algorithm - How do I use MATLAB to solve this PDE -

i have following question on practice exam:

i need use matlab solve it. problem is, have not seen problem before , i'm struggling started.

i have 1x1 grid, split 10x10. know can calculate whole bottom row besides corners using 1/10 * x*2. know can calculate entire right row using (1/10)(1+t)^2. however, cannot figure out how enough points able fill in values entire grid. know must have partial derivatives given in problem, i'm not quite sure come play (especially u_x equation). can me start here?

i don't need whole solution. once have enough points can write matlab program solve rest. really, think need x=0 axis solved, fill in middle of grid.

i have calculated bottom row, minus 2 corners, 0.001, 0.004, 0.009, 0.016, 0.025, 0.036, 0.049, 0.064, 0.081. , similarly, entire right row trival calculate using given boundry condition. can't piece go there.

edit: third boundry condition equation mistyped. should read:

u_x(0,t) = 1/5t, not u(0,t) = 1/5t

first realise equation have solve linear wave equation, , numerical scheme given can rewritten as

( u^(n+1)_m - 2u^n_m + u^(n-1)_m )/k^2 = ( u^n_(m-1) - 2u^n_m + u^n_(m+1) )/h^2 

where k time step , h delta x in space.

the reformulated numerical scheme makes clear left- , right-hand sides second order centred finite difference approximations of u_tt , u_xx respectively.

to solve problem numerically, however, need use form given because explicit update formula need implement numerically: gives solution @ time n+1 function of previous 2 times n , n-1. need start initial condition , march solution in time.

observe solution assigned on boundaries of domain (x=0 , x=1), values of discretized solution u^(n)_0 , u^(n)_10 known n (t=n*k). @ nth time step unknown vector [u^(n+1)_1, u^(n+1)_2, ..., u^(n+1)_9].

observe use update formula find solution @ n+1 step, requires knowledge of solution @ two previous steps. so, how start n=0 if need information two previous times? initial conditions come play. have solution @ n=0 (t=0), have u_t @ t=0. these 2 pieces of information combined can give both u^0 , u^1 , started.

i use following start-up scheme:

u^0_m = u(h*m,0)  // initial condition on u (u^2_m - u^0_m)/(2k) = u_t(h*m,0)  // initial condition on u_t 

that combined numerical scheme used n=1 gives need define linear system both u^1_m , u^2_m m=1,...,9.

to summarize:

--use start-up scheme find solution @ n=1 , n=2 simultaneously.

--from there on march in time using numerical scheme given.

if lost check out things like: finite difference schemes, finite difference schemes advection equations, finite difference schemes hyperbolic equations, time marching.

editing:

for boundary condition on u_x typically use ghost cell method:

• introduce ghost cell @ m=-1, i.e. fictitious (or auxiliary) grid point used deal boundary condition, not part of solution.

• the first node m=0 unknown vector, i.e. working [u_0 u_1 ... u_9].

• use left side boundary condition close system. specifically, writing down centered approx of boundary condition

  u^n_(1) - u^n_(-1) = 2*h*u_x(0,k*n)

• the above equation allows express solution on ghost node in terms on solution on internal, real node. therefore can apply time-marching numerical scheme (the 1 given) m=0 node. (the numerical scheme applied m=0 contain contributions m=-1 ghost node, have expressed in terms of m=1 node.)