I am working on this question:

Show that advection-diffusion equation $$ u_t=Du_{xx}+Au_x+Bu,\quad x\in\Bbb R, t>0 $$ where $A, B, D > 0$ are constants, can be transformed into heat equation for a function $v$ by choosing $$ u(x,t) = e^{\alpha x−\beta t}v(x,t). $$ Find $\alpha$ and $\beta$ to achieve this transformation.

I am unsure how to proceed: I have seen both the diffusion and heat equations, and learnt the methods of fundamental and similarity solution, but I'm confused by the different format of the equation to the standard diffusion equation: how does the $au_x+bu$ factor into the calculation?


  • 1
    $\begingroup$ Start by computing the derivatives in the advection-diffusion equation using the function u(x,t) you gave, leaving the unknown function v(x,t) in that general form. So you're going to need to use the product rule, as you have a product of functions. Then see if you can figure it out. $\endgroup$ – Anthony P Mar 10 at 18:40


$u(x,t) = e^{\alpha x−\beta t}v(x,t), \tag 1$


$u_t(x, t) = -\beta e^{\alpha x−\beta t}v(x,t) + e^{\alpha x−\beta t}v_t(x,t), \tag 2$

$u_x(x, t) = \alpha e^{\alpha x−\beta t}v(x,t) + e^{\alpha x−\beta t}v_x(x,t), \tag 3$

$u_{xx}(x, t) = \alpha^2 e^{\alpha x−\beta t}v(x,t) + \alpha e^{\alpha x−\beta t}v_x(x,t) + \alpha e^{\alpha x−\beta t}v_x(x,t) + e^{\alpha x−\beta t}v_{xx}(x,t)$ $= \alpha^2 e^{\alpha x−\beta t}v(x,t) + 2\alpha e^{\alpha x−\beta t}v_x(x,t) + e^{\alpha x−\beta t}v_{xx}(x,t); \tag 4$

we are given that $u(x, t)$ satisfies

$u_t=Du_{xx}+Au_x+Bu,\quad x\in\Bbb R, t>0; \tag 5$

we substitute in the formulas (1)-(4):

$-\beta e^{\alpha x−\beta t}v + e^{\alpha x−\beta t}v_t = D( \alpha^2 e^{\alpha x−\beta t}v + 2\alpha e^{\alpha x−\beta t}v_x + e^{\alpha x−\beta t}v_{xx})$ $+ A(\alpha e^{\alpha x−\beta t}v + e^{\alpha x−\beta t}v_x) + Be^{\alpha x−\beta t}v, \tag 6$

and cancel out

$e^{\alpha x−\beta t} \ne 0: \tag 7$

$-\beta v + v_t = D( \alpha^2 v + 2\alpha v_x + v_{xx}) + A(\alpha v + v_x) + Bv, \tag 8$

and gather $v$ and its derivatives into like terms:

$v_t = Dv_{xx} + (2D \alpha + A) v_x + (D\alpha^2 + A\alpha + (B + \beta))v, \tag 9$

which is reducible to a heat equation of the form

$v_t = D v_{xx} \tag{10}$

provided we may take $\alpha$ and $\beta$ satisfying

$2D\alpha + A = 0, \tag{11}$

$D\alpha^2 + A\alpha + B + \beta = 0; \tag{12}$

these two equations are easy to solve; (11) yields

$\alpha = -\dfrac{A}{2D}, \tag{13}$

and when this is substituted into (12):

$D \left ( -\dfrac{A}{2D} \right)^2 + A \left ( -\dfrac{A}{2D} \right ) + B + \beta = 0, \tag{14}$


$\dfrac{A^2}{4D} - \dfrac{A^2}{2D} + B + \beta = 0, \tag{15}$


$\beta = \dfrac{A^2}{4D} - B; \tag{16}$

these values of $\alpha$ (13) and $\beta$ (16) result in the heat equation (10) when inserted into (9).



Substituting $u = e^{\alpha x-\beta t}v$ into

$$ \left(\partial_t-D\partial_x^2-A\partial_x-B\right)u = 0 $$

we get

$$ \left(v_t-(A+2\alpha D)v_x-(B+\beta+\alpha(A+\alpha D))v-Dv_{xx}\right)e^{\alpha x - \beta t}=0 $$

now choosing $\alpha, \beta$

$$ A+2\alpha D = 0\\ B+\beta+\alpha(A+\alpha D) = 0 $$

and with $\alpha = -\frac{A}{2D}, \beta = \frac{A^2-4BD}{4D}$ we obtain

$$ v_t - D v_{xx}= 0 $$


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