I am trying to solve a first order differential equation with the condition that g ( y

Jonathan Miles 2022-07-01 Answered
I am trying to solve a first order differential equation with the condition that g ( y ) = 0 if y = 0:
a g ( c y ) + b g ( e y ) = α (1) g ( 0 ) = 0 ,
where parameters a,b,c,e are real nonzero constants; α is a complex constant; function g ( y ) : R C is a function mapping from real number y to a complex number. The goal is to solve for function g ( ). This is what I have done. Solve this differential equation by integrating with respect to y:
a c g ( c y ) + b e g ( e y ) = α y + β ,
where β is another complex constant. Plugging in y=0 and using the fact that g(0)=0, we have β = 0. Therefore, we have
a c g ( c y ) + b e g ( e y ) = α y .
The background of this problem is Cauchy functional equation, so my conjecture is one solution could be g ( y ) = γ y. Plugging in g ( y ) = γ y, I get γ = α a + b , which implies that one solution is g ( y ) = α a + b y. Then, I move on to show uniqueness. I define a vector-valued function h ( h 1 , h 2 ) T such that
h 1 ( y ) = a c g 1 ( c y ) + b e g 1 ( e y ) h 2 ( y ) = a c g 2 ( c y ) + b e g 2 ( e y ) ,
where g ( y ) g 1 ( y ) + i g 2 ( y ). Then, I rewrite this differential equation as
h ( y ) = α (2) h ( 0 ) = 0 ,
where α ( α 1 , i α 2 ) T . By the uniqueness theorem of first order differential equation, solution h(y) is unique. I have two questions. First, I think equation (1) and (2) should be equivalent. However, it seems that equation (1) can imply equation (2) but equation (2) may not imply equation (1). This is because h(0)=0 may imply either g 1 ( 0 ) = 0 , g 2 ( 0 ) = 0 or a c + b e = 0. Second, I have only proved that h(y) is unique. How should I proceed to show g(y) is also unique.
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (1)

jugf5
Answered 2022-07-02 Author has 18 answers
For people who are interested in my question, I got an answer from a friend. This is a smart way to show uniqueness of solution.
a g ( c y ) + b g ( e y ) = α
Define f ( ) g ( ):
a f ( c y ) + b f ( e y ) = α
View this equation as an infinite system of algebraic equations in unknowns (a,b). For example,
a f ( 0 ) + b f ( 0 ) = α a f ( c ) + b f ( e ) = α a f ( 2 c ) + b f ( 2 e ) = α
Since a and b are non-zero, this system of equations is solvable if and only if function f ( ) is a constant function. Thus, function g ( ) is a constant function which further implies that function g ( y ) = β y where β is a complex constant.
Did you like this example?
Subscribe for all access

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more