- #1
Frogeyedpeas
- 80
- 0
Consider the following linear functional operator:
$$Q_w[f(x)] = \lim_{h\rightarrow w} \lbrace \frac{f(x + h) - f(x)}{h} \rbrace $$
How does one solve the equation
$$a_0(x)Q_0[f(x)] = a_1(x)Q_1[f(x)]$$
Spelt out that is:
$$a_0(x)*f'(x) = a_1(x)(f(x+1) - f(x))$$
For the case of constant functions $a_0(x) = a_0$ and $a_1(x) = a_1$ the solution is simply found by assuming
$$f(x) =e^{Lx}$$
thereby implying:
$$a_0L e^{Lx} = a_1(e^{Lx}(e^{L} - 1))$$
which can be solved as
$$\frac{L}{1 - e^{L}} = =\frac{a_1}{a_0} $$
And L can be extracted through the use of Lambert-W function.But what about more general functions?
$$Q_w[f(x)] = \lim_{h\rightarrow w} \lbrace \frac{f(x + h) - f(x)}{h} \rbrace $$
How does one solve the equation
$$a_0(x)Q_0[f(x)] = a_1(x)Q_1[f(x)]$$
Spelt out that is:
$$a_0(x)*f'(x) = a_1(x)(f(x+1) - f(x))$$
For the case of constant functions $a_0(x) = a_0$ and $a_1(x) = a_1$ the solution is simply found by assuming
$$f(x) =e^{Lx}$$
thereby implying:
$$a_0L e^{Lx} = a_1(e^{Lx}(e^{L} - 1))$$
which can be solved as
$$\frac{L}{1 - e^{L}} = =\frac{a_1}{a_0} $$
And L can be extracted through the use of Lambert-W function.But what about more general functions?
