Question

1) Based on the finite difference method, what is the advanced scheme of the first, second and third derivatives with second order accuracy?

2) Consider the following equation:

d³y/dt³+2d²y/dt²+4dy/dt+8y=te²t

Where y(0) = 0 and y(3) = 28.2

By using the results from the session (1), solve the equation for t E [0 ; 3], with delta(t) = 0.3

Answer 1

1). We remind that the finite difference method is a class of numerical techniques for solving differential equations by approximating derivatives with finite differences. In order to achieve the second order accuracy, we consider the Taylor series expansion:

$f(x_0+h)=f(x_0)+\frac{f'(x_0)}{1!}h+\frac{f''(x_0)}{2!}h^2+\frac{f'''(x_0)}{3!}h^3+R_n(x),$

where $R_n(x)$ denotes the reminder term. Another expansion is:

$f(x_0-h)=f(x_0)-\frac{f'(x_0)}{1!}h+\frac{f''(x_0)}{2!}h^2-\frac{f'''(x_0)}{3!}h^3+\tilde{R}_n(x),$

where $\tilde{R}_n(x)$ is another reminder term. From the latter we find: $f(x_0+h)-f(x_0-h)=2f'(x_0)h+2h^3\frac{f'''(x_0)}{3!} +R_n(x)-\tilde{R}_n(x)$

and: $f(x_0+h)+f(x_0-h)=2f(x_0)+f''(x_0)h^2+R_n(x)+\tilde{R}_n(x)$

The latter yields: $f'(x_0)=\frac{f(x_0+h)-f(x_0-h)}{2h}-h^2\frac{f'''(x_0)}{3!} -\frac{(R_n(x)-\tilde{R}_n(x))}{2h}$; $f''(x_0)=\frac{f(x_0+h)-f(x_0-h)-2f(x_0)}{h^2}-\frac{(R_n(x)+\tilde{R}_n(x))}{h^2}$.

Thus, the approximate expressions are $f'(x_0) \approx\frac{f(x_0+h)-f(x_0-h)}{2h}$ and $f''(x_0)\approx\frac{f(x_0+h)+f(x_0-h)-2f(x_0)}{h^2}$.

We express $f'''(x_0)$ as: $f'''(x_0)\approx\frac{f''(x_0+h)-f''(x_0-h)}{2h}\approx\frac{1}{2h}\left(\frac{f(x_0+2h)+f(x_0)-2f(x_0+h)}{h^2}-\frac{f(x_0)+f(x_0-2h)-2f(x_0-h)}{h^2}\right)=\frac{f(x_0+2h)-2f(x_0+h)+2f(x_0-h)-f(x_0-2h)}{2h^3}$

2). We can substitute the approximations directly and receive the system for the values $f(x_0-2h)$, $f(x_0-h)$, $f(x_0)$, $f(x_0+h)$, $f(x_0+2h)$ at different points. On the other hand, we can solve the system directly. Namely, we substitute the expression $y=ce^{\lambda x}$, $c,\lambda\in{\mathbb{C}}$ into the homogenuous equation and receive: $ce^{\lambda x}(\lambda^3+2\lambda^2+4\lambda+8)=0$. The roots are: $\lambda=-2,2I,-2I$. Thus, the general solution of the homogenuous system is: $y=c_1e^{-2x}+c_2e^{2Ix}+c_3e^{-2Ix}$, where $c_1,c_2,c_3\in{\mathbb{C}}$ . We check, whether the following function $y_p=\alpha xe^{2x}+\beta e^{2x}$, $\alpha,\beta\in{\mathbb{C}}$, is a solution of the initial equation. The derivatives has the form: $y_p'=2\alpha x e^{2x}+(\alpha+2\beta) e^{2x}$, $y_p''=4\alpha x e^{2x}+(4\alpha+4\beta) e^{2x}$, $y_p'''=8\alpha x e^{2x}+(12\alpha+8\beta) e^{2x}$. We substitute it into equation and receive: $(32\alpha x+(12\alpha+8\alpha+4\alpha+8\beta+8\beta+16\beta))e^{2x}=(32\alpha x+(24\alpha+32\beta))e^{2x}=xe^{2x}$

From the latter we receive: $\alpha=\frac{1}{32}$, $\beta=-\frac{3}{4}\alpha=-\frac{3}{128}$ Thus, $y_p=\frac{1}{128}(4x-3)e^{2x}$. The solution (without initial conditions) has the form: $y=c_1e^{-2x}+c_2e^{2Ix}+c_3e^{-2Ix}+\frac{1}{128}(4x-3)e^{2x}$. We set $y(0)=0$ and $y(3)=28.2$ and receive a system for coefficients $c_1,c_2,c_3.$ Thus, we get the solution of the original system with initial conditions.