Difference Quotient and 2nd Order PDEs

In summary, the conversation discusses the approximation of a second order partial derivative using a difference equation. The equation presented in the book is compared to a different approach using central differences, which is found to be more intuitive and easier to understand. The concept of central differences is explained and a link is provided for further understanding.
  • #1
egsmith
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I am trying to match a result in one of my textbooks. To assist with one of their arguments they are approximating a 2nd order PDE by using a difference quotient and they show the approximation as follows:

(d^2u[x,t])/(dx^2) =~ (1/h^2)(u[x+h,t]-2u[x,t]+u[x-h,t])


When I actually use the equation it makes sense, i.e. it is positive when u[x] is concave up and negative when u[x] is concave down. However when I try to derive it I get to a part that doesn't make sense to me geometrically (I am an engineer by trade so geometric arguments appeal to me).

I'll change the function so there is only one independent variable and start with the first order difference quotient.

f[x,h,t] = du[x,t]/dx =~ (u[x+h,t] - u[x,t])/h

This makes sense because it is the slope of the secant (which approaches the tangent as h gets small) at u[x,t]. I am going to rename this result to f so future equations look less cluttered.

I was hoping I could just use recursion like this:

d^2u[x,t]/dx^2 = f[f[x,h,t],h,t]

But you need to know a lot about u[x,t] to have this result be as useful as the equation presented by the book.

Using a slightly different approach, +/- h in a hope to get things to cancel, I can almost match the approximation from the book.

d^2u[x,t]/dx^2 =~ (f[x,h,t] - f[x,-h,t])/2h =
1/(2h^2)(u[x+h,t]-2u[x,t]+u[x-h,t])

So I am half their result. However it seems, to me anyway, that the denominator should be 2h because I use the points at x+/-h to form the base of the triangle with the secant as the hypo. (This makes no difference in the textbook argument since all they are concerned with is sign and the 2 in question only scales.)

After some though I got even more confused because I am not really using the points at (x+/-h,t). I am not even using the point at (x,t). I am using the slope of the secants in front of and behind u[x,t]. From here my geometric intuition died and I came up with some bizarre stuff which really just simplified back to the recursion approach.

Can someone explain to me, or point me to a reference on, the proper way of deriving the approximation of a second order partial derivative using the difference equation? Perhaps the result from the book is some sort of average?
 
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  • #2
If you use central differences, it is a little easier. I'll ignore t, since, for this part, it has no role.

(1) du(x)/dx=~(u(x+h/2)-u(x-h/2))/h
(2) du(x+h/2)/dx=~(u(x+h)-u(x))/h
(3) du(x-h/2)/dx=~(u(x)-u(x-h))/h

Insert (2) and (3) into (1) and get

d/dx(du(x)/dx) (=d2u/dx2) =~ (u(x+h)-2u(x)+u(x-h))/h2
 
  • #3

FAQ: Difference Quotient and 2nd Order PDEs

What is a difference quotient?

A difference quotient is a mathematical expression that calculates the rate of change of a function over a certain interval. It is calculated by taking the difference between two points on the function and dividing it by the difference in their corresponding inputs.

How is a difference quotient used in 2nd order partial differential equations?

In 2nd order partial differential equations, the difference quotient is used to approximate the partial derivatives of a function. This allows us to solve the equation and find the function that satisfies the given conditions.

What is the difference between a 1st order and 2nd order PDE?

A 1st order partial differential equation involves only first derivatives of the function, while a 2nd order PDE involves second derivatives. This means that the solution to a 2nd order PDE will have more information about the function compared to a 1st order PDE.

What are some real-life applications of 2nd order PDEs?

2nd order PDEs are used in many areas of science and engineering, such as fluid dynamics, heat transfer, and electromagnetism. They are also commonly used in financial mathematics to model options pricing and in image processing for edge detection and noise reduction.

How are 2nd order PDEs solved?

There are various methods for solving 2nd order PDEs, including separation of variables, the method of characteristics, and the use of integral transforms. The specific method used will depend on the type of PDE and the boundary conditions given in the problem.

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