Numerical approximation: Forward diffrerence method

In summary, the conversation is about finding difference equations for an approximate solution to a heat equation using forward and backward difference methods. The person is stuck on finding an expression for the second derivative using the forward difference method, and is looking for help. Another person suggests using a defined operator for the first partial derivative with respect to x, and mentions that the resulting equation using this operator deviates only slightly from central differences.
  • #1
oddiseas
73
0

Homework Statement



[t]=-U+k[xx] u(x,0)=U(L,0)=0 u(x,0)sin(pix/L)


Write down difference equations for the approximate solution of this problem using the following methods:

1)forward difference
2)backward difference
3)crank nicholson

Homework Equations



I can do part 3, but i am stuck on the first two methods. I can find an exppression for the partial derivative of t, but the second derivative using the forward difference from a taylor approximation is 0 isn't it?

F(x+dx,t)=f(x,t)+[f][x](x,t)dx+[f][xx](x,t)(dx)^2

which if u solve for f''=0

The Attempt at a Solution

 
Physics news on Phys.org
  • #2
I'm not sure about the Taylor expansion (well, I do think that the relevant equations could be derived that way too, but that's not the usual method, I think).

I'd define the forward difference operator,
[tex]\partial U_i = \frac{U_{i+1}-U_i}{h}[/tex]
and use it successively. Similarly for the backward difference. Does this help?
 
  • #3
This is the definition for the first partial derivative with respect to x , using the forward difference method.I am trying to figure out if it is possible to find an expression for the second derivative with respect to x using the forward difference method. Because when i try i get zero. Usually for a heat equation i use forward difference in time and central difference for spatial, but this question specifically asks for a forward difference representation for the entire problem, so if anyone can help, because i am stuck.
 
  • #4
Yeah, I know, that's why I mentioned using the operator I defined in my post several times in a row (i.e. what is [itex]\partial\partial U_i[/itex]). The result you get that way deviates from central differences only by a small amount.
 

1. What is numerical approximation?

Numerical approximation is a method used in mathematics and science to estimate the value of a function or equation using a numerical approach, rather than an analytical one. It involves using numerical methods, such as the forward difference method, to approximate the value of a function at a given point.

2. What is the forward difference method?

The forward difference method is a numerical method used to approximate the derivative of a function at a given point, by using the slope of a nearby point. It involves calculating the difference between the function values at two points and dividing by the difference between the x-values of those points.

3. How does the forward difference method work?

The forward difference method works by using the slope of a nearby point to approximate the derivative of a function at a given point. This is done by calculating the difference between the function values at two points and dividing by the difference between the x-values of those points. The smaller the difference between the points, the more accurate the approximation will be.

4. What are the advantages of using the forward difference method?

One advantage of using the forward difference method is that it is relatively easy to implement and requires minimal computational resources. It is also a useful tool for approximating derivatives of functions that are difficult or impossible to find analytically. Additionally, it can provide a good estimate of the derivative at a specific point, even if the function is not smooth or continuous.

5. What are the limitations of the forward difference method?

One limitation of the forward difference method is that it can only provide an approximation of the derivative at a specific point, rather than an exact value. It also relies on the choice of step size, and a very small step size may result in rounding errors. Additionally, the accuracy of the approximation can vary depending on the function being approximated and the point at which the derivative is being estimated.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
649
  • Calculus and Beyond Homework Help
Replies
11
Views
688
  • Calculus and Beyond Homework Help
Replies
2
Views
754
  • Calculus and Beyond Homework Help
Replies
1
Views
663
  • Calculus and Beyond Homework Help
Replies
1
Views
633
  • Calculus and Beyond Homework Help
Replies
1
Views
797
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
869
  • Calculus and Beyond Homework Help
Replies
1
Views
599
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
Back
Top