Usbrth

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

Thank you very much for your consideration.

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
$$
u_t = fracpartialpartial t t^-alpha v = -alpha t^-alpha-1v-t^-alpha sum_i=1^n fracpartial eta_ipartial tfracpartial vpartial eta_i
$$
$$
u_t = -alpha t^-alpha-1v-beta t^-alpha-beta-1 sum_i=1^n x_i fracpartial vpartial eta_i
$$
which follows from the product and chain rules (the last of which will be used extensively below). The Laplacian of $u^gamma$ is
$$
Delta u^gamma = t^-gamma alpha sum_i=1^n fracpartial^2 v^gammapartial x_i^2 = t^-gammaalpha sum_i=1^n fracpartialpartial x_i fracpartial v^gammapartial x_i.
$$
We have
$$
fracpartialpartial x_i v^gamma = fracpartial eta_ipartial x_i fracpartialpartial eta_i v^gamma = gamma t^-beta v^gamma-1fracpartial vpartial eta_i
$$
and
$$
fracpartial^2partial x_i^2 v^gamma = fracpartialpartial x_i gamma t^-beta v^gamma-1fracpartial vpartial eta_i = fracpartial eta_ipartial x_i fracpartialpartial eta_i gamma t^-beta v^gamma-1fracpartial vpartial eta_i
$$
$$
fracpartial^2partial x_i^2 v^gamma = gamma t^-2beta fracpartial vpartial eta_i fracpartialpartial eta_i v^gamma-1 + v^gamma-1 fracpartialpartial eta_i fracpartial vpartial eta_i =
gamma t^-2beta left[ (gamma-1)v^gamma-2 left(fracpartial vpartial eta_i right)^2+ v^gamma-1 fracpartial^2 vpartial eta_i^2 right].
$$
Therefore,
$$
Delta u = gamma t^-gamma alpha -2beta sum_i=1^n left[ (gamma-1)v^gamma-2 left(fracpartial vpartial eta_i right)^2+ v^gamma-1 fracpartial^2 vpartial eta_i^2 right].
$$
Substituting in the equation, we have
$$
-alpha t^-alpha-1v-beta t^-alpha-beta-1 sum_i=1^n x_i fracpartial vpartial eta_i = gamma t^-gamma alpha -2beta sum_i=1^n left[ (gamma-1)v^gamma-2 left(fracpartial vpartial eta_i right)^2+ v^gamma-1 fracpartial^2 vpartial eta_i^2 right],
$$
simplifying,
$$
-alpha v-beta sum_i=1^n eta_i fracpartial vpartial eta_i = gamma t^1+alpha(1-gamma) -2beta sum_i=1^n left[ (gamma-1)v^gamma-2 left(fracpartial vpartial eta_i right)^2+ v^gamma-1 fracpartial^2 vpartial eta_i^2 right]
$$
(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
$$
2beta-alpha(1-gamma)=1.
$$
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
$$
sum_i=1^n left[ frac12 eta_i fracpartial vpartial eta_i + gamma(gamma-1)v^gamma-2 left(fracpartial vpartial eta_i right)^2+ gamma v^gamma-1 fracpartial^2 vpartial eta_i^2 right] = 0.
$$
If we let $v=v(eta_1+eta_2+cdots)$ and define $H=sum_i eta_i$ such that $v=v(H)$, we have
$$
fracpartial vpartial eta_i = fracpartial vpartial eta_j = fracd vd H = v'
$$
for all $i,j$, and now $v$ must satisfy
$$
frac12 H v' + gamma(gamma-1)v^gamma-2 v'^2 + gamma v^gamma-1 v''= 0.
$$

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
$$
v'' + fracH2 v'=0,
$$
whose solution is $v=mathrmerf(H/2)$ (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
$$
u = mathrmerf left( fracx2sqrtt right),
$$
which is the solution to the classical problem of heat conduction in an infinite bar.

Summary: choosing $alpha=0$, $beta=1/2$ and imposing $v:R^nto R$ to have the form $v(y_1,y_2,cdots) = v(y_1+y_2+cdots)$, such that $u$ has the form
$$
u = vleft( fracx_1sqrtt + fracx_2sqrtt + cdots right),
$$
$v$ must satisfy
$$
frac12 H v' + gamma(gamma-1)v^gamma-2 v'^2 + gamma v^gamma-1 v''= 0,
$$
in which the prime means derivation in relation to $H$.