Second order linear differential operator

In summary, the conversation discusses a second order linear differential operator over an interval, a function defined on that interval, and another function with the property that the operator applied to it equals the first function. The conversation then proceeds to show that if a function is the sum of two functions and one of them satisfies the operator's equation, then the sum also satisfies it. Finally, it is shown that if a function satisfies the operator's equation, it can be decomposed as the sum of two functions, one of which satisfies the operator's equation and the other is the zero vector.
  • #1
Ready2GoXtr
75
0

Homework Statement


Suppose that L is a second order linear differential operator over the interval J, that f is a function defined on J, and that the function v has the property that

Lv = f on J

(a) Show that if y = u + v and that Lu = 0 on J, then Ly = f on J
(b) Show that if Ly = f on J, then y = u + v for some u such that Lu = 0 on J

Homework Equations


None that apply

The Attempt at a Solution


Well unfortunately this one I was unable to attempt because i am not even sure what it is asking, this professor I have tends to deviate from the book.
 
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  • #2
Ready2GoXtr said:

Homework Statement


Suppose that L is a second order linear differential operator over the interval J, that f is a function defined on J, and that the function v has the property that

Lv = f on J

(a) Show that if y = u + v and that Lu = 0 on J, then Ly = f on J
(b) Show that if Ly = f on J, then y = u + v for some u such that Lu = 0 on J

Homework Equations


None that apply

The Attempt at a Solution


Well unfortunately this one I was unable to attempt because i am not even sure what it is asking, this professor I have tends to deviate from the book.

(a) If L is linear, then L(u+v)=Lu+Lv. So if we have Lu=0, then

Ly=L(u+v)=Lu+Lv=0+Lv=Lv=f.

(b) Since L is linear, then there is a zero vector u with Lu=0 and 0+y=y. Choose u=0 and v=y. Then u+v=0+y=y.
 
  • #3
Thanks man I would never figure that out!
 

Related to Second order linear differential operator

1. What is a second order linear differential operator?

A second order linear differential operator is a mathematical operation that involves taking the second derivative of a function. It is typically represented by the notation L[y], where y is the function and L is the operator.

2. What is the difference between a linear and non-linear differential operator?

A linear differential operator is one in which the operator acts on the function in a linear manner, meaning that the function and its derivatives are raised to the first power. A non-linear differential operator, on the other hand, involves higher powers of the function and its derivatives, making it more complex to solve.

3. How is a second order linear differential operator used in real-world applications?

Second order linear differential operators are commonly used in physics and engineering to model systems that involve acceleration or oscillation, such as in mechanics, electromagnetics, and signal processing. They also have applications in economics, biology, and other fields.

4. What are the steps to solving a second order linear differential equation using a differential operator?

The first step is to rewrite the equation in the form L[y] = 0, where L is the differential operator. Then, find the characteristic equation by setting the coefficients of the function and its derivatives equal to zero. Next, solve for the roots of the characteristic equation to find the general solution. Finally, use initial conditions or boundary values to determine the specific solution.

5. Are there any limitations to using second order linear differential operators?

One limitation is that they can only be used to solve linear differential equations, meaning that the equation must involve the function and its derivatives raised to the first power. Non-linear equations require different methods of solving. Additionally, the operator may not always have a closed-form solution and numerical methods may need to be used instead.

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