# SJÄLVSTÄNDIGA ARBETEN I MATEMATIK

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1 SJÄLVSTÄNDIGA ARBETEN I MATEMATIK MATEMATISKA INSTITUTIONEN, STOCKHOLMS UNIVERSITET Matching Conditions for the Laplace and Squared Laplace Operator on a Y-graph av Jean-Claude Kiik No 1 MATEMATISKA INSTITUTIONEN, STOCKHOLMS UNIVERSITET, STOCKHOLM

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3 Matching Conditions for the Laplace and Squared Laplace Operator on a Y-graph Jean-Claude Kiik Självständigt arbete i matematik 15 högskolepoäng, Grundnivå Handledare: Pavel Kurasov 2014

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5 Abstract The Laplace operator and squared Laplace operator on a branching graph are studied. Boundary conditions connecting three functions at a vertex are determined. It is shown that in general the boundary conditions are linear relationships between derivatives of the same degree. Introduction The background to the study of quantum graphs, a graph considered together with an operator and certain boundary conditions, lies in the theory of partial differential equations. The main interest of the square Laplace operator can be seen to stem from the equations for free oscillations of a plate, see Lifshitz and Landau in [7]. The equation ρ 2 ζ t 2 + Eh 2 12(1 σ 2 ) 2 ζ = 0 is a partial differential equation in the time domain, where ρ is the density of matter, ζ the distance function, t time, E the Young s Modulus (the measure of stiffness of the elastic material), h the height of the plate, σ the Poisson ratio (the negative ratio of transverse to axial strain). Fourier transform can be used on the equation to go from a time domain to a frequency domain. Using a Fourier transform we get the following equation, where, ρk 2 u + Eh 2 12(1 σ 2 ) 2 u = 0 Setting ρ = Eh 2 12(1 σ 2 = 1, we get the following partial differential equa- ) tion for u, u = ζe kt dt. 2 u = k 2 u. This equation can be used to describe waves in rods. Our aim is to study the system of three rods sharing one clamp with the three ends free. This system is described by the square Laplacian 2 on a Y-graph with certain boundary conditions at the vertex. 1

6 A lot of focus in research is on inverse scattering problems, see [6], which will not be covered in this report. The boundary, or matching, conditions are what will be discussed in the following sections. The first section will state definitions of concepts concerning unbounded operators and quantum graphs. The second section will treat the negative Laplace operator on a Y-graph and the necessary boundary conditions for it to be symmetric and connected at a vertex. The third section will treat the case of a symmetric Y-graph, with respect to the three angles between the branches, and the square of the Laplace operator acting on it. The last section will treat the general case of the square Laplace operator, with different angles between each branch. Operators and Graphs A graph Γ consists of a finite set of edges E = {e j } and a set of vertices V = {v i }, where each edge lies between two vertices (v i, v k ). If two edges share a vertex they are said to be connected and if all edges share one vertex v the graph Γ is said to be a star graph. We will focus on a graph with three edges sharing one vertex, which we will call a Y-graph. In order to uniquely define this Y-graph each edge e j must have a length, making it a metric graph, Definition 1. A graph Γ is said to be a metric graph if each edge e j is identified by an interval of the real line, i.e. e j = (0, l j ] where 0 < l j. We therefore extend the so-called Y-graph to be a metric graph with each edge e j that can be identified with the interval [0, ) with all the left endpoints identified. The edges emanate outward from the origin and we say that the operators we are interested in are acting on functions f i (x), where x is a point along an edge. The Y-graph with each branch parametrised by a function f i, i = 1, 2,, can be seen in figure 1. So far we have seen the Y-graph of interest and the functions on each edge, but since we will be treating unbounded operators the domain of the operators needs to be defined. The space L 2 [a, b] is defined as Definition 2. The space L 2 [a, b] is the collection of real (complex) valued square integrable functions with the inner product between f and g defined as and the norm of f defined as f, g = b a f(x)g(x)dx b f 2 L 2 [a,b] = f, f = f(x) 2 dx <. a 2

7 Figure 1: Y-graph equipped with functions on each branch. We have defined three functions and their domains, one on each edge of the Y-graph, and therefore we consolidate the Hilbert space with, Definition. The space L 2 (Γ) on Γ consists of functions that are measurable and square integrable on each edge e j and such that f 2 L 2 (Γ) = f 2 L 2 (e j ) < e j E i.e., L 2 (Γ) is the orthogonal direct sum of spaces L 2 (e j ). The scalar product in this setting f, g = Γ f j (x)g j (x)dx. The operators acting on the functions will be the two differential operators, The operator of negative second differentiation - the Laplace operator: L II = d2 dx 2. The operator of fourth order differentiation: L IV = L 2 II = d4 dx 4. We also note that our operators are linear, i.e. fulfilling Definition 4. Let X, Y be two vector spaces. A function L that maps every vector u from the domain D(L) X into a vector v = Lu of Y is called a linear operator from X to Y if L preserves linear relations, that is, if L(α 1 u 1 + α 2 u 2 ) = α 1 Lu 1 + α 2 Lu 2 for all u 1, u 2 from D(L) and all scalars α 1, α 2.

8 A symmetric linear operator is defined as Definition 5. A linear operator that satisfies the relationship Lf, g = f, Lg is said to be symmetric if the condition holds for all f and g in the domain D(L). The triple of metric graph, operator and conditions at the vertex V of the Y-graph are what we define as a quantum graph. Definition 6. A quantum graph is a metric graph Γ equipped with an operator L acting on it, together with matching (vertex) conditions. Second Order Linear Differential Operator The three functions defined on the Y-graph with the Laplace operator acting on them are defined on separate edges and their boundary values at the origin are independent, with an angle of 2α between two of the edges, according to figure 2. Each edge is parametrised by a real parameter x and the corresponding functions are denoted by f j for j = 1, 2,. The boundary values f j (0) and f j (0) should be connected by certain linear conditions to be called matching conditions. The conditions should be chosen so that the operator is symmetric as stated in the previous section. Figure 2: Y-graph with angles 2α and π α. The Laplace operator L II = d2 dx 2 has the domain i=1 W 2 2 [0, ), consisting of all functions f j from L 2 [0, ) such that f j L2 [0, ). The subscript 4

9 II will be dropped for the remainder of this section and L will denote the previous L II. Let us calculate the boundary form of the operator L, Lf, g f, Lg = = = { 0 f j g j dx + { f jg j f j g j dx} f jg j dx + f jg j 0 {f j(0)g j (0) f j (0)g j (0)}. 0 f jg j dx} The question of which vertex conditions need be imposed upon a function from the domain of the operator in order for the sum to be identically 0 is the same as making the operator symmetric. Here we assume that both functions f and g satisfy the same conditions. The following two vertex conditions can be considered on the Y-graph equipped with the negative Laplacian, Example 1. The Dirichlet condition with f 1 (0) = f 2 (0) = f (0) = 0 satisfies the stated requirements and yields the boundary form equal to 0. Imposing the Dirichlet condition at the origin vertex on the Y-graph is an example of a decoupling condition, it disconnects the edges entering the vertex. Under this condition the operator would be the direct sum of the operator on each edge. The edges are in this case independent of each other. Example 2. The Robin condition f j (0) = h jf j (0), with h j R also satisfies the requirements and yields a boundary form equal to 0. This is yet another example of a decoupling condition. In a physical application this would have the same result as the previous example, namely that the system would fall apart at the origin due to the edges being independent of each other. A vertex condition for physical applications often require the functions to be continuous at the vertex. In this case the edges are not independent. This important vertex condition is referred to as the standard, Neumann or Kirchoff conditions. Example. The standard conditions are the following { f 1 (0) = f 2 (0) = f (0) f 1 (0) + f 2 (0) + f (0) = 0. (1) They are chosen so that the system is connected at the origin and the boundary form is identically 0, since 5

10 Lf, g f, Lg = {f j(0)g j (0) f j (0)g j (0)} = g(0) f j(0) f(0) g j (0) 0. An example of a more general vertex condition than the above can be considered to be the δ-type condition Example 4. The δ-type condition is considered to be the following, { f 1 (0) = f 2 (0) = f (0) f 1 (0) + f 2 (0) + f (0) = hf(0) = hδ [f] for h R. These conditions reduce to the standard conditions when h = 0. All of the above examples in this section can be generalised in the following manner, f 1 f 1 i(s I) f 2 = (S + I) f 2 f f where I denotes the identity matrix and S is a unitary matrix containing the information about the conditions. From this we can see that for S = I we get the Dirichlet condition and with S = I we obtain the Neumann condition. For the standard conditions we have S = Fourth Order Linear Differential Operator Symmetric Angle We now consider the operator of fourth order differentiation L IV = d4 dx 4 acting on a function on the metric Y-graph, with a symmetric angle of 2α between two edges, see Figure. For α = 0 the branches f 1 and f 2 would lie on the same branch and for α = π the branches f 1 and f 2 would lie along f. Therefore we restrict the angle α and require that α 0, α π. In this section and the following we will be treating the same operator and for 6

11 Figure : Y-graph with angles 2α and π α. simplicity the subscript IV will be dropped in these two sections, leaving L in place of L IV. The boundary form of the operator L may be evaluated with the aid of partial integration as follows Lf, g f, Lg = ( = = = 0 f (IV ) j g j dx ( f j g j 0 f j g j ) f j g (IV ) j dx f j g j dx f jg j 0 + f jg j 0 ( f j (0)g j (0) + f j (0)g j (0) f j(0)g j (0) + f j(0)g j (0) ). The boundary form above shows us that we need more vertex conditions than in the previous section to ensure that the operator L is symmetric, more precisely in total six are needed. Our aim is to describe conditions that are straight forward generalisations of the standard conditions described in the previous section. The first equation in ( 1), page 5, means that the function is continuous even at the vertex. We reuse this condition requiring f 1 (0) = f 2 (0) = f (0). Another additional condition can be obtained if we require that the tangential lines to the graph of f j lie in the same plane. We should get one condition. Let us discuss how to express this condition using derivatives of f j. We solve this problem by introducing the coordinate system R by considering our Y-graph lying in the xy-plane in R, according to figure 4. Let 7 0 f j g j dx )

12 Figure 4: Y-graph in R with angles 2α, π α and π α. a j be the tangent vector to the curves z = f j (x j ) considered in R. Further, let φ j denote the angles between the tangential lines f j and the xy-plane, it can be observed that tan φ j = f j (0). The vector a is parallel to the x-axis, therefore has zero second coordinate and can be chosen equal to, a = (1, 0, tan φ ) = (1, 0, f (0)). The tangential vectors a 1 and a 2 may then be chosen equal to, 8

13 a 1 = ( cos α, sin α, tan φ 1 ) = ( cos α, sin α, f 1(0)) a 2 = ( cos α, sin α, tan φ 2 ) = ( cos α, sin α, f 2(0)). We want the branches of our Y-graph to be dependent and therefore find a matching condition through examining the determinant of the vectors a j, cos α sin α f 1 (0) det cos α sin α f 2 (0) = sin α(f 1(0) + f 2(0) + 2 cos αf (0)) = f (0) The restrictions α 0, π, mean that the value of sin α is non-zero and can be divided in the last equality above. This gives us the matching condition relating the first derivatives of f j, f 1(0) + f 2(0) + 2 cos αf (0) = 0. So far we have the following two matching conditions, { f 1 (0) = f 2 (0) = f (0) = f(0) f 1 (0) + f 2 (0) + 2 cos αf (0) = 0 (2) To find the remaining matching conditions we use these together with the boundary form at the beginning of this section, but we write f, f, f instead of f (0), f (0), f (0) respectively for all derivatives below for simplicity, since it is understood that we are investigating at the vertex. We use the conditions in ( 2 ) in the third step below, 9

14 Lf, g f, Lg = ( ) = f j (0)g j (0) + f j (0)g j (0) f j(0)g j (0) + f j(0)g j (0) ( ) = f j g(0) + f 1 g 1 + f 2 g 2 + f g f 1g 1 f 2g 2 f g + f j(0) ( ) g j ( ) = f j g(0) + f 1 ( g 2 2 cos αg ) + f 2 g 2 + f g ( ( f 2 2 cos f )g 1 f 2g 2 f g + f(0) g j ( = ) f j g(0) + ( f 1 + f 2 )g 2 + ( 2 cos αf 1 + f )g ( ) f 2( g 1 + g 2 ) f ( 2 cos αg 1 + g ) + f(0) g j. Looking at the last expression above we search for conditions that connect together derivatives of the same order and make the boundary form equal to zero. We choose the following conditions as our final matching conditions, 2 cos αf 1 (0) = 2 cos αf 2 (0) = f (0) f j (0) = 0. We now have the following six matching conditions for the operator L on our Y-graph in figure, ) f 1 (0) = f 2 (0) = f (0) = f(0) f 1 (0) + f 2 (0) + 2 cos αf (0) = 0 2 cos αf 1 (0) = 2 cos αf 2 (0) = f (0) f j (0) = 0. () Under these conditions the fourth order operator L is symmetric (and even self-adjoint). Let us consider two special cases. 10

15 Example 5. In the case α = π 2, cos α = 0 and the Y-graph resembles the letter T. The boundary conditions in ( 2 ) become, { f 1 (0) = f 2 (0) = f (0) = f(0) f 1 (0) + f 2 (0) = 0 Under these conditions the boundary form of L becomes Lf, g f, Lg = ( = f j ) g(0) + ( f 1 + f 2 )g 2 + f g ( ) f 2( g 1 + g 2 ) f g + f(0) g j. We observe that two separate sets of matching conditions ensure that this boundary form equals 0, namely and f 1 (0) = f 2 (0) = f (0) = f(0) f 1 (0) + f 2 (0) = 0 f 1 (0) = f 2 (0) f (0) = 0 f j (0) = 0, f 1 (0) = f 2 (0) = f (0) = f(0) f 1 (0) + f 2 (0) = 0 f (0) = 0 f 1 (0) = f 2 (0) f j (0) = 0. Example 6. We consider the case when α = π, thus cos α = 1. In this 2 case the boundary conditions in ( ) become 11

16 f 1 (0) = f 2 (0) = f (0) = f(0) f 1 (0) + f 2 (0) + f (0) = 0 f 1 (0) = f 2 (0) = f (0) f j (0) = 0. The first two matching conditions coincide with the standard conditions in ( 1 ). This is expected since the Y-graph is symmetric with respect to the angle between each branch when α = π. General Angle Figure 5: Y-graph with angles α, β and γ. In this section we examine a case similar to the previous section with the operator L acting on the Y-graph, but with the exception that we let the angle between each branch be different. We denote the angles by α, β and γ according to figure 5 and require that α, β, γ 0. As before the boundary form for the operator L evaluates to Lf, g f, Lg = ( = f j (0)g j (0) + f j (0)g j (0) f j(0)g j (0) + f j(0)g j ). (0) As earlier we search for six conditions to make the boundary form equal to zero. The first condition of continuity is still applicable f 1 (0) = f 2 (0) = 12

17 Figure 6: Y-graph in R with angles α, β and γ. f (0) and ensures that the branches are connected at the vertex. We continue in much the same way as in the previous section, by noting that the balance equation means that the tangential lines to the graph of f j lie in the same plane, from which we should get one condition. We solve this problem again by introducing the coordinate system R and consider our Y-graph lying in the xy-plane in R, according to figure 6. Let a j be the tangent vector to the curves z = f j (x j ) considered in R. Further, let φ j denote the angles between the tangential lines f j and the xy-plane, it can be observed that tan φ j = f j (0). The vector a is parallel to the x-axis, therefore has zero second coordinate and can be chosen equal to, 1

18 a = (1, 0, tan φ ) = (1, 0, f (0)) The tangential vectors a 1 and a 2 may then be chosen equal to, a 1 = ( cos(π β), sin(π β), tan φ 1 ) = (cos β, sin β, f 1(0)) a 2 = ( cos(π α), sin(π α), tan φ 2 ) = (cos α, sin α, f 2(0)) We want the branches of our Y-graph to be dependent and therefore find a matching condition through examining the determinant of the vectors a j, cos β sin β f 1 (0) det cos α sin α f 2 (0) = 1 0 f (0) sin αf 1(0) + sin βf 2(0) f (0)(cos β sin α + sin β cos α = 0. Using the trigonometric reduction sin α cos β + cos α sin β = sin(α + β) we get the following matching condition, sin αf 1(0) + sin βf 2(0) sin(α + β)f (0) = 0. Further, noting that sin(α + β) = sin(2π γ) = sin γ this condition can be written as sin αf 1(0) + sin βf 2(0) + sin γf (0) = 0 Thus, so far we have the following two matching conditions, { f 1 (0) = f 2 (0) = f (0) sin αf 1 (0) + sin βf 2 (0) + sin γf (0) = 0 To find the remaining matching conditions we use a rearrangement of the conditions above f 1 (0) = f 2 (0) = f (0) f (0) = sin αf 1 (0) + sin βf 2 (0) sin γ together with the boundary form at the beginning of this section. In order to use the stated rearrangement we require that γ is non-zero and γ π so that sin γ 0. Below we write f, f, f instead of f (0), f (0), f (0) respectively for all derivatives for simplicity, since it is understood that we are investigating at the vertex. The boundary form of the operator L becomes, 14

19 Lf, g f, Lg = ( ) = f j (0)g j (0) + f j (0)g j (0) f j(0)g j (0) + f j(0)g j (0) ( ) = f j g(0) + f 1 g 1 + f 2 g 2 + f g ( f 1g 1 f 2g 2 f g + f(0) g j ( = ) ( f j g(0) + f 1 g 1 + f 2 g 2 + f sin αg 1 + sin ) βg 2 sin γ) f 1g 1 f 2g 2 ( sin αf 1 + sin βf 2 sin γ ( = f 1 ( g 1 ) ( f j g(0) + f 1 sin αg sin γ ) ( f 2 g 2 ) ) ( ) g + f(0) g j ) sin βf g 2 sin γ ( sin αf )g 1 sin γ + f 2 ) ( ) sin βg + f(0) g j. sin γ Looking at the last expression above we search for conditions that connect together derivatives of the same order and make the boundary form equal to zero. We choose the following conditions as our final matching conditions, f 1 f 2 (0) sin αf sin γ = 0 (0) sin βf sin γ = 0 f j (0) = 0 Since we have stated earlier that γ is non-zero and γ π we rearrange and, in total, have the following six matching conditions for the operator L on the Y-graph in figure 5 and 6, 15

20 f 1 (0) = f 2 (0) = f (0) sin αf 1 (0) + sin βf 2 (0) + sin γf (0) = 0 sin γf 1 (0) sin αf = 0 sin γf 2 (0) sin βf = 0 f j (0) = 0. There is an interesting symmetry between the angles of the Y-graph and the first derivatives, namely that the first derivatives of each branch is associated with the angle subtending the other two branches, the angle opposite the branch itself. We continue by considering some special cases. Let us examine the cases when the angles α and β are equal to π. Example 7. In the case α = π the value of sin α = 0 and we get a similar case to Example 5, where two different sets of conditions ensure that the boundary form is zero, and f 1 (0) = f 2 (0) = f (0) sin βf 2 (0) + sin γf (0) = 0 sin γf 1 (0) = 0 sin γf 2 (0) sin βf = 0 f j (0) = 0, f 1 (0) = f 2 (0) = f (0) sin βf 2 (0) + sin γf (0) = 0 sin γf 1 (0) = 0 sin γf 2 (0) sin βf = 0 f j (0) = 0. Example 8. In the case β = π the value of sin β = 0 and this case is similar to the previous example when α = π, giving the two different sets of matching conditions, 16

21 and f 1 (0) = f 2 (0) = f (0) sin αf 1 (0) + sin γf (0) = 0 sin γf 2 (0) = 0 sin γf 1 (0) sin αf = 0 f j (0) = 0, f 1 (0) = f 2 (0) = f (0) sin αf 1 (0) + sin γf (0) = 0 sin γf 1 (0) sin αf = 0 sin γf 2 (0) = 0 f j (0) = 0. Example 9. Further, let us examine the case when α, β, γ all are equal to. In this case 2π sin α = sin β = sinγ = sin 2π = 2. The matching conditions become Dividing by f 1 (0) = f 2 (0) = f (0) 2 f 1 (0) + 2 f 2 (0) + 2 f (0) = 0 2 f 1 (0) 2 f = 0 2 f 2 (0) 2 f = 0 f j (0) = 0. 2 gives the following matching conditions f 1 (0) = f 2 (0) = f (0) f 1 (0) + f 2 (0) + f (0) = 0 f 1 (0) = f 2 (0) = f (0) = 0 f j (0). These conditions are expected since they concur with the standard conditions in ( ) and with the results in Example 6. 17

22 Conclusion A final remark about the results achieved above can be made, namely that in all three cases observed, the standard conditions are a set of linear relationships between derivatives of the same degree. For the Laplace operator noting that the standard conditions are linear relationships is straight forward. The standard conditions for the square Laplace operator in the symmetric case can be realised to have a linear relationship since 2 cos α is finite. Linear relationships in the case of a general angle between each branch discussed in the fourth section can be realised since sin α, sin β and sin γ are all finite. The linearity of the matching conditions seems to extend to an operator of any even order of differentiation, or perhaps to any order of differentiation, but proving this would require further study. The symmetry in the matching conditions in the general case would perhaps also extend to a star graph with any finite, or perhaps infinite, number of branches, but this would also warrant further study. 18

23 Bibliography [1] G Berkolaiko, R Carlson, S Fulling, and P Kuchment, editors. Analysis on graphs and its applications, volume Vol 77. American Mathematical Society, Jan-Jun [2] G Berkolaiko and P Kuchment. Introduction to quantum graphs. American Mathematical Society, 201. [] P Exner, J Keating, P Kuchment, T Sunada, and A Teplyaev, editors. Quantum Graphs and their applications. American Mathematical Society, Jun [4] T Kato. Perturbation Theory for Linear Operators. Springer-Verlag, [5] P Kurasov. Quantum Graphs: Spectral Theory and Inverse Problems. Birkhäuser, [6] P Kurasov and F Stenberg. On the inverse scattering problem on branching graphs. J. Phys. A: Math. Gen , [7] L D Landau and E M Lifshitz. Theory of Elasticity. Pergamon Press, [8] V Maz ja. Sobolev Spaces. Springer-Verlag, [9] F Riesz and B. Sz-Nagy. Functional Analysis. Dover Publications, Inc.,

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