The Lumenary · First-year graduate

Partial differential equations: modern theory

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Date: ______________________

You will learn to

Work with test functions, distributions and weak derivatives, and with the Sobolev spaces $H^1(\Omega)$ and $H^1_0(\Omega)$
Prove the Lax–Milgram theorem and use it with the Poincaré inequality to prove existence, uniqueness and stability of weak solutions of elliptic boundary value problems
Prove the weak maximum principle and use energy methods for the heat and wave equations
Explain elliptic regularity (weak solutions with smooth data are smooth) and Hadamard's notion of well-posedness, with his example of an ill-posed problem
Explain why the finite element method converges, via Galerkin orthogonality and Céa's lemma

1.
Let $\varphi$ be a test function with $\varphi(x) = 2 + 3x + 5x^2$ for $|x| < 1$ . Compute $\langle \delta', \varphi \rangle$ , where $\delta'$ is the distributional derivative of the Dirac delta.
2.
On $\Omega = (-1, 1)$ , which functions belong to $H^1(\Omega)$ ? Select all that apply.
- A. $u(x) = |x|$
- B. $u(x) = \operatorname{sign}(x)$
- C. $u(x) = |x|^{3/4}$
- D. $u(x) = |x|^{1/4}$
3.
On $V = H^1_0(0, 1)$ (norm $\|u\|_{H^1}^2 = \int u^2 + u'^2$ ), which bilinear forms are coercive? Select all that apply.
- A. $B(u, v) = \int_0^1 u'v'$
- B. $B(u, v) = \int_0^1 uv$
- C. $B(u, v) = \int_0^1 u'v' - 20\int_0^1 uv$
- D. $B(u, v) = \int_0^1 u'v' + \int_0^1 u'v$
4.
Hadamard's example $u_n(x, y) = \sin(nx)\sinh(ny)/n^2$ shows that the Cauchy problem $\Delta u = 0$ , $u(x, 0) = g$ , $\partial_y u(x, 0) = h$ is ill-posed. Which of Hadamard's conditions fails?
- A.Continuous dependence on the data.
- B.Existence: there is no solution for these data.
- C.Uniqueness: there are two different solutions with the same data.
- D.None: the problem is well-posed but the solutions are large.
5.
In the proof of Lax–Milgram, $A$ is defined by $B(u, v) = \langle Au, v \rangle$ . Which argument shows that the range of $A$ is closed?
- A.Coercivity gives $\|Au\| \ge c\|u\|$ ; so if $Au_k \to w$ then $(u_k)$ is Cauchy, converges by completeness to some $u$ , and $w = Au$ .
- B.Every linear subspace of a Hilbert space is closed.
- C. $A$ is bounded, and the range of a bounded operator is closed.
- D.By the open mapping theorem, since $A$ is surjective.
6.
Suppose hcoer : ∀ u, C * ‖u‖ * ‖u‖ ≤ B u u and hu : ∀ v, B u v = ℓ v. Which conclusion is the stability estimate of Lax–Milgram (and is provable from these hypotheses)?
- A.C * ‖u‖ ≤ ‖ℓ‖
- B.‖u‖ ≤ C * ‖ℓ‖
- C.‖ℓ‖ ≤ C * ‖u‖
- D.C * ‖u‖ * ‖u‖ = ℓ u
7.
For a smooth solution of $u_t = u_{xx}$ on $(0, 1)$ with $u(0, t) = u(1, t) = 0$ , let $E(t) = \int_0^1 u^2\,dx$ . Using the sharp Poincaré inequality $\int_0^1 u_x^2 \ge \pi^2\int_0^1 u^2$ , which bound holds?
- A. $E(t) \le e^{-2\pi^2 t} E(0)$
- B. $E(t) \le e^{-\pi^2 t} E(0)$
- C. $E(t) = E(0)$ for all $t$
- D. $E(t) \le e^{2\pi^2 t} E(0)$
8.
Solve $-u'' = f$ on $(0, 1)$ , $u(0) = u(1) = 0$ , by finite differences: with $h = 1/N$ and $x_i = ih$ , the unknowns $u_1, \dots, u_{N-1}$ satisfy $\frac{-u_{i-1} + 2u_i - u_{i+1}}{h^2} = f(x_i)$ with $u_0 = u_N = 0$ . Write solve_poisson(f, n) returning the list $[u_0, u_1, \dots, u_N]$ , solving the tridiagonal system by forward elimination and back substitution (the Thomas algorithm).
Computational — Practice

Answer key — Partial differential equations: modern theory

1.
$\langle \delta', \varphi \rangle = -\langle \delta, \varphi' \rangle = -\varphi'(0) = -3$ . The sign comes from integration by parts: for a differentiable $u$ , $\int u' \varphi = -\int u \varphi'$ , and the distributional derivative is defined so that this identity always holds.
2.
$|x|^a \in H^1(-1, 1)$ iff $2(a - 1) > -1$ , i.e. $a > \frac12$ : so (a) ( $a = 1$ ) and (c) ( $a = \frac34$ ) are in, (d) ( $a = \frac14$ ) is out. (b) has a jump; its distributional derivative $2\delta$ is not in $L^1_{\mathrm{loc}}$ . The threshold $a > \frac12$ matches the embedding $H^1 \subseteq C^{0,1/2}$ in one dimension.
3.
(a) and (d) are coercive. In (d) the convection term vanishes on the diagonal because $\int_0^1 (u^2/2)' = 0$ for $u$ vanishing at both ends, so $B(u, u) = \int u'^2 \ge \frac12\|u\|_{H^1}^2$ (Poincaré with constant $1$ ). (b) fails because $L^2$ cannot control derivatives; (c) fails because $\int u'^2 \ge \pi^2 \int u^2$ is sharp (equality for $\sin \pi x$ ), and $\pi^2 < 20$ .
4.
The data tend to $0$ uniformly ( $|\sin(nx)/n| \le 1/n$ ), but $u_n(\pi/(2n), 1) = \sinh(n)/n^2 \to \infty$ (formalised as hadamard_blowup). Since zero data give $u = 0$ , the map from data to solutions is not continuous: the third condition fails. Elliptic equations call for boundary value problems, not Cauchy problems.
5.
From $c\|u\|^2 \le B(u, u) = \langle Au, u \rangle \le \|Au\|\|u\|$ we get $\|Au\| \ge c\|u\|$ . If $Au_k \to w$ , then $\|u_k - u_m\| \le c^{-1}\|Au_k - Au_m\|$ , so $(u_k)$ is Cauchy and converges to some $u$ ; by continuity $Au = w$ . Hence $\operatorname{ran} A$ is closed. With density this gives surjectivity, and with injectivity (also from $\|Au\| \ge c\|u\|$ ) bijectivity.
6.
Option (a). $C\|u\|^2 \le B(u, u) = \ell(u) \le |\ell(u)| \le \|\ell\|\|u\|$ ; if $u \ne 0$ divide by $\|u\|$ , and if $u = 0$ the claim is $0 \le \|\ell\|$ . For $-\Delta u = f$ this is the energy estimate $\|u\|_{H^1_0} \le C^{-1}\|f\|_{H^{-1}}$ : continuous dependence on the data.
7.
$E'(t) = 2\int_0^1 u u_t = 2\int_0^1 u u_{xx} = -2\int_0^1 u_x^2 \le -2\pi^2 E(t)$ , so $(e^{2\pi^2 t}E)' \le 0$ and $E(t) \le e^{-2\pi^2 t}E(0)$ . The mode $e^{-\pi^2 t}\sin(\pi x)$ attains equality, so the rate is sharp. Applied to the difference of two solutions with the same data, it proves uniqueness.
8.
```
def solve_poisson(f, n):
    h = 1.0 / n
    off = -1.0 / h ** 2
    diag = [2.0 / h ** 2] * (n - 1)
    rhs = [f((i + 1) * h) for i in range(n - 1)]
    for i in range(1, n - 1):
        m = off / diag[i - 1]
        diag[i] -= m * off
        rhs[i] -= m * rhs[i - 1]
    u = [0.0] * (n - 1)
    u[-1] = rhs[-1] / diag[-1]
    for i in range(n - 3, -1, -1):
        u[i] = (rhs[i] - off * u[i + 1]) / diag[i]
    return [0.0] + u + [0.0]
```
The central difference quotient is exact for quadratics, so $f = 2$ reproduces $x(1 - x)$ at the nodes. For $f = -6x$ the exact solution is $x^3 - x$ ; the error is $O(h^2)$ (here actually zero at the nodes up to rounding, a special feature of this one-dimensional problem). The matrix is an M-matrix whose inverse has positive entries, so $f \ge 0$ forces $u \ge 0$ : a discrete maximum principle.