Startseite Mathematik Stable approximation of unbounded matrix operators for the simultaneous inversion of source terms and initial values in time-fractional Black–Scholes equation
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Stable approximation of unbounded matrix operators for the simultaneous inversion of source terms and initial values in time-fractional Black–Scholes equation

  • Shuang Yu und Hongqi Yang EMAIL logo
Veröffentlicht/Copyright: 18. November 2025

Abstract

This paper is devoted to identifying source term and initial value simultaneously in a time-fractional Black–Scholes equation, which is an ill-posed problem. The inverse problem is transformed into a system of unbounded operator equation system, and conditional stability is established under certain source conditions. We propose a stable approximation method to solve the problem, error estimates by rules of a priori and a posteriori regularization parameter selection are derived respectively. Numerical experiments are presented to validate the effectiveness of the proposed regularization method.

Funding statement: This work is supported in part by the Key-Area Research and Development Program of Guangdong Province (No. 2021B0101190003), by Guangdong Province Key Laboratory of Computational Science at the Sun Yat-sen University (No. 2020B1212060032), by the National Natural Science Foundation of China under grant 11571386, and by Innovation Team Project of Regular Universities in Guangdong Province (No. 2025KCXTD037).

A Supplement to Example 2

To better understand the spatial characteristics of the source term and initial value function, we will separately consider the source inversion problem (BSIP f ) and the initial value inversion problem (BSIP φ ).

Since problem (6.1) is a linear problem, based on the superposition principle, the solution u ( x , t ) of problem (1.3) can be written as the sum of the solutions u 1 ( x , t ) and u 2 ( x , t ) of the following two subproblems:

(A.1) { D t α 0 C u 1 ( x , t ) = - m 2 u 1 ( x , t ) + 2 u 1 ( x , t ) x 2 + f ( x ) , ( x , t ) ( 0 , 1 ) × ( 0 , 1 ) , u 1 ( 0 , t ) = u 1 ( 1 , t ) = 0 , t ( 0 , 1 ) , u 1 ( x , 0 ) = 0 , x ( 0 , 1 ) ,

and

(A.2) { D t α 0 C u 2 ( x , t ) = - m 2 u 2 ( x , t ) + 2 u 2 ( x , t ) x 2 , ( x , t ) ( 0 , 1 ) × ( 0 , 1 ) , u 2 ( 0 , t ) = u 2 ( 1 , t ) = 0 , t ( 0 , 1 ) , u 2 ( x , 0 ) = φ ( x ) , x ( 0 , 1 ) .

We consider the following two inverse problems:

(BSIP$\boldsymbol{f}$) (The inverse source problem in time-fractional Black–Scholes equation).

Given data y 1 ( x ) := u ( x , t 1 ) , find the source term f in time-fractional Black–Scholes equation (A.1).

Using the separation of variables method, we can get the solutions of problem (A.1) as follows:

u 1 ( x , t ) = n = 1 + 1 - E α , 1 ( - ( m 2 + ( n π ) 2 ) t α ) m 2 + ( n π ) 2 f n sin ( n π x ) ,

and by the terminal conditions y 1 ( x ) := u ( x , t 1 ) this yields

f ( x ) = n = 1 + m 2 + ( n π ) 2 1 - E α , 1 ( - ( m 2 + ( n π ) 2 ) t 1 α ) y 1 , n sin ( n π x ) .

That is, f = B 1 y 1 , where

B 1 y 1 = n = 1 + m 2 + ( n π ) 2 1 - E α , 1 ( - ( m 2 + ( n π ) 2 ) t 1 α ) y 1 , n sin ( n π x ) .

In other words, the solution f of the inverse problem is obtained from the data y 1 via the operator B 1 defined on functions y 1 in the set

𝒟 ( B 1 ) = { y 1 L 2 [ 0 , 1 ] : n = 1 ( n π ) 4 y 1 , n 2 < , y 1 , n = ( y 1 , sin ( n π x ) ) } .

(BSIP$\boldsymbol{\varphi}$) (The inverse initial value in time-fractional Black–Scholes equation).

Given data y 2 ( x ) := u ( x , t 2 ) , find the initial value φ in time-fractional Black–Scholes equation (A.2).

Similarly, we can obtain the solution

u 2 ( x , t ) = n = 1 + E α , 1 ( - ( m 2 + ( n π ) 2 ) t α ) φ n sin ( n π x )

for problem (A.2) and then deduce

φ ( x ) = n = 1 + 1 E α , 1 ( - ( m 2 + ( n π ) 2 ) t 2 α ) y 2 , n sin ( n π x ) ,

based on the terminal conditions y 2 ( x ) := u ( x , t 2 ) . Define

B 2 y 2 = n = 1 + 1 E α , 1 ( - ( m 2 + ( n π ) 2 ) t 2 α ) y 2 , n sin ( n π x ) ,

where B 2 is the linear operator on L 2 [ 0 , 1 ] with domain

𝒟 ( B 2 ) = { y 2 L 2 [ 0 , 1 ] : n = 1 e 2 ( n π ) 2 y 2 , n 2 < , y 2 , n = ( y 2 , sin ( n π x ) ) } .

Remark 2.

It seems that the e 2 ( n π ) 2 in 𝒟 ( B 2 ) is not apparent. On one hand, according to Lemma 1, the Mittag-Leffler function is an extension of exponential functions. On the other hand, since E α , 1 ( - ( m 2 + ( n π ) 2 ) t 2 α ) is complex, it is difficult to provide an expression of 1 / E α , 1 ( - ( m 2 + ( n π ) 2 ) t 2 α ) with respect to exp ( ) . However, what we can assert is that the instability of the inverse initial value problem is caused by the factor e n 2 . Here, we provide the example (see [22])

E 1 / 4 , 1 ( - x ) = 2 x π 0 e t 2 erfc ( - t 2 ) x 2 + t 2 𝑑 t ,

where erfc ( ) represents the complementary error function

erfc ( x ) = 2 π x e - t 2 𝑑 t .

Alternatively, if we set α = 1 , then we have

f ( x ) = n = 1 e ( m 2 + ( n π ) 2 ) t 2 y 2 , n sin ( n π x ) .

As a result of inverse problems (BSIP f ) and (BSIP φ ), we can conclude that, in comparison between factors e n 2 and n 2 , the initial value function is more sensitive than the source term when there are disturbances or noise.

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Received: 2025-01-10
Revised: 2025-10-21
Accepted: 2025-11-09
Published Online: 2025-11-18

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