BOUNDARY VALUE PROBLEMS ON A HALF SIERPINSKI GASKET

Transcription

1 BOUNDARY VALUE PROBLEMS ON A HALF SIERPINSKI GASKET WEILIN LI AND ROBERT S. STRICHARTZ Abstract. We study boundary value problems for the Laplacian on a domain Ω consisting of the left half of the Sierpinski Gasket (), whose boundary is essentially a countable set of points X. For harmonic functions we give an explicit Poisson integral formula to recover the function from its boundary values, and characterize those that correspond to functions of finite energy. We give an explicit Dirichlet to Neumann map and show that it is invertible. We give an explicit description of the Dirichlet to Neumann spectra of the Laplacian with an exact count of the dimensions of eigenspaces. We compute the exact trace spaces on X of the L 2 and L domains of the Laplacian on. In terms of the these trace spaces, we characterize the functions in the L 2 and L domains of the Laplacian on Ω that extend to the corresponding domains on, and give an explicit linear extension operator in terms of piecewise biharmonic functions.. Introduction The Laplacian on the Sierpinski Gasket was first constructed as a generator of a stochastic process, analogous to Brownian motion, by Kusuoka [6] and Goldstein []. An analytic method of constructing the Laplacian on the Sierpinski Gasket as a renormalized limit of graph Laplacians was later developed by Kigami [4]. With a well defined Laplacian, it is possible to study differential equations on the Sierpinski Gasket, although strictly speaking, these are not differential equations. Harmonic functions on the Sierpinski Gasket have been studied in detail and the Dirichlet problem on the entire gasket reduces to solving systems of linear equations and multiplying matrices. However, there has been little research into boundary value problems on bounded subsets of fractals, except for [8], [9] and [] that consider domains generated by horizontal cuts of the gasket. Hence we believe it is appropriate to begin our exploration by studying the Dirichlet problem on a boundary generated by a vertical cut along one of the symmetry lines of the gasket. This is the simplest example of a boundary given as a level set of a harmonic function. We hope our results give insight into more general techniques for solving the Dirichlet problem and other boundary value problems on more general domains. Most of our results are applications of Kigami s harmonic calculus on fractals to our half gasket. His theory includes many mathematical objects specific to the world of fractal analysis, such as renormalized graph energies, normal derivatives and renormalized graph Laplacians. We will present some notation as we proceed, The first author was supported in part by the National Science Foundation grant DMS The second author was supported in part by the National Science Foundation grant DMS

2 2 WEILIN LI AND ROBERT S. STRICHARTZ but for precise definitions and known facts (in particular the results that we call Proposition), see textbooks [] and []. The Sierpinski Gasket, denoted, is the unique nonempty compact set satisfying = 2 j=0 F j where F j are contractive mappings given by F j x = (x+q j )/2 and q j are the vertices of an equilateral triangle. Following convention, the boundary of is defined to be V 0 = {q 0, q, q 2 }. Hence boundary in our language differs from the standard topological definition of boundary. Using the mappings F j, we can iteratively generate a set of vertices V m where m depends on the number of times we apply F j. From V m, we can find a graph approximation Γ m. See Figure. for an illustration. Notice how the boundary points {q j } are oriented and we keep this orientation for the entire paper. q 0 q 0 q q 2 q 0 q q 2 q 0 q q 2 q q 2 Figure.. Left to right: Γ 0, Γ, Γ 2, Γ of We work on the domain Ω, which can be defined in terms of the level sets of a harmonic function. Let h s be the skew symmetric harmonic function with boundary values (h s (q 0 ), h s (q ), h s (q 2 )) = (0,, ). Then Ω = {x \ V 0 : h s (x) > 0} and Ω = q 0 q X where X = {x \ V 0 : h s (x) = 0}. We write Ω = Ω Ω. Figure.2 provides an illustration of Ω, which is precisely the left half of including the points on the symmetry line. In the figure, we labeled the points x m = F0 m F 2 q and y m = F0 m q. Note that X = {x m }, so each x m is important for obvious reasons. Each y m is important topologically because the removal of any y m turns Ω into a disconnected set. We also labeled the open sets Y m = F0 m F ( \ V 0 ). Note that Y m = {x m, y m, y m } and we write Y m = Y m Y m. Y m is classified as a cell because a cell is defined to be the image of under any compositions of contractive mappings F j. Thus Ω = Y m, which is an almost disjoint union. Although Ω is not globally self-similar because Ω cannot be written as a union of smaller copies of itself, it is locally self-similar because each Y m is a fractal. The retention of this local property is extremely important for our analysis because any result regarding also holds for Y m with a proper normalization factor.

3 BOUNDARY VALUE PROBLEMS ON A HALF SIERPINSKI GASKET q 0 y 4 y x 4 Y y 2 x Y 2 y x 2 Y q x Figure.2. A decomposition of Ω In the later sections, we will be interested in restriction and extension operators. Hence, we need to label points on the other half of the gasket. Let z m and Z m the reflections of y m and Y m respectively across the symmetry line containing X. Then = m (Y m Z m ) is an almost disjoint union and this decomposition will be useful in the later sections. We begin by studying the Dirichlet problem on Ω: u = 0 on Ω, (.) u(q ) = a 0 on Ω, u(x m ) = a m on Ω, where denotes the (Kigami) Laplacian with respect to the standard measure, u : Ω R is the unknown, and {a m } m=0 is the boundary data. Notice that we do not prescribe boundary data at q 0 even though q 0 Ω. This is by preference and is inconsequential because for almost the entire paper, we will assume {a m } converges. We will refer to (.) as the BVP. In Section 2, we construct a solution to the BVP using the harmonic extension algorithm, which we explain in that section. The space of C(Ω) solutions to the BVP is one-dimensional, but in general, the solutions blow up at q 0. We show that if the boundary data converges, then we can find a C(Ω) solution that is unique in this function space. In Section, we study the graph energy of the C(Ω) solution to the BVP. Although its energy is complicated, the culminating theorem presents an equivalence between finite energy and the normalized summability of the the boundary data. In fact, finiteness depends only on how quickly the data converges and not on the limiting value. In Section 4, we show that given stronger assumptions on the boundary data, we can obtain the existence of normal derivatives on Ω. In particular, we are interested in the behavior of the normal derivatives on X. The normal derivatives of the C(Ω) solution on X can be found in terms of the boundary data. This

4 4 WEILIN LI AND ROBERT S. STRICHARTZ relationship allows us to define a Dirichlet to Neumann map and we show that this map is invertible. In Section, we discuss both Dirichlet and Neumann eigenfunctions on Ω. For more information on eigenvalues and eigenfunctions on fractals, see [2] and [0]. There are no new eigenfunctions on Ω, but for a fixed eigenvalue, its multiplicity on Ω is different from its multiplicity on. For each eigenfunction, we count the dimension of its eigenspace. Section 6 and Section 7 are closely related to each other. We define a restriction operator that maps a function to its restriction to and normal derivatives on X. We characterize the function spaces dom L 2 () and dom L () in terms of the restriction operator. Using this result, we provide necessary and sufficient conditions for extending functions in dom L 2 (Ω) and dom L (Ω) to biharmonic functions in dom L 2 () and dom L () respectively. Section 8 acts as an appendix and in this section, we prove numerous lemmas about Green s functions and special types of sequences and series. Since these results are used in multiple sections and are purely technical lemmas, we have decided to place them in its own section. While the sequence and series lemmas may not be new, we have not found them in previously published work. 2. Solution to the Boundary Value Problem We begin this section by discussing the graph energy. The energy plays a central role in fractal analysis on because other objects such as harmonic functions, normal derivatives and the Laplacian, are defined in terms of the energy. Given a fixed value of m and a real valued function u on, the (renormalized) graph energy of level m is E m (u) = ( ) m [u(x) u(y)] 2, x my where x m y means x and y are in the same cell of level m. The graph energy of u is E(u) = lim m E m (u), allowing the value +. Given boundary conditions, we define a harmonic function to be the unique function that minimizes the graph energy subject these constraints. Additionally, our suggestive use of the word harmonic is justified: harmonic functions as minimizers of energy are equivalent to functions that satisfy the differential equation u = 0. The Laplacian is defined in Section 4. The simplest tool for constructing harmonic functions subject to boundary conditions is the harmonic extension algorithm. For a function u defined on V m, we can define its harmonic extension to V m+ as follows. Let {v j } be the three boundary points of a cell with {u(v j )} given. Then the harmonic extension of u to the three new points is shown in Figure 2.. It is not difficult to see that given u on V m, this is the unique extension that minimizes the graph energy at level m +. We can apply the harmonic extension algorithm infinitely many times and the resulting function on V = m V m will be harmonic. It is not difficult to see that functions generated by the harmonic extension algorithm must be continuous. Furthermore, V is dense in and so for continuous functions, it suffices to define

5 BOUNDARY VALUE PROBLEMS ON A HALF SIERPINSKI GASKET them on a dense subset. Thus, we say a harmonic function is determined by its boundary values. u(v 0 ) 2u(v 0)+2u(v )+u(v 2) 2u(v 0)+u(v )+2u(v 2) u(v ) u(v 2 ) u(v 0)+2u(v )+2u(v 2) Figure 2.. Harmonic Extension Algorithm We can use the harmonic extension algorithm to construct a solution to the BVP. Any harmonic function on Y m is determined by its values on Y m. Since Ω = m Y m, any harmonic function on Ω is determined by its value at the points {x m } and {y m }. In the following lemma, we see that there are additional constraints we must take into account. Lemma 2.. Fix m 2. Let u be a continuous piecewise harmonic function with boundary data given by (.). Then u(y m ) = 0 if and only if (2.) u(y m ) = 6 u(y m ) u(y m 2) a m a m. Proof. Consider the level m approximation of Y m Y m. The value of u at the midpoint of y m and y m 2 and the midpoint of y m and x m are determined by the harmonic extension algorithm, shown in Figure 2.2. If u(y m ) = 0, then u satisfies the mean value property at y m. Thus, u(y m ) is the average of its four neighboring points in V m and simplifying that equation yields (2.). Conversely, if (2.) holds, then it is straightforward to check that u(y m ) = 0. Theorem 2.2. For every choice of convergent boundary data {a m }, there is a one dimensional space of C(Ω) solutions to the BVP. Given a parameter λ, the solution to the BVP u λ is the harmonic extension of u λ (x m ) = a m, u λ (y ) = λ and (2.2) where and u λ (y m ) = m F m (λ) + m G m(λ), ( F m (λ) = λ a 0 a 8 4 m k=2 ( G m (λ) = λ + a 0 + a ) k a k ) m k a k. k=2

6 6 WEILIN LI AND ROBERT S. STRICHARTZ u(y m ) u(y m ) a m 2u(y m )+2u(y m 2)+a m 2u(y m )+u(y m 2)+2a m u(y m 2 ) a m Figure 2.2. Harmonic extension Proof. By Lemma 2., u λ must satisfy the recurrence (2.). The recurrence is linear, so we can formulate the equation in terms of matrices. Define [ ] [ ] A = and B =. Then the recurrence can be written as [ ] [ ] uλ (y m ) = B m a0 u λ (y m+ ) λ Solving the system, we find that u λ (y m ) = m ( 4 ) ( λ a 0 m + m 6 [ ] B m k ak A. a k+ ) k c k + ( ) ( m m λ + a k c k ) where c k = a k+ + a k. We want our formula in terms of a k rather than c k, so substituting m m k c k = 4 k a k + a m a m and m k c k = 8 k=2 m k=2 k a k + a m a m into the previous equation for u λ (y m ) yields (2.2). Extending these values by the harmonic extension algorithm uniquely yields a harmonic function u continuous on Ω. Since u λ is a linear combination of a m term and a / m term, u λ may blow up at q 0. Naturally, we ask whether we can find a λ such that u λ is continuous on Ω.,

7 BOUNDARY VALUE PROBLEMS ON A HALF SIERPINSKI GASKET 7 Lemma 2.. Suppose u λ C(Ω) satisfies the BVP for convergent {a m }. Then u λ C(Ω) if and only if (2.) lim m u λ(y m ) = lim m u λ(x m ). Proof. Suppose u λ C(Ω) solves the BVP. Then u λ is continuous at q 0, which is equivalent to (2.). Conversely, it is easy to see that q 0 is the only point at which u λ can be discontinuous. Then (2.) implies u λ is continuous at q 0, which shows that u λ C(Ω). Theorem 2.4. If a m 0 as m, then the function u given by the harmonic extension of u(x m ) = a m, ( ) (2.4) u(y ) = a 0 + a + 8 k a k, and (for m 2) (2.) u(y m ) = m ( a k a k k=2 ) m k a k k a m+k solves the BVP. Furthermore, this function is the unique solution in C(Ω). Proof. Substituting (2.4) into (2.2) yields (2.). By triangle inequality, ( ) u(y m ) m a k a k + 2 m k a k k a m+k. We claim that u(y m ) 0 as m. Clearly the first term tends to zero in the limit. The second term tends to zero because convergent sequences are bounded. Since both the boundary data and / m converge to zero, for all ε > 0, there exists M such that for all m M, we have a m < ε and / m < ε. For m M, we see that and m m k a k = m M k a k + k a m+k ε m k=m+ k = ε 2 ( ) k m a k C ε max a k k M + C 2 ε. Therefore u satisfies condition (2.) and by Lemma 2., u C(Ω). Since harmonic functions that are continuous up to the boundary satisfy the maximum principle [], uniqueness follows from the standard uniqueness argument for linear differential equations that satisfy the maximum principle. Corollary 2.. If a m A as m for some constant A, then the function u given by the harmonic extension of u(x m ) = a m, ( ) (2.6) u(y ) = a 0 + a + 8 k a k, k=2

8 8 WEILIN LI AND ROBERT S. STRICHARTZ and (for m 2) (2.7) u(y m ) = m ( a k a k ) m k a k k a m+k solves the BVP. Furthermore, this function is the unique solution in C(Ω). Proof. Consider the modified BVP u = 0 on Ω, (2.8) u(q ) = a 0 A on Ω, u(x m ) = a m A on Ω. Since a m A 0, the hypotheses of Theorem 2.4 are satisfied. Then there exists w C(Ω) that solves (2.8) and the formula for w(y m ) is given by (2.) under the map a k a k A. By construction, the function u = w + A solves the BVP with u C(Ω). The maximum principle implies that u is unique.. Energy Estimate In this section, we look to answer questions regarding the energy of the C(Ω) solution to the BVP. In particular, is the energy always finite and if not, can we characterize functions of finite energy in terms of a condition on the boundary data? Our main theorem shows that harmonic functions on Ω do not necessarily have finite energy and provides a simple characterization. Given a function u, we say u dome if and only if E(u) <. Following standard notation, dom 0 E is the space of functions that have finite energy and vanish on the boundary V 0. It is known that dome C() and in fact, is a dense subset. Suppose u is a piecewise harmonic function on Ω that is harmonic on each Y m with data given by (.). Then the energy of u restricted to Y m is constant after level m and is determined by u(y m ), u(y m ), and a m. It follows that ( ) m [ E(u) Ym = (u(ym ) u(y m )) 2 + (u(y m ) a m ) 2 + (u(y m ) a m ) 2], where it is understood that u(y 0 ) = u(q ) = a 0. Then E(u) is the sum of the energy of each cell, (.) ( ) m [ E(u) = (u(ym ) u(y m )) 2 + (u(y m ) a m ) 2 + (u(y m ) a m ) 2]. If we add the additional assumption that u C(Ω) solves the BVP, then an equation for E(u) as a function of {a m } can be obtained by substituting (2.6) and (2.7) into (.). However, E(u) is series of quadratic terms of series, which is too complicated to analyze directly. Instead, we estimate it. Lemma.. Suppose u C(Ω) solves the BVP with convergent {a m }. Then we have the energy estimate ( ) m ( ) m (a m+ a m ) 2 E(u) C 2 (a m+ a m ) 2 <. C

9 BOUNDARY VALUE PROBLEMS ON A HALF SIERPINSKI GASKET 9 Proof. We prove the lower bound first. By ignoring the first term of (.), we have ( ) m [ E(u) (u(ym ) a m ) 2 + (u(y m ) a m ) 2]. Using basic calculus, we find that u(y m ) = (/8)(a m+ + a m ) minimizes the previous series. Substituting this value of u(y m ) into the previous inequality, we obtain ( ) m 8 (a m+ a m ) 2 + (a a 0 ) 2 E(u) <. For the upper bound, consider the piecewise harmonic function w given by the harmonic extension of w(x m ) = w(y m ) = a m and w(q ) = a 0. Since u is a global harmonic function while w is a piecewise harmonic function, we have E(u) E(w). Note that E(w) is given by (.) because w is a piecewise harmonic function satisfying the boundary conditions. Then E(u) E(w) = which completes the proof. ( ) m 0 (a m+ a m ) (a a 0 ) 2 <, Theorem.2. Suppose u C(Ω) solves the BVP with convergent boundary data a m A. Then u dome if and only if (/) m/2 (a m A) l 2 <. Additionally, we have the upper bound E(u) C (/) m/2 (a m A) l 2. Proof. Suppose u C(Ω) solves the BVP with convergent boundary data a m A. Lemma. says that E(u) < if and only if (/) m/2 (a m+ a m ) l 2 <. Applying Lemma 8.9 yields the desired statement. 4. Normal Derivatives Although the normal derivative and the (standard) Laplacian on are defined independently, they are closely connected via the Gauss-Green formula. For a continuous function u, its normal derivative at q j V 0, denoted n u(q j ), is defined to be (4.) n u(q j ) = lim m ( ) m [ 2u(qj ) u(fj m q j+ ) u(fj m q j ) ]. We say n u(q j ) exists if the above limit exists. In the special case u is harmonic, we have the simplified formula (4.2) n u(q j ) = 2u(q j ) u(q j ) u(q j+ ). The formula for the normal derivative of a harmonic function at a boundary point of a cell is similar to the above formula, except we require a renormalization factor depending on the level. A junction point is a boundary point of two adjacent cells of the same level, and the normal derivative with respect to the cells will differ by a minus sign. If we need to distinguish between the two normal derivatives at a junction point, we use either (, ), (, ) or (, ), corresponding to the geometrical notion of a normal derivative.

10 0 WEILIN LI AND ROBERT S. STRICHARTZ Proposition 4.. Suppose u dom. Then at each junction point, the local normal derivatives exist and n u + n u = 0. This is called the matching condition for normal derivatives. The Laplacian of a function is defined in terms of its weak formulation. First, we define the (symmetric) bilinear form of the energy: given functions u, v and integer m, the bilinear form of the energy is E m (u, v) = ( ) m [u(x) u(y)][v(x) v(y)]. x my has a unique symmetric self-similar probability measure that we denote dx. Then the Laplacian can be defined as follows. Suppose u dome and f is continuous. Then we say u dom with u = f if E(u, v) = f(x)v(x) dx for all v dom 0 E (functions in dome vanishing on V 0 ). Since E(u, v) = E(v, u), subtracting the Gauss-Green formula from its transposed version yields the symmetric Gauss-Green formula (4.) ( uv u v) dx (v n u u n v) = 0. V 0 The following result relates the normal derivatives of a function with its Laplacian. Proposition 4.2 (Gauss-Green). Suppose u dom. Then n u exists on V 0 and the Gauss-Green formula, E(u, v) = uv dx + v n u, V 0 holds for all v dome. For the remainder of this section, we assume u C(Ω) solves the BVP with convergent boundary data. Naturally, we are interested in analyzing the behavior of n u(x) for x Ω. For all points in Ω except q 0, the formulas for the normal derivatives of u are given by (4.2). Using this equation, with the appropriate normalization factor, the normal derivative of u at y m with respect to the cell Y m is (4.4) n u(y m ) = ( ) m [2u(y m ) u(y m ) a m ]. Similarly, the normal derivative of u at x m with respect to Y m is ( ) m (4.) n u(x m ) = [2a m u(y m ) u(y m )]. However (4.2) does not give us the equation for n u(q 0 ) because u is only defined on Ω. But we can define n u(q 0 ) in a natural way. Lemma 4.. If u dom (), then (4.6) n u(q 0 ) = 2 lim m nu(y m ).

11 BOUNDARY VALUE PROBLEMS ON A HALF SIERPINSKI GASKET Proof. Write u = u s + u a where u s and u a are the symmetric and skew-symmetric parts of u respectively. Since u a F m 0 () = O(/m ), we have n u a (q 0 ) = 2 lim m nu a (y m ) = 0. For the symmetric part, consider the triangle T m with boundary points {q 0, y m, z m } and the harmonic function v on T m with v(q 0 ) = v(y m ) = v(z m ) =. Applying the symmetric Gauss-Green formula (4.) for u s and v, we find that n u s (q 0 ) + n u s (y m ) + n u s (z m ) = u s dx. T m Notice that n u s (y m ) = n u s (z m ) by symmetry. Using the normal derivative matching condition of u at q 0, we see that n u s (y m ) = n u s (q 0 ). Making these substitutions and taking the limit m, we find that 2 lim nu s (y m ) n u s (q 0 ) = lim u s dx = 0, m m T m because u is bounded and the measure of T m tends to zero in the limit. Motivated by this lemma, we define n u(q 0 ) for u defined on Ω by (4.6). In the special case that u C(Ω) solves the BVP with convergent data, then [ ( ) 0 (4.7) n u(q 0 ) = lim m m 7 k a k ( ) m ] 2 m k a k, 7 k=m+ which we obtained by substituting (2.7) into the definition of n u(q 0 ). Notice that (4.2) implies that for harmonic functions defined on, its normal derivative exists everywhere. However, this is not true for harmonic functions on Ω because the limit in (4.7) may not exist. It is straightforward to see that if a m = A + A 2 (/) m + o((/) m ), then the limit in (4.7) exists and equals a constant times A 2. It is not clear whether or not the converse of this statement holds. To find the normal derivatives on X in terms of the boundary data, we substitute (2.7) into (4.), which yields (4.8) η m = ( ) ( m a m 2 7 ) ( k a m+k m 6a m k a k 4 7 ) k a k, where η m = n u(x m ). We can think of (4.8) as a Dirichlet to Neumann map on X because it maps the Dirichlet boundary data to the corresponding normal derivatives. Define the infinite vectors η a / η =. η i, a =. a i and a 0 = 6a 0. / i,...

12 2 WEILIN LI AND ROBERT S. STRICHARTZ and the infinite matrices L = Diag[(/) i ] and K with entries i if i = j, j K i,j = i 4 27 j 8 i if i < j, j j 4 27 i 8 i if i > j. j Then (4.8) can be written as η = 6 7 L(I K)a + a 0. Since we assumed {a m } converges and u C(Ω), we see that {a m }, {u(y m )} l. Then (4.) implies (/) m η m l <. For this reason, for a real number r, we define the space l r, = {{c m }: r m c m l < }. Then we define the Dirichlet to Neumann map D N : l l /, given by D N a = 6 7 L(I K)a + a 0. Theorem 4.4. The Dirichlet to Neumann map is invertible. Proof. We see that D N is a composition of L: l l /, with I K: l l plus a translation. The translation is not important and obviously L is invertible because it is diagonal. It is well known that I K is invertible if and only if ρ(k) <, where ρ(k) is the spectral radius of K. The sum of the entries of the i-th row is i K i,j = K i,i + K i,j + j= Consequently, j= K = sup i j= j=i+ K i,j < i j= K i,j < j + j= j j= j i + i j j=i+ =. Since K is a positive matrix, the Perron-Frobenius Theorem for positive matrices states that ρ(k) K. Thus, ρ(k) <, which shows that I K is invertible... Eigenfunctions The exact spectral asymptotics on the whole gasket and the structure of the spectrum has been analyzed previously [2]. Motivated by that result, we discuss eigenvalues and eigenfunctions on the half gasket. Observe that: () A Dirichlet eigenfunction on Ω extends by odd reflection to a Dirichlet eigenfunction on and conversely. (2) A Neumann eigenfunction on Ω extends by even reflection to a Neumann eigenfunction on and conversely.

13 BOUNDARY VALUE PROBLEMS ON A HALF SIERPINSKI GASKET Thus there are no new eigenvalues on Ω because odd eigenfunctions on are Dirichlet eigenfunctions on Ω and even eigenfunctions on are Neumann eigenfunctions on Ω. Hence we count the number of even and odd eigenfunctions on. On, there are #V m = ( m+ + )/2 vertices on level m, of which m + lie on q 0 X and three are boundary points V 0. The eigenfunctions with eigenvalue λ C 0 m for a specific choice of C 0 are born on level k m and are in one-to-one correspondence with the graph eigenfunctions on V m, so there are ( m+ + )/2 Neumann eigenfunctions and ( m+ )/2 Dirichlet eigenfunctions. Thus on Ω, #{Neumann eigenfunctions with λ C 0 m } = ( m+ ) + + m +, 2 2 #{Dirichlet eigenfunctions with λ C 0 m } = ( m+ ) m, 2 2 because the m + vertices on q 0 X contribute even functions to the Neumann count while the m vertices on X do not contribute to the Dirichlet count. Note that the correction terms m + and m are of the order log m. This is consistent with the observation that Ω is zero dimensional. We can be more specific about individual multiplicities of eigenvalues on Ω. For a set U, define the functions N(U) = #{Neumann eigenfunctions on U}, D(U) = #{Dirichlet eigenfunctions on U}. () 0-series (constant eigenfunctions) have multiplicity N(Ω) = and D(Ω) = 0. (2) 2-series only show up in the Dirichlet spectrum on, but they are all even so they are absent from the Dirichlet spectrum of Ω. Thus, N(Ω) = 0 and D(Ω) = 0. () -series are entirely Neumann eigenfunctions on that are born on level 0 with multiplicity 2. Then N(Ω) = and D(Ω) = 0. (4) -series are born on level k where k for Dirichlet eigenfunctions and k 2 for Neumann eigenfunctions. If S k denotes the number of cycles of level less than k, then on, we find that N() = S k and D() = S k + 2. For a cycle that lies on X, the eigenfunction is odd, so that contributes to D() but not to N(). See Figure. for an example of such a function. Note that any unlabeled point means the function is defined to be zero at that point. Additionally, of the two extra Dirichlet eigenfunctions on, exactly one is odd, as shown in Figure.2. The number of cycles of level n is n and exactly one of these lies on X. So there are ( n +)/2 odd eigenfunctions and ( n )/2 even eigenfunctions. Thus k N(Ω) = 2 (n ) = ( k ) k 2 2 and D(Ω) = k n= n= 2 (n + ) + = ( k ) + + k. 2 2 () 6-series on are born on level k where k for Neumann eigenfunctions and k 2 for Dirichlet eigenfunctions. We know that N() = #V k and D() = #V k. Neumann eigenfunctions are obtained by giving arbitrary

14 4 WEILIN LI AND ROBERT S. STRICHARTZ Figure.. Odd eigenfunction on Γ 2 Figure.2. Another odd eigenfunction on Γ 2 values on the points in V k, while Dirichlet eigenfunctions are obtained by giving arbitrary values on the points V k \ V 0. To find the multiplicities on Ω, we just have to count the even eigenfunctions and the odd eigenfunctions. Hence N(Ω) = ( k ) + + k and D(Ω) = ( k ) k Trace Theorem Consider the restriction map R given by Ru = {(u(x m ), n u(x m ))}, where u is some function defined on some set containing X. That is, R maps u to its function values on X and its normal derivatives on X. In this section, we determine the image of dom L 2 () and dom L () under R. We say that u dom L 2 () if u is continuous on and u L 2 (), and analogously for u dom L (). To simplify notation, we define the following spaces. Define the Lipschitz space Lip = {{c m }: there exists M such that c m+ c m M for all m}. The norm on Lip/Constants is c m Lip = inf M where the infimum is taken over all M satisfying the previous condition. It follows directly from the definition of Lip that {c m } Lip if and only if there exists M such that c m c n M m n for all m and n.

15 BOUNDARY VALUE PROBLEMS ON A HALF SIERPINSKI GASKET We define the following trace spaces: T = { {(a m, η m )}: a m = A + A 2 (/) m + a m, m a m l, m η m Lip < }, T 2 = { {(a m, η m )}: a m = A + A 2 (/) m + a m, (2/) m/2 a m l 2, m/2 η m l 2 < }, with their respective norms {(a m, η m )} T = A + A 2 + m a m l + m η m Lip, {(a m, η m )} 2 T 2 = A 2 + A (2/) m/2 a m 2 l 2 + m/2 η m 2 l 2. Clearly both trace norms satisfy the triangle inequality. Note that the defined norm T2 makes T 2 a Hilbert Space with the obvious inner product. Similarly, we define norms on dom L () and dom L 2 () by u doml () = u L () + u L (), u 2 dom L 2 () = u 2 L 2 () + u 2 L 2 (). In the above definition, we could have replaced 2 L term with 2 2 L, but that would not be a Hilbert Space norm. As suggested by the notation, our goal is to prove that R maps dom L () and dom L 2 () to their corresponding trace spaces. In Section 7, we will show that the mapping is onto. Theorem 6. (Trace Theorem). () The restriction operator R: dom L () T is bounded and Ru T C u L () + C 2 u L (). (2) The restriction operator R: dom L 2 () T 2 is bounded and Ru T2 C u L () + C 2 u L 2 (). The proof of the theorem is technical and rather long, so we split the proof into multiple lemmas. Our primary tool will be the Green s formula. Given any function u on for which u exists, we can write (6.) u(x) = G(x, y) u(y) dy + h(x), where G(x, y) is the Green s function (the definition is given in Section 8.) and h is the harmonic function with boundary conditions h V0 = u V0. We will use the Green s function to relate an arbitrary function to its restriction to X and its normal derivatives on X. The derivations are digressive, so we have placed these computations into their own section. The important formulas and inequalities are given by (8.4), (8.), and (8.7). Note that the definition of the function Ψ m is given in (8.). Since it is easy to check the conditions for the harmonic function h in (6.), let us do that first. Lemma 6.2. If h is harmonic, then Rh T and Rh T 2 with (6.2) (6.) Rh T = u(q 0 ) + 2 u(q ) + u(q 2 ) 2u(q 0 ) + 2 u(q ) u(q 2 ), Rh T2 = u(q 0 ) + 2 u(q ) + u(q 2 ) 2u(q 0 ) u(q ) u(q 2 ).

16 6 WEILIN LI AND ROBERT S. STRICHARTZ Proof. If h is harmonic, then h is a linear combination of the constant function, the skew-symmetric harmonic function (with respect to X) and the symmetric harmonic function (with respect to X). Then u(q 0 ) 0 u(q ) = A + A 2 + A u(q 2 ) 0, where the coefficients are the coefficients A, A 2, and A are the weights of the constant, symmetric and skew-symmetric functions respectively. Solving the system for A, A 2, A in terms of u V0, we find A = u(q 0 ), A 2 = 2 (u(q ) + u(q 2 ) 2u(q 0 )), and A = 2 (u(q ) u(q 2 )). On X, we see that () a constant function is constant with zero normal derivative. (2) a skew-symmetric harmonic function is zero with normal derivative A / m. () a symmetric harmonic function has values A 2 (/) m with zero normal derivative. Then h(x m ) = A + A 2 (/) m and n h(x m ) = A / m. In the following lemma, we prove the bulk of the dom L () case. Proving the lemma directly from the Green s formula would be difficult, so we employ the following indirect method. For the function values of u dom L () on the vertical boundary, we prove an intermediate statement about the linear combination u(x m+ ) u(x m ). We consider the linear combination u(x m+ ) u(x m ) because the troublesome m Ψ k(, 2, 2) term of (8.) cancels out in the linear combination G(x m+, y) G(x m, y). Then the intermediate result, coupled with a lemma from Section 8.2, will give us the desired statement, except for a few estimates which we prove without much trouble. Likewise, for the normal derivatives of u dom L () on the vertical boundary, we prove an intermediate statement about the linear combination η m+ η m because the troublesome m k Ψ k (0,, ) term in (8.7) disappears in the linear combination. The intermediary result, combined with the proper lemma from Section 8.2 and more bounding, yields the desired normal derivative estimate. Lemma 6.. If u dom L () with u = 0 on V 0, then Ru T and (6.4) Ru T C u L (). Proof. Suppose u dom L () with Ru = {(a m, η m )}. Using the Green s formula (Proposition 8.) on a m+ a m and the equation for G(x m, y) given by (8.), after some simplification, we obtain a m+ a m = ( ) m [Ψ m+ (,, ) Ψ m (,, )] u dy 0 Then applying inequality (8.4) yields ( ) m a m+ a m u L Ψ m+ (,, ) Ψ m (,, ) dy 0 C u L m.

17 BOUNDARY VALUE PROBLEMS ON A HALF SIERPINSKI GASKET 7 Rearranging the above inequality yields m (a m+ a m ) l C u L. Lemma 8.6 implies that a m = A(/) m + a m, where A = lim m (/) m a m and m a m l m (a m+ a m ) l. The previous two inequalities immediately yield (6.) m a m l C u L. It follows from the Green s formula and standard bounding methods that ( ) m ( ) m a m G(x m, y) u dy C u L + C 2 u L m. Since A = lim m (/) m a m, the above implies that (6.6) A C u L. We use a similar technique to prove the desired statement about the normal derivatives. Using the equation for η m given by (8.7) to compute η m+ η m, we obtain η m+ η m = [ Ψ m+ (,, ) + Ψ m (,, )] u dy ϕ m+ + ϕ m, 0 where ϕ m was defined in the lemma. Then η m+ η m C u L Ψ m+ (,, ) Ψ m (,, ) dy + ϕ m+ ϕ m C u L m, where we used (8.4) and (8.) to bound the first and second terms respectively. Rearranging, we find that m (η m+ η m ) l C u L. The above estimate allows us to apply Lemma 8.7 which gives us The previous two inequalities imply m η m Lip = m (η m+ η m ) l. (6.7) m η m Lip C u L. Finally, combining our inequalities (6.), (6.6) and (6.7), we see that Ru T = A + m a m l + m η m Lip C u L. Since a m = A(/) m + a m and Ru <, we conclude that Ru T. In the following lemma, we prove the majority of the dom L 2 () statement of the Trace Theorem. We use an indirect approach similar to that of the proof for the dom L () case, except the statements are considerably harder to prove. Proving the lemma directly from the Green s formula without proving the intermediary result would be extremely difficult, mainly because the Cauchy-Schwarz inequality is too wasteful for the type of estimate we desire. The outline of the proof is similar to that of Lemma 6.. For u dom L 2 (), we prove intermediary results about the linear combinations a m+2 8a m+ + a m and η m+ 6η m+ +η m, where as usual, a m = u(x m ) and η m = n u(x m ). These linear combinations are written as linear combinations of integrals, but the primary

18 8 WEILIN LI AND ROBERT S. STRICHARTZ integrand of each linear combination is supported on a set not containing q 0. This support allows us give a more precise estimate, thereby limiting the wastefulness of Cauchy-Schwartz. Then applying results from Section 8. and some more bounding will give us the desired statements. Lemma 6.4. If u dom L 2 () with u = 0 on V 0, then Ru T 2 and (6.8) Ru T2 C u L2 (). Proof. Suppose u dom L 2 () with Ru = {(a m, η m )}. Using the Green s formula (Proposition 8.) on a m+2 8a m+ + a m and the equation for G(x m, y) given by (8.), after much computation, we obtain a m+2 8a m+ + a m = ( ) m G m u dy, where we defined G m (y) = 0 [9Ψ m+2(,, ) 20Ψ m+ (, 0, 0) + 2Ψ m (,, )]. We show that G m is supported on D m = Y m Y m+ Y m+2 Z m Z m+ Z m+2. Since G m is a linear combination of harmonic splines, we see that G m vanishes on Y m Z m for m < m. Using the harmonic extension algorithm, notice that 2Ψ m (,, )(y m+2 ) = 2Ψ m (,, )(z m+2 ) = 9, 20Ψ m+ (, 0, 0)(y m+2 ) = 20Ψ m+ (, 0, 0)(z m+2 ) = 0, 9Ψ m+2 (,, )(y m+2 ) = 9Ψ m+2 (,, )(z m+2 ) = 9. Thus G m (y m+2 ) = G m (z m+2 ) = 0 and consequently, G m vanishes on Y m Z m for m > m + 2, which proves that G m is supported on D m. Taking advantage of the support of G m, we can write ( ) m a m+2 8a m+ + a m = G m u dy, D m Applying Cauchy-Schwarz and inequality (8.4) on the above equation yields ( ) m a m+2 8a m+ + a m 2 C u 2 L 2 (D m). 2 By definition of D m and linearity of the integral, we have (6.9) m+2 u 2 L 2 (D = m) k=m u 2 L 2 (Y k Z k ), u 2 L 2 () = u 2 L 2 (Y k Z k ). Using the upper bound on a m+2 8a m+ + a m 2 and the above two equations, we obtain (2/) m/2 (a m+2 8a m+ + a m ) l 2 C u L2 (). This estimate allows us to apply Lemma 8.0. Thus a m = A + A 2 (/) m + a m, where A = lim m a m, A 2 = lim m (/) m (a m A ), and (2/) m/2 a m l 2 C (2/) m/2 (a m+2 8a m+ + a m ) l 2.

19 BOUNDARY VALUE PROBLEMS ON A HALF SIERPINSKI GASKET 9 The above two inequalities immediately yield (6.0) (2/) m/2 a m l 2 C u L2 (). We claim that A = 0 and A 2 C u L2 (). Applying Cauchy-Schwarz to the Green s formula for a m, we find that that ( ) m a m C u L2() + C 2 u L 2(). m/2 The above inequality implies that A = 0 and (6.) A 2 C u L 2 (). We use a similar argument to prove the estimate on the normal derivatives. Using Lemma 8. to compute η m+2 6η m+ + η m, we see that η m+2 6η m+ + η m = Φ m u dy (ϕ m+2 6ϕ m+ + ϕ m ), where we defined Φ m = 0 [ Ψ m+2(,, ) + 0Ψ m+ (8,, ) 2Ψ m (,, )]. We show that Φ m has support on D m as well. Since Φ m is a linear combination of harmonic splines, Φ m vanishes on Y m Z m for m < m. Using the harmonic extension algorithm, we have 2Ψ m (,, )(y m+2 ) = 2Ψ m (,, )(z m+2 ) =, 0Ψ m+ (8,, )(y m+2 ) = 0Ψ m+ (8,, )(z m+2 ) = 2, Ψ m+2 (,, )(y m+ ) = Ψ m+2 (,, )(z m+ ) =. Thus, Φ m (y m+2 ) = Φ m (z m+2 ) = 0 and consequently, Φ m vanishes on Y m Z m for m > m + 2. Using the compact support of Φ m, we can write η m+2 6η m+ + η m = Φ m u dy (ϕ m+2 6ϕ m+ + ϕ m ), D m It is straightforward to find an upper bound on the linear combination of ϕ m terms. Using Cauchy-Schwarz and inequality (8.2), we obtain η m+2 6η m+ + η m 2 C ( ϕ m ϕ m+ 2 + ϕ m 2) C u 2 L 2 (D m) m. Using Cauchy-Schwarz and inequality (8.4), we find that 2 Φ m u dy u 2 L 2 (D m) Φ m 2 dy C u 2 L 2 (D m) D m D m m. Combining the above two inequalities and (6.9) yields (6.2) m/2 (η m+2 6η m+ + η m ) l 2 C u L2 (). The hypothesis of Lemma 8. is satisfied, so we have η m = m A + η m with (6.) m/2 η m C (η 2 η ) 2 + C 2 m (η m+2 6η m+ + η m ) l 2. However, applying Cauchy-Schwarz to (8.7) yields η m C u L2 (). m/2

20 20 WEILIN LI AND ROBERT S. STRICHARTZ This forces A = 0 and so η m = η m. Note that the above bound provides the upper bound (η 2 η ) 2 C u 2 L 2 (). Combining this inequality with (6.2) and (6.) yields (6.4) m/2 η m 2 C u 2 L 2 (). Finally, using (6.0), (6.) and (6.4), we see that Ru 2 T 2 = A 2 + A (2/) m/2 a m 2 l + 2 m/2 η m 2 l 2 C u 2 L 2 (). Since a m = A 2 (/) m + a m and Ru 2 T 2 <, we conclude that Ru T 2. Finally, we have the necessary results to prove the Trace Theorem. Proof of the Trace Theorem. Suppose u dom L () or u dom L 2 (), and Ru = {(a m, η m )}). Let h be the harmonic function determined by the boundary values h V0 = u V0. Let w = u h, and note that w = u and w = 0 on V 0. The Green s formula states that u(x) = h(x) + G(x, y) w(y) dy. () Suppose u dom L (). Using triangle inequality on u = w + h, the estimate (6.2) applied to h, and the estimate (6.4) applied to w, we find that Ru T u(q 0 ) + 2 u(q )+u(q 2 ) 2u(q 0 ) + 2 u(q ) u(q 2 ) +C u L (). (2) Suppose u dom L 2 (). Using triangle inequality on u = w + h, (6.) applied to h, and (6.8) applied to w, we find that Ru T2 u(q 0 ) + 2 u(q )+u(q 2 ) 2u(q 0 ) u(q ) u(q 2 ) +C u L 2 (). 7. Extension Operators In this section, we present two different extension theorems. The first extension will be a right inverse to the restriction map R. The second extension will map solutions to differential equations on the half-gasket to a well-behaved function on the whole gasket. The ideas behind the two extensions are similar, but with different computations and formulas. In order to construct the desired extensions, we will require the following result. If will give us the exact conditions under which a piecewise function is in the domain of the Laplacian. Proposition 7. (Gluing Theorem). Let u and f be defined by gluing pieces {u j } and {f j } (j = 0,, 2), with u j = f j on F j. Then u dom with u = f if and only if f j (F i q j ) = f j (F j q i ) (i j) holds for {u j } and {f j } (so u and f are continuous) and the matching conditions on normal derivatives hold at the three points.

21 BOUNDARY VALUE PROBLEMS ON A HALF SIERPINSKI GASKET The Inverse Operator to R. We seek a linear extension operator E that is a right inverse of the restriction operator R. The desired extension will satisfy E : T dom L () and E : T 2 dom L 2 (). In order to construct this extension operator, we study piecewise biharmonic functions. Biharmonic functions satisfy the differential equation 2 u = 0 and in particular, biharmonic functions satisfying u = C for some constant C is a four-dimensional space on. One way to specify a constant Laplacian function on is to specify the value of the function on V 0 and the constant. Lemma 7.2. Suppose u = C on some cell of level m with boundary points p 0, p, p 2. Then the outward normal derivative of u at p j is ( ) m (7.) n u(p j ) = [2u(p j ) u(p j+) u(p j )] + C m+. Proof. Let v be the harmonic function on the cell with the boundary values v(p j ) = and v(p j+ ) = v(p j ) = 0. Since v is harmonic on a cell of level m, using (4.) with the proper normalization, we have n v(p j ) = 2(/) m while n v(p j+ ) = n v(p j ) = (/) m. Applying the symmetric Gauss-Green formula (4.), we obtain the desired formula. Lemma 7.. Given any sequences {a m } and {η m }, there exist a piecewise biharmonic function u on and sequences {C m} and {C m } such that Ru = {(a m, η m )}, u = C m on Y m, u = C m on Z m, and the normal derivative matching conditions hold at {x m }, {y m }, and {z m }. Proof. We construct two functions u and u 2 such that u (x m ) = a m but n u (x m ) = 0, while u 2 (x m ) = 0 but n u(x m ) = η m. Then the sum u = u + u 2 will satisfy Ru = {(a m, η m )}. Of course, we must do this carefully so that u satisfies the other claimed properties. Consider the symmetric piecewise biharmonic function u satisfying u (x m ) = a m, u (y m ) = u (z m ) = (/8)(a m+ + a m ), and u = D m on Y m Z m with ( ) D m = m (a m+ 8a m + a m ). 8 This information determines u on Y m Z m because as mentioned earlier, a constant Laplacian function is determined by its boundary values and the value of its Laplacian. Consequently, u is determined everywhere because = m (Y m Z m ). Using (7.) to compute the normal derivatives of u at x m, y m and z m, it is straightforward to check that n u (x m ) = 0 and the normal derivative matching conditions hold. Consider the skew-symmetric piecewise biharmonic function u 2 satisfying the conditions u 2 (x m ) = 0, u 2 (y m ) = (/8)(/) m (η m+ + η m ), u 2 (z m ) = u 2 (y m ), u 2 = E m on Y m and u 2 = E m on Z m, where ( ) E m = m (η m+ 6η m + η m ). 8 Again, these constraints determine u 2 everywhere on. Writing down the normal derivatives of u 2 at x m, y m and z m using (7.), we see that n u 2 (x m ) = η m and the normal derivative matching conditions hold.

22 22 WEILIN LI AND ROBERT S. STRICHARTZ Then the function u = u + u 2 satisfies u(x m ) = a m, n u(x m ) = η m, u(y m ) = 8 (a m+ + a m ) ( ) m (η m+ + η m ), 8 (7.2) u(z m ) = 8 (a m+ + a m ) + ( ) m (η m+ + η m ), 8 u = C m on Y m and u = C m on Z m where (7.) ( ) ( ) C m = m (a m+ 8a m + a m ) m (η m+ 6η m + η m ), 8 8 ( ) ( ) C m = m (a m+ 8a m + a m ) + m (η m+ 6η m + η m ). 8 8 Because normal derivatives add linearly, u satisfies the normal derivative matching conditions at x m, y m and z m. As a result of the above lemma, we can define the extension operator E which maps two sequences {(a m, η m )} to the function u given in the lemma. This operator is well defined because the process described by the lemma generates exactly one function for each pair of sequences. Additionally, it is not difficult to see that E is a linear operator. Theorem 7.4. There exist a bounded linear extension map E : T dom L () and E : T 2 dom L 2 () with R E = Id. Proof. Suppose {(a m, η m )} T and let u = E{(a m, η m )}. In order to apply the Gluing Theorem, we need to check that u is continuous. It suffices to check for continuity at q 0 because u is clearly continuous everywhere else. In order to show that u is continuous at q 0, we need to show that lim m u(x m ) = lim m u(y m ) = lim m u(z m ). Since {(a m, η m )} T, we have a m = A + A 2 (/) m + a m with m a m l < and m η m Lip <. Then (7.2) reads ( ) m A 2 + ( ) m (η m+ + η m ), u(y m ) = A + 4 u(z m ) = A + 4 ( 8 (a m+ + a m) 8 ) m A (a m+ + a m) + 8 ) m (η m+ + η m ). Taking the limit m in the above equations, we see that A = lim m u(y m ) = lim m u(z m ) = lim m a m, which verifies the continuity of u at q 0. Recall that Lemma 7. tells us that u satisfies the normal derivative matching conditions at {x m }, {y m } and {z m }. Thus the hypotheses of the Gluing Theorem are satisfied, so the theorem implies that u is well defined. We need to show that u L (). Observe that (7.) reads ( ) ( ) C m = m (a m+ 8a m + a 8 m ) m (η m+ 6η m + η m ), 8 ( ) ( ) C m = m (a m+ 8a m + a 8 m ) + m (η m+ 6η m + η m ). 8 (

23 BOUNDARY VALUE PROBLEMS ON A HALF SIERPINSKI GASKET 2 Using Lemma 8.7 to obtain an upper bound on the normal derivative terms in C m and C m, we find that u L C m l + C m l M m a m l + M 2 m η m Lip. Therefore, E : T dom L (). Suppose {(a m, η m )} T 2 and let u = E{(a m, η m )}. Again, we need to check that u is continuous at q 0 in order to apply the Gluing theorem. By definition of T 2, we have a m = A + A 2 (/) m + a m with (2/) m/2 a m l 2 < and m/2 η m l 2 <. Then a m 0 and η m 0. By the same argument for the T case, u is continuous at q 0, hence continuous everywhere. By Lemma 7., u satisfies the normal matching conditions at {x m }, {y m } and {z m }. Then u is well defined by the Gluing Theorem. Finally, u L 2 () because u 2 L = C m 2 + C m 2 ( ) m 2 2 m M a m 2 + M 2 m η m 2. Therefore, E : T 2 dom L 2 () Extensions of Solutions to Differential Equations on Ω. The material presented in this section is motivated by the classical theory of extending functions with u L p on a nice domain in Euclidean space R n to functions with the same property on R n. We ask: () Given u dom L (Ω), does there exist an extension u dom L ()? (2) Given u dom L 2 (Ω), does there exist an extension u dom L 2 ()? We present two motivating examples before we proceed to the main extension results. Theorem 7.. If u is a nonconstant harmonic function on Ω, then its even reflection is not in dom. Proof. Suppose, for the purpose of contradiction, even reflection extends to a function u dom (). Then u = 0 on, and n u(x m ) = 2u(x m ) u(y m ) u(y m ), n u(x m ) = 2u(x m ) u(z m ) u(z m ). Since u is assumed non-constant, both normal derivatives are nonzero. However, n u(x m ) = n u(x m ), so the normal derivative matching condition at x m does not hold. Therefore, u dom (). Theorem 7.6. Suppose u C(Ω) solves the BVP with a 0 = C and a m = (2/)(/) m (C + C 2 ) for some constants C, C 2. Then there exists a harmonic extension of u. Proof. Consider the harmonic function u on determined by the boundary values u(q 0 ) = 0, u(q ) = C and u(q 2 ) = C 2. Simple computation shows that u(x m ) = (2/)(/) m (C +C 2 ). Thus, u = u on Ω and u = 0, which shows that u is indeed a harmonic extension. In special cases, such as the one presented in the previous result, there exists a harmonic extension. In general, the desired extension will not be harmonic because the space of harmonic functions on is a three dimensional space so finding a

24 24 WEILIN LI AND ROBERT S. STRICHARTZ harmonic extension u of u satisfying the infinite number of conditions u(x m ) = a m is unlikely. For that reason, we look for a piecewise biharmonic extension. In fact, this motivates our study of piecewise biharmonic functions to begin with. To prove the existence of an extension, we need the analogue of Lemma 7.. Lemma 7.7. Suppose u dom L (Ω) or u dom L 2 (Ω). Then there exist a sequence {C m } and a piecewise biharmonic function u on satisfying u = u on Ω, u = C m on Z m, and the normal derivative matching conditions hold at {x m } and {z m }. Proof. For convenience, we write a m = u(x m ) and η m = n u(x m ). Consider the function u = u on Ω, (7.4) u(z m ) = 8 (a m+ + a m ) + ( ) m (η m+ + η m ), 8 and u = C m on Z m where ( ) ( ) (7.) C m = m (a m+ 8a m + a m ) + m (η m+ 6η m + η m ). 8 8 For the same reason as before, these constraints completely determine u on Z m. Hence we have defined a function u on. We claim that the normal matching conditions hold at x m and z m. Using (7.), ( ) m n u(x m ) = [2u(x m ) u(z m ) u(z m )] + C m m+, ( ) m n u(z m ) = [2u(z m ) u(z m ) u(x m )] + C m m+, ( ) m+ n u(z m ) = [2u(z m ) u(z m+) u(x m+)] + C m+ m+2. It is straightforward to check that our formulas for u(x m ), u(z m ), and C m imply the matching conditions hold at {x m } and {z m }. The lemma allows us to define an extension operator. Let E Ω be the extension operator that maps a function u dom L (Ω) or u dom L 2 (Ω) to the function E Ω u on as given in the lemma. This operator is well defined because for each u, there is exactly one E Ω u. It is clear that E Ω is linear and that E Ω u is continuous except possibly at q 0. Theorem 7.8. Suppose u dom L (Ω). If Ru T, then E Ω u dom L () and (E Ω u) L () u L (Ω) + C Ru T. The Trace Theorem implies the converse: if E Ω u dom L (), then Ru T. Proof. Suppose u dom L (Ω) and Ru = {(a m, η m )} T. By definition of T, we have a m = A + A 2 (/) m + a m with m a m l < and m η m Lip <. We need to check that E Ω is continuous at q 0. Observe that (7.4) becomes ( ) m E Ω u(z m ) = A + A (a m+ + a m) + ( ) m (η m+ + η m ). 8

25 BOUNDARY VALUE PROBLEMS ON A HALF SIERPINSKI GASKET 2 Taking the limit in the above equation, we see that A = lim m a m = lim m E Ω u(z m ). This proves that E Ω u is continuous. By Lemma 7.7, the matching conditions for u at {x m } and {z m } are satisfied. This allows us to apply the Gluing Theorem, and so (E Ω u) exists. To prove that E Ω u dom L (), observe that and by Lemma 8.7, m (a m+ 8a m + a m ) l 6 m a m l m (η m+ 6η m + η m ) l 6 m η m Lip. Using the above inequalities and the equation for C m given by (7.), we find that (E Ω u) L (Ω ) = max m C m M m a m l + M 2 m η m Lip. Then by triangle inequality, (E Ω u) L () u L (Ω) + M m a m l + M 2 m η m Lip, which completes the proof. Theorem 7.9. Suppose u dom L 2 (Ω). If Ru T 2, then E Ω u dom L 2 () and (E Ω u) 2 L 2 () u 2 L 2 (Ω) + C Ru 2 T 2. The Trace Theorem implies the converse: if E Ω u dom L 2 (), then Ru T 2. Proof. Suppose u dom L 2 (Ω) and Ru = {(a m, η m )} T 2. By definition of T 2, we know that a m = A + A 2 (/) m + a m with (2/) m/2 a m l 2 < and m/2 η m l 2 <. Then a m 0 and η m 0. Using these limits, the same argument given in the proof of Theorem 7.8 shows that E Ω u is continuous. Again, Lemma 7.7 guarantees the matching conditions for u at {x m } and {z m } hold. The Gluing Theorem implies (E Ω u) is well defined. To see why E Ω u dom L 2 (), we first see that (E Ω u) 2 L 2 (Ω ) = ( ) m 2 m C m 2 M a m 2 + M 2 m η m 2. Since (E Ω u) 2 L 2 () = u 2 L 2 (Ω) + (E Ωu) 2 L 2 (Ω ), using the above inequality gives us ( ) m 2 (E Ω u) 2 L 2 () u 2 L 2 (Ω) + M a m 2 + M 2 m η m 2. We can interpret Theorem 7.8 and Theorem 7.9 by the following: Ru T is the minimal condition for extending an arbitrary function in dom L (Ω) to a function in dom L () and Ru T 2 is the minimal condition for extending an arbitrary function in dom L 2 (Ω) to a function in dom L 2 (). A function belonging to dom L 2 (Ω) or dom L (Ω) is naturally a solution to the differential equation u = f for f L 2 or f L respectively. Solutions to this differential equation can be found using Theorem 8.2. As a special case of E Ω, we can extend harmonic functions u on Ω provided that Ru T 2 or Ru T. Recall that the solution to this differential equation was