I want to solve the PDE
$$\frac{\partial^2u}{\partial t\,\partial x} = \frac{\partial^2u}{\partial x^2},\quad -\infty < x < \infty,$$
$$u(x,0) = e^{-\lvert x\rvert}.$$
without the final answer involving integrals. The problem has a mixed partial derivatives, how can I handle that?
$$\frac{\partial^2u}{\partial t\,\partial x} = \frac{\partial^2u}{\partial x^2}$$ Let $\quad v=\frac{\partial u}{\partial x}$ $$\frac{\partial v}{\partial t}=\frac{\partial v}{\partial x}$$ This is a first order linear PDE which general solution is wellknown : $$v(x,t)=f(x+t)$$ where $f$ is an arbitrary function. $$u=\int f(x-t)dx=F(x+t)+g(t)$$ $F$ and $g$ are arbitrary functions. They have to be determined according to the condition $u(x,0) = e^{-\lvert x\rvert}.$ $$u(x,0)=F(x)+g(0)=e^{-\lvert x\rvert}$$ $$F(x)=e^{-\lvert x\rvert}-g(0)$$ The function $F$ is determined. We put it into the above equation $u=F(x+t)+g(t)$ where the argument of $F$ is $(x+t)$. $$u(x,t)=e^{-\lvert x+t\rvert}+g(t)-g(0)$$ This is the solution satisfying both the PDE and the solution. Thus the solution is not unique since they are infinity many functions $g(t)$. The specified condition is not suffisant to determine a unique solution.