Consider the PDE $$c_t + u c_x - Dc_{xx} = 0$$ If I consider a semi-infinite domain $0 < x < \infty$ with the following boundary conditions $$c(0,t) = M\delta(t), c(\infty, t) = 0$$ and initial condition $$c(x,0) = 0,$$ there is an exact solution
I got this by taxing a Laplace transform in time to get $$s\hat{c} + u\hat{c}_x -D\hat{c}_{xx} = 0$$ subject to $\hat{c}(x=0) = M, \hat{c}(x = \infty) = 0$. Solving this we get $\hat{c} = M \exp{((u-\sqrt{u^2+4sD})/2D)}$ which after inversion gives $$ c = \frac{M x}{\sqrt{4\pi D t^3}} \exp{\left(-\frac{(x-ut)^2}{4Dt}\right)}$$ However if I change the initial and boundary conditions to $$c(0,t) = 0, c(\infty, t) = 0$$ and $$c(x,0) = M\delta(x),$$ am stuck.
Physically, it seems to me both PDEs are describing the same physical process -- that of a pulse injection at the instant $t=0$ at the location $x=0$.
Then why am I unable to solve the second equation for an exact solution if I follow the same technique of Laplace transforms that enabled me to solve the first one ?
You have the spatial ODE
$$s\hat{c}(s,x)+u\hat{c}_x(s,x)-D\hat{c}_{xx}(s,x)=M\delta(x).$$
This is a second order linear ordinary differential equation with constant coefficients. The solution, for any fixed $s$, is the particular solution satisfying $\hat{c}(s,0)=0,\hat{c}_x(s,0)=M$. This is given by $\frac{M}{\lambda_2 - \lambda_1}(e^{\lambda_2 x}-e^{\lambda_1 x})$ where $\lambda_1,\lambda_2$ are the roots of $-D\lambda^2+u\lambda+s=0$. (Note that $\lambda_1,\lambda_2$ depend on $s$.) If these are in fact the same i.e. $u^2+4Ds=0$, then that ratio should be understood as a limit as $\lambda_1 \to \lambda_2$. (Physically this should never happen, since $u$ should be real, $D$ should be positive, and $s$ should have positive real part.) The solution to the original problem is then the inverse Laplace transform of this.