i need help please help me nonlinear partial differential equations let  $$u(x,t)=t^{-l}w(t^{-l/n}x)$$ where $w(y)=u(y,1)$ $y\in R^n$ $k=t^{-l/n}$ $l=\frac{n}{2+(m-1)n}$
prove that $u$ is a solution of $$\partial_t u-\Delta u^m=0$$ if and only if $w$ satisfies $$\Delta w^m(y)+\frac{l}{n}<y,\nabla w(y)>+lw(y)=0$$ where $y\in R^n$
 A: We know that $u(x,t)=t^{-l}w(t^{-l/n}x)$,
that $\partial_t u-\Delta u^m=0\tag 1$
and that $l=\dfrac{n}{2+(m-1)n}$ or $-lm-2l/n=-l-1\tag 2$
Computing the partial wrt $t$
$\partial_tu=-lt^{-l-1}w(y)-t^{-l}(l/n)\langle t^{-l/n-1}x,\nabla w(y)\rangle=$
$=-t^{-l-1}\left(lw(y)+(l/n)\langle t^{-l/n}x,\nabla w(y)\rangle\right)=$
$=-t^{-l-1}\left(lw(y)+(l/n)\langle y,\nabla w(y)\rangle\right)$
We are working in fact with the laplacian applied to two different functions, $u(x,t)$ and $w(y)$. For the first,
$$\Delta u(x,t)=\sum_1^n\partial_{x_ix_i}u$$
and for the second,
$$\Delta w(y)=\sum_1^n\partial_{y_iy_i}u$$
We have too, $\partial_{x_i} y_i=t^{-l/n}\;;i=1,\cdots,n$ With these considerations, we can calculate the laplacian for $u^m$
$$\partial_{x_ix_i} u^m=\partial_{x_i}(\partial_{x_i}t^{-lm}w^m)=$$
$$=t^{-lm}\partial_{x_i}(mw^{m-1}\partial_{y_i}w\partial_{x_i}y_i)=t^{-lm}t^{-l/n}\partial_{x_i}(mw^{m-1}\partial_{y_i}w)=$$
$$=t^{-lm-l/n}(mw^{m-1}\partial_{y_iy_i}w\partial_{x_i}y_i+m(m-1)w^{m-2}(\partial_{y_iy_i}w)^2\partial_{x_i}y_i)=$$
$$=t^{-lm-l/n}(t^{-l/n}mw^{m-1}\partial_{y_iy_i}w+t^{-l/n}m(m-1)w^{m-2}(\partial_{y_iy_i}w)^2)=$$
$$=t^{-lm-2l/n}(mw^{m-1}\partial_{y_iy_i}w+m(m-1)w^{m-2}(\partial_{y_iy_i}w)^2)=$$
$$\Delta u^m(x,t)=t^{-lm-2l/n}\sum_1^n(mw^{m-1}\partial_{y_iy_i}w+m(m-1)w^{m-2}(\partial_{y_iy_i}w)^2)$$
Now the laplacian for $w^n$
$$\partial_{y_iy_i}w^n=\partial_{y_i}(\partial_{y_i}w^n)=\partial_{y_i}(nw^{n-1}\partial_{y_i}w)=$$
$$=nw^{n-1}\partial_{y_iy_i}w+n(n-1)w^{n-2}(\partial_{y_i}w)^2$$
$$\Delta w^n=\sum_1^n(nw^{n-1}\partial_{y_iy_i}w+n(n-1)w^{n-2}(\partial_{y_i}w)^2)$$
The relation between the laplacians is, taking into account $(2)$,
$$\Delta u^m=t^{-tm-2t/n}\Delta w^m=t^{-l-1}\Delta w^m$$
Putting all in $(1)$ we get the identity,
$$\partial_tu-\Delta u^m=-t^{-l-1}\left(lw(y)+(l/n)\langle y,\nabla w(y)\rangle\right)-t^{-l-1}\Delta w^m=0$$
Or,
$$lw(y)+(l/n)\langle y,\nabla w(y)\rangle+\Delta w^m=0$$
I couldn't prove the assertion because the exponent is $m$ and not $n$. Nevertheles, I think it's false and the assertion holds with $m$ instead of $n$ for the exponent.
