# Sherman-Morrison-Woodbury Formula for Linear Integrodifferential Equations

Authors: Feng Wu
Journal: Mathematical Problems in Engineering
 (2016)

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## Abstract

The well-known Sherman-Morrison-Woodbury formula is a powerful device for calculating the inverse of a square matrix. The paper finds that the Sherman-Morrison-Woodbury formula can be extended to the linear integrodifferential equation, which results in an unified scheme to decompose the linear integrodifferential equation into sets of differential equations and one integral equation. Two examples are presented to illustrate the Sherman-Morrison-Woodbury formula for the linear integrodifferential equation.


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## Body

## 1. Introduction

This paper is devoted to the study of the linear integrodifferential equation (IDE):(1)Lvθ,x+kθ,x∫ΘθQθ,x,s,Dvs,xds=fθ,x,x∈Ω,θ∈Θ,Θθ⊆Θwith the homogeneous boundary conditions:(2)Bvθ,x=0,x∈∂Ω,Cvθ,x=0,θ∈∂Θ,where(3)θ=θ1θ2⋯θmT∈Θ,x=x1x2⋯xnT∈Ω,Lvθ,x, Bvθ,x, Cvθ,x, and Dvθ,x are all linear differential operators. Qθ,s,Dvs,x is a linear function with respect to Dvθ,x. Θθ⊆Θ is dependent on the variableθ. vθ,x is the unknown function of two types of variables, that is, θ and x. However, the integration in (1) is with respect to θ, without involving x.The IDE can be found to describe various kinds of phenomena such as Bose-Einstein condensates [1], wave-power hydraulics [2], heterogeneous heat transfer [3], slow erosion of granular flow [4], and wind ripple in the desert [5]. Solving this equation has been one of the interesting tasks for researchers. To date, there are many numerical methods such as finite element method [6] and finite difference method [7] that can be used to obtain an approximate solution for this problem. There are also some analytical methods such as the classical polynomial expansion method [8], the operational approach [9], the homotopy perturbation method [10], the separation of variables technique [11], and the resolvent method [12]. When we employ these analytical methods to solve the IDE, one common feature is that the IDE is converted to a series of pure differential equations and pure integral equations. In this article we propose a new technique for (1), that is, the Sherman-Morrison-Woodbury (SMW) formula [13]. The SMW formula is a powerful device for calculating the inverse matrix of a square matrix, which will be described briefly in Section 2. In Section 3, we extend the SMW formula to solve the IDE. We find that the SMW formula provides a unified procedure to decompose the linear IDE into several differential equations and one pure integral equation. In Section 4 two simple examples are used to explain this scheme.


## 2. Sherman-Morrison-Woodbury Formula

The SMW formula was developed for calculating the inverse matrix of a square matrix. Suppose that we have already obtained, by herculean effort, the inverse matrixA-1 of a square matrix A. Then a small change, for example, ΔA=abT, in A happens, and we need to calculate the inverse matrix A+ΔA-1. SMW formula is a powerful device for this problem.Lemma 1 (Sherman-Morrison-Woodbury formula).
IfΔA=abT, the inverse matrix of A+ΔA is(4)A+ΔA-1=A-1-A-1aI+bTA-1a-1bTA-1.Proof.
Refer to [13].It can be observed from (4) that calculating A+ΔA-1 is converted to calculating I+bTA-1a-1. In the next section we will develop the IDE version of the SMW formula.


## 3. SMW Formula for the IDE

In this section we extend the SMW formula to the IDE.Definition 2.
The functionws,c;θ,x is the Green function of the following differential equation (5), if it satisfies(5)Lws,c;θ,x=δx-cδθ-s,x∈Ω,θ∈Θ,Bws,c;θ,x=0,x∈∂Ω,Cws,c;θ,x=0,θ∈∂Θ,where(6)s=s1s2⋯smT∈Θ,c=c1c2⋯cnT∈Ω,δx-c=∏i=1nδxi-ci,δθ-s=∏j=1mδθi-si.δx is the Dirac function.Lemma 3.
For the following differential equation:(7)Luθ,x=fθ,x,x∈Ω,θ∈Θ,Buθ,x=0,x∈∂Ω,Cuθ,x=0,θ∈∂Θ,its solution is(8)uθ,x=∫Θ∫Ωws,c;θ,xfs,cdsdc.Proof.
Substituting (8) into (7), we have (9)Luθ,x=L∫Θ∫Ωws,c;θ,xfs,cdsdc=∫Θ∫Ωδx-cδθ-sfs,cdsdc=fθ,x,Buθ,x=B∫Θ∫Ωws,c;θ,xfs,cdsdc=∫Θ∫ΩBws,c;θ,xfs,cdsdc=0,Cuθ,x=C∫Θ∫Ωws,c;θ,xfs,cdsdc=∫Θ∫ΩCws,c;θ,xfs,cdsdc=0.So the proof of the lemma is completed.Note that in the proof we assume that the linear differential operatorsL, B, and C and the integral operator∫Θ∫Ω· are commutative. The assumption holds throughout the paper.Theorem 4.
The solution of (5) can be written as(10)vθ,x=uθ,x-pθ,x,where uθ,x satisfies(11)Luθ,x=fθ,x,x∈Ω,θ∈Θ,Buθ,x=0,x∈∂Ω,Cuθ,x=0,θ∈∂Θ,pθ,x satisfies(12)Lpθ,x=kθ,xgθ,x,x∈Ω,θ∈Θ,Bpθ,x=0,x∈∂Ω,Cpθ,x=0,θ∈∂Θ,and gθ,x satisfies(13)gθ,x+∫Θ∫Ωqs,c;θ,xks,cgs,cdsdc=∫ΘθQθ,x,s,Dus,xdsin which(14)qs,c;θ,x=∫ΘθQθ,x,α,Dws,c;α,xdα.Proof.
It can be observed easily in terms of (11) and (12) that (10) satisfies the boundary conditions defined by (2). Substituting uθ,x-pθ,x into the left side of (1) and using (11) and (12), we have (15)Lvθ,x+kθ,x∫ΘθQθ,x,s,Dvs,xds=Luθ,x-Lpθ,x+kθ,x∫ΘθQθ,x,s,Dus,xds-kθ,x∫ΘθQθ,x,s,Dps,xds=fθ,x-kθ,xgθ,x+kθ,x∫ΘθQθ,x,s,Dus,xds-kθ,x∫ΘθQθ,x,s,Dps,xds.In terms of Lemma 3, the solution of (12) can be written as(16)ps,x=∫Θ∫Ωwα,c;s,xkα,cgα,cdαdc.Using (14) and (16) to the last term in the right side of (15) yields(17)∫ΘθQθ,x,s,Dps,xds=∫ΘθQθ,x,s,D∫Θ∫Ωwα,c;s,xkα,cgα,cdαdcds=∫Θ∫Ω∫ΘθQθ,x,s,Dwα,c;s,xdskα,cgα,cdαdc=∫Θ∫Ωqα,c;θ,xkα,cgα,cdαdc=∫Θ∫Ωqs,c;θ,xkα,cgs,cdsdc.Substitute (17) into (15) and then use (29) to the result(18)Lvθ,x+kθ,x∫ΘθQθ,x,s,Dvs,xds=Luθ,x-Lpθ,x+kθ,x∫ΘθQθ,x,s,Dus,xds-kθ,x∫ΘθQθ,x,s,Dps,xds=fθ,x-kθ,xgθ,x+∫Θ∫Ωqs,c;θ,xks,cgs,cdsdc-∫ΘθQθ,x,s,Dus,xds=fθ,x.So the proof of the theorem is completed.Note that in the proof we assume that the linear differential operatorsQ, Dand the integral operator∫Θ∫Ω·are commutative. The assumption holds throughout the paper.Equation (10) can be seen as the IDE version of the Sherman-Morrison-Woodbury formula. In order to explain this point more clearly, we introduce some notations.Definition 5.
The following linear operators are defined by(19)L∘vθ,x=Lvθ,x,S∘vθ,x=∫ΘθQθ,x,s,Dvs,xds.Definition 6.
The inverse linear operator ofL is defined by(20)L-1∘fx,θ=ux,θ=∫Θ∫Ωws,c;θ,xfs,cdsdc,where ws,c;θ,x is the Green function defined by (5).
Using (19) and (20) to (1) yields(21)Lvθ,x+kθ,x·S∘vθ,x=fθ,x,x∈Ω,θ∈Θ.Lemma 7.
Equation (29) can be rewritten as(22)gθ,x+S∘L-1∘kθ,x·gθ,x=S∘L-1fθ,x.Proof.
In terms of (19), we have (23)∫Ω∫Θqs,c;θ,xfs,cdsdc=∫Ω∫Θ∫ΘθQθ,x,α,Dws,c;α,xdαfs,cdsdc=∫ΘθQθ,x,α,D∫Ω∫Θws,c;α,xfs,cdsdcdα=∫ΘθQθ,x,α,DL-1fx,θdα=S∘L-1∘fx,θ,Substituting (23) into (5) yields (22).We denote the solutiongθ,xof (22) by(24)gθ,x=1+S∘L-1∘kθ,x-1S∘L-1fθ,x.Theorem 8 (Sherman-Morrison-Woodbury formula).
The solution of (1) can be written as (25)vθ,x=L-1fθ,x-L-1kθ,x1+S∘L-1∘kθ,x-1S∘L-1fθ,x.Proof.
In terms of (20), we have (26)uθ,x=L-1∘fθ,x,pθ,x=L-1∘kθ,x·gθ,x.Combining (26) with (24) gives (27)vθ,x=uθ,x-pθ,x=L-1fθ,x-L-1kθ,xgθ,x=L-1fθ,x-L-1kθ,x1+S∘L-1∘kθ,x-1S∘L-1fθ,x.So the proof of the theorem is completed.It can be observed clearly that (25) is the SMW formula for the IDE. Theorems 4 and 8 show that solving the IDE (1) could be converted to solving the integral equation (13) or (29).According to Lemma3, the solution of (11) can be written as(28)us,x=∫Θ∫Ωwα,c;s,xfα,cdαdc.Using (28) to the right-hand side of (13) yields(29)gθ,x+∫Θ∫Ωqs,c;θ,xks,cgs,cdsdc=∫Ω∫Θqs,c;θ,xfs,cdsdc.Multiply both sides of (29) bykθ,xand rearrange the equation:(30)kθ,xgθ,x+∫Θ∫Ωqs,c;θ,xks,cgs,c-fs,cdsdc=0.Let(31)hθ,x=fθ,x-kθ,xgθ,xthen (30) can be rewritten as(32)hθ,x+∫Θ∫Ωqs,c;θ,xhs,cdsdc=fs,c.Noting (10)–(12) and (31), we have the following corollary.Corollary 9.
The solution of (1) satisfies (33)Lvθ,x=hθ,x,x∈Ω,θ∈Θ,Bvθ,x=0,x∈∂Ω,Cvθ,x=0,θ∈∂Θ,where hθ,x is defined by (32).In terms of Theorem4, we can obtain the following corollary.Corollary 10.
Ifkθ,x=kx is independent of the variable θ and Qθ,x,s,Dvs,x=Qx,s,Dvs,x is also independent of the variable θ, then the solution is(34)vθ,x=L-1∘fθ,x-L-1∘kx·gx,where gx satisfies(35)gx+∫Θ∫Ωqs,c;xkcgcdsdc=∫Ω∫Θqs,c;xfs,cdsdcin which (36)qs,c;x=∫ΘQx,α,Dws,c;α,xdα.Corollary10 shows that if kθ,x and Qx,s,Dvs,x are independent of the variable θ, solving the IDE is converted to solving the integral equation (35) which only involves the variable x. Hence, in this case, the problem is reduced by using Corollary 10.


## 4. Examples

In this section we present two simple examples. These examples are considered to illustrate the SMW formula for the IDE.Example 1.
We consider the following linear IDE firstly:(37)∂xxvθ,x+∫01e-θ∂xxvθ,xdθ=π2sin⁡πxeθ+1,vθ,0=vθ,1=0.According to Theorem 4, the solution of (37) can be written as (38)vθ,x=uθ,x-pθ,x,where uθ,x satisfies(39)∂xxuθ,x=π2sin⁡πxeθ+1,uθ,0=uθ,1=0,pθ,x satisfies(40)∂xxpθ,x=gx,pθ,0=pθ,1=0,and gx satisfies(41)gx+∫01qc,xgcdc=∫01e-θ∂xxuθ,xdθin which(42)qc,x=∫01e-θ∂xxwc,θ,xdθ=∫01e-θδx-cdθ=δx-c1-e-1.Solving (39) gives(43)uθ,x=-sin⁡πxeθ+1.Substituting (42) into (41) yields(44)gx+1-e-1gx=π2sin⁡πx2-e-1.In terms of (44), we have(45)gx=π2sin⁡πx.Once gx is obtained, (40) can be solved:(46)pθ,x=-sin⁡πx.So the exact solution is (47)vθ,x=uθ,x-pθ,x=-sin⁡πxeθ-sin⁡πx+sin⁡πx=-sin⁡πxeθ.Example 2.
Consider the following linear IDE:(48)v˙t+∫0πcos⁡t+svsds=1,v0=0.In terms of Corollary 9, (48) can be converted to the following equations:(49)v˙t=ht,v0=0,where ht satisfies(50)ht+∫0+∞qτ;thτdτ=1in which(51)qτ;t=∫0πcos⁡t+swτ;sds=sin⁡t+π-sin⁡t+τ,τ≤π,0,τ>π.Substituting (51) into (50) yields(52)ht-sin⁡t∫0πhτdτ-sin⁡t∫0πcos⁡τhτdτ-cos⁡t∫0πsin⁡τhτdτ=1.Solving (52) yields(53)ht=1+sin⁡t-8π4+π2+cos⁡t8-2π24+π2.Combining (53) with (49) yields(54)vt=t+8π4+π2cos⁡t-1+8-2π24+π2sin⁡twhich is the exact solution.


## 5. Conclusion

In this work, we extend the SMW formula for the linear IDE. By using the SMW formula, the linear IDE can be decomposed into several linear differential equations and one pure integral equation. The SMW formula could be a powerful tool to solve the linear IDE. Combining the SMW formula with some analytical methods developed for the integral equation may provide the analytical approach for the linear IDE. The SMW formula can also be combined with the numerical methods for solving the linear IDE numerically.



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