Backward Euler's Method

Question

This question was asked in CSIR. please help me to find out correct choice

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Let $y(t)$ satisfy the differential equation $$y'=\lambda y;y(0)=1$$. Then the backward Euler method for $n\geq 1$ and $h>0$ $$\frac{y_n-y_{n-1}}{h}=\lambda y_n;\quad y(0)=1$$ yields

1. a first order approximation to $e^{\lambda nh}$

2. a polynomial approximation to $e^{\lambda nh}$

3. a rational function approximation to $e^{\lambda nh}$

4. a Chebyshev polynomial approximation to $e^{\lambda nh}$

Answer 1

The correct answer is (1).

Let us prove it:

An the solution is to apply Taylor series. Let $y_n$ be the approximation point or the center of series, because it is occurs twice per numerical scheme.

$y(x)=y(c)+\frac{1}{1!}(x-c)y'(c)+...+\frac{1}{n!}(x-c)^n y^{(n)}(x_0)+...$

Let $c:=x_{n};\quad x:=x_{n-1};\quad y_n:=y(x_n);\quad y_{n-1}:=y(x_{n-1});\quad h:=x_n-x_{n-1}$

$y_{n-1} = y_{n} - h y'(x_{n})+\frac{h^2}{2}y''(x_{n})-\frac{h^3}{6}y'''(x_{n})+...$

$\frac{y_{n} - y_{n-1}}{h} = y'(x_{n}) - \frac{h}{2}y''(x_{n})+\frac{h^2}{6}y'''(x_{n})-...$

We may show that residual depends on step $h$ linearly: $0=\frac{y_{n} - y_{n-1}}{h}-y_n= \underbrace{(y'(x_{n})-\lambda y_n)}_{=0} - \frac{h}{2}y''(x_{n})+\frac{h^2}{6}y'''(x_{n})-...=O(h) $

So it is first order approximation.

You can find more info about finite difference method on wiki.

Answer 2

Since the explicit solution of that recursion is $$ y_n=\Bigl(1-\frac{λnh}{n}\Bigr)^{-n} $$ you get for $nh=t$ fixed a well-known approximation of the exponential $e^{λt}$. Using the logarithm series you can find the approximation order.

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2. and 3. are ambiguous since they do not specify the variable of the function, i.e., what is considered fixed in those expressions.