Quadratic spline collocation for Volterra integral equations D. - PDF document

Quadratic spline collocation for Volterra integral equations D. Saveljeva University of Tartu, Estonia

Consider Volterra integral equation � t y ( t ) = 0 K ( t, s, y ( s )) ds + f ( t ) , t ∈ [0 , T ] , (1) with given functions f : [0 , T ] → I R , K : S × I R → I R and set S = { ( t, s ) | t ∈ [0 , T ] , s ∈ [0 , t ] } . Take a mesh 0 = t 0 < t 1 < . . . < t N = T , denote h i = t i − t i − 1 , i = 1 , . . . , N . The method uses collocation points τ i = t i − 1 + ch i , i = 1 , . . . , N , with c ∈ (0 , 1]. There are imposed the following collocation conditions � τ i u ( τ i ) = 0 K ( τ i , s, u ( s )) ds + f ( τ i ) , i = 1 , . . . , N, and initial values u (0) = y (0) , u ′ (0) y ′ (0) . = We replaced one of initial conditions by the bound- ary condition u ′′ ( t N − 1 − 0) = u ′′ ( t N − 1 + 0) .

We have (1) ⇔ y = Ky + f with � t ( Kv )( t ) = 0 K ( t, s, v ( s )) ds, v ∈ C [0 , T ] . Then our method are u = P N Ku + P N f, where the projections P N are determined by ( P N f )(0) = f (0) , ( P N f )( τ i ) = f ( τ i ) , i = 1 , . . . , N, ( P N f ) ′′ ( t N − 1 − 0) = ( P N f ) ′′ ( t N − 1 + 0) , P N : C [0 , T ] → C [0 , T ] .

Let E be Banach spaces. Suppose we have an equation u = Ku + f, (2) where K ∈ K ( E, E ) and f ∈ E . Let it be given a sequence of approximating operators P N ∈ L ( E, E ), N = 1 , 2 , . . . . Consider also equations u N = P N Ku N + P N f. (3) General convergence theorem. Suppose u = Ku only if u = 0 and P N u → u for all u ∈ E in the process N → ∞ . Then 1. (2) has the unique solution u ∗ ; 2. there is N 0 such that for N ≥ N 0 , (3) has the unique solution u ∗ N ; N → u ∗ as N → ∞ ; 3. u ∗

4. there are C 1 , C 2 , C 3 > 0 such that C 1 � P N u ∗ − u ∗ � ≤ � u ∗ N − u ∗ � ≤ C 2 � P N u ∗ − u ∗ � (4) and N − P N u ∗ � ≤ C 3 � K ( P N u ∗ − u ∗ ) � . � u ∗ (5) P N u → u ∀ u ∈ E ⇒ (4) and (5); ⇓ � P N K − K � → 0 ⇒ (4); ⇓ P N K → K compactly ⇒ (4); ⇓ I − P N K → I − K stably or regularly ⇒ (4) and N − P N u ∗ � ≤ const � P N K ( P N u ∗ − u ∗ ) � . � u ∗

Results: c ∈ (0 , 1) In the space C [0 , T ] for quasi-uniform mesh we get P N → I as N → ∞ . C 1 [0 , T ] In the space for mesh with h N /h N − 1 = O (1) it holds P N → I as h max → 0. c = 1 C 1 [0 , T ] In the space C [0 , T ] and it holds P N = O ( N ). There is no compact convergence P N K → K , hence, there is no convergence � P N K − K � → 0 in both spaces. There is regular convergence I − P N K → I − K in the space C [0 , T ]. There is regular convergence in the Conjecture: space C 1 [0 , T ].

| u N ( τ i ) − y ( τ i ) | = | u N ( τ i ) − P N y ( τ i ) | ≤ � u N − P N y � ≤ ≤ const � P N K ( P N y − y ) � Results in superconvergence: Suppose that K and ∂ K /∂s are con- Theorem. tinuous on { ( t, s ) | 0 ≤ s ≤ t ≤ T } and y ′′′ ∈ Lip 1. Then, for c = 1 / 2, it holds 1 ≤ i ≤ N | u N ( t i − 1 + h/ 2) − y ( t i − 1 + h/ 2) | = O ( h 4 ) . max Theorem. Suppose that K , ∂ K /∂s , ∂ K /∂t , ∂ 2 K /∂t 2 and ∂ 3 K /∂t 3 are continuous on { ( t, s ) : 0 ≤ s ≤ t ≤ T } . Suppose also the function t �→ K ( t, t ) is twice continuously differentiable on [0 , T ] and y ′′′ ∈ Lip 1. Then, for c = 1, 0 ≤ i ≤ N | u N ( t i ) − y ( t i ) | = O ( h 4 ) max in the case of uniform mesh. Conjecture: There is superconvergence in the col- location points in case of c = O ( h 2 ).

Numerical examples: We considered a test equation � t y ( t ) = λ 0 y ( s ) ds + f ( t ) , t ∈ [0 , 1] , (6) with function f ( t ) = 1 2((1 − λ ) sin t + (1 + λ ) cos t + (1 − λ ) e t ) , and the exact solution y ( t ) = (sin t + cos t + e t ) / 2. Another test equation is � t y ( t ) = 0 ( t − s ) y ( s ) ds + sin t, t ∈ [0 , 1] , with the exact solution y ( t ) = (2 sin t + e t − e − t ) / 4. We calculated the error of the method � u − y � ∞ approximately as 0 ≤ k ≤ 10 | u ( t n − 1 + k 10 h ) − y ( t n − 1 + k max max 10 h ) | . 1 ≤ n ≤ N

Quadratic spline collocation for test equation (6): f ( t ) = sin t + e t λ = − 1 , N 4 16 64 1 . 13 · 10 − 3 2 . 21 · 10 − 5 3 . 63 · 10 − 7 c = 1 51 61 63 2 . 54 · 10 − 2 1 . 87 · 10 − 3 1 . 21 · 10 − 4 13 16 16 4 . 41 · 10 − 3 8 . 84 · 10 − 5 1 . 46 · 10 − 6 c = 0 . 5 50 61 63 2 . 54 · 10 − 2 1 . 87 · 10 − 3 1 . 21 · 10 − 4 13 16 16 7 . 71 · 10 − 3 1 . 62 · 10 − 4 2 . 70 · 10 − 6 c = 0 . 1 48 60 63 7 . 11 · 10 − 2 5 . 81 · 10 − 3 3 . 85 · 10 − 4 12 15 16 f ( t ) = (3 cos t − sin t − e t ) / 2 λ = 2 , N 4 16 64 1 . 88 · 10 − 3 4 . 02 · 10 − 5 6 . 82 · 10 − 7 c = 1 47 59 66 2 . 64 · 10 − 2 2 . 52 · 10 − 3 1 . 74 · 10 − 4 10 14 16 2 . 15 · 10 − 3 4 . 94 · 10 − 5 8 . 43 · 10 − 7 c = 0 . 7 44 59 62 4 . 10 · 10 − 2 3 . 40 · 10 − 3 2 . 27 · 10 − 4 12 15 16 7 . 24 · 10 − 3 1 . 56 · 10 − 4 2 . 62 · 10 − 6 c = 0 . 1 46 60 63 6 . 99 · 10 − 2 5 . 79 · 10 − 3 3 . 85 · 10 − 4 12 15 16 t � Numerical results for y ( t ) = ( t − s ) y ( s ) ds + sin t : 0 N 8 32 128 4 . 81 · 10 − 5 9 . 20 · 10 − 7 1 . 51 · 10 − 8 c = 1 52 61 63 1 . 95 · 10 − 4 3 . 71 · 10 − 6 6 . 07 · 10 − 8 c = 0 . 5 53 61 63 3 . 56 · 10 − 4 6 . 84 · 10 − 6 1 . 12 · 10 − 7 c = 0 . 1 52 61 63

Numerical results in superconvergence for test equation (6): f ( t ) = sin t + e t λ = − 1 , N 4 16 64 1 . 2 · 10 − 4 6 . 3 · 10 − 7 2 . 6 · 10 − 9 c = 1 186 239 252 8 . 1 · 10 − 5 4 . 5 · 10 − 7 1 . 9 · 10 − 9 c = 0 . 5 177 236 252 c = 10 − 6 1 . 6 · 10 − 5 8 . 0 · 10 − 8 3 . 3 · 10 − 10 203 244 245 f ( t ) = (3 cos t − sin t − e t ) / 2 λ = 2 , N 4 16 64 1 . 8 · 10 − 4 1 . 1 · 10 − 5 5 . 3 · 10 − 8 c = 1 161 207 249 1 . 2 · 10 − 4 6 . 8 · 10 − 7 3 . 1 · 10 − 9 c = 0 . 5 173 221 247 c = 10 − 6 9 . 3 · 10 − 5 6 . 2 · 10 − 7 2 . 7 · 10 − 9 150 227 251