Example 1: Let us consider the heat equation on the half-line (0, ∞) with homogeneous boundary conditions: If the solution to the above mixed initial/boundary value problem exists, then we know that it must be unique from an application of the maximum principle. In terms of the heat conduction, one can think of u in (1.1) as the temperature in an infinite rod, one end of which is kept at a constant zero temperature. The initial temperature of the rod is then given by u₀(x). Our goal is to solve the IBVP (1.1), and derive a solution formula, much like what we did for the heat IVP on the whole line. But instead of constructing the solution from scratch, it makes sense to try to reduce this problem to the IVP on the whole line, for which we already have a solution formula. This is achieved by extending the initial data u₀(x) to the whole line. We have a choice of how exactly to extend the data to the negative half-line, and one should try to do this in such a fashion that the boundary condition of (1.1) is automatically satisfied by the solution to the IVP on the whole line that arises from the extended data. This is the case, if one chooses the odd extension of φ(x), which we describe next. By definition, a function ψ(x) is odd, if ψ(−x) = −ψ(x). But then plugging in x = 0 into this definition, one gets ψ(0) = 0 for any odd function. Recall also that the solution u(x, t) to the heat IVP with odd initial data is itself odd in the x variable. This follows from the fact that the sum [u(x, t) + u(−x, t)] solves the heat equation and has zero initial data, hence, it is the identically zero function by the uniqueness of solutions. Then, by our above observation for odd functions, we would have that u(0, t) = 0 for any t > 0, which is exactly the boundary condition of (1.1)

This shows that if one extends u₀(x) to an odd function on the whole line, then the solution with the extended initial data automatically satisfies the boundary condition of (1.1). Let us then define \[ \mathfrak{\phi_{\mathrm{odd}}}(x) = \begin{cases} u_0 (x), &\quad\mbox{for }\ x> 0 , \\ - u_0 (-x), &\quad\mbox{for }\ x< 0 , \\ 0, &\quad\mbox{for }\ x = 0. \end{cases} \] It is clear that φ_odd is an odd function, since we defined it for negative x by reflecting the u₀(x) with respect to the vertical axis, and then with respect to the horizontal axis. This procedure produces a function whose graph is symmetric with respect to the origin, and thus it is odd. One can also verify this directly from the definition of odd functions. Now, let v(x, t) be the solution of the following IVP on the whole line \[ \begin{cases} v_t &= \alpha\,v_{xx} , \qquad -\infty < x < \infty , \quad 0 < t < \infty , \\ v(x,0) &= \mathfrak{\phi_{\mathrm{odd}}}(x) , \qquad -\infty < x < \infty . \end{cases} \tag{1.2} \] From previous section we know that the solution to (1.2) is given by the formul \[ v(x,t) = \int_{-\infty}^{+\infty} S(x-\xi , t)\,\mathfrak{\phi_{\mathrm{odd}}}(\xi )\,{\text d}\xi , \quad t > 0, \] where the kernel is \[ S(x,t) = \frac{1}{\sqrt{4\pi\alpha t}}\, e^{- x^2/(4\alpha t)} \] Restricting the x variable to only the positive half-line produces the function \[ u(x,t) = v(x,t) \big\vert_{x\ge 0} . \] We claim that this u(x, t) is the unique solution of IBVP (1.1). Indeed, u(x, t) solves the heat equation on the positive half-line, since so does v(x, t). Furthermore, \begin{align*} v(x,t) &= \int_0^{\infty} S(x-\xi , t) \,\mathfrak{\phi_{\mathrm{odd}}}(\xi )\,{\text d}\xi + \int_{-\infty}^0 S(x-\xi , t) \,\mathfrak{\phi_{\mathrm{odd}}}(\xi )\,{\text d}\xi \\ &= \int_0^{\infty} S(x-\xi , t) \,u_0 (\xi )\,{\text d}\xi - \int_{-\infty}^0 S(x-\xi , t) \,u_0 (-\xi )\,{\text d}\xi . \end{align*} Making the change of variables ξ ↦ −ξ in the second integral on the right hand-side, and flipping the integration limits gives \[ v(x,t) = \int_0^{\infty} S(x-\xi , t) \,u_0 (\xi )\,{\text d}\xi - \int_{-\infty}^0 S(x+\xi , t) \,u_0 (\xi )\,{\text d}\xi \] Using definition of ϕ_odd and the above expression for v(x, t), as well as the expression of the heat kernel S(x, t), we can write the solution formula for the IBVP (1) as \[ u(x,t) = \frac{1}{\sqrt{4\pi\alpha t}} \int_0^{\infty} \left[ e^{(x-\xi )^2 /(4\alpha t)} - e^{(x+\xi )^2 /(4\alpha t)} \right] u_0 (\xi )\,{\text d}\xi . \tag{1.3} \] The method used to arrive at this solution formula is called the method of odd extensions or the reflection method. We can make a physical sense of formula (1.3) by interpreting the integrand as the contribution from the point ξ minus the heat loss from this point due to the constant zero temperature at the endpoint.

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