I am trying to compute the Laplace transform of a PDE, solve it in the complex Laplace space, and then compute the inverse transform. However, my PDE involves products of functions. If we consider $f$ the function to solve and $y$ the variable in the physical space ($y\ge0$), products are :
$$ f(y) H(y-y_0) $$ and $$\delta_{y,y_0} \partial_y f (y) $$
Where $H$ is the Heaviside function and $\delta$ the Dirac impulsion. So far, it looks like the Laplace transform of both products would be :
$$\frac{1}{2 i \pi} \lim_{T\rightarrow+\infty} \int_{a-iT}^{a+iT} \mathscr{L} \left( f \right) \left( \sigma \right) \frac{e^{- y_0 (p-\sigma)}}{p-\sigma} d\sigma $$ and $$ \frac{1}{2 i \pi} \lim_{T\rightarrow+\infty} \int_{b-iT}^{b+iT} \left( \sigma \mathscr{L} \left( f \right) \left( \sigma \right) -f(0) \right) e^{- y_0 (p-\sigma)} d\sigma$$
First, it seems that if the function $f$ and its derivatives are vanishing quickly enough at infinity (and f has no singularity), then the choice of $a$ and $b$ is arbitrary. Please correct me if this is erroneous.
However, I can hardly solve the PDE with such integrals involving limits. Any idea to simplify them would be appreciated. For the sake of completeness, the PDE in the physical space is given below.
$$ \partial_{yy} f + \alpha \delta_{y,y_0} \partial_y f + \beta H (y-y_0) f + \gamma f = 0 $$
EDIT : the term $\alpha \delta_{y,y_0} \partial_y f (y)$ might be replaced by $\alpha \partial_y f (y_0)$ and therefore would not require any Laplace transform. However, the first derivative of $f$ is not continuous at $y_0$
EDIT2 : one way around could be to use Taylor expansion when estimating $f(y) H(y-y_0)$, so that we can express $f(y)$ as a function of $f$ and all its derivatives taken at $y-y_0$. However, not all derivatives of $f$ are continuous.
Lets consider the cases $y<y_0$ and $y>y_0$
$\partial_{yy} f + \gamma f = 0 \,\,(y<y_0)$
$\partial_{yy} f + (\beta+\gamma) f = 0 \,\,(y>y_0)$
with solutions:
$f_< = a e^{\sqrt{-\gamma}y} +b e^{-\sqrt{-\gamma}y}$
$f_> = c e^{\sqrt{-\beta-\gamma}y} +d e^{-\sqrt{-\beta-\gamma}y}$
lets call $\rho = \sqrt{-\gamma} ;\sigma = \sqrt{-\beta-\gamma}$
if we demand continuity and $f'(y_0)=0$ at $y=y_0$ we can "ignore" the delta function (and its prefactor $\alpha$). This implies 3 equations for $a,b,c,d$:
$\begin{pmatrix} e^{\rho y_0}&e^{-\rho y_0}&-e^{\sigma y_0}&-e^{-\sigma y_0}&\\ \rho e^{\rho y_0}&-\rho e^{-\rho y_0}&0&0&\\ 0&0&\sigma e^{\sigma y_0}&-\sigma e^{-\sigma y_0}&\\ \end{pmatrix} \begin{pmatrix} a\\ b\\ c\\ d\\ \end{pmatrix} =\begin{pmatrix} 0\\ 0\\ 0\\ \end{pmatrix} $
Adding a normalization $a^2+b^2+c^2+d^2=1$ we have 4 equations for 4 unknowns.
Note that $f''$ need not be continuous at $y=y_0$