# IVP for linear first-order PDE $3u_x + 4u_y + 5u_z =0$

I solved an old problem (I don't remember if I have already posted this problem: forgive me, if so)

$$\begin{cases} 3u_x + 4u_y + 5u_z =0\\ u(1,y,0)=2y-6 \end{cases}$$

I quite easily obtained the general solution:

$$u(x,y,z)=\Phi(\zeta_1,\zeta_2)=\Phi(4x-3y,4z-5y)$$

Then I had to evaluate the particular solution:

$$u(1,y,0)=\Phi(\zeta_1,\zeta_2)=\Phi(4-3y,-5y)=2y-6$$

Observing the parameters passed, I had

$$\begin{cases} \zeta_1=4-3y\\ \zeta_2=-5y \end{cases}$$

Expliciting the independent variables as function of constants $$\zeta_i$$:

$$\begin{cases} x \text{ undefined}\\ \displaystyle y=\frac{1}{3}(4-\zeta_1)=-\frac{1}{5}\zeta_2\\ z \text{ undefined} \end{cases}$$

I started to build the particular solution:

$$u(1,y,0)=\Phi(\zeta_1,\zeta_2)=2y-6$$

and replaced $$y$$ with $$\zeta_i$$ expressions

$$u_1(x,y,z)=\Phi_1(\zeta_1,\zeta_2)=2\left(\frac{1}{3}(4-\zeta_1)\right)-6$$

or else

$$u_2(x,y,z)=\Phi_2(\zeta_1,\zeta_2)=2\left(-\frac{1}{5}\zeta_2\right)-6$$

Making the last replacements, I had $$u_1(x,y,z)=\Phi_1(\zeta_1,\zeta_2)=2\left(\frac{1}{3}(4-(4x-3y))\right)-6 = -\frac{8}{3}x+2y-\frac{10}{3}$$

or else

$$u_2(x,y,z)=\Phi_2(\zeta_1,\zeta_2)=2\left(-\frac{1}{5}(4z-5y)\right)-6=2y-\frac{8}{5}z-6$$

Both solutions verify the problem and I noticed (in a similar way like in one of my previous posts) that

$$u(x,y,z) = \frac{1}{2} u_1(x,y,z) + \frac{1}{2}u_2(x,y,z)$$

and I started wondering if, for example

$$u(x,y,z) = \alpha u_1(x,y,z) + (1-\alpha)u_2(x,y,z)$$

with $$0 \leq \alpha \leq 1$$.

Question: is there any way to represent the particular solution that include even the two forms obtained?

• I think the boundary conditions have to be given for a surface for the solution to be unique, so, it's not s surprise you got a whole family of solutions depending on $\alpha$ – Rafa Budría Apr 15 at 6:48

$$\begin{cases} 3u_x + 4u_y + 5u_z =0\\ u(1,y,0)=2y-6 \end{cases}$$ I agree with your general solution : $$u(x,y,z)=\Phi(4x-3y,4z-5y)$$ Among this set of solutions, a sub-set is : $$u(x,y,z)=F(4x-3y)+G(4z-5y)$$ with $$F$$ and $$G$$ arbitrary functions.

Among this sub-set of solutions there is an even smaller sub-set made of linear functions $$F(X)=\alpha X$$ and $$G(X)=\beta X$$ : $$u(x,y,z)=\alpha(4x-3y)+\beta(4z-5y)$$ with $$\alpha$$ and $$\beta$$ arbitrary constants.

CONDTION : $$\quad u(1,y,0)=2y-6=\alpha(4-3y)+\beta(0-5y)$$ .

$$(2+3\alpha+5\beta)y-6-4\alpha=0 \quad\implies\quad \alpha=-\frac32$$ and $$\beta=\frac12$$ .

Thus a particular solution is : $$\boxed{u(x,y,z)=-\frac32(4x-3y)+\frac12(4z-5y)}$$ Of course they are an infinity of other particular solutions which satisfy both the PDE and the specified condition since the wording of the question doesn't specify enough conditions to make the solution unique.

NOTE :

If we are not looking for the general solution, but only for a particular solution on linear form, a more direct method is simpler :

Let $$u(x,y,z)=ax+by+cz$$

$$3u_x + 4u_y + 5u_z =0 = 3a+4b+5c$$

$$u(1,y,0)=2y-6=a+by$$

$$\begin{cases} -6=a\\ 2=b\\ 3a+4b+5c=0 \end{cases} \quad\implies\quad \begin{cases} a=-6\\ b=2\\ c=2 \end{cases}$$ $$\boxed{u(x,y,z)=-6x+2y+2z}$$ Which is the same as $$u(x,y,z)=-\frac32(4x-3y)+\frac12(4z-5y)$$ .

• Thanks JJacquelin! I should have thought about the last direct method....it is a good simple idea! :) – Clyde A. Jansen Apr 17 at 17:26