Answer to Question #175524 in Differential Equations for Arsene Dongmo

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.