How can I solve this 1st order non-homogeneous PDE?

Question

I've been trying to solve the initial value problem: $$u_t+2u_x-3u_y+4u_z+u=e^{xyz}$$ $$u(x,y,z,0)=\sin(xyz)$$ I've tried to approach the problem by characteristics' method but the non-homogeneous condition prevented me from completing the solution. Basically, I've found: $$u(x,y,z,t)=e^{-t}\sin\left((x-2t)(y+3t)(z-4t)\right)+e^{-t}\int_0^t e^{s+(2s+A)(-3s+B)(4s+C)}ds$$ where after the integral's resolution I'd replace $A=x-2t$, $B=y+3t$ and $C=z-4t$. If you have any ideia of how to solve this deadlock, I'd be forever grateful.

Answer 1

$$u_t+2u_x-3u_y+4u_z=-u+e^{xyz}$$ Charpit-Lagrange characteristic ODEs : $$\frac{dt}{1}=\frac{dx}{2}=\frac{dy}{-3}=\frac{dz}{4}=\frac{du}{-u+e^{xyz}}$$ A first characteristic equation from solving $\frac{dt}{1}=\frac{dx}{2}$ : $$x-2t=c_1$$ A second characteristic equation from solving $\frac{dt}{1}=\frac{dy}{-3}$ : $$y+3t=c_2$$ A third characteristic equation from solving $\frac{dt}{1}=\frac{dz}{4}$ : $$z-4t=c_3$$ A fourth characteristic equation from solving $\frac{dt}{1}=\frac{du}{-u+e^{xyz}}=\frac{du}{-u+e^{(2t+c_1)(-3t+c_2)(4t+c_3)}}$ :

$\frac{du}{dt}+u=e^{(2t+c_1)(-3t+c_2)(4t+c_3)}$

$$e^tu-\int_0^t e^{\tau+(2\tau+c_1)(-3\tau+c_2)(4\tau+c_3)}d\tau=c_4$$

The general solution of the PDE on the form of implicit equation $c_4=F(c_1,c_2,c_3)$ is : $$e^tu-\int_0^t e^{\tau+(2\tau+(x-2t))(-3\tau+(y+3t))(4\tau+(z-4t))}d\tau=F\big((x-2t),(y+3t),(z-4t)\big)$$ $F$ is an arbitrary function of three variables. $$\boxed{u(x,y,z,t)=e^{-t}\int_0^t e^{\tau+(2(\tau-t)+x)(-3(\tau-t)+y)(4(\tau-t)+z)}d\tau+e^{-t}F\big((x-2t),(y+3t),(z-4t)\big)}$$ CONDITION : $u(x,y,z,0)=\sin(xyz)=0+e^0F\big((x+0),(y+0),(z+0)\big)$ $$F(x,y,z)=\sin(xyz)$$ The function $F$ is determined. We put it into the above general solution : $$\boxed{u(x,y,z,t)=e^{-t}\int_0^t e^{\tau+(2(\tau-t)+x)(-3(\tau-t)+y)(4(\tau-t)+z)}d\tau+e^{-t}\sin\big((x-2t)(y+3t)(z-4t)\big)}$$ The integral cannot be expressed in terms of a finite number of standard functions. A closed form would require special functions not standardised up to now.

NOTE :

The above integral is a particular case of $\quad\int e^{P(X)}dX\quad$ where $P(X)$ is a polynolmial.

Case of first degree (Involving elementary function): $$\int e^{a_0+a_1X}dX=\frac{1}{a_1}e^{a_0+a_1X}+C$$

Case of second degree (Involving a special function erfi) https://mathworld.wolfram.com/Erfi.html : $$\int e^{a_0+a_1X+a_2X^2}dX=\frac{\sqrt{\pi}}{2\sqrt{a_2}}e^{a_0-\left(a_1^2/(4a_2)\right)}\text{erfi}\left(\frac{a_1+2a_2X}{2\sqrt{a_2}}\right)+C$$

Case of third degree $\int e^{a_0+a_1X+a_2X^2+a_3X^3}dX$ :

Presently there is no special function standardised for this kind of integral. Thus stop wasting time to find a closed form in general (except for some particular values of the coefficients).