Hello,


Can anyone help me see how one can get from the following integral


$$\lambda\int\_{\beta=0}^{y}\int\_{\alpha=x}^{Q-\beta}e^{-\mu(\alpha-x)}f(\alpha,\beta)d\alpha d\beta+k\int\_{\alpha=s}^{x}f(\alpha,y)d\alpha$$
$$ = $$
$$\lambda\int\_{\beta=0}^{y}\int\_{\alpha=s}^{Q-\beta}e^{-\mu(\alpha-s)}f(\alpha,\beta)d\alpha d\beta+k\int\_{\alpha=s}^{x}f(\alpha,0)d\alpha$$
to this second order partial differntial equation $\frac{\partial^{2}}{\partial x\partial y}f(x,y)-\mu\frac{\partial}{\partial y}f(x,y)-\frac{\lambda}{k}\frac{\partial}{\partial x}f(x,y)=0$,
where $\mu,k,\lambda,Q,s$ are all constants.



Writing an integral equation as a partial differential equation
Let $U =$
$$
\lambda \int\_{0}^{y} \int\_{x}^{Q - \beta} \operatorname{e} ^{\bigl(-\mu (\alpha - x)\bigr)} f (\alpha,\beta) d \alpha d \beta - \lambda \int\_{0}^{y} \int\_{s}^{Q - \beta} \operatorname{e} ^{\bigl(-\mu (\alpha - s)\bigr)} f (\alpha,\beta) d \alpha d \beta +
k \int\_{s}^{x} f (\alpha,y) d \alpha - k \int\_{s}^{x} f (\alpha,0) d \alpha
$$
so that our equation is $U=0$. Then compute
$$
0 = \frac{\partial^{3} U}{\partial y \partial x^{2}} - \mu \frac{\partial^{2} U}{\partial y \partial x} = 
-\lambda \frac{\partial f (x,y)}{\partial x} + k \frac{\partial^{2} f (x,y)}{\partial y \partial x} - \mu k \frac{\partial f (x,y)}{\partial y}
$$


