PDE with oscillating boundary conditions

  • Thread starter robl123
  • Start date
  • #1
robl123
2
0
Hi,

Say I have this pde:

[itex]u_t=\alpha u_{xx}[/itex]
[itex]u(0,t)=\sin{x}+\sin{2x}[/itex]
[itex]u(L,t)=0[/itex]

I know the solution for the pde below is v(x,t):

[itex]v_t=\alpha v_{xx}[/itex]
[itex]v(0,t)=\sin{x}[/itex]
[itex]v(L,t)=0[/itex]

And I know the solution for the pde below is w(x,t)

[itex]w_t=\alpha w_{xx}[/itex]
[itex]w(0,t)=\sin{2x}[/itex]
[itex]w(L,t)=0[/itex]

Would the complete solution be u(x,t)=v(x,t)+w(x,t)?
 

Answers and Replies

  • #2
VantagePoint72
821
34
Your first boundary conditions are screwed up: the left side is a function of t and the right side is a function of x. Assuming you fix that, then yes, functions satisfying pieces of the boundary conditions can be added together just as you've done. You can confirm thus just by substituting [itex]u[/itex] into the original PDE and using the linearity of differentiation.
 
  • #3
robl123
2
0
Thanks!

I was worried I was doing it wrong. I meant to write [itex]u(0,t)=\sin{t}+\sin{2t}[/itex] , ...
 

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