# Fourth-order homogeneous ODE

I've encountered the following fourth-order homogeneous differential equation while reading some lecture notes on the dynamic analysis of beams:

$\Phi(s)'''' - \beta^2 \Phi(s)'' - \alpha^4 \omega^2 \Phi(s) = 0$,

(where I've copied coefficients and exponents as found to specify whether they are positive or negative).

The notes also indicate this solution:

$\Phi(s)=C_1 \cos(\lambda s) + C_2 \sin(\lambda s) + C_3 \cosh(\mu s) + C_4 \sinh(\mu s)$,

where:

$\lambda=\sqrt{\sqrt{\frac{\beta^4}{4} + \alpha^4 \omega^2} -\frac{\beta^2}{2}}$ and

$\mu=\sqrt{\sqrt{\frac{\beta^4}{4} + \alpha^4 \omega^2} +\frac{\beta^2}{2}}$.

Since it's been several years since the last time I handled differential equations, and anyway I've never had to go beyond second-order ones, what I'm looking for would be for someone to check whether the presented solution is correct, and if possible to explain the steps necessary to find it. Some resources on the topic (preferably accessible to a civil engineering student with some years between him and Calculus courses) would also be much appreciated.

Thank you very much in advance.

Here's an elementary derivation, using the operator notation, which should be clear enough from the first glance. We rewrite the equation using a differential form of the derivatives:

$$\frac{d^4\Phi}{ds^4} - \beta^2 \frac{d^2\Phi}{ds^2} - \alpha^4 \omega^2 \Phi = 0$$

Now we present everything that happens to $\Phi$ as an operator acting on the function from the left:

$$\left(\frac{d^4}{ds^4} - \beta^2 \frac{d^2}{ds^2} - \alpha^4 \omega^2 \right)\Phi = 0$$

'Multiplication' for operators means applying the same operator several times. So we can actually introduce an explicit operator notation here:

$$\left(\frac{d^4}{ds^4} - \beta^2 \frac{d^2}{ds^2} - \alpha^4 \omega^2 \right)=\left( \hat{D}^4 - \beta^2 \hat{D}^2 - \alpha^4 \omega^2 \right)$$

Now we pretend it's just a quadratic function, which we can factor in the usual way:

$$\left( \hat{D}^2 - \frac{1}{2} \left(\beta^2+ \sqrt{\beta^4+4\alpha^4 \omega^2} \right) \right) \left( \hat{D}^2 - \frac{1}{2} \left(\beta^2- \sqrt{\beta^4+4\alpha^4 \omega^2} \right) \right)$$

Here I used the usual formulas for quadratic equations.

Now we get back to our equation:

$$\left( \hat{D}^2 - \frac{1}{2} \left(\beta^2+ \sqrt{\beta^4+4\alpha^4 \omega^2} \right) \right) \left( \hat{D}^2 - \frac{1}{2} \left(\beta^2- \sqrt{\beta^4+4\alpha^4 \omega^2} \right) \right) \Psi=0$$

This can be valid if we introduce a solution in the following form:

$$\Psi=\Psi_I+\Psi_{II}$$

Where:

$$\left( \hat{D}^2 - \frac{1}{2} \left(\beta^2+ \sqrt{\beta^4+4\alpha^4 \omega^2} \right) \right) \Psi_I=0$$

$$\left( \hat{D}^2 - \frac{1}{2} \left(\beta^2- \sqrt{\beta^4+4\alpha^4 \omega^2} \right) \right) \Psi_{II}=0$$

Notice that we can actually factor both of the equations above in the same way by using the formula $a^2-b^2=(a-b)(a+b)$, and setting:

$$\Psi_I=\Psi_1+\Psi_2$$

$$\Psi_{II}=\Psi_3+\Psi_4$$

Where:

$$\left( \hat{D} - \sqrt{\frac{1}{2} \left(\beta^2+ \sqrt{\beta^4+4\alpha^4 \omega^2} \right) } \right) \Psi_1=0$$

$$\left( \hat{D} + \sqrt{\frac{1}{2} \left(\beta^2+ \sqrt{\beta^4+4\alpha^4 \omega^2} \right) } \right) \Psi_2=0$$

$$\left( \hat{D} - \sqrt{\frac{1}{2} \left(\beta^2- \sqrt{\beta^4+4\alpha^4 \omega^2} \right) } \right) \Psi_3=0$$

$$\left( \hat{D} + \sqrt{\frac{1}{2} \left(\beta^2- \sqrt{\beta^4+4\alpha^4 \omega^2} \right) } \right) \Psi_4=0$$

Now we get back to:

$$\hat{D}=\frac{d}{ds}$$

We obtain four 1st order differential equations:

$$\frac{d\Psi_1}{ds} - \sqrt{\frac{1}{2} \left(\beta^2+ \sqrt{\beta^4+4\alpha^4 \omega^2} \right) } \Psi_1=0$$

$$\frac{d \Psi_2}{ds} + \sqrt{\frac{1}{2} \left(\beta^2+ \sqrt{\beta^4+4\alpha^4 \omega^2} \right) } \Psi_2=0$$

$$\frac{d \Psi_3}{ds} - \sqrt{\frac{1}{2} \left(\beta^2- \sqrt{\beta^4+4\alpha^4 \omega^2} \right) } \Psi_3 =0$$

$$\frac{d \Psi_4}{ds} + \sqrt{\frac{1}{2} \left(\beta^2- \sqrt{\beta^4+4\alpha^4 \omega^2} \right) } \Psi_4=0$$

The solutions all have the form of the exponential function. However, because:

$$\beta^2 \leq \sqrt{\beta^4+4\alpha^4 \omega^2}$$

We should change the signs in the last two radicals and use the imaginary unit:

$$\frac{d \Psi_3}{ds} - i \sqrt{\frac{1}{2} \left( \sqrt{\beta^4+4\alpha^4 \omega^2}-\beta^2 \right) } \Psi_3 =0$$

$$\frac{d \Psi_4}{ds} + i \sqrt{\frac{1}{2} \left( \sqrt{\beta^4+4\alpha^4 \omega^2}-\beta^2 \right) } \Psi_4=0$$

Now we introduce:

$$\lambda=\sqrt{\frac{1}{2} \left( \sqrt{\beta^4+4\alpha^4 \omega^2}-\beta^2 \right) }$$

$$\mu=\sqrt{\frac{1}{2} \left( \sqrt{\beta^4+4\alpha^4 \omega^2}+\beta^2 \right) }$$

And obtain, finally:

$$\frac{d\Psi_1}{ds} - \mu \Psi_1=0$$

$$\frac{d\Psi_2}{ds} + \mu \Psi_2=0$$

$$\frac{d\Psi_3}{ds} - i \lambda \Psi_3=0$$

$$\frac{d\Psi_4}{ds} + i \lambda \Psi_4=0$$

Which gives us the final solution:

$$\Psi=A_1 e^{-\mu s}+A_2 e^{\mu s}+A_3 e^{-i \lambda s}+A_4 e^{i \lambda s}$$

This is the same as the solution in the notes, if you remember the definitions for trigonometric and hyperbolic functions in terms of the exponential function.

$$\cosh x = \frac{1}{2} \left(e^x+e^{-x} \right)$$

$$\sinh x = \frac{1}{2} \left(e^x-e^{-x} \right)$$

$$\cos x = \frac{1}{2} \left(e^{ix}+e^{-ix} \right)$$

$$\sin x = \frac{1}{2i} \left(e^{ix}-e^{-ix} \right)$$