The "Vectors" Series — 5. Plane Vector Differential Equations and Complex Numbers

《向量》系列——5.平面向量微分方程与复数 苏剑林 · 2010-10-03 · Mathematics differential equation vector complex number

Let's start by considering a complex differential equation

$$\dot{z}=f(z,t)\tag{1}$$If we let $z=x+yi,f(z,t)=f(x+yi,t)=g(x,y,t)+i*h(x,y,t)$, then this equation corresponds to

$$\begin{aligned}\dot{x}=g(x,y,t) \\ \dot{y}=h(x,y,t)\end{aligned}$$

This shows that, to some extent, a two-variable differential equation is equivalent to a complex differential equation. more

Let's look at an example first: solve the system of equations

$\dot{x}=x^2-y^2,\dot{y}=2xy$, with initial value $t=0,x=1,y=1$.

If we let $z=x+yi$, then the equation is in fact equivalent to $\dot{z}=z^2$, with initial condition $t=0,z=1+i$. Using the method for solving a single-variable differential equation, we obtain

$z=-\frac{1}{t+C}$, and from the initial condition we get $C=1/2 (i-1)$, i.e.

$$x+yi=z=-\frac{1}{t+1/2 (i-1)}=\frac{1/2-t}{(1/2-t)^2+(1/2)^2}+\frac{1/2 i}{(1/2-t)^2+(1/2)^2}$$

By the condition for equality of complex numbers, we obtain:

$$\begin{aligned}x=\frac{1/2-t}{(1/2-t)^2+(1/2)^2} \\ y=\frac{1/2}{(1/2-t)^2+(1/2)^2}\end{aligned}$$

As we can see, this method is quite quick — it's an important route for turning the unfamiliar into the familiar. The obvious advantage of converting a two-variable real differential equation into a single-variable complex differential equation is elimination: rather than calling this merely a change of form, it's better described as an "elimination method." We've also previously discussed the equivalence between complex numbers and two-dimensional vectors. Hence, for differential equations involving two-dimensional vectors, we can always treat the vector as a "number" (a complex number) when solving (as long as the vector equation doesn't involve the "magnitude"). And solving a complex differential equation is, to a large extent, the same as solving a real differential equation. Many of the equations that arise in practical problems (especially in physics and mechanics) are essentially ideal two-dimensional vector equations. Since a vector is equivalent to a complex number, why not take advantage of solving it as a complex differential equation, which is quick and convenient?

Below is a practical application from kinematics:

A projectile is fired from the ground with speed $v_0$ at an angle θ to the ground, and air resistance is q times the velocity (q being a constant). Find the trajectory of the projectile.

Let the mass of the projectile be m, the initial velocity be $\vec{v}_0$, the gravitational acceleration be $\vec{g}$, and the trajectory be $\vec{r}$. We can then write down

$\ddot{\vec{r}}=-\frac{q}{m}\dot{\vec{r}}+\vec{g}$, and letting $-\frac{q}{m}=k$, we get

$$\ddot{\vec{r}}=k\dot{\vec{r}}+\vec{g}$$

Integrating once, and substituting the initial condition, we obtain

$$\dot{r}=k\vec{r}+\vec{g}t+\vec{v}_0$$

Since a vector is equivalent to a complex number, we can treat the vector as a complex function, so the above equation is a first-order differential equation. Using what we know about solving first-order differential equations, we get

$$\vec{r}=(\frac{\vec{g}}{k^2}+\frac{\vec{v}_0}{k})(e^{kt}-1)-\frac{\vec{g}t}{k}$$

Let $\vec{r}=x+yi,\vec{g}=-gi,\vec{v}_0=v_0 cos\theta+i*v_0 sin\theta$ (equivalent to decomposing the vector orthogonally), and substituting this into the equation above, by the condition for equality of complex numbers, we obtain

$$\begin{aligned}x=(\frac{v_0 \cos\theta}{k})(e^{kt}-1) \\ y=(\frac{v_0 \sin\theta}{k}-\frac{g}{k^2})(e^{kt}-1)+\frac{g t}{k} \\ k=q/m\end{aligned}$$