Question

(b) For a given function u(t), explain how the derivative of u(t) with respect to t can be approximated on a uniform grid with grid spacing ∆t, using the one-sided forward difference approximation

du/dt ≈ ui+1 − ui/∆t , where ui = u(ti). You should include a suitable diagram explaining your answer.

(c) Using the one-sided forward difference approximation from part (b) and Euler’s method, calculate the approximate solution to the initial value problem

du/dt + t cos(u) = 0, subject to u(0) = −0.2,

at t = 0.4, on a uniform grid with spacing ∆t = 0.1.

Answer 1

SUBPART B

Let us assume a curve for the function u(t) and plot out a few points first.

Now as we see, from the graph where is unit increment along the horizontal axis or the step size. It has a very small value.

The derivative of a curve at a point is also the slope of the tangent to the curve at that particular point. At point A, the derivative of the function is denoted as

Now if the step size is really less, then the curve can be approximated to coincide with the straight line AB. In other words, the point B is approximated to lie almost on the curve. In such a case, we can write

Thus if we know the value of the function at a point and the derivative equation of the curve, we can use them to approximate a data set for the original curve. Equation 1 is the forward difference algorithm while the equation 2 is the Euler algorithm. So essentially they are the same thing.

SUBPART C

Given equation is

Thus using the forward difference method and the Eulers algorithm we can write

where and

Using these values, let us calculate the values at the iteration numbers

Iteration (i)	u(i)	t(i)
0	-0.2	0
1	-0.2	0.1
2	-0.2098	0.2
3	-0.2293	0.3

Thus the value at t =0.4s as approximated by the Euler / forward difference method is u = -0.2585

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