Solve the following PDE, $$ u_t(x,t)=ku_{xx}(x,t)-bu(x,t)$$ where $b>0$, with boundary conditions $$u(0,t)=u(c,t)=0 $$

My attempt

Assume $u(x,t)=X(x)T(t)$ and plugging into the diffirential equation,



Then solving $X(x)$ gives the ODE


with boundary conditions,


How to go from here?


Setting up the characteristic equation, $$r^2+\dfrac{\lambda-b}{k}=0$$ gives, $$r=i\mkern1mu\sqrt{\dfrac{\lambda-b}{k}}$$ which gives a general solution, $$X(x)= c_1\cos\bigg(\sqrt{\dfrac{\lambda-b}{k}}x\bigg)+c_2\sin\bigg(\sqrt{\dfrac{\lambda-b}{k}}x\bigg)$$ And from here her I don't know

  • $\begingroup$ Well, did you solve $kX''(x)−(b+\lambda)X(x)=0$? $\endgroup$ – Hyperplane Oct 2 '17 at 22:05
  • $\begingroup$ and once you have the general solution to the ODE for $X(x)$, what has to be true of $\lambda$ to make your solution satisfy the boundary conditions? $\endgroup$ – Brian Borchers Oct 2 '17 at 22:06
  • $\begingroup$ Would I have to consider the cases $\lambda =b$, $\lambda > b$, and $\lambda <b$ ? $\endgroup$ – Alex Oct 3 '17 at 0:01
  • $\begingroup$ Now you apply your boundary condition. What does it tell you about the constants $c_1, c_2$ and $\lambda$? $\endgroup$ – Hyperplane Oct 3 '17 at 9:41

You can rewrite the equation as $$ u_t(x,t)+bu(x,t)=ku_{xx}(x,t) $$ Multiplying by $e^{bt}$ gives the more standard form: $$ (e^{bt}u)_{t}=(e^{bt}u)_{xx} $$ Therefore $v(x,t)=e^{bt}u(x,t)$ satisfies $$ v_{t}=kv_{xx} \\ v(0,t)=0=v(c,t). $$ Using separation of variables gives a solution $$ v(x,t) = \sum_{n=1}^{\infty}A_ne^{-n^2\pi^2kt/c^2}\sin(n\pi x/c). $$ The constants $A_n$ are determined through orthogonality from the initial data $v(x,0)$. Then $u(x,t)=e^{-bt}v(x,t)$.


There are a discrete infinity of values of lambda that correspond to the number of half sine waves that can fit in the interval. The general solution is a sum over all of them. The coefficients are found from the initial condition using orthogonalty of the sines.


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