How to solve this 1-dimensional heat equation

Question

Find the explicit forumula for the solution of Cauchy problem on ${\mathbb{R}^ + } \times \mathbb{R}$

$\left\{ \begin{gathered} {u_t} - k{u_{xx}} + b{u_x} + cu = 0 \\ u\left( {0,x} \right) = g\left( x \right) \\ \end{gathered} \right.$

I started by substituting $v\left( {t,x} \right) = u\left( {t,x} \right){e^{ct}}$ which gives $\left\{ \begin{gathered} {v_t} - k{v_{xx}} + b{v_x} = \left( {{u_t} + cu - k{u_{xx}} + b{u_x}} \right){e^{ct}} = 0 \\ v\left( {0,x} \right) = g\left( x \right) \\ \end{gathered} \right.$

however, I don't know what to do next to get rid of the $b{v_x}$ term.

EDIT: $k,b,c$ are constants

Answer 1

Case $1$: $k=0$ and $b=0$

Then $u_t+cu=0$

$\dfrac{du}{dt}=-cu$

$\dfrac{du}{u}=-c~dt$

$\int\dfrac{du}{u}=\int-c~dt$

$\ln u(t,x)=-ct+f(x)$

$u(t,x)=F(x)e^{-ct}$

$u(0,x)=g(x)$ :

$F(x)=g(x)$

$\therefore u(t,x)=g(x)e^{-ct}$

Case $2$: $k=0$ and $b\neq0$

Then $u_t+bu_x+cu=0$

$u_t+bu_x=-cu$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$

$\dfrac{dx}{ds}=b$ , letting $x(0)=x_0$ , we have $x=bs+x_0=bt+x_0$

$\dfrac{du}{ds}=-cu$ , letting $u(0)=f(x_0)$ , we have $u(t,x)=f(x_0)e^{-cs}=f(x-bt)e^{-ct}$

$u(0,x)=g(x)$ :

$f(x)=g(x)$

$\therefore u(t,x)=g(x-bt)e^{-ct}$

Case $3$: $k\neq0$

Then $u_t-ku_{xx}+bu_x+cu=0$

Let $u(t,x)=T(t)X(x)$ ,

Then $T'(t)X(x)-kT(t)X''(x)+bT(t)X'(x)+cT(t)X(x)=0$

$T'(t)X(x)=(kX''(x)-bX'(x)-cX(x))T(t)$

$\dfrac{T'(t)}{T(t)}=\dfrac{kX''(x)-bX'(x)-cX(x)}{X(x)}=-\dfrac{4k^2s^2+b^2+4kc}{4k}$

$\begin{cases}\dfrac{T'(t)}{T(t)}=-\dfrac{4k^2s^2+b^2+4kc}{4k}\\kX''(x)-bX'(x)+\dfrac{4k^2s^2+b^2}{4k}X(x)=0\end{cases}$

$\begin{cases}T(t)=c_3(s)e^{-\frac{t(4k^2s^2+b^2+4kc)}{4k}}\\X(x)=\begin{cases}c_1(s)e^{\frac{bx}{2k}}\sin xs+c_2(s)e^{\frac{bx}{2k}}\cos xs&\text{when}~s\neq0\\c_1xe^{\frac{bx}{2k}}+c_2e^{\frac{bx}{2k}}&\text{when}~s=0\end{cases}\end{cases}$

$\therefore u(t,x)=\int_0^\infty C_1(s)e^{\frac{2bx-t(4k^2s^2+b^2+4kc)}{4k}}\sin xs~ds+\int_0^\infty C_2(s)e^{\frac{2bx-t(4k^2s^2+b^2+4kc)}{4k}}\cos xs~ds$

$u(0,x)=g(x)$ :

$\int_0^\infty C_1(s)e^{\frac{bx}{2k}}\sin xs~ds+\int_0^\infty C_2(s)e^{\frac{bx}{2k}}\cos xs~ds=g(x)$

$e^{\frac{bx}{2k}}\int_0^\infty C_1(s)\sin xs~ds=g(x)-e^{\frac{bx}{2k}}\int_0^\infty C_2(s)\cos xs~ds$ or $e^{\frac{bx}{2k}}\int_0^\infty C_2(s)\cos xs~ds=g(x)-e^{\frac{bx}{2k}}\int_0^\infty C_1(s)\sin xs~ds$

$\mathcal{F}_{s,s\to x}\{C_1(s)\}=g(x)e^{-\frac{bx}{2k}}-\mathcal{F}_{c,s\to x}\{C_2(s)\}$ or $\mathcal{F}_{c,s\to x}\{C_2(s)\}=g(x)e^{-\frac{bx}{2k}}-\mathcal{F}_{s,s\to x}\{C_1(s)\}$

$C_1(s)=\mathcal{F}^{-1}_{s,x\to s}\left\{g(x)e^{-\frac{bx}{2k}}\right\}-\mathcal{F}^{-1}_{s,x\to s}\{\mathcal{F}_{c,s\to x}\{C_2(s)\}\}$ or $C_2(s)=\mathcal{F}^{-1}_{c,x\to s}\left\{g(x)e^{-\frac{bx}{2k}}\right\}-\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_1(s)\}\}$

$\therefore u(t,x)=\int_0^\infty \mathcal{F}^{-1}_{s,x\to s}\left\{g(x)e^{-\frac{bx}{2k}}\right\}e^{\frac{2bx-t(4k^2s^2+b^2+4kc)}{4k}}\sin xs~ds-\int_0^\infty\mathcal{F}^{-1}_{s,x\to s}\{\mathcal{F}_{c,s\to x}\{C_2(s)\}\}e^{\frac{2bx-t(4k^2s^2+b^2+4kc)}{4k}}\sin xs~ds+\int_0^\infty C_2(s)e^{\frac{2bx-t(4k^2s^2+b^2+4kc)}{4k}}\cos xs~ds~\text{or}~\int_0^\infty C_1(s)e^{\frac{2bx-t(4k^2s^2+b^2+4kc)}{4k}}\sin xs~ds+\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\left\{g(x)e^{-\frac{bx}{2k}}\right\}e^{\frac{2bx-t(4k^2s^2+b^2+4kc)}{4k}}\cos xs~ds-\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_1(s)\}\}e^{\frac{2bx-t(4k^2s^2+b^2+4kc)}{4k}}\cos xs~ds$

Answer 2

By using the substitution $v\left( {t,x} \right) = u\left( {t,x} \right){e^{\left( {c + \frac{{{b^2}}} {{4k}}} \right)t - \frac{b} {{2k}}x}}$ , the problem becomes $\left\{ \begin{gathered} {v_t} - k{v_{xx}} = 0 \\ v\left( {0,x} \right) = g\left( x \right){e^{ - \frac{b} {{2k}}x}}\\ \end{gathered} \right.$

which has the solution $v\left( {t,x} \right) = \frac{1} {{\sqrt {4\pi kt} }}\int_{ - \infty }^{ + \infty } {{e^{ - \frac{{{{\left( {x - y} \right)}^2}}} {{4kt}}}}g\left( y \right){e^{ - \frac{b} {{2k}}y}}} dy$, giving $u\left( {t,x} \right) = {e^{ - \left( {c + \frac{{{b^2}}} {{4k}}} \right)t + \frac{b} {{2k}}x}}\frac{1} {{\sqrt {4\pi kt} }}\int_{ - \infty }^{ + \infty } {{e^{ - \frac{{{{\left( {x - y} \right)}^2}}} {{4kt}}}}g\left( y \right){e^{ - \frac{b} {{2k}}y}}} dy$