PDE9

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Transform this equation: $u_{t}=\nu u_{xx} + \lambda u_x + \mu u\,$ into the standard heat equation: $v_t=v_{xx}\,$

Assume a solution in the form $u(x,t) = e^{\alpha x + \beta t}v(x,t)\,$

Find its derivatives and plug it into the orginal DE:

$u_t = \beta e^{\alpha x + \beta t} v + e^{\alpha x + \beta t} v_t\,$

$u_x = \alpha e^{\alpha x + \beta t} v + e^{\alpha x + \beta t} v_x\,$

$u_{xx} = \alpha^2 e^{\alpha x + \beta t} v + \alpha e^{\alpha x + \beta t} v_x + \alpha e^{\alpha x + \beta t} v_x + e^{\alpha x + \beta t} v_{xx}\,$

$\beta v + v_t = \nu(\alpha^2 v + 2 \alpha v_x + v_{xx} ) + \lambda( \alpha v + v_x ) + \mu v\,$

Collect derivatives of $v\,$ and find the appropriate values of $\alpha\,$ and $\beta\,$ to make the equation look like the heat equation.

$v(\beta - \nu \alpha^2 - \lambda\alpha - \mu) + v_t - v_x(2\alpha\nu+\lambda) - v_{xx}(\nu)=0\,$

$\beta - \nu \alpha^2 - \lambda\alpha-\mu=0\,$

$2\alpha\nu+\lambda = 0\,$

$\lambda = -2 \alpha \nu\,$

$\alpha = \frac{-\lambda}{2\nu}\,$

$\beta-\nu\left(\frac{-\lambda}{2\nu}\right)^2 - \lambda\left(\frac{-\lambda}{2\nu}\right) - \mu = 0\,$

$\beta - \frac{-\lambda^2}{4\nu} + \frac{2\lambda^2}{4\nu} - \mu = 0\,$

$\beta = \mu - \frac{\lambda^2}{4\nu}\,$

So the solution is

$u(x,t) = e^{\frac{-\lambda}{2\nu}x + \left(\mu-\frac{\lambda^2}{4\nu}\right)t}v(x,t)\,$

Taking derivatives and plugging back in gives this equation:

$v_t = \nu v_{xx}\,$

Partial Differential Equations

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