Physical meaning of PDE When studying PDE, I want to ask if any physical meaning for the following PDE:
$$\frac{\partial W}{\partial u}=\frac{\partial^{2} W}{\partial x^{2}}$$ such that $t>0, 0<x<\infty$ and $W(0,x)=\delta_{0}(x)$. Standard book has physical meaning if $-\infty<x<\infty$ (Placing a unit impulse at $x=0$ and let it diffuse and recording the temperature). In the example given, can we use the same interpretation? What makes me confuse is that the domain defined. Usually, we do not define at $x=0$ and $t=0$ when I read the books. So I guess $W(u,x)=0$ since $\delta_{0}(x)=0$ if $x\neq 0$. But I can think of $W(u,x)\neq 0$. $\frac{1}{\sqrt{4\pi u}}e^{-\frac{x^{2}}{4u}}$ is a solution
 A: The heat-type equation you are studying is the differential formulation (i.e. a formulation involving pointwise values of the quantities involved, like spatial derivatives) of the general balance equation, and the $\delta(x)$ represents a source term located in $x=0$ and independent of $t$. To see this, let's analyze the $n$-dimensional case and then particularize the results for $n=1$.
Let $G\subset\mathbb{R}^n$ a domain for which some form of the divergence theorem holds, for example a Caccioppoli set, and suppose that this set is characterized by a time-dependent quantity $Q(t)$. 
The quantity $Q$ can vary respect time only only under the action of sources lying in the interior of $G$ or by flowing across the boundary $\partial G$ of $G$: this is the heuristic content of the balance equation
$$
\frac{\mathrm{d}Q(t)}{\mathrm{d}t}=\int_{\partial G}\boldsymbol{q}(x,t)\cdot\boldsymbol{n}_x\mathrm{d}x+\int_G s(x,t)\,\mathrm{d}x\tag{1}\label{1}
$$
where


*

*$\boldsymbol{q}(x,t)$ is the flux density vector of the quantity $Q(t)$ across $\partial G$ at the point $x$

*$\boldsymbol{n}_x$ is the inner (measure theoretic, for a Caccioppoli set) normal at the point $x$.

*$s(x,t)$ is a general inner term (representing for example processes of absorption/generation of the given quantity taking place within $G$). 


The further step in the deduction is to assume a particular form for $Q(t)$ and for the flux $\boldsymbol{q}(x,t)$. This is precisely the meaning of the following equations:
$$
Q(t)=\int_G c(x)\rho(x) w(x,t)\,\mathrm{d}x,\tag{2}\label{2}
$$
where


*

*$c(x)$ is a function which depends only on the properties of the medium which fills $G$ (if we are dealing with heat, it is called heat capacity),

*$\rho(x)$ is a function which depends only on the density of the medium (it is commonly called mass density),


and
$$
\boldsymbol{q}(x,t)=\boldsymbol{\hat{K}}(x)\nabla w(x,t),\tag{3}\label{3}
$$
where $\boldsymbol{\hat{K}}(x)$ is tensor that again takes care of the possibly anisotropic behavior of the flux. 
Now, by using equations \eqref{1}, \eqref{2}, \eqref{3} and the Gauss-Green-Ostrogradsky (divergence) theorem we obtain
$$
\begin{split}
\frac{\mathrm{d}Q(t)}{\mathrm{d}t} &=\int_G c(x)\rho(x)\frac{\partial w(x,t)}{\partial t}\,\mathrm{d}x\\
&=\int_{\partial G}\boldsymbol{q}(x,t)\cdot\boldsymbol{n}_x\mathrm{d}x+\int_G s(x,t)\,\mathrm{d}x\\
&=\int_{\partial G}\boldsymbol{\hat{K}}(x)\nabla w(x,t)\cdot\boldsymbol{n}_x\mathrm{d}x+\int_G s(x,t)\,\mathrm{d}x\\
&=\int_{G}\big[\nabla\cdot\big(\boldsymbol{\hat{K}}(x)\nabla w(x,t)\big)+ s(x,t)\big]\mathrm{d}x
\end{split}
$$
and then
$$
\int_G\left[ c(x)\rho(x)\frac{\partial w(x,t)}{\partial t}-\nabla\cdot\big(\boldsymbol{\hat{K}}(x)\nabla w(x,t)\big)-s(x,t)\right]\mathrm{d}x=0
$$
The above equation integral equation is satisfied if and only if
$$
c(x)\rho(x)\frac{\partial w(x,t)}{\partial t}=\nabla\cdot\big(\boldsymbol{\hat{K}}(x)\nabla w(x,t)\big) + s(x,t)\tag{4}\label{4}
$$
Equation \eqref{4} is the general linear (divergence form) parabolic equation: if we assume $c(x)\equiv\rho(x)\equiv1$, $\boldsymbol{\hat{K}}(x)\equiv\boldsymbol{1}$ and $s(x,t)=\delta(x)\times\delta(t)=\delta(x)\delta(t)$ (the tensor products of Dirac deltas respect to the variables $x\in\mathbb{R}^n$ and $t\in\mathbb{R}$) we obtain the standard heat equation:
$$
\begin{split}
\frac{\partial w(x,t)}{\partial t}&=\Delta w(x,t)+\delta(x)\delta(t)\quad n>1\\
\frac{\partial w(x,t)}{\partial t}&=\frac{\partial^2w(x,t)}{\partial x^2}+\delta(x)\delta(t)\quad n=1
\end{split}
$$
If you need to specify only the presence of a time independent term, basically considering only $\delta(x)$ or, the other way out, only a time-dependent term, you can put
$$
s(x,t)=
\begin{cases}
\delta(x) & x\in\mathbb{R}^n\\
\delta(t) & t\in\mathbb{R}
\end{cases}
$$
therefore deltas respect to single variables or their tensor products can be interpreted as source terms in the balance equation \eqref{1}.
Notes


*

*Many (if not all) evolution equations of mathematical physics can be deduced from the general balance equation \eqref{1} (for example the wave equation), just by properly choosing the defining equations \eqref{2} and \eqref{3}, which therefore play the rôle of constitutive axioms of a theory. In particular \eqref{3} is the Fourier-Duhamel law of heat conduction.

*A deduction of the one-dimensional heat equation (without considering the source term), however conceptually identical to the deduction shown above, is offered by Cannon (1984) (§1.1, pp. 13-15) and by Widder (1978) (§2 and §3, pp. 1-5). I prefer the one above since it don't requires the knowledge of any special (however elementary) physical concepts, being valid also for the general linear diffusion equations and other parabolic PDEs.
[1] Cannon, J. R. (1984), The One-Dimensional Heat Equation, Encyclopedia of Mathematics and Its Applications, 23 (1st ed.), Reading etc.: Addison-Wesley Publishing Company, pp. XXV+483, ISBN 978-0-521-30243-2, MR 0747979, Zbl 0567.35001
[2] Widder, D. V. (1978), The Heat Equation, Pure and Applied Mathematics 67, Academic Press, pp. xiv+267, ISBN 0-12-748540-6, MR0466967, Zbl 0322.35041. 
A: This equation describes e.g. heat diffusion.
Your initial condition describes a unit source at $(0, 0)$.
The complex version is e.g. the Schrödinger equation.
