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Question, solve the given PDE :

$$ \frac{\partial C}{\partial t} = a \frac{\partial^2 C}{\partial x^2} -kC $$ where a, k are constants.

boundary conditions : $C= C_0$ at $x=0$, $C=0 $ at $x =$ infinitum

initial condition : $C=0$ at $t=0$

My attempt :

$$C(x,t) = X(x)~T(t)$$ $$ C = X~T$$ $$ \tag{1}\boxed{ \frac{\partial C}{\partial t} = T^{\prime} X,~ \space \frac{\partial C}{\partial x}= X^{\prime}T, ~ \frac{\partial^2 C}{\partial x^2} = X^{\prime \prime } T}$$ replacing the partial derivatives into the Original equation : $$ T^{\prime}X = a ~X^{\prime \prime}T-k(X~T) $$ $$ X (T^{\prime} +KT)=a ~X^{\prime \prime}T $$ $$\frac{T^{\prime}}{T} = a ~\frac{ X^{\prime \prime}}{X} -k$$ $$ \tag{2}\boxed{\frac{T^{\prime}}{T} = J, ~\frac{ X^{\prime \prime}}{X} -k = J}$$ equating each side to a constant $J = -\lambda^2$ $$ \tag{3}\boxed{\frac{d T}{d t} = -\lambda^2 T ,~ \frac{d^2 X}{d x^2} = \left (\frac{-\lambda^2 + k}{a} \right) X}$$ solving each differential equation : $$T(t) = Ae^{- \lambda t} , ~ ~~~X(x) = B ~e^{\left (x \sqrt{\frac{ k - \lambda^2}{a}} \right)} + C ~e^{\left (-x \sqrt{\frac{ k - \lambda^2}{a}} \right) } $$ Note : D = AB and E = AC

Corrected equation : $$\tag{4} \boxed{{C(x,t) = \left(D \ e^ {x \sqrt{\frac{k-\lambda^2}{a}}} + E e^ {-x \sqrt{\frac{k-\lambda^2}{a}}} \right) e^{-\lambda t}}}$$

confused with Boundary conditions and initial condition, could you guys help please

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  • $\begingroup$ Have you plugged in your solution into the equation and conditions to check it? I guess not, because as it is now you have: 1) $C(0,t)=0 \ne C_0$ 2) $C(\infty,t) \sim \sin{x} \ne 0$ $\endgroup$
    – Rolazaro Azeveires
    May 29, 2021 at 23:22
  • $\begingroup$ @RolazaroAzeveires hey appreciate the response. I did plug in the condition values after evaluating $C(x,t)$. I will do it again $\endgroup$
    – Gaurang
    May 30, 2021 at 0:03
  • $\begingroup$ @RolazaroAzeveires how is $C(\infty,t)$~$\sin x$? And I think first I should follow the boundary conditions and then initial. I might be wrong on that, do correct me $\endgroup$
    – Gaurang
    May 30, 2021 at 0:10
  • $\begingroup$ @RolazaroAzeveires also $C = C_0$ at $x=0$ That’s a boundary condition $\endgroup$
    – Gaurang
    May 30, 2021 at 0:19
  • $\begingroup$ Hi. You asked to check your solution, I pointed that your solution does not fit your problem in, at least, two cases: your solution does not obey the boundary conditions. Neither at $x=0$ nor $x\rightarrow\infty$ Your method looks fine but something is wrong. I would redo it, knowing that I was losing the boundary conditions on the way. $\endgroup$
    – Rolazaro Azeveires
    May 30, 2021 at 11:03

1 Answer 1

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Using the Laplace transform

$$ sC(s,x) - a C_{xx}(s,x)+ k C(s,x) = c(0,x) $$

here $c(0,x) = 0$ so we follow with

$$ (s+k)C(s,x) - a C_{xx}(s,x)=0 $$

now solving for $x$ we have

$$ C(s,x) = \phi_1(s) e^{\sqrt{\frac{s+k}{a}}x}+\phi_2(s) e^{-\sqrt{\frac{s+k}{a}}x} $$

and due to the boundary conditions $\phi_1(s) = 0,\phi_2(s) = c_0$ so

$$ C(s,x) = c_0e^{-\sqrt{\frac{s+k}{a}}x} $$

and finally

$$ c(t,x) = \mathcal{L}^{-1}\left[C(s,x)\right]=\frac{c_0 x e^{-\left(\frac{x^2}{4 a t}+k t\right)}}{2 \sqrt{a\pi } t^{3/2}},\ \ \ \frac {x}{\sqrt{a}}\gt 0 $$

NOTE

The answer above considers the boundary conditions at $x = 0$ as initial conditions (Dirac impulses). To consider a step at $x = 0$ with amplitude $c_0$ then the boundary conditions should be $\phi_1(s) = \frac{c_0}{s}$ and then

$$ C(s,x) = \frac{c_0}{s}e^{-\sqrt{\frac{s+k}{a}}x} $$

with inverse Laplace transform

$$ c(t,x) = \frac{1}{2} c_0 e^{x \left(-\sqrt{\frac{k}{a}}\right)} \left(2-\text{erfc}\left(\frac{2 t \sqrt{a k}-x}{2 \sqrt{a t}}\right)+e^{2 x \sqrt{\frac{k}{a}}} \text{erfc}\left(\frac{2 t \sqrt{a k}+x}{2 \sqrt{a t}}\right)\right) $$

Follows a plot for $k=a=1, c_0=2,\ \ 0\le x\le 1, \ \ 0 \le t\le 2$.

enter image description here

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  • $\begingroup$ Hey thanks for the response. How did you use the boundary condition could you explain please, And verify wheather my last step $(4)$ is correct or not? $\endgroup$
    – Gaurang
    May 31, 2021 at 14:29
  • $\begingroup$ Boundary condition for time version* $\endgroup$
    – Gaurang
    May 31, 2021 at 14:35
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    $\begingroup$ The $c(t,x)$ can be obtained from a transform table. $\endgroup$
    – Cesareo
    May 31, 2021 at 15:40
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    $\begingroup$ Yes. This is the reason behind the choice for $\phi_1, \phi_2$ $\endgroup$
    – Cesareo
    Jun 1, 2021 at 6:43
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    $\begingroup$ We can use the initial value theorem from Laplace which states. $c(0,x) = \lim_{s\to \infty}s C(s,x)$ giving $0$. $\endgroup$
    – Cesareo
    Jun 1, 2021 at 14:59

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