Let $u(x,t) $ be a solution of $u_t=ku_{xx}$. Show that the following facts hold.

a. For constants $a, x_0 $ and $t_0$, the function $v(x,t) =u(ax-x_0,a^{2}t-t_0)$ satisfies $v_t=kv_{xx}$

b. For any constant $k'$ the function $v(x,t)=u(x,(\frac{k'}{k}t)$ satisfies $v_t=k'v_{xx}$

c. The function $v(x,t) = t^{\frac{-1}{2}}e^{\frac{-x^2}{4kt}} \cdot u( \frac{x}{t},\frac{-1}{t})$ satisfies $v_t=kv_{xx}$

I am beyond confused with this problem. This is what I'm thinking...first I need to solve for $u(x,t)$ so I can take the partial derivatives for $v(x,t)$. The question is do I solve it by itself or use the fundamental source solution for the heat equation $u_t=ku_{xx}$ which is $u(x,t) = \frac{1}{\sqrt{4 \pi kt}}e^{\frac{-x^2}{4kt}}$ for $ t > 0, -\infty < x < \infty $

I'm going to try $a$ since there are constants everywhere and by taking the derivative of a constant I get $0$. So, for $u(ax-x_0,a^{2}t-t_0)$ my $u_t =a^2$ and $u_x=a$, but then if I take the derivative with respect to x again, I will get 0 and that's obviously not going to work because $v_t \neq kv_{xx}$ and that's not good.

Do I need to take the product and chain rule for $u_{xx}$? Suppose I need to. Then $u_x = au$ and $u_{xx} = au_x$ because I am using the product rule, so I have $au_x+u(0) \rightarrow au_x$ Since I can substitute $u_x$ now I will have $a(au)$ which becomes $a^2u$.

Alright.. but what happens to the u?

If I keep it as is... I am close

$a^2=ua^2$. Unless we make $u=k$? If I don't have the pesky $k$ I'm done, but I need it .


Part a:

$$ v = u(x',t') $$ where $$ x' = ax - x_0 \\t' = a^2t - t_0 $$ Therefore $$ v_x = \frac{\partial{u}}{\partial{x'}}\frac{\partial{x'}}{\partial{x}} +\frac{\partial{u}}{\partial{t'}}\frac{\partial{t'}}{\partial{x}} =u_{x'}a + 0 $$ Similarly: $$ v_{xx} = a^2u_{x'x'} \\v_{t} = a^2u_{t'} $$ and the result follows using $u_{t'}=ku_{x'x'}$.

The other parts of the question can be done in a similar way.

  • $\begingroup$ so I just take the partial derivatives separately? Like for t, we have $a^2t-t_0$, so $v_t = a^2u_t+t(0)$ Since $t_0$ is already a constant, it would be pointless for me to type 0+0. hmmm... Do I take the product and chain rule of this? I see $v_x = (u_x)(x_x)+(u_t)(t_x)$ @MartinG $\endgroup$ – usukidoll Sep 28 '14 at 11:22
  • $\begingroup$ I recommend you keep the dash notation, otherwise it's easy to get confused as to the meaning of $u_x$ etc. Part b can be done by the same method as part a. Part c requires the product rule as well as the chain rule for partial derivatives. $\endgroup$ – MartinG Sep 28 '14 at 11:30
  • $\begingroup$ so hmmmm... I need the chain rule and the product rule right? but....do I plug in $ v_{t} = a^2u_{t'}$ and $v_{xx} = a^2u_{x'x'}$ into $u_{t'}=ku_{x'x'}$? @MartinG $\endgroup$ – usukidoll Sep 28 '14 at 11:32
  • $\begingroup$ Yes, or rather the other way round: $v_{t} = a^2u_{t'} = a^2ku_{x'x'} = kv_{xx}$ $\endgroup$ – MartinG Sep 28 '14 at 11:58

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