18.102: NOTES ON DIRICHLET PROBLEM ON AN INTERVALRICHARD MELROSEAbstract. Here are my notes for the lecture on November, about the Dirichletproblem for d2/dx2on an interval.Compactness of integration.Theorem 1. The integration operator on [a, b] :(1) L2([a, b]) 3 u 7−→ Au(x) =Zxau(s)ds ∈ L2([a, b])is a compact operator.Proof. We need to show that A maps bounded sets into precompact sets, and todo this it suffices to show that for any sequence {un} with kunk ≤ 1 in L2([a, b]),vn= Aunhas a convergent subsequence in L2([a, b]). In fact we may assume thatun* u is weakly convergent and then we will show that vn→ v = Au is stronglyconvergent. So, observe first that vn∈ C([a, b]). In fact(2) |Aw(x) − Aw(y)| = |Zyxw(s)ds| ≤ |x − y|12kwkL2.This shows that the sequence vn= Aunis (uniformly) equicontinuous (each vnisuniformly continuous since it is continuous on a compact set). Morever(3) vn(x) =Zbaχ[a,x]un(s)ds → v(x) for each x ∈ [a, b]since χ[a,x]∈ L2([a, b]) and un* u. In fact it follows that vn→ v uniformly on[a, b] by combining (3) and (2) since given  > 0 we may choose n so large that(4) |vn(x) − v(x)| ≤ /3 ∀ x = a + m(b − a)/N, m ∈ {0, . . . , N}, N < 2/4.Then(5) |v(x) − vn(x)| ≤ |v(x) − v(x0)| + |v(x0) − vn(x0| + |vn(x0) − vn(x)| < where x0is one of the points in (4).Now vn→ v uniformly imples vn→ v in L2([a, b]), proving the compactness ofA. Let me first apply this to a ‘trivial’ problem. Suppose we are interested in theDirichlet problem(6)d2u(x)dx2+ V (x)u(x) = f(x) on [0, 1], u(0) = u(1) = 0.By the ‘trivial case’ I mean V ≡ 0. Then of course we can solve (6) by integration,assuming for instance that f is continuous (but this works for f ∈ L2([0, 1]). Thus,12 RICHARD MELROSEintegrating twice gives a solution of the differential equation(7) Bf(x) =Zx0Zt0f(s)dsdt.From the discussion above we see immediately that B : L2([0, 1]) −→ L2([0, 1]) iscompact. Clearly Bf (0) = 0 but on the other hand(8) Bf(1) =Z10Zt0f(s)dsdt =Z10(1 − s)f(s)ds = B1fneed not vanish. However there are solutions to the homogeneous equation, (6)with f ≡ 0, namely any linear function. Of course we don’t want to mess up thefact that Bf(0) = 0 so we should only add cx to this. Choosing c correctly, namely(9) Af(x) = Bf(x) − (B1f)xensures that u(x) = Af(x) satisfies (6) (at least if f is continuous) including theboundary conditions.Proposition 1. The operator A in (9) is a compact and self-adjoint as an operatoron L2([0, 1]).Proof. Since x is a fixed function and B1f is a constant, the extra term in (9) isfinite rank. Thus the compactness of A follows from that of B – in fact B is thecomposite of two compact operators.To see the self-adjointness of A we just need to compute its adjoint! First wecan change the order of integration to write(10) Bf(x) =Zx0(x − s)f(s)ds.Then, computing the adjoint of B, by again changing the order of integration,(11)hBf, φi =Z10Zx0(x − s)f(s)dsφ(x)dx =Z10Z1s(x − s)f(s)φ(x)dxds=Z10f(s)(B∗φ)(s)dxds = hf, B∗φi, where(B∗φ)(x) =Z1x(s − x)φ(s)ds =Z10(s − x)φ(s)ds +Zx0(x − s)φ(s)ds. Similarly,(12)h(B1f)x, φi =Z10Z10(1 − s)f(s)dsxφ(x)dx=Z10f(x)(1 − x)Z10sφ(s)dsdx = hf, (B1φ)xi +Z10f(x)Z10(s − x)φ(s)dsdx.Thus(13) A∗= (B −(B1f)x)∗= A.Thus A is indeed a compact self-adjoint operator. Let’s compute its eigenvalues,using the uniqueness of solutions to ODEs (should I prove this?)DIRICHLET 3First the non-zero eigenvalues. If Au = λu with λ 6= 0 then we can writeu = λ−1Au. Since Au is C1and vanishes at x = 0 and x = 1, so is and does u.Moverover if u is C1then Au is C2and hence so is u. In fact this argument showsthat u is C∞but we do not need to check that, since by differentiating we see that(14)d2udx2= λ−1u, u(0) = u(1) = 0.So, now we can use the uniqueness of solutions to this ODE. There is a uniquesolution of the Cauchy problem, with u(0) = 0 and u0(0) = 1. In fact this functionis sin(x/√−λ). So λ = −T2< 0 necessarily and then for sin(T ) = 0 we must haveT = πn, n ∈ N. Thus the general solution to Au − λu with λ 6= 0 is(15) u(x) = A sin(πnx), λ =−1π2n2, n ∈ N.Now, we have to think about the null space of A, Au = 0. Differentiation onceshows that(16)Zx0u(s)ds = 0 a.e.We already know that this implies that u = 0 in L2([0, 1]). Thus the null space istrivial.Now, applying the spectral theorem, we conclude that(17) φn(x) = sin(πnx), n ∈ N is a complete orthonormal basis of L2([0, 1]).Department of Mathematics, Massachusetts Institute of TechnologyE-mail address: