Two questions about PDE

Question

I deal with two problems:

1. $\frac{\partial v}{\partial y} - 2t\frac{\partial v}{\partial t} + 3v = e^{3y-t}$, $v(y,0)=e^y$.
2. $\frac{\partial v}{\partial y} - 2t\frac{\partial v}{\partial t}+3v=e^{3y-t}$, $v(0,t)=e^t$.

Here is my approach.

1. I found no solution. Let $y:=y(s,\tau)$, $t:=t(s,\tau)$ such that $$\frac{\partial y}{\partial s} = 1,\quad \frac{\partial t}{\partial s} = -2t,\quad \frac{\partial v}{\partial s} + 3v = e^{3y-t},$$ with initial conditions $y(0,\tau) = \tau$, $t(0,\tau)=0$, and $v(0,\tau)=e^{\tau}$. Note that $\frac{\partial t}{\partial s}=-2t$ give us $t(s,\tau)=e^{-2s}C_1(\tau)$. Since $t(0,\tau)=0$, we have $C_1(\tau)=0$ which implies $t(s,\tau)=0$, contradiction.
2. With same approach, we can solve that $y(s,\tau) = s$, $t(s,\tau)=e^{-2s}\tau$. Hence, we remain to solve that $\frac{\partial v}{\partial s} + 3v= e^{3y-t} = e^{3s-e^{-2s}\tau},$ and I have no idea how to solve it.

I am not sure why problem 1 doesn't have a solution and why problem 2 is due to the integration issue.

Answer 1

$$\frac{\partial v}{\partial y} - 2t\frac{\partial v}{\partial t} = -3v+e^{3y-t}$$ Charpit-Lagrange characteristic ODEs : $$\frac{dy}{1}=\frac{dt}{-2t}=\frac{dv}{-3v+e^{3y-t}}=ds$$ Note: This is equivalent to : $\frac{dy}{ds} = 1,\quad \frac{dt}{ds} = -2t,\quad \frac{dv}{ds} + 3v = e^{3y-t}$

A first characteristic equation comes from solving $\frac{dy}{1}=\frac{dt}{-2t}$ : $$t\:e^{2y}=c_1$$ $e^y=(\frac{c_1}{t})^{1/2}\quad\implies\quad e^{3y-t}=(\frac{c_1}{t})^{3/2}e^{-t}$

A second characteristic equation comes from solving $\frac{dt}{-2t}=\frac{dv}{-3v+e^{3y-t}}=\frac{dv}{-3v+(\frac{c_1}{t})^{3/2}e^{-t}}$ :

$\frac{dv}{dt}=-\frac{1}{2t}\left(-3v+(\frac{c_1}{t})^{3/2}e^{-t}\right)\quad\implies\quad \frac{dv}{dt}-\frac{3}{2t}v=-\frac{(c_1)^{3/2}}{2t^{5/2}}e^{-t}$

Solving this first order linear ODE involves the special function Ei : https://mathworld.wolfram.com/ExponentialIntegral.html The solution of the ODE is : $$v=\frac{1}{12}\left(\frac{c_1}{t}\right)^{3/2}\bigg((t^2-t+2)e^{-t}+t^3\text{Ei}(-t)\bigg)+c_2t^{3/2}$$ The general solution of the PDE on implicit form $c_2=F(c_1)$ is : $$v=\frac{1}{12}\left(\frac{te^{2y}}{t}\right)^{3/2}\bigg((t^2-t+2)e^{-t}+t^3\text{Ei}(-t)\bigg)+F(te^{2y})t^{3/2}$$ $$\boxed{v(y,t)=\frac{1}{12}e^{3y}\bigg((t^2-t+2)e^{-t}+t^3\text{Ei}(-t)\bigg)+t^{3/2}F(te^{2y})}$$ $F$ is an arbitrary function.

Or equivalently : $$v(y,t)=\frac{1}{12}e^{3y}\bigg((t^2-t+2)e^{-t}+t^3\text{Ei}(-t)\bigg)+e^{-3y}G(te^{2y})$$ $G$ is an arbitrary function (related to F).

$\text{CASE of boundary condition}$ $v(0,t)=e^t$ $$v(0,t)=\frac{1}{12}\bigg((t^2-t+2)e^{-t}+t^3\text{Ei}(-t)\bigg)+t^{3/2}F(t)=e^t$$ $$F(t)=t^{-3/2}e^t-\frac{1}{12}t^{-3/2}\left((t^2-t+2)e^{-t}+t^3\text{Ei}(-t)\right)$$ The function $F$ is determined. We put it into the above general solution of course not with argument $t$ but with argument $te^{2y}$. $$v(y,t)=\frac{1}{12}e^{3y}\bigg((t^2-t+2)e^{-t}+t^3\text{Ei}(-t)\bigg)+t^{3/2}\bigg((te^{2y})^{-3/2}e^{te^{2y}}-\frac{1}{12}(te^{2y})^{-3/2}\left(((te^{2y})^2-te^{2y}+2)e^{-(te^{2y})}+(te^{2y})^3\text{Ei}(-te^{2y})\right) \bigg)$$ After simplification : $$v(y,t)=\frac{1}{12}e^{3y}\bigg((t^2-t+2)e^{-t}+t^3\text{Ei}(-t)\bigg)+e^{-3y+te^{2y}}-\frac{1}{12}\left((t^2e^{4y}-te^{2y}+2)e^{-3y-te^{2y}}+t^3e^{3y}\text{Ei}(-te^{2y})\right) $$

$\text{CASE of initial condition}$ $v(y,0)=e^y$

$v(y,0)=\frac16 e^{3y}+Ce^{-3y}\neq e^y$ . No function $F$ or $G$ are found.

This is in aggrement with other arguments for no solution.

Answer 2

Your approach is correct.

For problem 1, the $y$-axis is a characteristic curve, so you should expect things to go wrong when you try to assign values to $v$ there.

For problem 2, you can in principle solve the ODE through multiplication by the integrating factor $e^{3s}$, but then you will need to calculate $$ \int e^{6s} e^{-\tau e^{-2s}} \, ds , $$ which has no elementary antiderivative if $\tau \neq 0$. (The substitution $u=e^{-2s}$ gives something which can be expressed in terms of the exponential integral Ei, according to Mathematica.)

What's the context? Did you get these problems as exercises in some course? In that case, it's probably best to ask your teacher to clarify.