Week 9: Existence and interior regularity

Existence of weak solutions (continued)

We can actually obtain more information on the values of μ\mu for which the boundary-value problem of finding uH01,2(Ω)u\in H_{0}^{1,2}(\Omega) such that Lμu=fL_{\mu}u=f admits a unique solution by making use of an important property of compact linear mappings. We recall the following spectral theorem.

Theorem 9.1 (spectral theorem for compact operators). Suppose K:XXK:X\rightarrow X is a compact operator on a Banach space XX and let

σ(K)=C\{ξC:KξI is bijective}\sigma(K)=\mathbb{C}\backslash\{\xi\in\mathbb{C}:K-\xi I\ \textup{is bijective}\}

denote the spectrum of KK. We have the following properties:

Remark 9.2. Note that if λC\lambda \in\mathbb{C} is an eigenvalue of KK, then it is automatically in σ(K)\sigma(K).

Theorem 9.3 (third existence theorem). There exists an at most countable set ΣR\Sigma\subset\mathbb{R} such that whenever λΣ\lambda\notin\Sigma, there exists a unique weak solution uH01,2(Ω)u\in H_{0}^{1,2}(\Omega) to the boundary-value problem

Lu=λu+f on ΩuΩ0.(###)\begin{aligned}Lu&=\lambda u + f\ \textup{on }\Omega\tag{\#\#\#}\\ \left.u\right|_{\partial\Omega}&\equiv 0.\end{aligned}

If Σ\Sigma is infinite, then Σ={λk}k=1\Sigma=\{\lambda_{k}\}_{k=1}^{\infty} for a nondecreasing sequence λk\lambda_{k}\rightarrow{\infty}.

Proof. Let μ>max{C,0}\mu>\max\{C,0\} be as in the first existence theorem. By that theorem, the boundary value problem (###) possesses a unique solution if λμ-\lambda\geq\mu, i.e. if λ+μ0\lambda + \mu \leq 0, so suppose λ+μ>0\lambda+\mu>0. By the Fredholm alternative, this boundary-value problem has a unique weak solution if and only if λ\lambda is not an eigenvalue of LL, i.e.

Lv=λvv=0Lv=\lambda v \Rightarrow v=0

for vH01,2(Ω)v\in H_{0}^{1,2}(\Omega). Fixing μ>0\mu>0 satisfying the condition of the first existence theorem, this in turn is equivalent to the condition that

Lμv=(λ+μ)vv=0,L_{\mu}v = (\lambda + \mu)v\Rightarrow v=0,

or Kv=μμ+λvv=0Kv=\frac{\mu}{\mu+\lambda}v\Rightarrow v=0, i.e. μμ+λ\frac{\mu}{\mu+\lambda} is not an eigenvalue of KK. Since KK is compact, the spectral theorem tells us that the eigenvalues of KK form an at most countable set {νk}k=1\{\nu_{k}\}_{k=1}^{\infty} with 00 as the only possible cluster point; therefore, λ\lambda is not an eigenvalue of LL iff μμ+λ{νk}λ{μνkμ}\frac{\mu}{\mu+\lambda} \notin\{\nu_{k}\}\Leftrightarrow\lambda\notin\{\frac{\mu}{\nu_{k}}-\mu\}. The sequence {μνkμ}\{\frac{\mu}{\nu_{k}}-\mu\} tends to \infty if it possesses a cluster point so that we may reörder it to obtain a nondecreasing sequence {λk}\{\lambda_{k}\} with λk\lambda_{k}\rightarrow\infty.

Remark 9.4. The set Σ\Sigma of all (real) eigenvalues λ\lambda of LL is called the spectrum of LL.

Remark 9.5. The equation in (###) is sometimes referred to as Helmholtz's equation.

We may also establish the boundedness of the inverse of L+λIL+\lambda I (λΣ\lambda\notin\Sigma) as a mapping into L2(Ω)L^{2}(\Omega) (cf. remarks following first existence theorem).

Theorem 9.6. Let λΣ\lambda\notin\Sigma. There exists a constant CC depending on λ\lambda, Ω\Omega and LL such that whenever fL2(Ω)f\in L^{2}(\Omega) and uu solves (###),

u2Cf2.||u||_{2}\leq C||f||_{2}.

Proof. Suppose that this is not the case so that there exist sequences {uk}k=1H01,2(Ω)\{u_{k}\}_{k=1}^{\infty}\subset H_{0}^{1,2}(\Omega) and {fk}k=1L2(Ω)\{f_{k}\}_{k=1}^{\infty}\subset L^{2}(\Omega) in L2(Ω)L^{2}(\Omega) such that Luk=λuk+fkLu_{k}=\lambda u_{k} + f_{k} and uk2kf2||u_{k}||_{2}\geq k||f||_{2} for all kNk\in\mathbb{N}. By dividing through by uk2||u_{k}||_{2}, we may assume that uk2=1||u_{k}||_{2}=1 and fk21k||f_{k}||_{2}\leq \frac{1}{k} for all kk so that fk0f_{k}\rightarrow 0. Now, by the energy estimates and Hölder's inequality, we actually have that

uk1,22C(Ω,λ,L)(BLλI(uk,uk)+uk2)C(Ω,λ,L)(fk22+2uk22)||u_{k}||_{1,2}^{2}\leq C(\Omega,\lambda,L)\left(B_{L-\lambda I}(u_{k},u_{k}) + ||u_{k}||_{2} \right)\leq C(\Omega,\lambda,L)\left(||f_{k}||_{2}^{2} +2||u_{k}||_{2}^{2} \right)

so that {uk}\{u_{k}\} is bounded in H01,2(Ω)H_{0}^{1,2}(\Omega). By the Banach-Alaoǧlu theorem and Rellich-Kondrachov, we may pass to a subsequence, again denoted {uk}\{u_{k}\}, such that ukuu_{k}\rightarrow u in L2(Ω)L^{2}(\Omega) for some uH01,2(Ω)u\in H_{0}^{1,2}(\Omega), and ukuu_{k}\rightharpoondown u in H01,2(Ω)H_{0}^{1,2}(\Omega). In particular, u2=limkuk2=1||u||_{2}=\lim_{k\rightarrow\infty} ||u_{k}||_{2}=1 so that u0u\neq 0 and for all vH01,2(Ω)v\in H_{0}^{1,2}(\Omega),

BLλI(u,v)=limkBLλI(uk,v)=limk<fk,v>=0B_{L-\lambda I}(u,v)=\lim_{k\rightarrow\infty}B_{L-\lambda I}(u_{k},v)=\lim_{k\rightarrow\infty}\left<f_{k},v\right>=0

so that Lu=λuLu = \lambda u on Ω\Omega. But this implies that λΣ\lambda\in\Sigma!

Interior regularity

We now turn our attention to the regularity of solutions to the equation Lu=fLu=f under the assumptions stated in Week 8. We first show that if uW1,2(Ω)u\in W^{1,2}(\Omega) solves this equation and aijC1(Ω)a_{ij}\in C^{1}(\Omega), then uu is actually twice weakly differentiable.

Theorem 9.7. Suppose uW1,2(Ω)u\in W^{1,2}(\Omega) solves the equation

Lu=f on ΩLu=f\ \textup{on}\ \Omega

weakly, i.e. for all vH01,2(Ω)v\in H_{0}^{1,2}(\Omega)

BL(u,v)=ΩfvB_{L}(u,v)=\int_{\Omega}f\cdot v

and LL satisfies the hypotheses given in Week 8. Suppose furthermore that for all i,j{1,,n}i,j\in\{1,\dots,n\}, aijC1(Ω)a_{ij}\in C^{1}(\Omega). Then uWloc2,2(Ω)u\in \wloc^{2,2}(\Omega) and whenever UΩU\Subset\Omega, the inequality

u2,2,UC(f2,Ω+u2,Ω)||u||_{2,2,U}\leq C\left(||f||_{2,\Omega} + ||u||_{2,\Omega}\right)

holds, where C>0C>0 is a constant depending only on UU, Ω\Omega and LL.

Proof. First note that we may rewrite the weak form of Lu=fLu=f as

Ωaijiujv=Ωf~v,(*)\int_{\Omega}a_{ij}\partial_{i}u\cdot\partial_{j}v = \int_{\Omega}\tilde{f}\cdot v,\tag{*}

where f~:=fbiiucuL2(Ω)\tilde{f}:=f-b^{i}\partial_{i}u-cu\in L^{2}(\Omega). We now choose an appropriate vv and exploit ellipticity to 'differentiate' the equation Lu=fLu=f. Fix open sets UVΩU\Subset V\Subset \Omega and a χC(Rn)\chi\in C^{\infty}(\mathbb{R}^{n}) such that 0χ10\leq \chi\leq 1, χU1\left.\chi\right|_{U}\equiv 1 and supp χV\supp\chi\subset V. For h0h\neq 0 sufficiently small and k{1,,n}k\in\{1,\dots,n\} fixed, let

v=δkh(χ2δkhu).v=-\delta_{k}^{-h}(\chi^{2}\cdot \delta^{h}_{k}u).

We now estimate both sides of (*) appropriately, making free use of the properties of the 'difference quotient operator' δkh\delta^{h}_{k} (product rule and integration by parts specifically). First note that

LHS=aijiuδkh(j(χ2δkhu))=δkh(aijiu)j(χ2δkhu)=aijiδkhujδkhuχ2+2δkhaijiuδkhuχjχ+2aijiδkhuχjχδkhu+δkhaijiujδkhuχ2.\begin{aligned} LHS &=-\int a_{ij}\partial_{i}u\cdot \delta_{k}^{-h}(\partial_{j}(\chi^{2}\cdot \delta_{k}^{h}u))\\ &=\int \delta^{h}_{k}(a_{ij}\cdot \partial_{i}u)\cdot \partial_{j}(\chi^{2}\cdot \delta^{h}_{k}u) \\ &=\int a_{ij}\cdot \partial_{i}\delta_{k}^{h}u \cdot \partial_{j}\delta_{k}^{h}u\cdot \chi^{2} + 2\int\delta_{k}^{h}a_{ij}\cdot \partial_{i}u\cdot \delta_{k}^{h}u\cdot \chi\cdot \partial_{j}\chi + 2\int a_{ij}\cdot \partial_{i}\delta_{k}^{h}u\cdot \chi \cdot \partial_{j}\chi\cdot \delta_{k}^{h}u + \int\delta_{k}^{h}a_{ij}\cdot \partial_{i}u\cdot \partial_{j}\delta_{k}^{h}u\cdot \chi^{2} .\end{aligned}

By the ellipticity condition on aija_{ij} (i.e. aijvivjλ0v2a_{ij}v^{i}v^{j}\geq \lambda_{0}|v|^{2}), we may estimate the first term from below:

aijiδkhujδkhuχ2λ0Dδkhu2χ2.\int a_{ij}\cdot \partial_{i}\delta_{k}^{h}u \cdot \partial_{j}\delta_{k}^{h}u\cdot \chi^{2}\geq \lambda_{0}\int|\D\delta_{k}^{h}u|^{2}\chi^{2}.

Since aijC1(Ω)a_{ij}\in C^{1}(\Omega) and all of these integrands are supported on the compact set supp χV\supp\chi\subset V, we may find a c0>1c_{0}>1 such that

((aij)+(δkhaij)+Dχ)c0(|(a_{ij})| + |(\delta_{k}^{h}a_{ij})|+|\D\chi|)\leq c_{0}

on this set so that we may bound the sum of the last two terms from below, using Hölder's inequality, the Peter-Paul inequality with ε=λ06c02\eps=\frac{\lambda_{0}}{6c_{0}^{2}}, and (the proof of) Lemma 6.3 to control the L2L^{2} norm of the difference quotient of uu:

2aijiδkhuχjχδkhu+δkhaijiujδkhuχ22c02Dδkhuδkhuχc0DuDδkhuχ3c02(Dδkhu2χ2)1/2(VDu2)1/2λ02Dδkhu2χ29c042λ0VDu2.\begin{aligned}&2\int a_{ij}\cdot \partial_{i}\delta_{k}^{h}u\cdot \chi \cdot \partial_{j}\chi\cdot \delta_{k}^{h}u + \int\delta_{k}^{h}a_{ij}\cdot \partial_{i}u\cdot \partial_{j}\delta_{k}^{h}u\cdot \chi^{2} \\ &\geq-2c_{0}^{2}\int|\D\delta_{k}^{h}u|\cdot |\delta_{k}^{h}u|\cdot \chi -c_{0}\int|\D u|\cdot |\D\delta_{k}^{h}u|\cdot \chi\\&\geq -3c_{0}^{2}\left(\int|\D\delta_{k}^{h}u|^{2}\chi^{2} \right)^{1/2}\left(\int_{V}|\D u|^{2} \right)^{1/2}\\&\geq -\frac{\lambda_{0}}{2}\int|\D\delta_{k}^{h}u|^{2}\chi^{2} - \frac{9c_{0}^{4}}{2\lambda_{0}}\int_{V}|\D u|^{2}.\end{aligned}

Finally, we take care of the second integral:

2δkhaijiuδkhuχjχ2c02Duδkhuχ2c02VDu2. 2\int\delta_{k}^{h}a_{ij}\cdot \partial_{i}u\cdot \delta_{k}^{h}u\cdot \chi\cdot \partial_{j}\chi \geq -2c_{0}^{2}\int|\D u|\cdot |\delta_{k}^{h}u|\cdot \chi \geq -2c_{0}^{2}\int_{V}|\D u|^{2}.

Altogether, LHSλ02Dδkhu2χ2c02(2+9c022λ0)VDu2LHS\geq \frac{\lambda_{0}}{2}\int|\D\delta_{k}^{h}u|^{2}\chi^{2} - c_{0}^{2}\left(2+\frac{9c_{0}^{2}}{2\lambda_{0}} \right)\int_{V}|\D u|^{2}. We now bound the right-hand side of (*) from above, again using Hölder's inequality and Lemma 6.3:

RHS=f~δkh(χ2δkhu)(f+βDu+γu)δkh(χ2δkhu)(f2,V+βDu2,V+γu2,V)D(χ2δkhu)2,V.\begin{aligned}RHS&=-\int\tilde{f}\cdot \delta_{k}^{-h}(\chi^{2}\cdot \delta_{k}^{h}u)\\ &\leq \int(|f| + \beta |\D u| + \gamma |u|)\cdot |\delta_{k}^{-h}(\chi^{2}\delta_{k}^{h}u)|\\ &\leq (||f||_{2,V} + \beta ||\D u||_{2,V} + \gamma||u||_{2,V})\cdot ||\D(\chi^{2}\delta_{k}^{h}u)||_{2,V}.\end{aligned}

By the product rule, D(χ2δkhu)=χ2Dδkhu+2χDχδkhu\D(\chi^{2}\delta_{k}^{h}u)=\chi^{2}\cdot \D \delta_{k}^{h}u + 2\chi \cdot \D\chi\cdot \delta_{k}^{h}u so that by the triangle inequality,

D(χ2δkhu)2,V(Dδkhu2χ2)1/2+2c0Du2,V;||\D(\chi^{2}\delta_{k}^{h}u)||_{2,V}\leq \left(\int |\D\delta_{k}^{h}u|^{2}\chi^{2} \right)^{1/2} + 2c_{0}||\D u||_{2,V};

Using Peter-Paul with ε\eps and using the elementary inequality (j=1Nαi)rC(N,r)j=1Nαir(\sum_{j=1}^{N}\alpha_{i})^{r}\leq C(N,r)\sum_{j=1}^{N}\alpha_{i}^{r}, we arrive at

RHSC04ε(f2,V2+(β2+γ2)u1,2,V2)+C0εDδkhu2χ2+4c02C0εDu2,V2.RHS \leq \frac{C_{0}}{4\eps}\cdot \left(||f||_{2,V}^{2} + (\beta^{2}+\gamma^{2})||u||_{1,2,V}^{2} \right) + C_{0}\eps\int|\D\delta_{k}^{h}u|^{2}\chi^{2} + 4c_{0}^{2}C_{0}\eps||\D u||_{2,V}^{2}.

Choosing ε=λ04C0\eps =\frac{\lambda_{0}}{4C_{0}}, we deduce that the inequality LHSRHSLHS\leq RHS implies that

λ04Dδkhu2χ2C1(f2,V2+u1,2,V2),\frac{\lambda_{0}}{4}\int |\D\delta_{k}^{h}u|^{2}\chi^{2}\leq C_{1}\cdot \left(||f||_{2,V}^{2} + ||u||_{1,2,V}^{2} \right),

for some constant C1>0C_{1}>0 depending only on UU, VV, Ω\Omega and LL, and Dδkhu2χ2UDδkhu2\int|\D\delta_{k}^{h}u|^{2}\chi^{2}\geq \int_{U}|\D \delta_{k}^{h}u|^{2}. Thus,

δkhu1,2,U2=Uδkhu2+UDδkhu24C1λ0f2,V2+(1+4C1λ0)u1,2,V2,||\delta_{k}^{h}u||_{1,2,U}^{2} = \int_{U}|\delta_{k}^{h}u|^{2} + \int_{U}|\D\delta_{k}^{h}u|^{2} \leq \frac{4C_{1}}{\lambda_{0}}||f||_{2,V}^{2} + (1+\frac{4C_{1}}{\lambda_{0}})||u||_{1,2,V}^{2},

which implies a bound on δkhu1,2,U||\delta_{k}^{h}u||_{1,2,U} independent of hh. By Corollary 6.6, we therefore have that uW2,2(U)u\in W^{2,2}(U), and we immediately obtain the bound

u2,2,UC2(f2,V+u1,2,V),(**)||u||_{2,2,U} \leq C_{2}\cdot\left(||f||_{2,V} + ||u||_{1,2,V} \right), \tag{**}

C2C_{2} having the same dependencies as C1C_{1}. To obtain the desired estimate, we now choose an ηC0(Ω)\eta\in C_{0}^{\infty}(\Omega) such that 0η10\leq \eta\leq 1 and ηV1\left.\eta\right|_{V}\equiv 1. Setting v=η2uv=\eta^{2}\cdot u in (*), we see that the left-hand side may be bounded from below as

aijiuj(η2u)=aijiujuη2+2aijiuujηηλ0Du2η22c02Duuηλ02Du2η2c02λ0Ωu2,\begin{aligned}\int a_{ij}\partial_{i}u\partial_{j}(\eta^{2}u)&=\int a_{ij}\cdot\partial_{i}u\cdot\partial_{j}u\cdot\eta^{2} + 2\int a_{ij}\cdot\partial_{i}u\cdot u \cdot \partial_{j}\eta\cdot \eta\\ &\geq \lambda_{0}\int|\D u|^{2}\eta^{2}-2c_{0}^{2}\int|\D u|\cdot |u|\cdot \eta\\ &\geq \frac{\lambda_{0}}{2}\int|\D u|^{2}\eta^{2}-\frac{c_{0}^{2}}{\lambda_{0}}\int_{\Omega}|u|^{2}, \end{aligned}

where we have used the Peter-Paul inequality with ε=λ04c02\eps=\frac{\lambda_{0}}{4c_{0}^{2}}. On the other hand, the left-hand side of (*) may be estimated as

f~η2ufu+βDuuη+γu2(12+γ+β2λ0)Ωu2+λ04Du2η2+12Ωf2,\begin{aligned}\int\tilde{f}\cdot \eta^{2}u&\leq \int|f|\cdot |u| + \beta|\D u|\cdot|u|\cdot \eta + \gamma|u|^{2}\\ &\leq (\frac{1}{2} + \gamma+\frac{\beta^{2}}{\lambda_{0}})\int_{\Omega}|u|^{2} + \frac{\lambda_{0}}{4}\int|\D u|^{2}\eta^{2} + \frac{1}{2}\int_{\Omega}|f|^{2},\end{aligned}

where we have again used the Peter-Paul inequality. Altogether, after rearranging terms, we obtain the inequality

Du2η24λ0(12f22+(12+γ+β2λ0+c02λ0)u22).\int|\D u|^{2}\eta^{2} \leq \frac{4}{\lambda_{0}}\left(\frac{1}{2}||f||_{2}^{2} + (\frac{1}{2}+\gamma + \frac{\beta^{2}}{\lambda_{0}} + \frac{c_{0}^{2}}{\lambda_{0}})||u||_{2}^{2} \right).

Since Du2η2VDu2\int|\D u|^{2}\eta^{2}\geq \int_{V}|\D u|^{2}, plugging the resulting inequality into (**) yields the desired estimate.