- #1

robl123

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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)?