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I did the dpe:
$$\frac{\partial u}{\partial y} = \frac{\partial ^2 u}{\partial x^2} - 4u$$

$0 < x < \pi $

With boundary conditions: $\begin{array}{l} u(0,y) = u(\pi ,y) = 0 \\ u(x,0) = {x^2} - \pi x \\ \end{array}$

I used method separation of variables to solve the problem, and i got the solution is:

$$u(x,y) = \sum\limits_{n = \text{even}} \frac{- 8}{\pi n^3} \sin (nx) e^{ - (4 + n^2)y} $$

However, I feel that the solution is not fit with the boundary conditions Could you give some hint??

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Using separation of variables $$ u = XY \\ u_y = XY' \\ u_{xx} = X''Y \\ XY' = X''Y - XY \\ \frac {Y'}Y = \frac {X''}X - 4 $$ Since LHS and RHS are functions of different variables, they must be equal to some constant $$ \frac {Y'}Y = \frac {X''}X - 4 = -\lambda^2-4 \\ \frac {Y'}Y = -\lambda^2-4 \\ Y = Y_0 e^{-(\lambda^2+4) y} \\ \frac {X''}X = -\lambda^2 \\ X = X_{01} \cos( \lambda x) + X_{02} \sin (\lambda x) \\ u = Y_0 e^{-(\lambda^2+4) y} \left (X_{01} \cos(\lambda x) + X_{02} \sin (\lambda x) \right ) $$ Applying first and second BCs $$ u(0, y) = Y_0 e^{-(\lambda^2+4) y} X_{01} = 0 \\ u(\pi, y) = Y_0 e^{-(\lambda^2+4) y} X_{02} \sin (\lambda \pi) = 0 $$ From first equation $X_{01} = 0$, since if $Y_0$ one gets trivial solution. From second $$ \lambda \pi = \pi n, \quad n \in \mathbb Z \\ \lambda = n\\ u = \sum_{n = 1}^\infty Y_0 e^{-(n^2+4) y} X_{01n} \sin nx $$ Coefficients $X_{01n}$ can absorb $Y_0$, so without loss of generality, one can say $$ u = \sum_{n = 1}^\infty X_{01n} e^{-(n^2+4) y} \sin nx $$ Now apply third BC $$ u(x, 0) = \sum_{n = 1}^\infty X_{01n} \sin nx = x^2 - \pi x $$ So, basically you need to find LHS expansion of $x^2-\pi x$. $$ b_n = \frac 2\pi \int_0^\pi (x^2 - \pi x) \sin n x dx = -\frac 4 {n^3 \pi} (-1+\cos n \pi) = \frac 8{n^3 \pi},\quad n \text{ is odd} $$ so $$ u = \frac 8{\pi^3} \sum_{n = odd} \frac {e^{-(n^2+4) y} \sin nx}{n^3} $$

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  • $\begingroup$ Incidentally, all the work you have shown is in OP's question. And the actual issue of the question is covered by the link to nowhere. $\endgroup$ – Artem May 13 '13 at 11:27
  • $\begingroup$ @Artem you're right, I accidentally deleted one extra symbol. As for the answer, all is needed to find one single integral. $\endgroup$ – Kaster May 13 '13 at 11:49
  • $\begingroup$ @Kaster: Thank you very much! Your solution and mine is:$u(x,y) = \sum\limits_{n = even} {\frac{{ - 8}}{{\pi {n^3}}}\sin (nx){e^{ - (4 + {n^2})y}}} $, Are they equal? $\endgroup$ – MacArthur Nguyen May 13 '13 at 14:28
  • $\begingroup$ @MacArthurNguyen I don't think so, one of use made a mistake. I'm not really sure it's not me. Would be great if you go through of both our solutions and try to find faulty one. $\endgroup$ – Kaster May 13 '13 at 14:38
  • $\begingroup$ @Kaster could you please check my question (almost same as this) math.stackexchange.com/questions/849181/… $\endgroup$ – square_one Jun 27 '14 at 7:07

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