In order to illustrate the significance of the fact that even the boundary conditions for a differential equation are to be accurately approximated when difference methods are used, we examine the differential equation y″ = y, with boundary conditions y′(0) = 0, y(1) = 1, which has the solution

Question

In order to illustrate the significance of the fact that even the boundary conditions for a differential equation are to be accurately approximated when difference methods are used, we examine the differential equation

[latex]y^{\prime \prime}=y,[/latex]

with boundary conditions [latex]y^{\prime}(0)=0, y(1)=1[/latex], which has the solution [latex]y(x)=\frac{\cosh x}{\cosh (1)}[/latex].

We put [latex]x_{n}=n h[/latex], assume that [latex]1 / h[/latex] is an integer and use the difference approximation

[latex]y_{n}^{\prime \prime} \approx\left(y_{n+1}-2 y_{n}+y_{n-1}\right) / h^{2} .[/latex]

Two different representations for the boundary conditions are

(1) symmetric case : [latex]y_{-1}=y_{1} ; y_{N}=1, N=1 / h[/latex],

(2) non-symmetric case

[latex]y_{0}=y_{1}, y_{N}=1.[/latex]

(a) Show that the error [latex]y(0)-y_{0}[/latex] asymptotically approaches [latex]a h^{2}[/latex] in the first case, and [latex]b h[/latex] in the second case, where [latex]a[/latex] and [latex]b[/latex] are constants to be determined.

(b) Show that the truncation error in the first case is [latex]O\left(h^{2}\right)[/latex] in the closed interval [latex][0,1][/latex].

Accepted Answer

(a) Substituting the second order difference approximation into the differential equation, we get the difference equation

[latex]y_{n+1}-2\left(1+\frac{h^{2}}{2} ight) y_{n}+y_{n-1}=0.[/latex]

The characteristic equation is given by

[latex] \xi^2-2\left(1+\begin{array}{c} h^2 \ 2 \end{array} ight) \xi+1=0 [/latex]

[latex] \begin{aligned} ext{with roots}\qquad\xi_{1 h} & =1+\frac{h^2}{2}+\left[\left(1+\frac{h^2}{2} ight)^2-1 ight]^{1 / 2} \ & =1+\frac{h^2}{2}+h\left[1+\frac{h^2}{4} ight]^{1 / 2}=1+h+\frac{h^2}{2}+\frac{h^3}{8}+\ldots \ & =e^h\left(1-\frac{1}{24} h^3+O\left(h^4 ight) ight), \ \xi_{2 h} & =1+\frac{h^2}{2}-\left[\left(1+\frac{h^2}{2} ight)^2-1 ight]^{1 / 2}=1-h+\frac{h^2}{2}-\frac{h^3}{8}+\ldots \ & =e^{-h}\left(1+\frac{1}{24} h^3+O\left(h^4 ight) ight) . \end{aligned} [/latex]

The solution of the difference equation may be written as

[latex]y_{n}=C_{1} e^{n h}\left(1-\frac{n}{24} h^{3}+O\left(h^{4} ight) ight)+C_{2} e^{-n h}\left(1+\frac{n}{24} h^{3}+O\left(h^{4} ight) ight)[/latex]

where [latex]C_{1}[/latex] and [latex]C_{2}[/latex] are arbitrary parameters to be determined with the help of the discretization of the boundary conditions.

(1) Symmetric case : We have [latex]y_{-1}=y_{1}[/latex]. Hence, we obtain

[latex] \begin{aligned} & C_{1} e^{h}\left(1-\frac{1}{24} h^{3}+O\left(h^{4} ight) ight)+C_{2} e^{-h}\left(1+\frac{1}{24} h^{3}+O\left(h^{4} ight) ight) \ = & C_{1} e^{-h}\left(1+\frac{1}{24} h^{3}+O\left(h^{4} ight) ight)+C_{2} e^{h}\left(1-\frac{1}{24} h^{3}+O\left(h^{4} ight) ight) \end{aligned} [/latex]

We get [latex]C_{1}=C_{2}[/latex].

Next, we satisfy [latex]y_{N}=1, N h=1[/latex].

[latex]y_{N}=C_{1} e\left(1-\frac{1}{24} h^{2}+O\left(h^{3} ight) ight)+C_{2} e^{-1}\left(1+\frac{1}{24} h^{2}+O\left(h^{3} ight) ight)=1 .[/latex]

Since [latex]C_{1}=C_{2}[/latex], we obtain

[latex] \begin{aligned} C_{1} & =\frac{1}{\left[2 \cosh (1)-\left(h^{2} / 12 ight) \sinh (1)+O\left(h^{3} ight) ight]} \ & =\frac{1}{2 \cosh (1)}\left[1-\frac{h^{2} \sinh (1)}{24 \cosh (1)}+O\left(h^{3} ight) ight]^{-1} \ & =\frac{1}{2 \cosh (1)}\left[1+\frac{h^{2} \sinh (1)}{24 \cosh (1)}+O\left(h^{3} ight) ight] \end{aligned} [/latex]

The solution of the difference equation becomes

[latex] \begin{aligned} & y_{n}=C_{1}\left[2 \cosh x_{n}-\frac{x_{n} h^{2}}{12} \sinh x_{n} ight] . \ & y_{0}=2 C_{1}=\frac{1}{\cosh (1)}+\frac{h^{2}}{24} \frac{\sinh (1)}{\cosh ^{2}(1)}+O\left(h^{3} ight) . \end{aligned} [/latex]

We get from the analytic solution [latex]y(0)=1 / \cosh (1)[/latex].

Hence, we have [latex]y(0)-y_{0}=a h^{2}[/latex],

where [latex]a=-\frac{1}{24} \frac{\sinh (1)}{\cosh ^{2}(1)}=-0.020565 . [/latex]

(2) Non-symmetric case: [latex]y_{0}=y_{1}, y_{N}=1[/latex].

Satifying the boundary conditions, we obtain

[latex]C_{1}+C_{2}=C_{1} e^{h}+C_{2} e^{-h}+O\left(h^{3} ight),[/latex]

or [latex]\left(e^{h}-1 ight) C_{1}=\left(1-e^{-h} ight)=C_{2}+O\left(h^{3} ight),[/latex]

or [latex] \left[h+\frac{h^2}{2}+O\left(h^3 ight) ight] C_1=\left[h-\frac{h^2}{2}+O\left(h^3 ight) ight] C_2 [/latex]

or [latex] C_1=\left\{\left(1+\frac{h}{2} ight)^{-1}\left(1-\frac{h}{2} ight)+O\left(h^2 ight) ight\} C_2=\left[1-h+O\left(h^2 ight) ight] C_2 ext {. } [/latex]

[latex] y_N=C_1 e+C_2 e^{-1}+O\left(h^3 ight) ext {, } [/latex]

or [latex] 1=\left[(1-h) e+e^{-1}+O\left(h^2 ight) ight] C_2 ext {. } [/latex]

Neglecting the error term, we obtain

[latex] \begin{aligned} C_{2} & =\frac{1}{2 \cosh (1)-h e}=\frac{1}{2 \cosh (1)}\left[1-\frac{h e}{2 \cosh (1)} ight]^{-1} \ & =\frac{1}{2 \cosh (1)}\left[1+\frac{h e}{2 \cosh (1)} ight], \end{aligned} [/latex]

where the [latex]O\left(h^{2} ight)[/latex] term is neglected.

[latex] C_{1}=(1-h) C_{2}=\frac{1}{2 \cosh (1)}\left[1-h+\frac{h e}{2 \cosh (1)} ight] [/latex]

where the [latex]O\left(h^{2} ight)[/latex] term is neglected.

[latex] \begin{aligned} ext{Thus, we have}\qquad y_{0} & =C_{1}+C_{2} \ & =\frac{1}{2 \cosh (1)}\left[2-h+\frac{h e}{\cosh (1)} ight]=\frac{1}{2 \cosh (1)}\left[2+\frac{h \sinh (1)}{\cosh (1)} ight] \ & =\left[\frac{1}{\cosh (1)}+\frac{h \sinh (1)}{2 \cosh ^{2}(1)} ight] \ y(0)-y_{0} & =\frac{1}{\cosh (1)}-y_{0}=-\frac{h}{2}\left(\frac{\sinh (1)}{2 \cosh ^{2}(1)} ight)=12 a h . \end{aligned} [/latex]

We have [latex] b h=12 a h, \quad ext { or } \quad b=12 a=-0.24678 .[/latex]

(b) Hence, from [latex](a)[/latex], in the symmetric case the truncation error is [latex]O\left(h^{2} ight)[/latex] while in the nonsymmetric case it is [latex]O(h)[/latex].

Question 5.69: In order to illustrate the significance of the fact that eve......

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