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Consider the boundary-value problem

$$\frac{∂u}{∂t} - \frac{\partial^2u}{∂x^2} = 0, \quad x\in[0,2], t\in[0,\infty) $$

$$u(0,t) = u(2,t)=0$$ $$u(x,0)=x(x-1)(x-2)$$

Show that $u(x, t) = −u(2 − x, t),\ \forall x ∈ [0, 2], t \in [0, \infty)$.

NB: You are expected to achieve the result without actually finding the solution $u(x, t)$ here.

I think I need to use the result that the solution of this problem is unique, I've tried a few things but seem to get to dead ends. Would appreciate any help.

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Consider the transformation $X=2-x$, and let $U(X,t):=u(X,t)=u(2-x,t)$.

Then it is easy to see that

  • $\frac{\partial U}{\partial t}-\frac{\partial^2 U}{\partial X^2}=\frac{\partial u}{\partial t}-\frac{\partial^2u}{\partial x^2}=0$
  • $U(0,t)=u(2,t)=0$
  • $U(2,t)=u(0,t)=0$
  • $U(X,0)=u(2-x,0)=-(2-x)(1-x)x=-u(x,0)$

So $u(x,t)$ and $-U(X,t)$ satisfy the same equation and conditions; hence by uniqueness, they are the same.

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Let $v(x,t):=-u(2-x,t)$. Then show:

  1. $\frac{∂v}{∂t} - \frac{∂^2v}{∂x^2} = 0,$

and

  1. $v(0,t)=0, v(2,t)=0, v(x,0)=x(x-1)(x-2).$

Since the boundary-value problem has a unique solution $u$, we get

$u(x,t)=v(x,t)=-u(2-x,t).$

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