Approximate Derivatives and Differential Equations Solutions

Q. 1
Solution:
Given the data we determine approximate derivatives of function f(x) at
x=2.
From the data it can be deduced that h=0.1
Difference schemes to be employed:
f ′(x) ≈ f (x + h) − f (x − h)
2h
f ′′(x) ≈ f (x + h) + f (x − h) − 2f (x)
h2
x f (x) f ′(x) f ′′(x)
1.9 2.981206427 x − h
2.0 3.323350970 x 3.717023853−2.981206427
2∗0.1
3.717023853+2.981206427−2∗3.323350970
0.1∗0.1
= 3.67908713 = 5.151088
2.1 3.717023853 x + h
Answer: Approximately, f ′(2) = 3.67908713 and f ′′(2) = 5.151088

Q. 2
Solution:
Differential equation to be is solved is the 1-d heat equation:
∂u
∂t (x, t) = ∂2u
∂x2 (x, t) (1)
We begin by assuming a solution of the form:
u(x, t) = X(x)T (t)
Substituting u(x,t) in (1) yields two separate ODE’s:
X′′(x) + λX(x) = 0 (2)
T ′(t) + λT (t) = 0 (3)
where λ is a constant.
Equation (2) is a Laplace equation with general solution:
X(x) = A cos √λx + B sin √λx
The boundary conditions: ∂u
∂x(0, t) = 0
∂u
∂x(2, t) = 0
imply: A = 0 and sin √λ = 0 or λ = (nπ/2)2, where n=1,2,3, . . .
∴
X(x) = bn sin (nπx
2 ) n = 1, 2, 3, . . .
Solving equation (2) gives the solution:
T (t) = cne−(nπ/2)2t n = 1, 2, 3 . . .
Combining them and applying the principle of superposition gives the full
solution:
u(x, t) = X(x)T (t) =
∞∑
n=1
Bn sin (nπx
2 )e−(nπ/2)2t

where, Bn = bncn. This constant can be determined from the intial condition:
∂u
∂t (x, 0) = x4 − 8x2
which implies, ∞∑
1
Bn sin (nπx
2 ) = x4 − 8x2
This is the sine Fourier series for the function: f (x) = x4 − 8x2 for which the
constant Bn is given as:
Bn = 2
2
∫ 2
0
(x4 − 8x2) sin (nπx
2 ) dx (4)
So the exact solution to the DE is:
u(x, t) =
∞∑
n=1
Bn sin (nπx
2 )e−(nπ/2)2t
with Bn as determined in (4)