Solution of the non-linear Heat Equation

Question

How to find $v$ such that $$u(x,t)=t^{-\alpha}v(xt^{-\beta})$$ is the solution of the non-linear Heat equation : $$u_t-\Delta(u^{\gamma})=0$$where $\frac{n-2}{n}<\gamma<1$ , $x$ $\in R^n$ and $t$ $>$ $0$.

Thank you very much for your consideration.

Answer 1

We want to determine $\alpha$, $\beta$ and $v$ such that $$ u(t,x_1,x_2,\cdots) = t^{-\alpha} v(x_1 t^{-\beta},x_2 t^{-\beta},\cdots) $$ is solution of the heat equation, with $v:R^n \to R$. It is convenient to define the similarity variables $\eta_i = x_i t^{-\beta}$. Therefore, $v=v(\eta_1,\eta_2,\cdots)$. Derivating $u$ in relation to $t$, we have $$ u_t = \frac{\partial}{\partial t} t^{-\alpha} v = -\alpha t^{-\alpha-1}v-t^{-\alpha} \sum_{i=1}^n \frac{\partial \eta_i}{\partial t}\frac{\partial v}{\partial \eta_i} $$ $$ u_t = -\alpha t^{-\alpha-1}v-\beta t^{-\alpha-\beta-1} \sum_{i=1}^n x_i \frac{\partial v}{\partial \eta_i} $$ which follows from the product and chain rules (the last of which will be used extensively below). The Laplacian of $u^\gamma$ is $$ \Delta u^\gamma = t^{-\gamma \alpha} \sum_{i=1}^n \frac{\partial^2 v^\gamma}{\partial x_i^2} = t^{-\gamma\alpha} \sum_{i=1}^n \frac{\partial}{\partial x_i} \frac{\partial v^\gamma}{\partial x_i}. $$ We have $$ \frac{\partial}{\partial x_i} v^\gamma = \frac{\partial \eta_i}{\partial x_i} \frac{\partial}{\partial \eta_i} v^\gamma = \gamma t^{-\beta} v^{\gamma-1}\frac{\partial v}{\partial \eta_i} $$ and $$ \frac{\partial^2}{\partial x_i^2} v^\gamma = \frac{\partial}{\partial x_i} \gamma t^{-\beta} v^{\gamma-1}\frac{\partial v}{\partial \eta_i} = \frac{\partial \eta_i}{\partial x_i} \frac{\partial}{\partial \eta_i} \gamma t^{-\beta} v^{\gamma-1}\frac{\partial v}{\partial \eta_i} $$ $$ \frac{\partial^2}{\partial x_i^2} v^\gamma = \gamma t^{-2\beta} \frac{\partial v}{\partial \eta_i} \frac{\partial}{\partial \eta_i} v^{\gamma-1} + v^{\gamma-1} \frac{\partial}{\partial \eta_i} \frac{\partial v}{\partial \eta_i} = \gamma t^{-2\beta} \left[ (\gamma-1)v^{\gamma-2} \left(\frac{\partial v}{\partial \eta_i} \right)^2+ v^{\gamma-1} \frac{\partial^2 v}{\partial \eta_i^2} \right]. $$ Therefore, $$ \Delta u = \gamma t^{-\gamma \alpha -2\beta} \sum_{i=1}^n \left[ (\gamma-1)v^{\gamma-2} \left(\frac{\partial v}{\partial \eta_i} \right)^2+ v^{\gamma-1} \frac{\partial^2 v}{\partial \eta_i^2} \right]. $$ Substituting in the equation, we have $$ -\alpha t^{-\alpha-1}v-\beta t^{-\alpha-\beta-1} \sum_{i=1}^n x_i \frac{\partial v}{\partial \eta_i} = \gamma t^{-\gamma \alpha -2\beta} \sum_{i=1}^n \left[ (\gamma-1)v^{\gamma-2} \left(\frac{\partial v}{\partial \eta_i} \right)^2+ v^{\gamma-1} \frac{\partial^2 v}{\partial \eta_i^2} \right], $$ simplifying, $$ -\alpha v-\beta \sum_{i=1}^n \eta_i \frac{\partial v}{\partial \eta_i} = \gamma t^{1+\alpha(1-\gamma) -2\beta} \sum_{i=1}^n \left[ (\gamma-1)v^{\gamma-2} \left(\frac{\partial v}{\partial \eta_i} \right)^2+ v^{\gamma-1} \frac{\partial^2 v}{\partial \eta_i^2} \right] $$ (see that a $\eta$ appeared in the second term of LHS). We can make sure that the initial choice of $u$ works if we can get rid of the $t$ in the RHS. It happens if we choose $\alpha$ and $\beta$ such that $$ 2\beta-\alpha(1-\gamma)=1. $$ There are infinitely many pairs of $\alpha$ and $\beta$ satisfying this relation; one convenient choice is $\alpha=0$ and $\beta=1/2$. This choice is convenient because 1) the expression for $u$ is more simple; 2) it echoes the fact that for the classic (linear) heat equation, we also have $\alpha=0$ and $\beta=1/2$ for certain boundary conditions; and 3) the first term on the LHS of the equation vanishes, which allows a simpler form of $v$.

Assuming those values for $\alpha$ and $\beta$, our equation reduces to $$ \sum_{i=1}^n \left[ \frac{1}{2} \eta_i \frac{\partial v}{\partial \eta_i} + \gamma(\gamma-1)v^{\gamma-2} \left(\frac{\partial v}{\partial \eta_i} \right)^2+ \gamma v^{\gamma-1} \frac{\partial^2 v}{\partial \eta_i^2} \right] = 0. $$ If we let $v=v(\eta_1+\eta_2+\cdots)$ and define $H=\sum_i \eta_i$ such that $v=v(H)$, we have $$ \frac{\partial v}{\partial \eta_i} = \frac{\partial v}{\partial \eta_j} = \frac{d v}{d H} = v' $$ for all $i,j$, and now $v$ must satisfy $$ \frac{1}{2} H v' + \gamma(\gamma-1)v^{\gamma-2} {v'}^2 + \gamma v^{\gamma-1} v''= 0. $$

We can check that we did all right if we apply the equation for the classical linear one-dimensional case (i.e., $\gamma=1$). In that case, the equation reduces to $$ v'' + \frac{H}{2} v'=0, $$ whose solution is $v=\mathrm{erf}(H/2)$ (see that we can set any value for the integration constants). Since the problem is one-dimensional, $H=\eta_1=x/t^{-1/2}$, and the solution is $$ u = \mathrm{erf} \left( \frac{x}{2\sqrt{t}} \right), $$ which is the solution to the classical problem of heat conduction in an infinite bar.

Summary: choosing $\alpha=0$, $\beta=1/2$ and imposing $v:R^n\to R$ to have the form $v(y_1,y_2,\cdots) = v(y_1+y_2+\cdots)$, such that $u$ has the form $$ u = v\left( \frac{x_1}{\sqrt{t}} + \frac{x_2}{\sqrt{t}} + \cdots \right), $$ $v$ must satisfy $$ \frac{1}{2} H v' + \gamma(\gamma-1)v^{\gamma-2} {v'}^2 + \gamma v^{\gamma-1} v''= 0, $$ in which the prime means derivation in relation to $H$.