75042 Partial differential equations

Problem set #8 — suggested solutions

#8.1: Aaarggh! Misprint! You were supposed to show

∫_{∂Ω} u\frac{∂u}{∂ν} dS ≥ 0, ∫_{∂Ω} \frac{∂u}{∂ν} dS = 0

(these are boundary integrals) where u is harmonic and suitably smooth at the boundary. The second one is almost trivial. This is just the divergence theorem:

∫_{∂Ω} \frac{∂u}{∂ν} dS = ∫_{∂Ω} ν · Du dS = ∫_{Ω} div Du dx = ∫_{Ω} Δu dx = 0

The first one is just partial integration. I remind you that this is nothing but the divergence theorem applied to a product: Since div(uDu) = Du · Du + u div Du = |Du|^{2} + uΔu = |Du|^{2} (when u is harmonic) we find

∫_{∂Ω} u\frac{∂u}{∂ν} dS = ∫_{∂Ω} ν · uDu dS = ∫_{Ω} div(uDu) dx = ∫_{Ω} |Du|^{2} dx ≥ 0

and, by the way, the inequality must be strict unless u is constant (on each component of Ω).

#8.2: We use the linearity. If u, v solve the given problem then their difference w = u − v will solve

Δw = 0 in Ω, \frac{∂w}{∂ν} = 0 on ∂Ω.

Clearly, any constant function satisfies this; so the best we can ever hope to prove is that two solutions of the original problem differ by a constant. And we can indeed prove this, for if w is not constant then by the final remark in the solution of #8.1 above, we have

∫_{∂Ω} w \frac{∂w}{∂ν} dS > 0

which is clearly impossible since ∂w/∂ν = 0 on ∂Ω. (One sometimes comes by mixed boundary value problems, in which u is given on part of the boundary and ∂w/∂ν on the rest. Clearly the same proof works there, and in this case we do get uniqueness, provided only that u is given on a nonempty subset of the boundary.)

The compatibility condition ∫_{∂Ω} g dS = 0 is necessary for the existence of a solution because of the second formula of #8.1.

#8.3: The problem cannot have more than one solution. For if u and v solve the problem, let w = u − v. Then w = 0 on ∂Ω. If w is not identically zero, then either w > 0 somewhere or w < 0 somewhere (or both). In the former case, w has a positive maximum at some interior point of Ω. At this point x, then, Δw(x) = Δu(x)−Δv(x) = u(x)^{3} −v(x)^{3} > 0 because w(x) > 0 ⇔ u(x) > v(x). But elementary calculus tells us that we cannot have Δw(x) > 0 at a maximum point; no fancy PDE theory is required at all. (The argument works equally well if we replace u^{3} by f(u) where f is any strictly increasing function.)

#8.4: Clearly u(x, t, a, b) = a · x − tH(a) + b gives us u_{t} = H(a) and Du = a (the D operator here is assumed to only act on the x variables), so u is an integral of the Hamilton-Jacobi equation u_{t} − H(Du) = 0.

To show it is a complete integral, we must show the (n + 2) × (n + 1) matrix

[\begin{matrix}u_{a_{1}} & … & u_{a_{n}} & u_{b} \\ u_{x_{1}a_{1}} & … & u_{x_{1}a_{n}} & u_{x_{1}b} \\ ⋮ & ... & ⋮ & ⋮ \\ u_{x_{n}a_{1}} & … & u_{x_{n}a_{n}} & u_{x_{n}b} \\ u_{ta_{1}} & … & u_{ta_{n}} & u_{tb} \end{matrix}] = [\begin{matrix}x_{1} − tH_{p_{1}}(a) & … & x_{n} − tH_{p_{n}}(a) & 1 \\ 1 & … & 0 & 0 \\ ⋮ & ... & ⋮ & ⋮ \\ 0 & … & 1 & 0 \\ H_{p_{1}} & … & H_{p_{n}}(a) & 0 \end{matrix}]

75042 PDE/1996: Problem set #8 Page 2

has rank n + 1. (This matrix is transposed relative to what the lecture note does, by the way.) The block consisting of rows 2, …, n + 1 and columns 1, …, n is an identity matrix. Clearly the rows 1, …, n + 1 are linearly independent, so the matrix has the necessary rank.

#8.5: I rename (x_{1}, x_{2}) to (x, y) for convenience.

(a) We solve u_{x} + u_{y} = u^{2} with initial data u(x, 0) = g(x). The characteristic equations are

ẋ = 1, ẏ = 1, ż = z^{2}

and their solution is x = t, y = t + A, z = 1/(B − t). (There should have been an integration constant in the solution for x, but I threw it away because I am free to choose the starting point of a parametrization of the characteristics. Of course, I can only throw away one integration constant this way.) Eliminate t to describe the characteristic curves as

y = x + A, z = \frac{1}{B − x}.

A characteristic lying on the graph of the solution u must fit the initial data u(x, 0) = g(x), so that z = g(x) when (x, y, z) is on the characteristic with y = 0. This happens when x = −A, so we must have

\frac{1}{B − (−A)} = g(−A) and therefore B = −A + \frac{1}{g(−A)}.

Thus, passing back to the general form for the characteristic curve we get A = y − x, B = x − y + 1/g(x − y), and so

u(x, y) = z = \frac{1}{B − x} = \frac{g(x − y)}{1 − yg(x − y)}

which does indeed solve the problem.

(b) We solve xu_{y} − yu_{x} = u with initial data u(x, 0) = g(x). The characteristic equations are

ẋ = −y, ẏ = x, ż = z

with solutions x = A cos t, y = A sin t, z = Be^{t}. Here again we have thrown away a useless integration constant — to be added to t — in the expressions for x and y.

We note that the projected characteristics are circles in the (x, y) plane, and each of these cross the initial curve y = 0 twice. Since |z| grows as we travel along a characteristic with increasing t, the initial data cannot be satisfied everywhere. If, however, we restrict our attention to the right half plane x > 0 there is no trouble at all.

We eliminate t from our equations for the characteristic and get x^{2} + y^{2} = A^{2} and z = B exp ( arctan(y/x)). The initial data implies that we should have z = g(x) when y = 0. At this point on the characteristic x = A, so we get Thus B = g(A). In general, then, our solution becomes

u(x, y) = z = g (\sqrt{x^{2} + y^{2}}) exp ( arctan \frac{y}{x}).

#8.6: Let it wait until next week. This problem set was too big, perhaps. To compensate, I’ll make the next one too small.

Problem set #9 — for next week

#9.1: Evans problem 5 (p. 152).

#9.2: Evans problem 6 (p. 152).