Solve 1D Wave Equation (Hyperbolic PDE)

I have the following equation:

where f = 2q, q is a function of both x and t. I have the initial condition:

where sigma = 1/8, x lies in [-1,1]. There is also a boundary condition that q(-1) = q(+1).

How do I solve this (get the function q(x,t), or at least q(x) at some particular time t) in Matlab? Can it be done using ODE45? Or do I have to use some PDE solver? Thank you!

[Bruno Luong]

In my book, this equation is a transport equation or convection. It's not an hyperbolic PDE (or wave equation) which is a second order equation.

[Torsten]

It's called hyperbolic conservation equation:

http://php.math.unifi.it/users/cimeCourses/2009/01/200912-Notes.pdf

Best wishes

Torsten.

Answer 1

[Torsten]

Solve the system of ordinary differential equations

df(i)/dt = -2*(f(i)-f(i-1))/(x(i)-x(i-1))  (1 <= i <= n)
f(i) @ t=0 = exp(-x(i)^2/(2*sigma^2)) (?)

with 

f(0) = f(n-1) and x(i) = -1+2/(n-1)*(i-1)

using "ode15s".

Best wishes

Torsten.

Answer 2

[Bruno Luong]

In my book, this equation is a transport equation or convection. It's not an hyperbolic PDE (or wave equation) which is a second order equation.

One can solve it by characteristics equation, meaning look for a curve x(t) such that dx/dt = 2.

x(t) = x(t=0) + 2*t.

Now if you define

p(t) = q(t,x(t))

then dp/dt = dq/dt + dq/dx*dx/dt = dq/dt + 2 * dq/dx = dq/dt + df/dx = 0.

Meaning p(t) is constant on the characteristic curve x(t).

q(t,x) = p(t,x0+2*t) = p(0,x0) = q(0,x-2*t) = exp((x-2*t)^2/(2*sigma))

provide (x-2*t) is in (-1,1), the place where the caracteristics lines cut the interval t=0, x in (-1,1).

The rest of the domain you can fill with anything that respects q(1)=q(-1), it doesn't matter.

[Torsten]

I think you'll have to periodically continue the function given at t=0 and advect this function with velocity +2. So if the periodically continued function is called q_per(x) (now defined on all of IR), the solution of the PDE is q(x,t) = q_per(x-2*t).