In this post we derived the wave equation for a one-dimensional object moving in two-dimensional space (namely, a string vibrating up and down). This time we’re going to derive a family of solutions for this one-dimensional wave equation.
So the equation we derived previously looked like this: . This equation is often rewritten as: , where . We want to find solutions describing how the string looks at any point in space and time. Well, we know in general what a wave looks like. It looks more or less like a sine or cosine. Suppose our solution looks like this:
, where is the time component of the solution. In other words, we’re assuming that the space and time components of the wave are independent. Let’s differentiate:
Plugging these in, the PDE becomes:
Since we already have our space component, let’s focus on the time component:
This is just a second order ODE with constant coefficients. And it’s in quite a nice form. Which function’s derivative looks like the original function? That would, of course, be the exponential function. Let’s assume for a second that . Then , and the time component equation becomes: . Dividing out the exponentials, we get a simple quadratic equation: . This equation is called the “characteristic equation” (not to be confused with the equation of the characteristic curve! These are different things!). Solving, we get that .
Now we need to alter the ‘s to get rid of their imaginary parts.
Very quickly, the principle of superposition says that if are solutions of the linear equation , then (for any index between and and for any constants ) is also a solution of .
Let’s use this to get a family of real solutions. Let’s compute and , both of which are solutions:
This is convenient, because now we can just multiply to the first equation and to the second, and then add the two. That gives us the solution , where and can be any real number.
Finally, plugging back into the solution , we get: