Question

我有一个看起来像这样的符号功能

syms x y(x) h
fn(x) = y + (h^2*(diff(y(x), x) + 2))/2 + (h^5*diff(y(x), x, x, x, x))/120 + (h^3*diff(y(x), x, x))/6 + (h^4*diff(y(x), x, x, x))/24 + h*(2*x + y(x) - 1)

我想用其一阶导数替换y的导数的所有实例，即

subs(fn, sym('diff(y(x), x)'), dy)

其中dy已定义为

dy(x) = 2*x + y(x) - 1

结果如下：

ans(x) =

y + (h^2*(2*x + y(x) + 1))/2 + (h^5*diff(y(x), x, x, x, x))/120 + (h^3*diff(y(x), x, x))/6 + (h^4*diff(y(x), x, x, x))/24 + h*(2*x + y(x) - 1)

它取代了一阶导数，而不是高阶导数。我想要的是h^5字词(h^5*diff(dy(x), x, x, x)。有没有办法做到这一点？

我当前的方法非常hackish，涉及将sym转换为字符串，用dy替换一阶导数，然后转换回sym并评估将系列中每个项的顺序减少一，但是它必须是递归的，因为在每个阶段dy的衍生物然后被包含diff(y, ...)的东西取代。我希望有更清洁的方法来处理这个问题。

Answer 1

您需要记住，Matlab将diff(y,x)和diff(y,x,2)之类的内容视为不同的变量。它不知道如何将diff(y,x)替换为diff(y,x,2)，因为抽象函数的一般操作（一个没有明确定义的操作，例如y(x)）是不明确的

这样的事情如何从相反的一端执行替换，从最高阶导数开始：

syms y(x) h
dy(x) = 2*x + y - 1
fn(x) = y + (h^2*(diff(y, x) + 2))/2 + (h^5*diff(y, x, 4))/120 + (h^3*diff(y, x, 2))/6 + (h^4*diff(y, x, 3))/24 + h*(2*x + y - 1)
fn2 = subs(fn, diff(y, x, 4), diff(dy, x, 3));
fn2 = subs(fn2, diff(y, x, 3), diff(dy, x, 2));
fn2 = subs(fn2, diff(y, x, 2), diff(dy, x));
fn2 = subs(fn2, diff(y, x), dy);

返回

fn2(x) =

y(x) + (h^2*(2*x + y(x) + 1))/2 + (h^3*(2*x + y(x) + 1))/6 + (h^4*(2*x + y(x) + 1))/24 + (h^5*(2*x + y(x) + 1))/120 + h*(2*x + y(x) - 1)

或者您最初可以将dy(x)作为抽象符号表达式：

syms y(x) dy(x) h
fn(x) = y + (h^2*(diff(y, x) + 2))/2 + (h^5*diff(y, x, 4))/120 + (h^3*diff(y, x, 2))/6 + (h^4*diff(y, x, 3))/24 + h*(2*x + y - 1)
fn2 = subs(fn, diff(y, x, 4), diff(dy, x, 3));
fn2 = subs(fn2, diff(y, x, 3), diff(dy, x, 2));
fn2 = subs(fn2, diff(y, x, 2), diff(dy, x));
fn2 = subs(fn2, diff(y, x), dy)

返回

fn2(x) =

y(x) + (h^4*diff(dy(x), x, x))/24 + (h^2*(dy(x) + 2))/2 + (h^5*diff(dy(x), x, x, x))/120 + (h^3*diff(dy(x), x))/6 + h*(2*x + y(x) - 1)

在MATLAB中替换符号导数

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