How do we write a sinusoidal solution to a 2nd order DE as a sum of exponentials raised to complex roots?

[zenterix]

Homework Statement

I have a doubt that has bugged me for a long time about sinusoidal solutions to linear 2nd order differential equations with constant coefficients.

Relevant Equations

I can solve differential equations, but I would like to understand how we get from a real solution of a 2nd order linear differential equation with constant coefficients to a form (which I imagine is a complex solution) that has exponentials raised to exponents that are the complex roots of the characteristic polynomial.

Can this be done?

Consider the differential equation

$$y''+ay'+by=0$$

We have analytical solutions for this equation.

There are three cases to consider based on the discriminant of the characteristic polynomial associated with the equation.

$$\Delta=a^2-4b$$

I just want to discuss the case where $$\Delta <0$$.

It can be shown that if we define $k=\frac{\sqrt{-\Delta}}{2}$ the general solution is

$$y(x)=e^{-ax/2}(c_1\sin{kx}+c_2\cos{kx})$$

So here is my question.
In the other two possible cases ($\Delta=0$ or $\Delta >0$) we can write the general solutions, respectively, as

$$y(x)=c_1e^{rx}+c_2xe^{rx}=c_1+c_2x$$

$$y(x)=c_1e^{r_1x}+c_2e^{r_2x}$$

Is there a way to express the general solution in the case $\Delta <0$ in a form similar to these?

For example, suppose $\Delta <0$. Then the two roots are $\frac{a\pm\sqrt{\Delta}}{2}$.

Let's start with the general solution

$$y(x)=e^{-ax/2}(c_1\sin{kx}+c_2\cos{kx})$$

$$=e^{-ax/2}A\cos{(kx-\phi)}$$

where $k=\frac{\sqrt{-\Delta}}{2}$, $A=\sqrt{c_1^2+c_2^2}$, and $\phi=\tan^{-1}{\frac{c_1}{c_2}}$.

This solution is the real part of a complex solution

$$y(x)=\Re(z(x))=\Re[e^{-ax/2}A(\cos{(kx-\phi)}+i\sin{(kx-\phi)})]$$

$$=\Re[e^{-ax/2}Ae^{i(kx-\phi)}]$$

$$=\Re[Ae^{-i\phi}e^{\frac{-a+\sqrt{-\Delta}}{2}x}]$$

By a similar derivation we can also show that

$$y(x)=\Re[Ae^{-i\phi}e^{\frac{-a-\sqrt{-\Delta}}{2}x}]$$

It seems that any linear combination of these two complex solutions is also a complex solution and, furthermore, the real part is a solution to the original real equation.

But I don't know what to conclude from this.

An alternative way is to do the following

$$y(x)=\Re(z(x))=\Re[e^{-ax/2}A(\cos{(kx-\phi)}+i\sin{(kx-\phi)})]$$

$$y(x)=\Re[e^{-ax/2}(A\cos{\phi}\cos{kx}+A\sin{\phi}\sin{kx}+i(A\sin{kx}\cos{\phi}-A\cos{kx}\sin{\phi}))]$$

where we used

$$A\sin{(kx-\phi)}=A(\sin{kx}\cos{\phi}-\cos{kx}\sin{\phi})$$

$$=(A\cos{\phi})\sin{kx}-(A\sin{\phi})\cos{kx}$$

Then

$$y(x)=\Re[e^{-ax/2}(A\cos{\phi}(\cos{kx}+i\sin{kx})+A\sin{\phi}(\sin{kx}-i\cos{kx})]$$

$$=\Re[e^{-ax/2}(A\cos{\phi}e^{ikx}-iA\sin{\phi}e^{ikx})]$$

 

[FactChecker]

Good point. The solutions for $\Delta \ge 0$ are not oscillating. Your solutions for the case of $\Delta \lt 0$ work for the equations with oscillating solutions. There is usually some cyclical trade-off between two physical (or mathematical) quantities which are represented by the real and imaginary parts of your solution. The oscillations can be stable (damped) or unstable.

 

[pasmith]

If y&#039;&#039; - 2py&#039; + qy = 0 has real cofficients and z is a complex-valued solution, then it is easy to check that the complex conjugate \bar{z} is also a solution. Since the ODE is linear, it follows that both (z + \bar{z})/2 = \Re(z) and (z - \bar{z})/(2i) = \Im(z) are linear combinations of solutions and therefore also solutions, which happen to be real-valued.

We can always write Ae^{r_1x} + Be^{r_2x} = \exp\left(\frac{ r_1 + r_2 }2 x\right)\left(<br /> C\cosh\left(\frac{r_1 - r_2}2 x\right) + D\sinh\left(\frac{r_1 - r_2}2x\right)\right) where C = A + B and D = A - B. The second formulation is sometimes preferable, since it makes it trivial to apply initial conditions on y and y&#039; at x = 0, and to apply an iniitial condition at x = x_0 we can simply replace x with x - x_0 throughout.

Since the cofficients of the ODE are real, r_1 and r_2 are either both real or a complex conjugate pair, and in the latter case we have <br /> e^{\Re(r)x}(C \cosh(i\Im(r)x) + D\sinh(i\Im(r)x) = e^{\Re(r)x}(C\cos(\Im(r)x) + iD\sin(\Im(r)x)) and the result will be real-valued provided C and iD are real. This requires that B = \bar{A}.