# Solve the boundary-value problem $u_t + u_{xx} = 0$

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.

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.

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

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