Question

Let ${\mathit{z}}_{\mathbf{1}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$and${\mathit{z}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$be two complex valued solution of initial valued problem$\frac{\mathbf{d}\mathbf{z}\mathbf{\left(}\mathbf{t}\mathbf{\right)}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{\lambda }\mathit{z}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$with$\mathit{z}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$where$\mathit{\lambda }$is a complex number. Suppose that${\mathit{z}}_{\mathbf{2}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{\ne }\mathbf{0}$for all$t$.

(a): Using the quotient rule (Exercise 37), show that the derivative oflocalid="1662091749567">z1(t)z2(t)is zero. Conclude thatlocalid="1662091758671">z1(t)=z2(t)for alllocalid="1662091797882">t.

(b): Show that the initial value problem initial valued problemlocalid="1662091766773">dz(t)dt=λz(t)withlocalid="1662091774693">z(0)=1has unique complex-valued solutionlocalid="1662091784628">z(t).

Short answer

(a) The derivative of$\frac{z_1\left(t\right)}{z_2\left(t\right)}$ is $0$.

(b)$z\left(t\right)=e^{\lambda t-iArg\left(z\right)}$

Step-by-step solution

(a) Step 1: Determine the derivative of the function z1(t)z2(t).

The derivative of the function$\frac{g\left(t\right)}{f\left(t\right)}$ with respect to $t$is $\frac{d}{dt}\left(\frac{g\left(t\right)}{f\left(t\right)}\right)=\frac{f\left(t\right)\frac{dg}{dt}-g\left(t\right)\frac{df}{dt}}{{\left[f\left(t\right)\right]}^2}A$.

Consider the complex valued functions $\frac{z_1\left(t\right)}{z_2\left(t\right)}$.

Using the definition, differentiate$\frac{z_1\left(t\right)}{z_2\left(t\right)}$ with respect to$t$ as follows.

$\frac{d}{dt}\left\{\frac{z_1\left(t\right)}{z_2\left(t\right)}\right\}=\frac{z_2\left(t\right)\frac{d{z}_1\left(t\right)}{dt}-z_1\left(t\right)\frac{d{z}_2\left(t\right)}{dt}}{{\left[z_2\left(t\right)\right]}^2}$

Simplify the equation ddt{z1(t)z2(t)}=z2(t)dz1(t)dt-z1(t)dz2(t)dt[z2(t)]2.

As $\frac{d}{dt}\left\{z\left(t\right)\right\}=\lambda z\left(t\right)$, substitute the values$\lambda z_1\left(t\right)$ for $\frac{d}{dt}\left\{z_1\left(t\right)\right\}$and$\lambda z_2\left(t\right)$ for $\frac{d}{dt}\left\{z_2\left(t\right)\right\}$in the equation $\frac{d}{dt}\left\{\frac{z_1\left(t\right)}{z_2\left(t\right)}\right\}=\frac{z_2\left(t\right)\frac{d{z}_1\left(t\right)}{dt}-z_1\left(t\right)\frac{d{z}_2\left(t\right)}{dt}}{{\left[z_2\left(t\right)\right]}^2}$as follows.

$\begin{array}{c}\frac{d}{dt}\left\{\frac{z_1\left(t\right)}{z_2\left(t\right)}\right\}=\frac{z_2\left(t\right)\frac{d{z}_1\left(t\right)}{dt}-z_1\left(t\right)\frac{d{z}_2\left(t\right)}{dt}}{{\left[z_2\left(t\right)\right]}^2}\\ =\frac{z_2\left(t\right)\left\{\lambda z_1\left(t\right)\right\}-z_1\left(t\right)\left\{\lambda z_2\left(t\right)\right\}}{{\left[z_2\left(t\right)\right]}^2}\\ =\frac{\lambda z_1\left(t\right)z_2\left(t\right)-\lambda z_1\left(t\right)z_2\left(t\right)}{{\left[z_2\left(t\right)\right]}^2}\\ \frac{d}{dt}\left\{\frac{z_1\left(t\right)}{z_2\left(t\right)}\right\}=0\end{array}$

Hence, the derivative of$\frac{z_1\left(t\right)}{z_2\left(t\right)}$ is$0$ .

(b)Step 3: Simplify the equation dz(t)dt=λz(t).

Consider the initial value problem$\frac{dz\left(t\right)}{d\left(t\right)}=\lambda z\left(t\right)$ with$z\left(0\right)=1$ .

Simplify the equation $\frac{dz\left(t\right)}{d\left(t\right)}=\lambda z\left(t\right)$as follows.

$\begin{array}{l}\frac{dz\left(t\right)}{d\left(t\right)}=\lambda z\left(t\right)\\ \frac{dz\left(t\right)}{z\left(t\right)}=\lambda d\left(t\right)\end{array}$

Determine the solution.

Take integration both side in the equation $\frac{dz\left(t\right)}{z\left(t\right)}=\lambda d\left(t\right)$as follows.

$\begin{array}{c}\frac{dz\left(t\right)}{z\left(t\right)}=\lambda d\left(t\right)\\ \int \frac{dz\left(t\right)}{z\left(t\right)}=\int \lambda d\left(t\right)\\ \ln z\left(t\right)=\lambda t+c\end{array}$

Substitute the value $0$for $t$in the equation$\ln z\left(t\right)=\lambda t+c$as follows.

$\begin{array}{c}\ln z\left(t\right)=\lambda t+c\\ \ln z\left(0\right)=\lambda \left(0\right)+c\\ \ln \left(1\right)=c\\ c=0\end{array}$

Substitute the value $0$for$c$in the equation $\ln z\left(t\right)=\lambda t+c$as follows.

$\begin{array}{l}\ln z\left(t\right)=\lambda t+c\\ \ln z\left(t\right)=\lambda t\end{array}$

As $\ln z\left(t\right)=\ln \left|z\left(t\right)\right|+iArg\left(z\right)$, Substitute the value$\ln \left|z\left(t\right)\right|+iArg\left(z\right)$for $\ln z\left(t\right)$in the equation $\ln z\left(t\right)=\lambda t$as follows.

$\begin{array}{c}\ln z\left(t\right)=\lambda t\\ \ln z\left(t\right)=\lambda t\\ \ln \left|z\left(t\right)\right|+iArg\left(z\right)=\lambda t\\ \ln \left|z\left(t\right)\right|=\lambda t-iArg\left(z\right)\end{array}$

Taking anti-log both side in the equation$\ln \left|z\left(t\right)\right|=\lambda t-iArg\left(z\right)$as follows.

$\begin{array}{c}\ln \left|z\left(t\right)\right|=\lambda t-iArg\left(z\right)\\ e^{\ln \left|z\left(t\right)\right|}=e^{\lambda t-iArg\left(z\right)}\\ z\left(t\right)=e^{\lambda t-iArg\left(z\right)}\end{array}$

Hence, the solution ofthe initial value problem$\frac{dz\left(t\right)}{d\left(t\right)}=\lambda z\left(t\right)$with$z\left(0\right)=1$is $z\left(t\right)=e^{\lambda t-iArg\left(z\right)}$.