Question 12.15: Let P(x, D) be the linear differential operator P(x,D) = ∑j=......

Let P(x, D) be the linear differential operator

P(x, D)=\overset{m}{\underset{j=0}{\sum}} a_j(x) D^j         (12.33)

where a_j \in C^{\infty}( \mathbf{R} ), j=0,1, \ldots, m, and D is the distributional derivation operator.
a) Prove that the solution of the following equation in \mathcal {D} ^{\prime}( \mathbf{R})

P(x, D) \delta * u=\left(\overset{m}{\underset{j=0}{\sum}} a_j(x) D^j \delta\right) * u=\delta

is the function

u(x)=H(x) \cdot v(x)

where H is the Heaviside function, while v \in C^m(R) is the solution of the initial value problem

P(x, D) v=0, \quad v(0)=v^{\prime}(0)=\ldots=v^{(m-2)}(0)=0, v^{(m-1)}(0)=1

b) Suppose that each a_j, j=0,1, \ldots, m, is a constant; then for the operator given by (12.33) we put simply P(D). Assume F is a distribution with support in [0,∞). Prove then that the solution of the equation

P(D) \delta * u=F

is the distribution

u=(H \cdot v) * F

where v is the function defined in part a).