Solve, using the Method of Characteristics, the equation $\frac{\partial \rho}{\partial t} + \frac{\partial \rho}{\partial x}=-\mu\rho$
for $x,t>0$ with the conditions $\rho(x,0)=f(x)$ and $\rho(0,t)=g(t)$.

I'm currently trying to solve above problem using method of characteristics. I used initial condition $\rho(x,0)=f(x)$ and obtained $|\rho(x,t)|=|f(x-t)|e^{- \mu t}$. But I need to find $\rho(x,t)$ and also I don't know how to use the boundary condition $\rho(0,t)=g(t)$. I'm posting this problem here to get a hint.Thanks in advance.


$$\frac{\partial \rho}{\partial t} + \frac{\partial \rho}{\partial x}=-\mu\rho$$ System of ODEs for the characteristics equations : $\quad \frac{dt}{1}=\frac{dx}{1}=\frac{d\rho}{-\mu\rho}$

First family of characteristic curves, from $\quad \frac{dt}{1}=\frac{dx}{1} \quad\to\quad t-x=c_1$

Second family of characteristic curves, from $\quad \frac{dt}{1}=\frac{d\rho}{-\mu\rho}\quad\to\quad \rho e^{\mu t}=c_2$

General solution on form of implicit equation : $\quad\Phi(t-x\:,\:\rho e^{\mu t})=0$

$\Phi$ is any differentiable function of two variables.

Or equivalently, on explicit form : $\quad\rho e^{\mu t} =F(t-x)$

$F(X)$ is any differentiable function where $X=t-x$.

$$\rho(x,t)=e^{-\mu t}F(t-x)$$

Case of condition : $\rho(0,t)=g(t)$ :

$\rho(0,t)=g(t)=e^{-\mu t}F(t-0)\quad$ which determines the function $F$ :

$F(X)=g(X)e^{\mu X}\quad$ that we put into the above general solution with $X=t-x$ $$\rho(x,t)=e^{-\mu t}g(t-x)e^{\mu(t-x)} = g(t-x)e^{-\mu x} $$

Case of condition : $\rho(x,0)=f(x)$ :

$\rho(x,0)=f(x)=e^{-\mu 0}F(-x)\quad$ which determines the function $F$ :

$F(X)=f(-x)\quad$ that we put into the above general solution with $X=t-x$ $$\rho(x,t)=e^{-\mu t}f(-(t-x))=f(x-t)e^{-\mu t} $$

If the two conditions are specified together and without information about the functions $f$ and $g$, it seems doubtful that further calculus be possible about the respective ranges of validity of the two above different functions $\rho(x,t)$ and the boundaries between them.

Comment :

If we don't know what are the functions $f$ and $g$ ( smooth or piecewise, related functions or not), in the general case $f(0)\neq g(0)$. Then there would be a contradiction between the conditions $\rho(0,0)=e^{-\mu 0}f(0-0)=f(0)$ and $\rho(0,0)=e^{-\mu 0}g(0-0)=g(0)$.

This implies that $\rho(x,t)$ would be a piecewise function like this, for example : $$\rho(x,t)=H(x-t)f(x-t)e^{-\mu t}+H(t-x)g(t-x)e^{-\mu x}$$ $H$ is the Heaviside step function.

In writing such a form of solution, we accept that the PDE be not valid on the line $x=t$ where the partial derivatives would be not continuous.

  • $\begingroup$ Thanks for your answer. It sounds good. $\endgroup$ – Mayuran Sriskandasingam Dec 7 '17 at 2:29
  • 1
    $\begingroup$ I added to my answer a comment about the case of piecewise solution. $\endgroup$ – JJacquelin Dec 7 '17 at 6:24
  • $\begingroup$ It makes more sense. Thanks again! $\endgroup$ – Mayuran Sriskandasingam Dec 7 '17 at 8:00

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.