How to solve the pde $\frac{\partial^3f}{\partial x\partial y\partial z}=f(x,y,z)$.

Question

I know just $f'=f$. What if the problem?

Looks interesting. Where is this PDE from? Maybe $f(x,y,z)=C\exp(x+y+z)$? Not sure about the uniqueness though. — , Nov 12 '21 at 01:48

score 2 · Answer 1 · answered Nov 12 '21 at 03:26

As AHusain mentions, the independence of $x,y,z$ suggests a solution of the form $f(x,y,z)=X(x)Y(y)Z(z)$. Since the PDE is first-order with respect to each variable, try a solution of the form $X(x)=C_1e^{a_1x},Y(y)=C_2e^{a_2y},Z(z)=C_3e^{a_3z}$.

Then: $f(x,y,z)=C_1e^{a_1x}C_2e^{a_2y}C_3e^{a_3z}=Ce^{a_1x+a_2y+a_3z}$.

Taking the partial derivative of this, we get

$\frac{\partial^3}{\partial x\partial y\partial z}[Ce^{a_1x+a_2y+a_3z}]= \frac{\partial^2}{\partial x\partial y}[a_3Ce^{a_1x+a_2y+a_3z}]= \frac{\partial}{\partial x}[a_2a_3Ce^{a_1x+a_2y+a_3z}]= a_1a_2a_3Ce^{a_1x+a_2y+a_3z}$.

Plugging this into our PDE, we get:

$a_1a_2a_3Ce^{a_1x+a_2y+a_3z}=Ce^{a_1x+a_2y+a_3z}$

$\to a_1a_2a_3=1$

Solving for $a_3$, we get $a_3=\frac{1}{a_1a_2}$

Thus any function of the form $f(x,y,z)=Ce^{ax+by+\frac{z}{ab}}$ should solve the PDE.

What are the general answer, It should contain three idependent function! — xldd, Nov 12 '21 at 06:32

score 2 · Answer 2 · answered Nov 12 '21 at 07:53

In order to complete the answer from ash4fun :

Any linear combination of particular solutions is solution of the PDE. $$f(x,y,z)=\sum_{\forall a} \sum_{\forall b} C_{a,b}e^{ax+by+\frac{1}{ab}z}$$ $C_{a,b}$ are arbitrary constants. Or instead of the above discrete form : $$\boxed{f(x,y,z)=\int\int \varphi(a,b)\:e^{ax+by+\frac{1}{ab}z}\:da\:db}$$ $\varphi(a,b)$ is any arbitrary function of two variables.

How to solve the pde $\frac{\partial^3f}{\partial x\partial y\partial z}=f(x,y,z)$.

2 Answers2