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- Apr 13, 2013

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I am looking at the following exercise:

Find the solution $u(t,x)$ of the problem

$$u_t-u_{xx}=2 \sin{x} \cos{x}+ 3\left( 1-\frac{x}{\pi}\right)t^2, t>0, x \in (0,\pi) \\ u(0,x)=3 \sin{x}, x \in (0,\pi) \\ u(t,0)=t^3, u(t, \pi)=0, t>0$$

At the suggested solution, it is stated that the boundaries are non-zero, so we want to set them equal to $0$.

We set $v=u-\left( 1-\frac{x}{\pi}\right) t^3-\frac{x}{\pi} \cdot 0 \Rightarrow v=u-t^3+t^3 \frac{x}{\pi}$.

Could you explain to me why we take this $v$? Is there a formula that we use that enables us to solve the problem?