
−(a(x)u0)0 = 0 x∈ [0, 1] (1.13)
u(0) = 1;
u(1) = 0;
where the diffusion coefficient a(x) assumes the values:
a(x) =
(a1 if 0≤ x < 0.5 a2 if 0.5 < x≤ 1
Since a(x) > 0 for each x ∈ [0, 1] is is an elliptic equation. In this case the solution can be obtained by first subdividing the domain interval in two halves and integrating the equation in each subinterval:
u(x) =
(u1(x) = c11x + c12 x∈ [0, 0.5) u2(x) = c21x + c22 x∈ (0.5, 1].
We can see that the solutions are defined in terms of four constants. We need thus four equations. Two are given by the boundary conditions, but the other two are still missing. One natural condition is the request that u(x) be continuous (at least C0([0, 1])) in the domain [0, 1]. The second condition can be determined by looking at the left-hand-side of eq. (1.13) and looking for existence requirement of this term. Before we discuss this requirement we note that we can define the “flux” of u(x) as q(x) =−a(x)u0(x). Hence, the requirement for the existence of the left-hand-side (as long as we do not use the product rule for the derivative of the flux) is that q(x) must be continuous for all x∈ [0, 1]
x u(x)
FIGURE 1.2: Solution of example 1.4 for a1 = 1 and a2 = 10.
(again the requirement here is q(x)∈ C0([0, 1]). This observation suggests the sought condition, that yield the following system of equations for the constants ci:
u1(0) = 0 u2(1) = 0;
u1(0.5) = u2(0.5) q1(0.5) = q2(0.5),
−a1(0.5)u01(0.5) =−a2(0.5)u02(0.5).
We note that the last condition physically means that the flux of the quantity u(x) that exits from the subdomain on the left of x = 0.5 enters the subdomain on the right of x = 0.5. It is a conservation statement. Solving the system, the solution becomes:
u(x) =
(1− a12a+a22x x∈ [0, 0.5]
2a1
a1+a2 − a12a+a12x x∈ [0.5, 1], shown in fig. 1.2 in the case a1 = 1 and a2 = 10.
Remark 1.5. The previous example shows that the differential equation with discontinuous coeffi-cients has a solution that is continuous but not differentiable: the gradient is discontinuous. On the other hand the flux is continuous, and thus more regular. We will use this fact in to properly define our numerical solution. This property, that can be also shown theoretically, is very important in ap-plications, and characterizes “conservation laws”. In other words, the partial differential eq. (1.13) represents the balance of the quantity u(x). This quantity can be thought of as mass, then the equation is a mass-balance equation, a temperature, in which case the equation is an energy conservation equa-tion, a fluid velocity, and then the equation is a force balance equation (first Newton law), etcetera.
The determination of the conservation properties of numerical discretization schemes is an active and important field of research in the case of highly variable diffusion coefficients.
We would like to remark that in the case of jumps in the diffusion coefficient we cannot use the product rule to expand the left-hand-side of eq. (1.13). In fact we cannot write the following:
−a(x)u00(x)− a0(x)u0(x) = 0
because both u00(x) and a0(x) do not exist for x = 0.5. However, the solution u(x) exists and is intuitively sound, i.e., without any singularity, although it does not possess a second derivative. Hence, the equation must be written exclusively as in eq. (1.13). In general, using the chain rule for derivative is numerically counterproductive even if the regularity of the mathematical objects allows it.
Example 1.6 (Poisson equation). Find u : [0, 1]7→ R such that:
−u00= f (x) x∈ [0, 1] (1.14)
u(0) = u(1) = 0 (1.15)
with
f (x) =
(1 if x = 0.5, 0 otherwise..
This is an elliptic equation. The solution if this problem can be found by means of Green’s functions and is given by:
u(x) = (1
4(1− x) if 0 ≤ x ≤ 0.5,
1
4(x− 1) if 0.5≤ x ≤ 1. (1.16)
This solution is continuous but it has a piecewise constant first derivative with a jump in x = 0.5.
Hence the second derivative u00(x) does not exists in the midpoint. This seems a contradiction as in this case the left-hand-side of eq. (1.14) does not exists for all x ∈ [0, 1]. However, the solution u(x) given in eq. (1.16) in terms of the integral of the Green’s function is mathematically sound.
Thus we need to define a more “forgiving” formulation, whose solution can have discontinuous first derivatives. This is the role of the so called “weak” formulation to be seen in the next sections.
Example 1.7. Transport equation.
Given a vector field of constant velocity λ > 0, find the function u = u(x, t) such that:
ut+ λux = 0, (1.17a)
u(x, 0) = f (x). (1.17b)
FIGURE 1.3: Left panel: characteristic lines for eq. (1.17a) in the (x, t) plane. Right panel: graph of the solutionu(x, t) at t = 0 and t = t1 > 0 in the (u, x) plane. The solution is a wave with shape given byf (x) (a line in this case) that propagates towards the right with speed λ.
The characteristic curve is a line in the plane (x, t) given by:
x− λt = const = ξ.
Along this line the original eq. (1.17a) becomes:
du dt = ∂
∂tu(ξ + λt, t) = λux+ ut = 0.
Hence, the solution u is constant along a characteristic curve and this constant is determined by the initial conditions eq. (1.17b):
u(x, t) = f (ξ) = f (x− λt). (1.18)
At a fixed time t1 the solution is given by the rigid translation of the initial condition (f (x)) by a quantity λt1, as shown in fig. 1.3 (right panel).
Example 1.8. Advection (or convection) and diffusion equation (ADE).
Find the function u(x, t) : [0, T ]× R 7→ R such that:
The solution is given by [3]:
u(x, t) = 1 where the function erfc is the complementary error function.