Approximation of solutions of DDEs under nonstandard assumptions via Euler scheme

Abstract

We deal with approximation of solutions of delay differential equations (DDEs) via the classical Euler algorithm. We investigate the pointwise error of the Euler scheme under nonstandard assumptions imposed on the right-hand side function f. Namely, we assume that f is globally of at most linear growth, satisfies globally one-side Lipschitz condition but it is only locally Hölder continuous. We provide a detailed error analysis of the Euler algorithm under such nonstandard regularity conditions. Moreover, we report results of numerical experiments.

Appendix. Analytical properties of solutions of ODEs and its Euler approximation

This section consists of some auxiliary results for solutions of ordinary differential equations and its Euler approximation that are used in the paper, especially for proving the main Theorem 1.

Lemma A1.1

Let us consider the following ODE

$$ z^{\prime}(t)=g(t,z(t)), \quad t\in [a,b], \quad z(a)=\xi, $$

(A.1)

where $-\infty <a<b<+\infty$, ξ ∈ ℝ^d and g : [a, b] × ℝ^d → ℝ^d satisfies the following conditions

• (G1) g ∈ C([a, b] × ℝ^d; ℝ^d).

• (G2) There exists $K\in (0,+\infty )$ such that for all (t, y) ∈ [a, b] × ℝ^d

  $$ \|g(t,y)\|\leq K(1+\|y\|). $$

• (G3) There exists H ∈ ℝ such that for all t ∈ [a, b], y₁, y₂ ∈ ℝ^d

  $$ \langle y_1-y_2, g(t,y_1)-g(t,y_2)\rangle \leq H \| y_1 - y_2 \|^2. $$

Then we have what follows.

• (i) The (A.1) has a unique solution $z\in C^{1}([a,b];\mathbb {R}^{d})$ such that

  $$ \sup\limits_{t\in[a,b]}\|z(t)\|\leq (\|\xi\|+K(b-a))e^{K(b-a)}, $$

  (A.2)

  and for all t, s ∈ [a, b]

  $$ \|z(t)-z(s)\|\leq \bar K |t-s|, $$

  (A.3)

  where $\bar K=K\Bigl (1+(\|\xi \|+K(b-a))e^{K(b-a)}\Bigr )$.

• (ii) Let us consider $u:[a,b]\to \mathbb {R}^{d}$ the solution of

  $$ u^{\prime}(t)=g(t,z(t)), \quad t\in [a,b], \quad u(a)=\zeta, $$

  (A.4)

  with $\zeta \in \mathbb {R}^{d}$. Then for all t ∈ [a, b] we have

  $$ \|u(t)-z(t)\|\leq e^{H_{+}(b-a)}\|\xi-\zeta\|. $$

  (A.5)

Proof

Since the right-hand side function g is continuous and it is of at most linear growth, Peano’s theorem guarantees existence of the C¹-solution (e.g., see Theorem 70.4, page 292 in [16]). Now, we show that the uniqueness follows from the one-sided Lipschitz assumption (G3). Namely, let us assume that (A.1) has two solutions z = z(t) and x = x(t) with the same initial value z(a) = x(a) = ξ. By (G3) we have for all t ∈ [a, b] that

$$ \begin{array}{@{}rcl@{}} &&\frac{d}{dt}\| z(t)-x(t)\|^{2} = 2 \left\langle z(t)-x(t), g(t, z(t))-g(t, x(t)) \right\rangle\\ &&\leq 2 H \| z(t)-x(t) \|^{2}\leq 2 H_{+} \| z(t)-x(t) \|^{2}. \end{array} $$

(A.6)

Let us consider the C¹-function $[a,b]\ni t\to \varphi (t) = \| z(t)-x(t)\|^{2}\in [0,+\infty )$, where φ(a) = 0. Integrating two sides of the preceding inequality we get

$$ \varphi(t) \leq 2 H_+ {\int}_a^t \varphi(s) ds. $$

Hence, from the Grownall’s lemma we obtain that φ(t) = 0 for all t ∈ [a, b], which in turns implies that z(t) = x(t) for all t ∈ [a, b].

Note that, by the assumption (G2), for all t ∈ [a, b] it holds

$$ \|z(t)\| \leq \|\xi\| + {{\int}_{a}^{t}} \|g(s, z(s))\| ds\leq \|\xi\| + K(b-a) + K {{\int}_{a}^{t}}\| z(s) \| ds. $$

(A.7)

Again by the Gronwall’s lemma we obtain for all t ∈ [a, b] that

$$ \|z(t)\| \leq (\|\xi\| + K(b-a)) e^{K(t-a)}. $$

(A.8)

This implies (A.2).

By (G2) and (A.2) we obtain for all t, s ∈ [a, b]

$$ \begin{array}{@{}rcl@{}} &&\|z(t) - z(s)\| = \|z(t \vee s) - z(t \wedge s)\| \leq {\int}_{t \wedge s}^{t \vee s} \|g(u, z(u))\| du\\ &&\leq K (1 + \sup_{t \in [a,b]} \|z(t)\|)(t \wedge s - t \vee s) \leq \bar{K} |t-s|, \end{array} $$

(A.9)

where $\bar {K}= K\Bigl (1+(\|\xi \|+K(b-a))e^{K(b-a)}\Bigr )$. Hence, the proof of (A.3) is completed.

As in (A.6) we get that for all t ∈ [a, b]

$$ \frac{d}{dt}\| z(t)-u(t)\|^{2} \leq 2H_{+}\|z(t)-u(t)\|^{2} $$

(A.10)

which implies that

$$ \|z(t)-u(t)\|^{2}\leq \|\xi-\zeta\|^{2}+2H_{+}{\int\limits_{a}^{t}}\|z(s)-u(s)\|^{2}ds. $$

(A.11)

By using Gronwall’s lemma we get (A.5). □

Lemma A.2

Let us consider the following ordinary differential equation

$$ z^{\prime}(t)=g(t,z(t)), \quad t\in [a,b], \quad z(a)=\xi, $$

(A.12)

where $-\infty <a<b<+\infty$, η ∈ ℝ^d and g : [a, b] × ℝ^d → ℝ^d satisfies the following conditions:

• (G1) g ∈ C([a, b] × ℝ^d; ℝ^d).

• (G2) There exists $K\in (0,+\infty )$ such that for all (t, y) ∈ [a, b] × ℝ^d

  $$ \|g(t,y)\|\leq K(1+\|y\|). $$

• (G3) There exists H ∈ ℝ such that for all t ∈ [a, b], y₁, y₂ ∈ ℝ^d

  $$ \langle y_1-y_2, g(t,y_1)-g(t,y_2)\rangle \leq H \| y_1 - y_2 \|^2. $$

• (G4) There exist L ≥ 0, $p \in \mathbb {N}$, α, β₁, β₂,…,β_p ∈ (0, 1], such that for all t₁, t₂ ∈ [a, b], y₁, y₂ ∈ ℝ^d

  $$ \|g(t_1,y_1)-g(t_2,y_2)\|\leq L\Bigl((1+\|y_1\|+\|y_2\|)\cdot |t_1-t_2|^{\alpha}+ {\sum}_{i=1}^p \|y_1-y_2\|^{\beta_i}\Bigr). $$

Let us consider the Euler method based on equidistant discretization. Namely, for n ∈ ℤ+, ${\Delta }\in [0,+\infty )$ we set h = (b − a)/n, t_k = a + kh, k = 0, 1,…,n, and let y₀ ∈ ℝ^d be any vector from the ball B(ξ,Δ) = {y ∈ R^d : ∥ξ − y∥≤Δ}. We take

$$ y_{k+1}=y_{k}+h \cdot g(t_{k},y_{k}), \quad k=0,1,\ldots,n-1. $$

(A.13)

Then the following holds

• (i) There exists $\tilde {C_{1}}=\tilde {C_{1}}(b-a,K)\in (0,+\infty )$ such that for all n ∈ ℤ+, ${\Delta } \in [0,+\infty )$, $\xi \in \mathbb {R}^{d}$ y₀ ∈ B(ξ,Δ) we have

  $$ \max\limits_{0\leq k\leq n}\|y_{k}\|\leq \tilde{C_{1}}(1+\|\xi\|)(1+{\Delta}). $$

  (A.14)

• (ii) There exists $\tilde {C_{2}}=\tilde {C_{2}}(b-a,L,K,H,p,\alpha , \beta _{1},\ldots ,\beta _{p})\in (0,+\infty )$ such that for all n ∈ ℤ+, ${\Delta } \in [0,+\infty )$, $\xi \in \mathbb {R}^{d}$ y₀ ∈ B(ξ,Δ) we have

  $$ \max\limits_{0\leq k\leq n}\|z(t_{k})-y_{k}\|\leq \tilde{C_{2}}(1+\|\xi\|)\Bigl({\Delta}+h^{\alpha} + \sum\limits_{i=1}^{p} h^{\beta_{i}}\Bigr). $$

  (A.15)

Proof

Fix n ∈ ℤ+, ${\Delta }\in [0,+\infty )$, $\xi \in \mathbb {R}^{d}$ and y₀ ∈ B(ξ,Δ). By the assumption (G2) we have that for all k = 0, 1,…,n − 1

$$ \|y_{k+1}\|\leq \|y_k\|+h \|g(t_k,y_k)\| \leq (1+hK)\|y_k\|+hK $$

and, since y₀ ∈ B(ξ,Δ), ∥y₀∥≤Δ + ∥ξ∥. Hence, by the discrete version of Gronwall’s lemma we get that for all k = 0, 1,…,n that

$$ \|y_k\|\leq e^{K(b-a)}({\Delta} + \|\xi\|) + e^{K(b-a)}-1\leq e^{K(b-a)}(1+\|\xi\|)(1+{\Delta}). $$

This proves (A.14) with $\tilde C_{1}= e^{K(b-a)}$.

We now prove (A.15). For k = 0, 1,…,n − 1 we consider the following local ODE

$$ z_{k}^{\prime}(t)=g(t,z_{k}(t)), \quad t\in [t_{k},t_{k+1}], \quad z_{k}(t_{k})=y_{k}. $$

(A.16)

By Lemma A1.1 there exists unique solution $z_{k}:[t_{k},t_{k+1}]\to \mathbb {R}^{d}$ of (A.16). From the assumption (G2) and by (A.14) we get for all t ∈ [t_k, t_k+ 1], k = 0, 1,…,n − 1 that

$$ \|z_k(t)\| \leq \tilde{C_1} (1+\|\xi\|)(1+{\Delta})+K(b-a)+K\int\limits_{t_k}^t \|z_k(s)\|ds. $$

The use of Gronwall’s lemma yields

$$ \max\limits_{0\leq k\leq n-1}\sup\limits_{t\in [t_{k},t_{k+1}]}\|z_{k}(t)\|\leq C_{2} (1+\|\xi\|)(1+{\Delta}), $$

(A.17)

where $C_{2}=e^{K(b-a)} \Bigl (\tilde {C_{1}}+K(b-a) \Bigr )$. By (G2) and (A.17) we get for all t ∈ [t_k, t_k+ 1], k = 0, 1,…,n − 1

$$ \begin{array}{@{}rcl@{}} &&\|z_{k}(t)-y_{k}\|\leq \int\limits_{t_{k}}^{t} \|g(s,z_{k}(s))\|ds\leq hK\Bigl(1+\sup\limits_{t\in [t_{k},t_{k+1}]}\|z_{k}(t)\|\Bigr)\\ &&\leq hK\Bigl(1+C_{2}(1+\|\xi\|)(1+{\Delta})\Bigr)\leq hC_{3}(1+\|\xi\|)(1+{\Delta}), \end{array} $$

(A.18)

with C₃ = K(1 + C₂). From Lemma A1.1 (ii) we arrive at

$$ \begin{array}{@{}rcl@{}} \|z(t_{k+1})-y_{k+1}\|&\leq& \|z(t_{k+1})-z_{k}(t_{k+1})\|+\|z_{k}(t_{k+1})-y_{k+1}\|\\ &\leq& e^{hH_{+}}\|z(t_{k})-y_{k}\|+\|z_{k}(t_{k+1})-y_{k+1}\| \end{array} $$

(A.19)

for k = 0, 1,…,n − 1. Using (G4), (A.14), (A.17) and (A.18) we get

$$ \begin{array}{@{}rcl@{}} &&\|z_{k}(t_{k+1})-y_{k+1}\|\leq\int\limits_{t_{k}}^{t_{k+1}}\|g(s,z_{k}(s))-g(t_{k},y_{k})\|ds\\ & \leq& L\int\limits_{t_{k}}^{t_{k+1}}\Bigl((1+\|z_{k}(s)\|+\|y_{k}\|)\cdot (s-t_{k})^{\alpha}+ \sum\limits_{i=1}^{p} \|z_{k}(s)-y_{k}\|^{\beta_{i}}\Bigr)ds\\ &\leq& Lh\left( (1+(\tilde{C_{1}}+C_{2})(1+\|\xi\|)(1+{\Delta}))\frac{1}{\alpha+1}\cdot h^{\alpha}\right.\\ &&\left.+ \sum\limits_{i=1}^{p} h^{\beta_{i}}C_{3}^{\beta_{i}}(1+\|\xi\|)^{\beta_{i}}(1+{\Delta})^{\beta_{i}}\right). \end{array} $$

(A.20)

It is easy to see that for all x ∈ ℝ₊ ∪{0}, ϱ ∈ (0, 1]

$$ (1+x)^{{\varrho}}\leq 2(1+x). $$

(A.21)

Hence

$$ \|z_{k}(t_{k+1})-y_{k+1}\|\leq C_{4} (1+\|\xi\|)(1+{\Delta}) h \left( h^{\alpha} + \sum\limits_{i=1}^{p} h^{\beta_{i}}\right), $$

(A.22)

where $\displaystyle {C_{4}=L\max \limits \Bigl \{\frac {1+\tilde {C_{1}}+C_{2}}{\alpha +1}, 4 \max \limits _{1\leq j \leq p}C_{3}^{\beta _{j}} \Bigr \}}$. From the above considerations we see that C₄ depends only on α, β₁,…,β_p, p, L, K, b − a. By (A.19) and (A.22) we get

$$ \|z(t_{k+1})-y_{k+1}\|\leq e^{hH_{+} } \cdot \| z(t_{k})-y_{k}\| +C_{4} (1+\|\xi\|)(1+{\Delta}) h \left( h^{\alpha} + \sum\limits_{i=1}^{p} h^{\beta_{i}}\right). $$

(A.23)

Let us denote

$$ e_{k}=z(t_{k})-y_{k}, \quad k=0,1,\ldots,n. $$

(A.24)

Of course ∥e₀∥ = ∥ξ − y₀∥≤Δ. Hence, we arrive at the following recursive inequality

$$ \|e_{k+1}\|\leq e^{hH_{+}}\|e_{k}\| + C_{4} (1+\|\xi\|)(1+{\Delta}) h \left( h^{\alpha} + \sum\limits_{i=1}^{p} h^{\beta_{i}}\right), $$

(A.25)

for k = 0, 1,…,n − 1. Applying the Gronwall’s lemma we get when H₊ > 0

$$ \begin{array}{@{}rcl@{}} &&\|e_{k}\| \leq e^{H_{+}(b-a)} {\Delta} + \frac{e^{H_{+}(b-a)}-1}{e^{hH_{+}}-1} C_{4} (1+\|\xi\|)(1+{\Delta}) h \left( h^{\alpha} + \sum\limits_{i=1}^{p} h^{\beta_{i}}\right)\\ &&\leq e^{H_{+}(b-a)} {\Delta} + \frac{e^{H_{+}(b-a)}-1}{H_{+}} C_{4} (1+\|\xi\|)(1+{\Delta}) \left( h^{\alpha} + \sum\limits_{i=1}^{p} h^{\beta_{i}}\right) \end{array} $$

(A.26)

for k = 0, 1,…,n, and when H₊ = 0

$$ \|e_{k}\| \leq {\Delta} + C_{4} (1+\|\xi\|)(1+{\Delta})(b-a) \left( h^{\alpha} + \sum\limits_{i=1}^{p} h^{\beta_{i}}\right) $$

(A.27)

for k = 0, 1,…,n. By elementary calculations we arrive at (A.15). □

The following fact is well-known, see, for example, pages 3–4 in [15].

Lemma A.3

For all ϱ ∈ (0, 1] and x, y ∈ ℝ it holds

$$ |x|^{{\varrho}}\leq 1+|x|, $$

(A.27)

and

$$ \left||x|^{{\varrho}}-|y|^{{\varrho}}\right|\leq |x-y|^{{\varrho}}. $$

(A.28)

Below we establish main properties of the functions (5.1) and (5.2).

Fact A.4

Let A, B, C ≥ 0, D ∈ ℝ, ϱ, γ ∈ (0, 1] and consider $f:[0,+\infty )\times \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ defined as follows¹

$$ f(t,y,z)=A-B\cdot \text{sgn} (y)\cdot |y|-C\cdot \text{sgn}(y)\cdot |y|^{{\varrho}}\cdot |z|^{\gamma} +D \cdot y \cdot |z|^{\gamma}. $$

Then the function f satisfies the assumptions (F1)–(F4).

Proof

Let us define h₁(y) = −sgn(y) ⋅|y| = −y, h₂(y) = −sgn(y) ⋅|y|^ϱ for all y ∈ ℝ. Then we can write that

$$ f(t,y,z)=A+B\cdot h_{1}(y)+C\cdot h_{2}(y)\cdot |z|^{\gamma} + D \cdot y \cdot |z|^{\gamma}. $$

(A.29)

Of course h₁ ∈ C(ℝ). Moreover,

$$ \lim\limits_{y\to 0-}h_{2}(y)=\lim\limits_{y\to 0-}(-y)^{{\varrho}}=0=h_{2}(0)=\lim\limits_{y\to 0+}h_{2}(y)=\lim\limits_{y\to 0+}(-1)\cdot y^{{\varrho}}, $$

(A.30)

therefore h₂ ∈ C(ℝ). In particular, this implies that $f \in C([0, \infty ) \times \mathbb{R} \times \mathbb{R} )$, so f satisfies (F1).

By Lemma A.3 we have |h₁(y)| = |y|≤ 1 + |y|, |h₂(y)| = |y|^ϱ ≤ 1 + |y| for all y ∈ ℝ. Hence, for $(t, y, z) \in [0,+\infty )\times \mathbb{R} \times \mathbb{R}$

$$ \begin{array}{@{}rcl@{}} |f(t,y,z)| &\leq& A+B\cdot |h_{1}(y)|+C\cdot |h_{2}(y)|\cdot |z|^{\gamma} + |D||y||z|^{\gamma} \\ &\leq& A+B (1+|y|)+C (1+|y|)(1+|z|) + |D|(1+|y|)(1+|z|) \\ &\leq& (A+B+C+|D|)(1+|y|)(1+|z|), \end{array} $$

and therefore f satisfies (F2).

For all $t\geq 0,z,y_{1},y_{2}\in \mathbb {R}$

$$ \begin{array}{@{}rcl@{}} &&(y_{1}-y_{2})\Bigl(f(t,y_{1},z)-f(t,y_{2},z)\Bigr) \\ &=& B (y_{1}-y_{2}) \Bigl(h_{1}(y_{1})-h_{1}(y_{2})\Bigr) +C\cdot |z|^{\gamma} (y_{1}-y_{2})\Bigl(h_{2}(y_{1})-h_{2}(y_{2})\Bigr)\\&& + D \cdot |z|^{\gamma} (y_{1}-y_{2})^{2} \end{array} $$

Since h₁, h₂ are decreasing, it holds for all y₁, y₂ ∈ ℝ, i = 1, 2 that (y₁ − y₂)(h_i(y₁) − h_i(y₂)) ≤ 0. Moreover B, C,|z|^γ ≥ 0, hence

$$ (y_1-y_2)\Bigl(f(t,y_1,z)-f(t,y_2,z)\Bigr) \leq D \cdot |z|^{\gamma} (y_1-y_2)^2 \leq |D| \cdot (1+|z|) (y_1-y_2)^2, $$

and f satisfies (F3).

For all y₁, y₂ ∈ ℝ we have that |h₁(y₁) − h₁(y₂)| = |y₁ − y₂|. We now justify that h₂ satisfies for all y₁, y₂ ∈ ℝ

$$ |h_{2}(y_{1})-h_{2}(y_{2})| \leq 2|y_{1} - y_{2}|^{{\varrho}}. $$

(A.31)

When y₁, y₂ < 0 or y₁, y₂ ≥ 0, by Lemma A1.3, we have

$$ |h_2(y_1)-h_2(y_2)| = \big| |y_1|^{{\varrho}} - |y_2|^{{\varrho}} \big| \leq |y_1 - y_2|^{{\varrho}} \leq 2 |y_1 - y_2|^{{\varrho}}. $$

For the case when y₁ < 0, y₂ ≥ 0 (the case y₁ ≥ 0, y₂ < 0 is analogous) we have

$$ |h_2(y_1)-h_2(y_2)| = ||y_1|^{{\varrho}}+y_2^{{\varrho}}|=|y_1|^{{\varrho}}+y_2^{{\varrho}}\leq 2|y_1-y_2|^{{\varrho}}, $$

since − y₁ > 0, |y₁ − y₂| = y₂ + (−y₁) ≥ y₂ ≥ 0, |y₁ − y₂| = y₂ + (−y₁) ≥−y₁ = |y₁|≥ 0, and $[0,+\infty )\ni x\to x^{{\varrho }}$ is increasing. Combining the facts above we obtain for all t₁, t₂ ≥ 0, $y_{1},y_{2},z_{1},z_{2}\in \mathbb {R}$

$$ \begin{array}{@{}rcl@{}} &&|f(t_{1},y_{1},z_{1})-f(t_{2},y_{2},z_{2})| \\ &\leq& B|h_{1}(y_{1})-h_{1}(y_{2})| + C\left||z_{1}|^{\gamma}\cdot h_{2}(y_{1}) - |z_{2}|^{\gamma}\cdot h_{2}(y_{2})\right| + |D|\cdot \left|y_{1} |z_{1}|^{\gamma} - y_{2} |z_{2}|^{\gamma}\right|\\ &\leq& B|y_{1}-y_{2}|+C\cdot |z_{1}|^{\gamma}\cdot |h_{2}(y_{1})-h_{2}(y_{2})|+C\cdot |h_{2}(y_{2})|\cdot \left||z_{1}|^{\gamma}-|z_{2}|^{\gamma}\right| \\ &&+ |D|\cdot |y_{2}| \cdot\left| |z_{1}|^{\gamma} - |z_{2}|^{\gamma}\right| + |D|\cdot |z_{1}|^{\gamma} \cdot |y_{1} - y_{2}| \\ &\leq& B|y_{1}-y_{2}|+2 C (1+|z_{1}|) |y_{1} - y_{2}|^{{\varrho}} +C(1+|y_{2}|)\cdot |z_{1}-z_{2}|^{\gamma} \\ &&+ |D|\cdot |y_{2}|\cdot |z_{1} - z_{2}|^{\gamma} + |D|\cdot (1+|z_{1}|)\cdot |y_{1} - y_{2}| \\ &\leq& \max\{B+|D|, 2C, C+|D|\} \left[ (1+|z_{1}|+|z_{2}|)(1+|y_{1}|+|y_{2}|)|t_{1}-t_{2}| \right.\\ &&+ (1+|z_{1}|+|z_{2}|)\cdot |y_{1}-y_{2}|+(1+|z_{1}|+|z_{2}|)\cdot |y_{1}-y_{2}|^{{\varrho}}\\ &&\left.+ (1+|y_{1}|+|y_{2}|)\cdot |z_{1}-z_{2}|^{\gamma} \right]. \end{array} $$

That ends the proof. □

The proof of the fact below is analogous to the proof of Fact A.4 and is omitted.

Fact A.5

Let A, B, C ≥ 0, D ∈ ℝ, ϱ, γ ∈ (0, 1] and define a function $f:[0,+\infty )\times \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ as follows

$$ f(t,y,z)=A-B\cdot \text{sgn} (y)\cdot |y|-C\cdot \text{sgn}(y)\cdot |y|^{{\varrho}}\cdot |z| +D \cdot y \cdot z. $$

Then the function f satisfies the assumptions (F1)–(F4).

The proof of the following fact is straightforward.

Fact A.6

For all $h \in (0,\frac {1}{2})$ it holds

$$ 0 < \frac{1}{1-h} \leq 1 + 2h \leq 2. $$