Finite-difference methods

[5/40%]
Show
LaTeX

Let h>0h > 0 be given. Consider the linear difference equation

y((n+2)h)−2y((n+1)h)+y(nh)h2=−y(nh),n=0,1,2,…. \frac{y((n+2)h) - 2y((n+1)h) + y(nh)}{h^2} = -y(nh), \quad n = 0, 1, 2, \ldots .

(Note the analogy with the differential equation y′′=−yy'' = -y.)

?
(1)

Find the general solution of the equation by trying suitable exponential substitutions.

(2)

Find the solution with y(0)=0y(0) = 0 and y(h)=hy(h) = h. Denote it by Sh(nh)S_h(nh), n=1,2,…n = 1, 2, \ldots.

(3)

Let xx be fixed and h=xnh = \frac{x}{n}. Show that

lim⁡n→∞Sxn(nxn)=sin⁡x. \lim _{n \to \infty } S_{\frac{x}{n}}\left(\frac{nx}{n}\right) = \sin x.

Divide the interval [0,1][0, 1] into five equal subintervals, and apply the finite difference method in order to approximate the solution of the two-point boundary value problem

y′′(t)=125t,y(0)=y(1)=0 y''(t) = 125 t, \quad y(0) = y(1) = 0

at the four interior grid points. Compare your approximate values at the grid points with the exact solution at the grid points. Note: You should not expect very accurate approximations with only four interior grid points.

?

Divide [0,1][0, 1] into n+1n+1 equal subintervals, and apply the finite difference approximation method to derive the linear system associated with the two-point boundary value problem

y′′(t)−y′(t)=f(t),y(0)=y(1)=0. y''(t) - y'(t) = f(t), \quad y(0) = y(1) = 0.
?

Divide [0,1][0, 1] into five equal subintervals, and approximate the solution to

y′′(t)−y′(t)=125t,y(0)=y(1)=0 y''(t) - y'(t) = 125 t, \quad y(0) = y(1) = 0

at the four interior grid points. Compare the approximations with the exact values at the grid points.

?

Consider the partial differential equation (Laplace equation)

∂2u∂x2+∂2u∂y2=0on[0,1]×[0,1] \frac{\partial ^{2} u}{\partial x^{2}}+\frac{\partial ^{2} u}{\partial y^{2}} = 0 \quad \text{on} \quad [0,1] \times [0,1]

with the boundary conditions

u(x,0)=1,u(x,1)=2,u(0,y)=1,u(1,y)=2. u(x,0) = 1, \qquad u(x,1) = 2, \qquad u(0,y) = 1, \qquad u(1,y) = 2.

Apply the central difference scheme

(∂2u∂x2)j,k≈uj−1,k−2uj,k+uj+1,k(Δx)2,(∂2u∂y2)j,k≈uj,k−1−2uj,k+uj,k+1(Δy)2 \left( \frac{\partial ^{2} u}{\partial x^{2}} \right)_{j,k} \approx \frac{u_{j-1,k}-2 u_{j,k}+u_{j+1,k}}{(\Delta x)^{2}}, \qquad \left( \frac{\partial ^{2} u}{\partial y^{2}} \right)_{j,k} \approx \frac{u_{j,k-1}-2 u_{j,k}+u_{j,k+1}}{(\Delta y)^{2}}

and then solve the linear equation. Consider the cases Δx=Δy=1/3\Delta x = \Delta y = 1/3 and Δx=Δy=1/4\Delta x = \Delta y = 1/4.

?

References

[deSi04] de Souza P.N., Silva J. Berkeley Problems in Mathematics. 3rd ed., Springer, 2004.
[Mey00] Meyer C.D. Matrix Analysis and Applied Linear Algebra. SIAM, 2000.
[StHa17] Steeb W., Hardy Y. Problems and Solutions in Introductory and Advanced Matrix Calculus. 2nd ed., World Scientific Publishing, 2017.