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I am trying to solve the following problem on $[0,\pi]$ through separation of variables:
$u_t=u_{xx}$
$u(x,0)=x^3\space (1)$
$u(0,t)=0\space (2)$
$u(\pi,t)=\pi^3\space (3)$
So far I have come to the conclusion that the separation constant $\lambda=-k^2<0$. This led me to the solution $u(x,0)=(\sum_{n=0}^\infty A_n\cos(k_nx)+\sum_{n=0}^\infty B_n\sin(k_nx))(\sum_{n=0}^\infty C_ne^{-k_n^2t})$.
From (1) it follows that $\sum_{n=0}^\infty A_n\cos(k_nx)+\sum_{n=0}^\infty B_n\sin(k_nx)=x^3$, which can be realised by constructing the Fourier Series of $x^3$, which is $\sum_{n=0}^\infty b_n\sin(nx)$. I computed $b_n$ but it would take up too much unnecessary space.
It follows from this that (1) is met if we set $A_n=0$, $B_n=b_n$ and $k_n=n$. Then we have $$u(x,0)=(\sum_{n=0}^\infty b_n\sin(nx))(\sum_{n=0}^\infty C_ne^{-n^2t})$$This indeed makes (2) true, but $u(\pi,t)$ also becomes $0$. I don't see how this can be resolved. Did I make an error earlier on?

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  • $\begingroup$ This is an inhomogeneous problem so you cannot use Separation of Variables to solve it. $\endgroup$
    – user610336
    Mar 12, 2019 at 18:55
  • $\begingroup$ So... if the problem specifically states that I should use the method of separation of variables, the correct solution is: "No solution can be obtained through this method."? $\endgroup$
    – user569579
    Mar 12, 2019 at 18:59
  • $\begingroup$ You may try method of shifting the data $\endgroup$
    – user610336
    Mar 12, 2019 at 19:04
  • $\begingroup$ That seems like a highly nontrivial method that is not within the scope of my course, but I will give it a try, thanks. $\endgroup$
    – user569579
    Mar 12, 2019 at 19:18
  • $\begingroup$ It should not be so hard because the BC of your problem is simple $\endgroup$
    – user610336
    Mar 12, 2019 at 19:21

1 Answer 1

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Let $v(x,t)=u(x,t)-\pi^2 x$. Then $v$ will satisfy $$ v_t = v_{xx} \\ v(x,0) = u(x,0)-\pi^2 x = x^3-\pi^2 x \\ v(0,t) = u(0,t)-\pi^2 0 = 0 \\ v(\pi,t) = u(\pi,t)-\pi^3 = 0. $$ This equation is solved by $v(x,t)=\sum_{n=1}^{\infty}A_n e^{-n^2\pi^2 t} \sin(n\pi x)$, where the constants $A_n$ are chosen to satisfy $\sum_{n=1}^{\infty}A_n \sin(n\pi x)=x^3-\pi^2 x$. The constants $A_n$ are uniquely determined by the mutual orthogonality of the $\sin(n\pi x)$ functions. Finally, $u(x,t)=v(x,t)+\pi^2 x$.

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  • $\begingroup$ That's a very neat solution, thanks. $\endgroup$
    – user569579
    Mar 13, 2019 at 10:00
  • $\begingroup$ @user569579 : It is a very neat trick that I learned in my course on PDEs. :) $\endgroup$ Mar 13, 2019 at 13:52

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