Solution to Differential Equation with Limit Boundary Condition

In summary, the conversation discusses finding a unique solution to a second order differential equation with given boundary conditions. The obtained solution is provided, and the process of applying the boundary conditions is explained. The conversation ends with the suggestion to substitute a simplified expression for D into the solution and take the limit as x approaches 0.
  • #1
a1234
78
6
Homework Statement
A second order differential equation was found to have a solution, y_2 (provided below). Apply the boundary conditions y(1) = 1 and lim of y as y approaches 0 = 0 to find the unique solution.
Relevant Equations
Original differential equation and obtained solution provided in the box below.
The original differential equation is:

1663934426962.png

My solution is below, where C and D are constants. I have verified that it satisfies the original DE.
1663934166996.png


When I apply the first boundary condition, I obtain that
1663934588145.png
, but I'm unsure where to go from there to apply the second boundary condition. I know that I should try to choose C such that the undefined contributions from both terms cancel out, but haven't found anything that does this.
 
Physics news on Phys.org
  • #2
I would use [tex]
\exp(\pm ikx^2/2) = \cos(kx^2/2) \pm i\sin(kx^2/2)[/tex] to write the solution in the form [tex]
y(x) = \frac{A\cos(kx^2/2) + B\sin(kx^2/2)}{\sqrt{x}}.[/tex]
 
  • Like
Likes topsquark
  • #3
a1234 said:
Homework Statement:: A second order differential equation was found to have a solution, y_2 (provided below). Apply the boundary conditions y(1) = 1 and lim of y as y approaches 0 = 0 to find the unique solution.
Relevant Equations:: Original differential equation and obtained solution provided in the box below.

The original differential equation is:

View attachment 314495
My solution is below, where C and D are constants. I have verified that it satisfies the original DE.
View attachment 314494

When I apply the first boundary condition, I obtain that View attachment 314496, but I'm unsure where to go from there to apply the second boundary condition. I know that I should try to choose C such that the undefined contributions from both terms cancel out, but haven't found anything that does this.
The right side of the equation for D simplifies to $$D = e^{-ik/2}(1 + \frac C {2ik})$$
I would substitute the above for D into the solution you found, and take the limit as x approaches 0.
 
  • Like
Likes a1234 and topsquark
  • #4
I was able to figure out this problem.
 
  • Like
Likes PhDeezNutz

FAQ: Solution to Differential Equation with Limit Boundary Condition

What is a differential equation?

A differential equation is an equation that relates an unknown function to its derivatives. It is commonly used to describe the change of a system over time or space.

What is a boundary condition?

A boundary condition is a set of constraints that are applied to a differential equation to determine a unique solution. It specifies the behavior of the unknown function at the boundaries of the domain.

What is a limit boundary condition?

A limit boundary condition is a type of boundary condition where the behavior of the unknown function is specified as it approaches a certain value or point in the domain.

How do you solve a differential equation with a limit boundary condition?

To solve a differential equation with a limit boundary condition, you need to first find the general solution to the differential equation. Then, you can use the limit boundary condition to determine the specific values of any arbitrary constants in the general solution, resulting in a unique solution.

What are some applications of solving differential equations with limit boundary conditions?

Solving differential equations with limit boundary conditions is useful in many fields, including physics, engineering, and economics. It can be used to model and predict the behavior of systems such as population growth, heat transfer, and electrical circuits.

Similar threads

Replies
1
Views
620
Replies
8
Views
2K
Replies
2
Views
862
Replies
4
Views
621
Replies
2
Views
1K
Replies
4
Views
2K
Back
Top