Verify this function is a solution of the heat equation

[docnet]

Homework Statement

verify this function is a solution of the heat equation

Relevant Equations

$\partial_t u -\Delta u=0$

I spent hours looking at this and cannot figure out where the error is. I'm wondering if there is an error before the boxed expression.

@Orodruin and @PeroK may I ask for your assistance?Consider a solution $u:[0,\infty)\times \mathbb{R}^n\rightarrow \mathbb{R}$ of the heat equation, ie. $\partial_t u-\Delta u=0$. For $\kappa>0$ and $V\in \mathbb{R}^n$ we define the Galilean boosts of $u(t,x)$ by $$u_\kappa=exp(-<x-t\kappa V,\kappa V>)\cdot u(t,x-2t\kappa V)$$
$$\Rightarrow exp(t\kappa^2 |V|^2-\kappa \sum x_iV_i )\cdot u(t,x-2t\kappa V)$$
To show $u_\kappa$ is again a solution of the heat equation, we operate on $u_\kappa$ with $\partial_t$ using the product and chain rules
$$\partial_t u_\kappa=\partial_t\big[ exp(t\kappa^2 |V|^2-\kappa \sum x_iV_i )\big]\cdot u(t,x-2t\kappa V)$$
$$+exp(t\kappa^2 |V|^2-\kappa \sum x_iV_i )\cdot \partial_t\big[u(t,x-2t\kappa V)\big]$$
$$=\kappa^2 |V|^2exp(-<x-t\kappa V,\kappa V>)\cdot u(t,x-2t\kappa V)$$$$+exp(-<x-t\kappa V,\kappa V>)\cdot \partial_tu\big[(t,x-2t\kappa V)\big]$$
and operate on $u_\kappa$ with $\Delta$ using the product and chain rules
$$\Delta u_\kappa=\Delta\big[ exp(t\kappa^2 |V|^2-\kappa \sum x_iV_i )\big]\cdot u(t,x-2t\kappa V)$$$$+exp(t\kappa^2 |V|^2-\kappa \sum x_iV_i )\cdot \Delta \big[u(t,x-2t\kappa V)\big]$$
$$= \kappa^2 |V|^2exp(-<x-t\kappa V,\kappa V>)\cdot u(t,x-2t\kappa V)$$$$+exp(-<x-t\kappa V,\kappa V>)\cdot \Delta\big[ u(t,x-2t\kappa V)\big]$$
and compute
$$\partial_t u_\kappa - \Delta u_\kappa$$
$$\Rightarrow \kappa^2 |V|^2exp(-<x-t\kappa V,\kappa V>)\cdot u(t,x-2t\kappa V)$$$$-\kappa^2 |V|^2exp(-<x-t\kappa V,\kappa V>)\cdot u(t,x-2t\kappa V)$$$$+exp(-<x-t\kappa V,\kappa V>)\cdot \partial_t\big[ u(t,x-2t\kappa V)\big]$$$$-exp(-<x-t\kappa V,\kappa V>)\cdot \Delta \big[u(t,x-2t\kappa V)\big]$$
The first two terms cancel and we have the resulting expression $$\Rightarrow exp(-<x-t\kappa V,\kappa V>)\Big[\partial_t \big[u (t,x-2t\kappa V)\big]-\Delta\big[ u(t,x-2t\kappa V)\big]\Big]$$
We conclude $u_\kappa$ satisfies the heat equation if the terms inside the parenthesis vanish to zero.
$$\partial_t[u(t,x-2t\kappa V)]=\partial_tu(t,x-2t\kappa V)\partial_tt+\partial_xu(t,x-2t\kappa V)\partial_t(x-2t\kappa V)$$$$=\partial_tu(t,x-2t\kappa V)\boxed{-2\kappa V\partial_xu(t,x-2t\kappa V)}$$
$$\Rightarrow \text{ the}\quad\partial_x\quad\text{term creates problems}$$
The $\Delta$ term
$$\partial_{x_i}[u(t,x-2t\kappa V)]=\partial_{x_i}u(t,x-2t\kappa V)\partial_{x_i}(x-2t\kappa V)=\partial_{x_i}u(t,x-2t\kappa V)$$
$$\partial_{x_i}[\partial_{x_i}u(t,x-2t\kappa V)]=\partial_{x_i}^2u(t,x-2t\kappa V)\partial_{x_i}(x-2t\kappa V)=\partial_{x_i}^2u(t,x-2t\kappa V)$$
sum over i
$$\Rightarrow \Delta u(t,x-2t\kappa V)$$
if the boxed term vanishes, we could conclude that by the heat equation
$$\partial_t u_\kappa -\Delta u_\kappa = 0$$
the Galilean boosts of $u_\kappa$ is also a solution of the heat equation.

 

[PeroK]

and operate on $u_\kappa$ with $\Delta$ using the product and chain rules
$$\Delta u_\kappa=\Delta\big[ exp(t\kappa^2 |V|^2-\kappa \sum x_iV_i )\big]\cdot u(t,x-2t\kappa V)$$$$+exp(t\kappa^2 |V|^2-\kappa \sum x_iV_i )\cdot \Delta \big[u(t,x-2t\kappa V)\big]$$
$$= \kappa^2 |V|^2exp(-<x-t\kappa V,\kappa V>)\cdot u(t,x-2t\kappa V)$$$$+exp(-<x-t\kappa V,\kappa V>)\cdot \Delta\big[ u(t,x-2t\kappa V)\big]$$

That operator has second derivatives, so you should have cross terms as well.

 

[docnet]

thanks @PeroK! for helping me see my mistake. I'm posting a new attempt for completeness.

Consider a solution $u:[0,\infty)\times \mathbb{R}^n\rightarrow \mathbb{R}$ of the heat equation, ie. $\partial_t u-\Delta u=0$. For $\kappa>0$ and $V\in \mathbb{R}^n$ we define the Galilean boosts of $u(t,x)$ by $$u_\kappa=exp(-<x-t\kappa V,\kappa V>)\cdot u(t,x-2t\kappa V)$$$$= exp(t\kappa^2 |V|^2-\kappa \sum x_iV_i )\cdot u(t,x-2t\kappa V)$$
Define the previous expression by the following shorthand notation
$$\Rightarrow AB$$
We compute the partial derivatives of $A$ and $B$
$$\partial_t A=\kappa^2|V|^2A$$
$$\partial_{x_i}A=-\kappa|V|A$$
$$\partial_{x_i}^2A=\kappa^2|V|^2A$$
$$\partial_tB=\partial_tB-2\kappa V\partial_xB$$
$$\partial_{x_i}B=\partial_{x_i}B$$
$$\partial_{x_i}\partial_{x_i}B=\partial_{x_i}^2B$$
To show $u_\kappa=AB$ is again a solution of the heat equation, we operate on $AB$ with $\partial_t$.
$$\partial_t[AB]= \partial_tA\cdot B+A\partial_tB=\kappa^2|V|^2AB+A(\partial_tB-2\kappa V\partial_xB)$$
and operate on $AB$ with $\partial_{x_i}$
$$ \partial_{x_i}[AB]=\partial_{x_i}A\cdot B+A\partial_{x_i}B=-\kappa|V|AB+A\partial_{x_i}B$$
and operate on $\partial_{x_i}[AB]$ with $\partial_{x_i}$ $$\partial_{x_i}^2[AB]=-\kappa|V|(\partial_{x_i}A\cdot B+A\partial_{x_i}B)+\partial_{x_i}A\partial_{x_i}B+A\partial_{x_i}^2B$$
$$=-\kappa|V|(-\kappa|V|AB+A\partial_{x_i}B)-\kappa|V|A\partial_{x_i}B+A\partial^2_{x_i}B$$
$$=\kappa^2|V|^2AB-2\kappa|V|A\partial_{x_i}B+A\partial^2_{x_i}B$$
sum over i
$$=\kappa^2|V|^2AB-2\kappa|V|A\cdot \partial_xB+A\Delta B$$
We compute
$$\partial_t[AB]-\Delta[AB]=A(\partial_tB-\Delta B)$$
by the PDE, the terms inside the parenthesis vanish to zero.
$$A\big(\partial_tu(t,x-t\kappa V)-\Delta u(t,x-t\kappa V)\big)=A\cdot 0$$
We conclude that $AB=u_\kappa$ is indeed a solution of the heat equation.