Satisfying first and second order PDEs given any two smooth functions

Question

I understand that this might be much simpler than the way I understand PDEs. At first, I tried to solve this problem using examples of functions $f$ and $g$, but apparently I must prove it for any smooth functions $f$ and $g$. I would appreciate your help.

Here goes:

• Given two smooth functions of form $f=f(y)$ and $g=g(z)$, prove that:
  1. $u(t,x) = f(x-t)$ is a solution of the PDE $\frac{\partial u}{\partial t}+\frac{\partial u}{\partial x}=0$.
  2. $u(t,x) = g(x+t)$ is a solution of the PDE $\frac{\partial u}{\partial t}-\frac{\partial u}{\partial x}=0$.
  3. $u(t,x) = f(x-t)+g(x+t)$ is a solution of the PDE $\frac{\partial^2 u}{\partial t^2}-\frac{\partial^2 u}{\partial x^2}=0$.

The functions $f$ and $g$ are supposed to be the same throughout 1. to 3. Any smooth functions $f$ and $g$. Without using specific Ansatz/sample functions, how do I expand each partial derivative of the function $u$?

Answer 1

This really is the chain rule. Taking the first example, maybe it would help to rewrite $x-t=w(x,t)$ and compute $$ u_x=\partial_xf(w(x,t))=f'(w(x,t))w_x(x,t) $$ so $u_x=f'(x-t)$ since $w_x=1$.

Similarly, $$ u_t=-f'(x-t) $$ So that $u_t+u_x=0$.

All we asked from $f$ in the above is that it be differentiable.

Answer 2

As an alternative way of thinking of the problem is by going through the solution of the PDE. The first two PDEs are first order linear PDEs that can be solved using the method of characteristics. Equip the first equation with the initial condition $u(t_0,x) = f(x)$ for $f$ smooth. Then, to solve the PDE, we basically run the solution backwards along characteristic curves until they intersect with the boundary of the domain (given by $t =0$).

Here, we will assume that $t_0=0$. The characteristics are fairly simple to find: we will assume that both $t$ and $x$ are functions of some unknown parameter $s$, so $u(t,x) = u(t(s),x(s))$, and $$ \frac{d}{ds}u = \partial_t u \frac{d}{ds}t(s) + \partial_x u \frac{d}{ds}x(s)= 0. $$

By comparing the above equation with the given PDE, we see immediately that $$ \frac{d}{ds}t(s) = \frac{d}{ds}x(s) =1, $$ so characteristic curves are given by $ t = s$ and $ x = s + x_0 = t+x_0 $. Along the characteristics, the unknown function $u$ satisfies $$ \frac{d}{ds} u(t(s),x(s)) = 0, $$ which has solution $ u(t(s),x(s)) = u(0,x_0) = u(0,x-t) = f(x-t)$.

The same strategy will work for the second PDE, once again assuming $t_0=0$, and the initial condition $u(0,x) =g(x)$. The characteristics are given by $t = s$ and $x= -s+x_0$ so that $x_0 = t+x$, and the solution is therefore $u(t,x) = u(0,x_0)=u(0,x+t)= g(x+t)$.

The final question can be obtained by "factoring" the second order partial differential equation to obtain $$ [\partial_t - \partial_x][\partial_t+\partial_x]u(x,t) = 0. $$ Full details here come from d'Alembert's solution to the wave equation ( https://en.wikipedia.org/wiki/Wave_equation )