Yes, although it is true that the documentation for pdepe describes it as a solver for parabolic systems, it can often obtain solutions to systems of hyperbolic PDE with no problems.
To convert your two-equation system to a form acceptable to pdepe, I defined two auxiliary variables to yield the following four-equation system
I made a few assumptions about the values of the constants, the initial, and boundary conditions and used the following MATLAB code to solve this system
function matlabAnswers_12_31_2020
nx=31;
x = linspace(0,1,nx);
nx2=ceil(nx/2);
t = linspace(0, 1, 50);
m = 0;
sol = pdepe(m,@pdefun,@pdeic,@pdebc,x,t);
u=sol(:,:,1);
v=sol(:,:,2);
figure; plot(x, u(end,:) ,x,v(end,:)); grid; legend('u', 'v');
title 'solution at final time';
figure; plot(t, u(:,nx2),t,v(:,nx2)); grid; legend('u', 'v');
title 'solution at the center as a function of time';
end
function [ c, f, s ] = pdefun ( x, t, U, dUdx)
a=1; b=1;
c=[1 1 1 1]';
f=[0 0 a^2*dUdx(1) a^2*b^2*dUdx(2)]';
s=[U(3) U(4) U(4) -U(3)]';
end
function u0 = pdeic(x)
u0=[sin(pi*x) sin(3*pi*x) 0 0]';
end
function [ pl, ql, pr, qr ] = pdebc ( xl, ul, xr, ur, t )
pl = ul;
ql = [0 0 0 0]';
pr = ur;
qr = [0 0 0 0]';
end
