# PDE wave equation general solution

Question: find the general solution of $$u_{xx} + 2u_{xt} - 20u_{tt} = 0$$

I'm bit confused how exactly the form of general solution $$u(x,t)$$ looks like. Because what I learn from the book is when wave equation is $$u_{tt} - c^2u_{xx} = 0$$. Then $$u(x,t) = f(x - ct) + g(x - ct)$$ where $$f$$ and $$g$$ are arbitrary function and $$c$$'s are two roots. However, I'm referring to few websites here and the equation $$u_{xx} +u_{xt} - 20u_{tt} = 0$$. This,https://www.math.cuhk.edu.hk/course_builder/1516/math4220/Solutions%20to%20assignment%202.pdf by factoring the equation into $$(\frac{\partial }{\partial x} - 4\frac{\partial }{\partial t})(\frac{\partial }{\partial x} + 5 \frac{\partial }{\partial t}) = 0$$ and obtained the general solution $$u(x,t) = f(x + \frac{1}{4}t) + g(x - \frac{1}{5}t)$$. I don't understand why $$c$$ became $$\frac{1}{c}$$ here.

In addition, this link http://individual.utoronto.ca/rifkind/teaching/PDE2015/Assignments/Assignment2/solution2.pdf using changing coordinate method to find the general solution $$u(x,t) = f(t - 5x) + g(4x + t)$$. These two solutions look equivalent to me, though I can't tell why at this moment.

Now, back to my question, it's bit trickier than the one I'm referring to. Since it can't factor in general way, but we can still 'factor' using quadratic formula to find the roots and expressed into

$$(D_{x}^2 + 2D_{xt} - 20D_{t}^2)u = 0$$

$$(D_x - (-1 + \sqrt{21})D_t)(D_x - (- 1 - \sqrt{21})D_t) = 0$$

$$(D_x + (1 - \sqrt{21})D_t)(D_x + ( 1 + \sqrt{21})D_t) = 0$$

I think I'm stuck on the very last part which is writing down the general solution $$u(x,t)$$.

Any suggestion?

First about the $$\frac{1}{c}$$

$$u_{tt} - c^2 u_{xx} = 0\\ u_{xx} - \frac{1}{c^2} u_{tt} = 0\\ (D_x + \frac{1}{c} D_t)(D_x - \frac{1}{c} D_t) u = 0\\$$

f and g are the cases when you satisfy one of the two parts of the product.

$$(D_x + \frac{1}{c} D_t) f(x-ct) = \dot{f}(x-ct)+\frac{1}{c}\dot{f}(x-ct)*(-c)\\ = \dot{f}(x-ct)-\dot{f}(x-ct)=0\\ (D_x - \frac{1}{c} D_t) g(x+ct) = \dot{g}(x+ct)-\frac{1}{c}\dot{g}(x+ct)*(c)\\ = \dot{g}(x+ct)-\dot{g}(x+ct)=0\\$$

It's not that c became $$\frac{1}{c}$$. It's about which goes with $$u_{xx}$$ and which goes with $$u_{tt}$$. You can see this by looking at units. x has units of meters. t has units of seconds. c has units of m/s. Here 5 is implicitly 5 s/m if the units are too make sense in $$\partial_x + 5 \partial_t$$, so you need $$c=\frac{1}{5} m/s$$ if you want something that would make sense in $$g(x-ct)$$

It does not matter if you can't factor it with rational numbers. Using the quadratic formula still counts as factoring.

Call it

$$(D_x + \alpha D_t)(D_x - \beta D_t) u = 0$$

Don't worry about what $$\alpha$$ and $$\beta$$ are yet.

Repeat just like with $$f$$ and $$g$$ above.

$$(D_x + \frac{1}{c} D_t) f(x-ct) = 0\\ (D_x + \alpha D_t) f(x- \frac{1}{\alpha} t) = 0\\ (D_x - \frac{1}{c} D_t) g(x+ct) = 0\\ (D_x - \beta D_t) g(x+\frac{1}{\beta}t) = 0\\$$

by renaming $$\frac{1}{c} \to \alpha$$ and $$\frac{1}{c} \to \beta$$ in the f and g cases respectively.

So the general solution is $$f(x- \frac{1}{\alpha} t) + g(x+\frac{1}{\beta}t)$$ for any functions $$f$$ and $$g$$.

Now substitute what $$\alpha$$ and $$\beta$$ are from using the quadratic formula.

$$\alpha = 1 - \sqrt{21}\\ \beta = -1 - \sqrt{21}\\$$