Sep 15, 2009 #1 aminfar Messages 1 Reaction score 0 I am new to non-linear PDEs. So I tried to solve it, but I stuck in the beginning. U^2_xU_t - 1 = 0 U(x, 0) = x
I am new to non-linear PDEs. So I tried to solve it, but I stuck in the beginning. U^2_xU_t - 1 = 0 U(x, 0) = x
Oct 5, 2009 #2 gato_ Messages 116 Reaction score 0 The equation admits a separation of variables, and the solution can be written as U(x,t)=f(x)+g(t). Inserting this, you get f'(x)^2=1/g'(t)=constant. The solution can then be checked to be U=x+t
The equation admits a separation of variables, and the solution can be written as U(x,t)=f(x)+g(t). Inserting this, you get f'(x)^2=1/g'(t)=constant. The solution can then be checked to be U=x+t