I have a question about the dilation Properties of the heat Equation. It is proven that each dilation $v(x, t) = u(\sqrt{ax}, at)$, by using the Chain rule, and treating a as a constant. That I understand. What I do not understand is the following: To "prove" that $u$ is actually only a function of


$a$ is chosen to be


where $k$ is the diffusion coefficient and $t$ is time. $k$ can be constant for one specific material, but $t$ has to change with time. How can $a$ be a function of $t$ at the same time that it is treated as a constant when proving that $v(x, t) = u(\sqrt{ax}, at)$? David

  • $\begingroup$ You don't understand "It is proven that each dilation v(x,t)=u(ax‾‾√,at), by using the Chain rule, and treating a as a constant". That is not a clear sentence. What do you prove? $\endgroup$ – abnry Feb 10 '15 at 2:41

Here's what you do. First, you suppose you have a solution to the IVP $$u(x,0) = \delta(x), \qquad u_t = u_{xx}.$$ The initial conditions here are important.

Next you show that for all $a > 0$ then $v(x,t) = a^{1/2} u(a^{1/2}x, at)$ is also a solution to the IVP. It is easy to verify it satisfies $v_t = v_{xx}$ but you need to be a little more careful showing that $v(x,0) = \delta(x)$. Here you use the identity $\delta(x) = \alpha \delta(\alpha x)$ for $\alpha > 0$.

Finally, you appeal to the uniqueness of the solution to conclude that $v(x,t) = u(x,t)$ or that $$u(x,t) = a^{1/2} \; u(a^{1/2}x, at)$$ for all $a > 0$. This is a functional identity. It is sound no matter how we got here. No matter what $a$ we chose on the right side we will still get $u(x,t)$. This means that in some sense $a$ must always cancel out. So we have no fear in substituting $a=1/t$ to get

$$u(x,t) = \frac{1}{\sqrt{t}} u(\frac{x}{\sqrt{t}},1) = \frac{1}{\sqrt{t}} f\left( \frac{x}{\sqrt{t}} \right)$$ where $f(\alpha)$ is some function.

  • $\begingroup$ Thank you for Your comment. What I meant by "It is proven that each dilation v(x,t)=u(ax‾‾√,at), by using the Chain rule, and treating a as a constant", is the following: $\endgroup$ – David Feb 10 '15 at 7:30
  • $\begingroup$ First: p= xa^0.5; q=at. We found out that (du^2/dx^2)=a*(du^2/dp^2), and that du/dt=a*du/dq. Then we said that if u(p,q) is a solution, du/dq has to be Equal to k*(du^2/dp^2). This was shown by inserting this into the heat Equation. In Our Equation we had that v(x,t)=u(sqrt(a)x,at). We did not have the a in front of the u-function. I don't understant how we can put a=1/4kt. The reason for doing this is to show that the solution to the heat Equation is actually only a function of x/sqrt(4kt). $\endgroup$ – David Feb 10 '15 at 8:12

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