Solve the partial differential equation dC/dt = D * d2C/dx2 - k*C.

Step-by-step solution

1. 1

   Read the physics in the two terms. $D\,C_{xx}$ is ordinary diffusion and $-kC$ is first-order decay: substance spreads out while also disappearing at a rate proportional to how much is present. The two mechanisms are independent, which suggests the solution should factor into a decay part times a diffusion part.

2. 2

   Make the substitution that peels off the decay. Set

   $C(x,t)=e^{-kt}u(x,t)$

   Then $C_{t}=e^{-kt}\left(u_{t}-ku\right)$ and $C_{xx}=e^{-kt}u_{xx}$, because the exponential does not depend on $x$.

3. 3

   Substitute and cancel. The equation becomes

   $e^{-kt}\left(u_{t}-ku\right)=De^{-kt}u_{xx}-ke^{-kt}u$

   Dividing by $e^{-kt}$ (never zero) gives $u_{t}-ku=Du_{xx}-ku$, and the $-ku$ terms cancel on both sides:

   $u_{t}=Du_{xx}$

   The decay term is gone exactly — not approximately — so every known solution of the heat equation transfers.

4. 4

   Solve the reduced problem by separation of variables. Writing $u=X(x)T(t)$ and dividing by $XT$:

   $\frac{T'}{T}=D\frac{X''}{X}=-\lambda$

   so $T(t)=e^{-\lambda t}$ and $X''+\dfrac{\lambda}{D}X=0$. For $\lambda>0$ this gives $X=A\cos\sqrt{\lambda/D}\,x+B\sin\sqrt{\lambda/D}\,x$; the admissible $\lambda$ are fixed by the boundary conditions.

5. 5

   Assemble the general separated solution. Undoing the substitution,

   $C(x,t)=e^{-kt}\sum_{n}\left(A_{n}\cos\sqrt{\tfrac{\lambda_{n}}{D}}x+B_{n}\sin\sqrt{\tfrac{\lambda_{n}}{D}}x\right)e^{-\lambda_{n}t}$

   with the coefficients $A_{n},B_{n}$ determined by the initial profile $C(x,0)$.

6. 6

   Sanity-check the limiting cases. If $k=0$ the substitution is trivial and the equation is the plain heat equation ✓. If $D=0$ the PDE reduces to $C_{t}=-kC$, whose solution $C=C_{0}e^{-kt}$ is exactly what the substitution predicts with $u$ constant ✓. Every mode decays at the combined rate $k+\lambda_{n}$, so decay and diffusion simply add in the exponent.