Duhamel's Method

Question

Using Duhamel's Method to solve the problem

$$ \begin{align}\begin{cases} c_{t} +v c_{x} = f(x,t)& x \in \mathbb{R} , t> 0 \\ c(x,0) = 0 \end{cases} \end{align} \tag{1}$$

Find an explicit formula when $f(x,t) = e^{-t}\sin(x)$

Attempt

The solution for the general problem

$$ \begin{align}\begin{cases} c_{t} +v c_{x} = f(x,t)& x \in \mathbb{R} , t> 0 \\ c(x,0) = g(x) \end{cases} \end{align} \tag{2}$$

is given by

$$ c(x,t) = g(x-vt) + \int_{0}^{t} f(x-v(t-s),s) ds \tag{3}$$

then we should have something like this

$$ f(x-v(t-s),s) = e^{-s}\sin(x-v(t-s)) \tag{4}$$

$$ c(x,t) = \int_{0}^{t} e^{-s}\sin(x-v(t-s)) ds \tag{5}$$

Is this right?

Answer 1

Your result is exact. The method of characteristics leads to the same result :

$$\frac{dt}{1}=\frac{dx}{v}=\frac{dc}{e^{-t}\sin(x)}$$ A first family of characteristic curves comes from $\frac{dt}{1}=\frac{dx}{v}$ $$x-vt=c_1$$ A second family of characteristic curves comes from $\frac{dt}{1}=\frac{dc}{e^{-t}\sin(x)}=\frac{dc}{e^{-t}\sin(c_1+vt)}$ $$c(x,t)=\int e^{-t}\sin(c_1+vt)dt+c_2$$ The condition $c(x,0)=0$ determines $c_2$ so that : $c(x,t)=\int_{s=0}^{s=t} e^{-s}\sin(c_1+vs)ds$ $$c(x,t)=\int_{0}^{t} e^{-s}\sin(x-vt+vs)ds$$