I am trying to solve a PDE that's order 1 in time $t\ge0$ and order 2 in space $x\ge0$. The solution $u(x,t)$ exists, is unique and possesses the following properties:
- $u(x,t)\ge0$ for all $x,t\ge0$;
- $u(x,0+)=\delta(x)$;
- $u_\infty(x)=\lim_{t\to+\infty}u(x,t)$ exists;
- $\int_{0}^\infty u(x,t)\,dx<\infty$ for all $t\ge0$.
Depending on the method used to solve the PDE, the function $u(x,t)$ ends up either having the form $$ u_1(x,t)=u_{\infty}(x)+\int_0^\infty e^{-t(1+y^2)}\frac{g(x,y)}{1+y^2}dy, $$ or the form $$ u_2(x,t)=\int_0^\infty e^{-t(1+y^2)}\frac{g(x,y)}{1+y^2}dy, $$ where $g(x,y)$ is a complicated mix of special functions. I'm trying to understand whether the two forms are equivalent (the uniqueness of the solutions implies they have to be). It's easy to see that $\partial_t u_1(x,t)=\partial_t u_2(x,t)=\partial_t u(x,t)=-\int_{0}^\infty e^{-t(1+y^2)} g(x,y)\,dy$. So the first form can be recovered using $$ u_1(x,t)=\int_{t}^\infty\partial_s u(x,s), $$ and property 3.
For the second form, the idea is to use $$ u_2(x,t)=\int_{0}^t\partial_s u(x,s), $$ and then somehow use property 1, but I am not sure how. Are the two forms of $u(x,t)$ really equivalent? I'm not particularly well greased in generalized functions.
The two forms "are equivalent" (that is, the two functions are the same) whether $$u_1(x,t)-u_2(x,t)=u_\infty(x)=0.$$ It seems this equality is valid because, taking $b>t$, we get \begin{align} u_1(x,t)&=\int_{t}^\infty\partial_s u(x,s)\ ds\\ &=\int_{t}^b \partial_s u(x,s)\ ds+\int_{b}^\infty\partial_s u(x,s)\ ds\\ &=\int_{t}^b \partial_s u(x,s)\ ds+\lim_{c\to\infty}\int_{b}^c\partial_s u(x,s)\ ds \end{align} so that \begin{align} u_\infty(x)&=\lim_{t\to\infty}u_1(x,t)\\ &=\lim_{t\to\infty}\int_{t}^b \partial_s u(x,s)\ ds + \lim_{c\to\infty}\int_{b}^c\partial_s u(x,s)\ ds\\&=\lim_{c\to\infty}\int_{c}^b \partial_s u(x,s)\ ds + \lim_{c\to\infty}\int_{b}^c\partial_s u(x,s)\ ds\\ &=\lim_{c\to\infty}\left[\int_{c}^b \partial_s u(x,s)\ ds+\int_{b}^c\partial_s u(x,s)\ ds\right]\\ &=\lim_{c\to\infty}\left[\int_{c}^b \partial_s u(x,s)\ ds-\int_{c}^b\partial_s u(x,s)\ ds\right]\\ &=\lim_{c\to\infty}0 \end{align}