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The problem is:

Solve the P.D.E. $U_t=U_{xx}$ with the following initial and boundary conditions: $$U(0,t)=U(\pi,t)=0,\quad\text{and}\quad U(x,0)=u_0-u_0\sin x$$

For the given fixed boundary conditions, I solved the equation for $U(x,0)=u_0$ and got $U_1(x,t)$, then solved it for $U(x,0)=-u_0\sin x$ and got $U_2(x,t)$ as the solution. Now I correctly got $U(x,t)=U_1(x,t)+U_2(x,t)$ as the solution to the original P.D.E..

Given that this technique worked for this problem, I wonder if I am allowed to do this when solving other P.D.E.s and whether it is a known method (for instance be known as the principle of superpositions).

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  • $\begingroup$ It is trivially right in this case. When you get the series $u(x,0)=\sum_{n=1}^{\infty}B_n\sin(nx)$, you go on to calculate $B_n=\frac{2}{\pi}\int_0^\pi u(x,0)\sin(nx)\mathrm{d}x$. What you are doing is to just decompose $u(x,0)$ as the sum of $u_1(x,0)=u_0$ and $u_2(x,0)=-u_0\sin x$, so $B_n$ is of course the sum of $B_{1n}=\frac{2}{\pi}\int_0^\pi u_1(x,0)\sin(nx)\mathrm{d}x$ and $B_{2n}=\frac{2}{\pi}\int_0^\pi u_2(x,0)\sin(nx)\mathrm{d}x$. $\endgroup$ Commented Nov 25 at 19:32
  • $\begingroup$ I might have made a typo in the integral. But you know what I mean. $\endgroup$ Commented Nov 25 at 19:47
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    $\begingroup$ It works when the equations are linear. $\endgroup$ Commented Nov 25 at 20:09

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