- #1

awholenumber

- 200

- 10

An equation containing the derivatives of one or more dependent variables, with respect to one or more independent variables, is said to be a differential equation

a question usually starts like this ...

find the function that gives , this instantaneous rate of change ...

BvU said:take ##f(x)= x^2-3## . Add initial condition f(0) = 0 so it yields a unique solution.

##x = 0 \rightarrow f(0) = 0##

##x = 0.1 \rightarrow f(0.1) = 0 + (0.01 - 3)* 0.1 = -0.299 \ ## exact solution ##f(0.1) = 0.001 - 3 = -0.299 \ ##

etc. etc.

we get:

So the exact solution is a function ##f(x) = x^3/3-3x## and the numerical approach is a table of aproximated function values.

(I took steps of 0.1 so the difference between exact and numerical becomes visible).

BvU said:this differential equation does have a condition: ##f(0) = 1##. You solve numerically by approaching f(0) as a linear function for a small step:$$f(0+\Delta x) \approx f(0) + \Delta x {dy\over dx}$$ when all you really know is $$ \lim_{h\downarrow 0} {f(0+h) - f(0) \over h } = {dy\over dx} \ .$$

When you take a few steps you get a sequence of approximated function values, but not a function (in the sense of an exact prescription what to do with x to get y).

So for ##\ f(x)= y\ \ \& \ \ f(0) = 1 \ ## you have found four function values, not four functions.

So the exact solution is a function ##f(x) = x^3/3-3x## and the numerical approach is a table of approximated function values.

When you take a few steps you get a sequence of approximated function values, but not a function (in the sense of an exact prescription what to do with x to get y).

is this what we use the numerical methods for ?? to simply get approximated function values??BvU said:Yes. Note that replacing ##\ dy\over dx\ ## by ##\ \Delta y\over \Delta x\ ## is an approximation...