1
$\begingroup$

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,

$$X(x)T'(t)=kX''(x)T(t)-bX(x)T(t)$$

$$\dfrac{T'(t)}{T(t)}=k\dfrac{X''(x)}{X(x)}-b=-\lambda$$

Then solving $X(x)$ gives the ODE

$$kX''(x)+(\lambda-b)X(x)=0$$

with boundary conditions,

$$X(0)=X(c)=0$$

How to go from here?

Edit

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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $u(x,t)=e^{-bt}v(x,t)$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.