Derivation of Euler integration#
Let us start by defining \(X = Kx\) and \(Y=Ky\). We can then write:
\[\begin{split}\begin{aligned}
f(a, b, X, Y) &= K^hf(a, b, x, y) \\
F &= K^h f
\end{aligned}\end{split}\]
We let \(f(a, b, X, Y) \equiv F\) and \(f(a, b, x, y) \equiv f\) for simplicity. The total differential of this expression is written as:
\[\begin{split}\begin{aligned}
&\left ( \frac{\partial F}{\partial a} \right)_{b, X, Y} da + \left ( \frac{\partial F}{\partial b} \right)_{a, X, Y} db + \left ( \frac{\partial F}{\partial X} \right)_{a, b, Y} dX + \left ( \frac{\partial F}{\partial Y} \right)_{a, b, X} dY = \\
& K^h\left ( \frac{\partial f}{\partial a} \right)_{b, x, y} da + K^h \left ( \frac{\partial f}{\partial b} \right)_{a, x, y} db + K^h \left ( \frac{\partial f}{\partial x} \right)_{a, b, y} dx + K^h \left ( \frac{\partial f}{\partial y} \right)_{a, b, x} dy + hK^{h-1} f dK
\end{aligned}\end{split}\]
The last term on the right hand side is from the chain rule. We also note that \(dX = Kdx + xdK\) and \(dY = Kdy + ydK\) from the product rule. We can now eliminate \(dX\) and \(dY\) and collect like terms (this is going to get a bit messy):
\[\begin{split}\begin{aligned}
&\left [ \left ( \frac{\partial F}{\partial a} \right)_{b, X, Y} - K^h\left ( \frac{\partial f}{\partial a} \right)_{b, x, y}\right ] da + \left [ \left ( \frac{\partial F}{\partial b} \right)_{a, X, Y} - K^h \left ( \frac{\partial f}{\partial b} \right)_{a, x, y} \right ] db\\
+ &\left [ K \left ( \frac{\partial F}{\partial X} \right)_{a, b, Y} - K^h \left ( \frac{\partial f}{\partial x} \right)_{a, b, y} \right ] dx + \left [ K \left ( \frac{\partial F}{\partial Y} \right)_{a, b, X} - K^h \left ( \frac{\partial f}{\partial y} \right)_{a, b, x} \right ] dy \\
+ &\left [ x \left ( \frac{\partial F}{\partial X} \right)_{a, b, Y} + y \left ( \frac{\partial F}{\partial Y} \right)_{a, b, X} - hK^{h-1} f \right ] dK = 0
\end{aligned}\end{split}\]
Since \(a\), \(b\), \(x\), and \(y\) are all independent variables, the only way that this expression is guaranteed to be 0 is if each coefficient of \(dx\), \(dy\), \(da\), and \(db\) are 0. We can thus write:
\[\begin{split}\begin{aligned}
\left ( \frac{\partial F}{\partial X} \right)_{a, b, Y} &= K^{h-1} \left ( \frac{\partial f}{\partial x} \right)_{a, b, y}\\
\left ( \frac{\partial F}{\partial Y} \right)_{a, b, X} &= K^{h-1} \left ( \frac{\partial f}{\partial y} \right)_{a, b, x} \\
x \left ( \frac{\partial F}{\partial X} \right)_{a, b, Y} + y \left ( \frac{\partial F}{\partial Y} \right)_{a, b, X} &= hK^{h-1}f
\end{aligned}\end{split}\]
Substituting the first two expressions into the third yields:
\[x \frac{\partial}{\partial x}[ f(a, b, x, y)] + y \frac{\partial}{\partial y} [f(a, b, x, y) ] = h f(a, b, x, y)\label{eq3.27}\]