derive general solution to first-order inhomogeneous pde

Question

Given $u_t+cu_x=b(x,t)$ with initial value $u(x,0)=\phi(x)$, I am asked to derive the following general solution $$u=\phi(x-ct)+\int_{0}^t b(x-c(t-\tau),\tau)d\tau$$

I am not sure how to proceed with this $b(x,t)$. I have solved such PDE where the explicit form of $b(x,t)$ given, but not in the general case. I tried the method of characteristics, $$\frac{dt}{1}=\frac{dx}{c}=\frac{du}{b(x,t)}$$ but got stuck. Hope someone could help! Thanks.

Answer 1

$$u_t+cu_x=b(x,t)$$ You have correctly written the Charpit-Lagrange system of ODEs : $$\frac{dt}{1}=\frac{dx}{c}=\frac{du}{b(x,t)}$$ A first characteristic equation comes from solving $\frac{dt}{1}=\frac{dx}{c}$ : $$x-ct=C_1\quad\implies\quad x=C_1+ct$$ A second characteristic equation comes from solving $dt=\frac{du}{b(x,t)}=\frac{du}{b\left(C_1+ct\:,\:t\right)}$ $$du-b\left(C_1+ct\:,\:t\right)dt=0$$ $$u-\int_0^t b\left(C_1+c\tau\:,\:\tau\right)d\tau=C_2$$

The general solution of the PDE $\quad C_2=\phi(C_1)\quad$ is : $$u-\int_0^t b\left((x-ct)+c\tau\:,\:\tau\right)d\tau=\phi(x-ct)$$

$\phi$ is an arbitrary function (to be determined according to some initial condition). $$u=\phi(x-ct)+\int_{0}^t b(x-c(t-\tau),\tau)d\tau$$

Answer 2

The general solution of equation $u_t+cu_x=b(x,t)$ is sum of general solution of homogenous equation $u_t+cu_x=0$ and the particular solution of inhomogenous equation $u_t+cu_x=b(x,t)$.

1. General solution of homogenous equation $u_t+cu_x=0$ can be obtained using method of characteristics. We compose the equation of characteristics $\frac{dt}{1}=\frac{dx}{c}$, and obtain its the first integral $x - c t$. So, general solution of homogenous equation is $u(x, t) = \psi(x - c t)$, where $\psi$ is a continuous-differentiable function.

2. The particular solution of equation $u_t+cu_x=b(x,t)$. We build it using extra condition $u(0, t) = 0$. It can be obtained using Fourier transform: $u(x, t) \overset{\mathcal{F}}{\rightarrow} \hat{u}(\omega,t)$ $$ \hat{u}(\omega,t) = \frac{1}{\sqrt{2 \pi}}\int\limits_{-\infty}^{+\infty} \exp(-i\omega x) u(x, t) dx, $$ $$ u(x,t) = \frac{1}{\sqrt{2 \pi}}\int\limits_{-\infty}^{+\infty} \exp(i\omega x)\hat{u}(\omega, t) d\omega. $$ For function $\hat{u}$ we get the Cauchy problem $\hat{u}_t+i \omega c \hat{u}=\hat{b}(\omega,t)$, $\hat{u}|_{t=0} = 0$. After solving it, we obtain formula $$ \hat{u}(\omega,t) = \int\limits_0^t \exp(i \omega c (\tau - t)) \hat{b}(\omega,\tau) d\tau. $$ So, $$ u(x,t) = \frac{1}{\sqrt{2 \pi}} \int\limits_{-\infty}^{+\infty} \left(\int\limits_0^t \exp(i \omega c (\tau - t)) \hat{b}(\omega,\tau) d\tau \right) \exp(i\omega x) d\omega, $$ or $$ u(x,t) = \int\limits_0^t \frac{d\tau}{\sqrt{2 \pi}} \int\limits_{-\infty}^{+\infty} \exp(i \omega c (\tau - t)) \hat{b}(\omega,\tau) \exp(i\omega x) d\omega. $$ After calculating it using $\exp(i \omega c (\tau - t)) \overset{\mathcal{F}^{-1}}{\rightarrow} \sqrt{2 \pi} \delta (x - c (t -\tau))$ and convolution theorem, we obtain

$$u(x, t)=\int_{0}^t b(x-c(t-\tau),\tau)d\tau.$$

3. So, the general solution of equation $u_t+cu_x=b(x,t)$ is $$ u(x, t)=\psi(x - c t) + \int_{0}^t b(x-c(t-\tau),\tau)d\tau, $$ where $\psi$ is a continuous-differentiable function. Substituting it into Cauchy condition $u(x, 0)=\phi(x)$ leads to $u(x, 0)=\psi(x)=\phi(x)$.

4. Thus, $$ u(x, t)=\phi(x - c t) + \int_{0}^t b(x-c(t-\tau),\tau)d\tau, $$ is the solution of your Cauchy problem.