欧拉方法

Often in the study of differential equations, a closed form solution is difficult if not impossible to find. In these cases, it is useful to be able to approximate the nature of these solutions. Euler's Method is one such method for doing this.

Euler's Method applies to differential equations of the form $\frac{dy}{dx}=f(x, y)$ given an initial point $(x_0, y_0)$. Then the value of $y(x)$ can be approximated in $n$ "steps" as follows:

Let $D=\frac{x-x_0}{n}$. For $0\le i\le n-1$, let: $$x_i=x_{i-1}+D$$ $$y_i=y_{i-1}+Df(x_{i-1}, y_{i-1})$$

Then $x_n=x$ and $y_n$ will be an approximation of $y(x)$.

Notice that at each step, $y_i$ is calculated using a tangent line approximation. In other words, between each $x_{i-1}$, $x_i$, we assume that line tangent to $y(x)$ at $x_{i-1}$ behaves similar to the graph itself. Generally speaking, as $n$ increases, this assumption becomes more valid since the gaps between successive $x_i$ decrease.

Example: Let $y(x)$ be a function of $x$ which satisfies $y'=\frac{1}{x+y}$ and $y(3)=1$. Approximate $y(5)$ with $2$ and with $4$ steps.

Let's do $2$ steps first.

The value is $D=\frac{5-3}{2}=1$.

We see that $x_1=x_0+D=3+1=4$, and also $y_1=y_0+1 \cdot (\frac{1}{x_0+y_0})=1.25$ so: $$(x_1, y_1)=(4, 1.25)$$ Repeating this: $$(x_2, y_2)\approx (5,1.4405)$$ And so $y(5)\approx 1.4405$ according to Euler's Method with $2$ steps.

Now try to approximate $y(5)$ using $4$ steps.

We have $D=0.5$, so the pairs $(x_i, y_i)$ are: $$(x_1, y_1)=(3.5, 1.125)$$ $$(x_2, y_2)=(4, 1.2331)$$ $$(x_3, y_3)=(4.5, 1.3287)$$ $$(x_4, y_4)=(5, 1.4144)$$ And so $y(5)\approx 1.4144$ according to Euler's Method with $4$ steps.

(Note: as expected, this second estimate is closer to the real value, which is around $1.3910$)

Euler's method applies to first-order ordinary differential equations of the form $y'=F(x, y)$, given an initial point $(x_0, y_0)$. Specifically, the value of $y(x)$ can be approximated in $n$ "steps" as follows: Let $D=\frac{x-x_0}{n}$. For $0\le i\le n-1$, let: $x_i=x_{i-1}+D$ and $y_i=y_{i-1}+D\times F(x_{i-1}, y_{i-1})$. Then $x_n=x$ and $y_n$ will be an approximation of $y(x)$. Notice that at each step, $y_i$ is calculated using a tangent line approximation: we are approximately (with step size $D$) "tracing along a solution curve in the slope field". Generally speaking, the approximation improves as $n$ increases and the step size $D$ decreases.