Understanding Linear and Nonlinear Operators in Differential Equations

In summary, The first question asks whether the operator Lu = du/dx + du/dy + 1 is linear or not. The second question asks whether the equations du/dt - d^2u/dx^2 + 1 = 0 and du/dt - d^2u/dx^2 = -1 are nonlinear, linear inhomogeneous, or linear homogeneous. The first equation is linear and non-homogeneous, while the second equation is linear and homogeneous. Your reasoning for the first question is correct, but there is a difference between the conditions for an operator to be linear and for an equation to be linear. To find the linear operator corresponding to a linear equation, the constant term must be separated.
  • #1
theneedtoknow
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These 2 questions are somewhat related:

Is the operator linear: Lu = du/dx + du/dy + 1

I said it's non-linear since L(u+v) = du/dx + dv/dx + du/dy + dv/dy + 1 = Lu + Lv - 1 =/= Lu + Lv

and then:
State whether each of the following equations is nonlinear, linear inhomogeneous, or linear homogeneous.

du/dt - d^2u/dx^2 + 1 = 0
which can be rewritten as du/dt - d^2u/dx^2 = -1
so now we have pretty much the same situation as above, except since it was set to zero, i can just bring the 1 on the other side and the resulting equation is linear and non-homogeneous Lu = -1

So then, is my reasoning for the first question wrong, that the operator (which is almost identical except for the order of derivatives) is actually linear? And then, why is my reasoning wrong?
 
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  • #2
theneedtoknow said:
So then, is my reasoning for the first question wrong, that the operator (which is almost identical except for the order of derivatives) is actually linear? And then, why is my reasoning wrong?

Your reasoning is fine, there's just a difference between the conditions for an operator or equation to be linear. Given a linear operator L, there's always an associated linear equation L u = c. If we want to find the linear operator corresponding to a general linear equation, we need to separate any nonzero constant term.
 

FAQ: Understanding Linear and Nonlinear Operators in Differential Equations

1. What is a simple linear operator?

A simple linear operator is a mathematical function that maps one vector space to another vector space in a linear manner. It is represented by a matrix and can be used to transform vectors in the input space to vectors in the output space.

2. How is a simple linear operator different from a general linear operator?

A simple linear operator is a special case of a general linear operator, where the input and output spaces are the same. In other words, the matrix representing a simple linear operator is a square matrix, whereas a general linear operator can have different dimensions for the input and output spaces.

3. What are the properties of a simple linear operator?

A simple linear operator has two main properties: linearity and homogeneity. Linearity means that the operator preserves addition and scalar multiplication, while homogeneity means that the output is directly proportional to the input.

4. How is a simple linear operator applied in scientific research?

Simple linear operators are commonly used in scientific research to model linear relationships between variables. They can also be used to transform data, perform dimensionality reduction, and solve differential equations.

5. What are some examples of simple linear operators?

Some examples of simple linear operators include the identity operator, which leaves the input unchanged, and the projection operator, which projects the input onto a subspace. Other examples include differentiation and integration operators, which are commonly used in calculus.

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