Solving this partial differential equation

Question

I'm stuck on a differential equation and would like help with it: $$ \frac{\partial}{\partial s} \Psi(s,x)+\frac{\partial}{\partial t}\Psi(t,x)=\frac{\partial}{\partial I}\Psi(I,x)$$

where $s+t=I.$

I took an introductory course in ordinary differential equations with some partial differential equations a few years ago. This might be a more difficult problem than I'm used to but I just want to see how someone would reason through a problem like this.

Answer 1

As @Multigrid said: if you consider $g(y) =\partial_y \psi(y)$ you have ( ignoring the $x$ which seems to be a red herring): $g(s)+g(t)=g(s+t)$ which more or less implies $g(y)=ay$ for some $a$.

Thus $\psi(y)= {1\over 2}a y^2 +b$ for some $a,b$

Answer 2

$$ \frac{\partial}{\partial s} \Psi(s,x)+\frac{\partial}{\partial t}\Psi(t,x)=\frac{\partial}{\partial I}\Psi(I,x) \tag 1$$ where $s+t=I.$

There is no derivative with respect to $x$, but only with respect to the first variable of the function $\Psi$. Thus the equation $(1)$ is not a PDE. It is an ODE where $x$ is considered as a parameter, just like any other parameters that might exist in the unknown function $\Psi$. $$\frac{d}{ds}\Psi(s)+\frac{d}{dt}\Psi(t)=\frac{d}{d(s+t)}\Psi(s+t)$$ $$\Psi'(s)+\Psi'(t)=\Psi'(s+t)$$ This is the same function taken for a unique variable say $X$ which name changes : $s,t,I$. Let $\Psi'(X)=\Phi(X)$ $$\Phi(s)+\Phi(t)=\Phi(s+t)$$ An obvious solution of this functional equation is : $\quad\Phi(X)=c\;X\quad$ since $cs+ct=c(s+t)$ . Integrating : $$\Psi(X)=\frac{c}{2} X^2+C$$ Coming back to equation $(1)$ the parameter $x$ appears into $c$ and $C$ which become functions of $x$ : $$\boxed{\Psi(X,x)=f(x)X^2+F(x)}$$ $f(x)$ and $F(x)$ are arbitrary functions.

CHECKING :

$\Psi(s,x)=f(x)s^2+F(x)\quad\implies\quad \frac{\partial}{\partial s} \Psi(s,x)=2f(x)s$

$\Psi(t,x)=f(x)t^2+F(x)\quad\implies\quad \frac{\partial}{\partial t} \Psi(t,x)=2f(x)t$

$\Psi(I,x)=f(x)I^2+F(x)\quad\implies\quad \frac{\partial}{\partial I} \Psi(I,x)=2f(x)I$

$$ \frac{\partial}{\partial s} \Psi(s,x)+\frac{\partial}{\partial t}\Psi(t,x)=2f(x)s+2f(x)=2f(x)(s+t)=2f(x)I=\frac{\partial}{\partial I}\Psi(I,x)$$ The original equation $\frac{\partial}{\partial s} \Psi(s,x)+\frac{\partial}{\partial t}\Psi(t,x)=\frac{\partial}{\partial I}\Psi(I,x)$ is satisfied.