Analysis of converting a DE into complex DE

[Hall]

In Lecture 7, Prof. Arthur Mattuck (MIT OCW 18.03) taught that the following equation
$$
y’ +ky = k \cos(\omega t)$$
can be solved by replacing cos⁡(ωt) by $e^{\omega t}$ and, rewriting thus,
$$
\tilde{y’} + k\tilde{y}= ke^{i \omega t}
$$
Where $\tilde{y} = y_1 + i y_2$. And the solution of the first equation is the real part of the solution of the second equation. The purpose to go from first equation to the second is to ease the process of solving.

I believed the transformation, and it also doesn’t seem very alien, but I just want to know more about it. Can Real/Complex analysis prove that the real part of the solution of the second equation is the solution of the first equation?

 

[topsquark]

In Lecture 7, Prof. Arthur Mattuck (MIT OCW 18.03) taught that the following equation
$$
y’ +ky = k \cos(\omega t)$$
can be solved by replacing cos⁡(ωt) by $e^{\omega t}$ and, rewriting thus,
$$
\tilde{y’} + k\tilde{y}= ke^{i \omega t}
$$
Where $\tilde{y} = y_1 + i y_2$. And the solution of the first equation is the real part of the solution of the second equation. The purpose to go from first equation to the second is to ease the process of solving.

I believed the transformation, and it also doesn’t seem very alien, but I just want to know more about it. Can Real/Complex analysis prove that the real part of the solution of the second equation is the solution of the first equation?

I'm not sure what you are asking.
$\tilde{y} = y_1 + i y_2 \implies \tilde{y}' = y_1' + i y_2'$

So what is the real part of $\tilde{y}' + k \tilde{y} = e^{it}$?

You get that the real part of the solution is what you need pretty much just by definition.

-Dan

 

[Hall]

I'm not sure what you are asking.
$\tilde{y} = y_1 + i y_2 \implies \tilde{y}' = y_1' + i y_2'$

So what is the real part of $\tilde{y}' + k \tilde{y} = e^{it}$?

You get that the real part of the solution is what you need pretty much just by definition.

-Dan

$$
\tilde{y’}+ k\tilde{y}= k e^{i \omega t}$$
Is basically a composition of two differential equations, real part and imaginary part. Why solving the complete equation solves the composite parts? And sifting the solution gives the solution of composite parts?

In physics it is an Axiom that if $\mathbf{F}= F_x \hat{i} + F_y\hat{j}$ is applied to an object, the effect of $F_x$ and $F_y$ can be analysed independently, they don’t affect each other’s domain. Same is the Axiom in projectile motions.

But mathematics doesn’t state any such Axiom.

 

[topsquark]

$$
\tilde{y’}+ k\tilde{y}= k e^{i \omega t}$$
Is basically a composition of two differential equations, real part and imaginary part. Why solving the complete equation solves the composite parts? And sifting the solution gives the solution of composite parts?

In physics it is an Axiom that if $\mathbf{F}= F_x \hat{i} + F_y\hat{j}$ is applied to an object, the effect of $F_x$ and $F_y$ can be analysed independently, they don’t affect each other’s domain. Same is the Axiom in projectile motions.

But mathematics doesn’t state any such Axiom.

Actually it does, but I can't think of the name of it. (The term "extension field" comes to mind but I'd have to look it up to be sure.) It comes from Algebra and is actually a Mathematical reason that we should suspect that motion in two perpendicular directions in independent. Though I'll admit that the extension field idea came later on.

But in any case, the resulting complex equation is linear and we can still solve it easily and under the same rules as for real variables. So the complex nature of the equation doesn't change anything about simply taking the real part at the end... everything is still nice and linear so you can decompose the real part without any troubles. There is no abstract mixing of complex and real solutions to complicate anything. (I don't think it would matter in any case but I don't have the Linear Algebra to prove that in general.)

-Dan

 

[pbuk]

I agree with @topsquark, there is no mystery here. But let me try it another way:

Why solving the complete equation solves the composite parts? And sifting the solution gives the solution of composite parts?

Edit: thanks to @pasmith for providing a correct explanation below - I have deleted mine.

Because if $z = h(t)$ is a solution to $z' = h'(z, t)$ then $x = \operatorname{Re}(z) = \operatorname{Re}(h(t))$ is a solution to $x' = \operatorname{Re}(z') = \operatorname{Re}(h'(z, t))$.

If that doesn't work for you, try: if $x + iy = f(x, t) + ig(y, t)$ is a solution to $x' + iy' = f'(x, t) + ig'(x, t)$ then $x = \operatorname{Re}(f(t) + ig(t)) = f(t)$ is a solution to $x' = \operatorname{Re}(x' + iy') = \operatorname{Re}(f'(x, t) + ig(x, t)) = f'(x, t)$ (provided $x, y, f, g$ are all real).

 

[pasmith]

In physics it is an Axiom that if $\mathbf{F}= F_x \hat{i} + F_y\hat{j}$ is applied to an object, the effect of $F_x$ and $F_y$ can be analysed independently, they don’t affect each other’s domain. Same is the Axiom in projectile motions.

But mathematics doesn’t state any such Axiom.

I think this only works if $\mathbf{F} = k(t)\dot{\mathbf{x}} + l(t)\mathbf{x}+ \mathbf{c}(t)$; otherwise you can't really analyse vertical and horizontal motion separately. And in this case instead of using vectors, you could use complex numbers: set $F = F_x + iF_y$ and $z = x + iy$.

This separation of real and imaginary parts works only when the ODE is linear with real coefficients. In this case $z' + kz$ is a linear operator, and if $$\begin{split} x' + kx &= f(t) \\ y' + ky &= g(t) \end{split}$$ then we can form the linear combination $z = x + iy$ and adding the first equation to $i$ times the second equation gives us $$z' + kz = f(t) + ig(t).$$ However this is really only useful when $$f(t) + ig(t) = e^{at}(\cos(bt) + i\sin(bt)) = e^{(a + ib)t}$$ because multiplying this by the integrating factor $e^{kt}$ gives us something which is easily integrated: $$\frac{d}{dt}(ze^{kt}) = \frac{1}{k + a + ib}\frac{d}{dt}e^{(k + a + ib)t} = \frac{k + a - ib}{(k + a)^2 + b^2}\frac{d}{dt}e^{(k + a + ib)t}$$ Now the real and imaginary parts of $z$ can be extracted and we have solved both $$\begin{split} x' + kx &= e^{at}\cos(bt) \\ y' + ky &= e^{at}\sin(bt)\end{split}$$ with only one integration (which if we'd stuck with real variables throughout would have been by parts).

Note that linearity is critical; if we had $z' + kz^2$ on the left hand side then this would not have worked: $$x^2 + iy^2 \neq x^2 - y^2 + 2ixy = z^2.$$

 

[pbuk]

Note that linearity is critical; if we had $z' + kz^2$ on the left hand side then this would not have worked: $$x^2 + iy^2 \neq x^2 - y^2 + 2ixy = z^2.$$

Oh good point - I'll edit my post.