How can one derive a partial differential equation from a function

Mathematics Asked by Bob Pen on December 3, 2020

The Problem

If you are given a nonlinear PDE

$$frac{partial c}{partial t} = Dnabla^2c + alpha c , , ,, vec{r} in Omega, ,, , tgt 0 , , , , (1)$$
where $D, alpha$ are constants.

And then you are introduced a function defined by:
$$phi(vec{r}, t) = c(vec{r}, t)e^{-alpha t} , , , , (2)$$
where $c$ satisfies $(1)$.

You are asked to derive the PDE for the function $phi$

How does one solve this? My attempt is given below, but I have no proper idea of what I am doing.

My attempt

I find a way to express $c$ in terms of $phi$:
$$c = phi e^{alpha t} , , , , (3)$$
Then I just insert $(3)$ into $(1)$ to get:
$$frac{partial(phi e^{alpha t})}{dt} = Dnabla^2phi e^{alpha t} + alpha phi e^{alpha t} , , , , (4)$$

Is this the solution, or are there more steps to the problem? If someone could give me a guiding hand I would appreciate it very much.

Continuation after assitance from user Ninad Munshi

$$e^{alpha t}frac{partialphi}{partial t}+ alpha phi e^{alpha t} = Dnabla ^2 phi e^{alpha t} + alpha phi e^{at} $$
Dividing by $e^{alpha t}$
$$frac{partialphi}{partial t}+ alpha phi = Dnabla ^2 phi + alpha phi $$
Taking $- alpha phi$ on both sides
$$frac{partialphi}{partial t} – Dnabla ^2 phi = 0 $$
$$frac{partialphi}{partial t} – Dfrac{partial^2phi}{partial x^2} = 0 $$

One Answer

Your new solution is right. You may recognize $phi=c e^{-alpha t}$ from the method of integrating factors for solving the ODE in $t$, $$ frac{dc}{dt}(t) - alpha c(t) = f(t).$$ Your exercise is essentially an application of this to the new setting of a certain PDE. It somehow worked out because $nabla^2 phi = (nabla^2 c)e^{-alpha t}$, by which I mean that this change of variables simplifies the equation from $(partial_t - alpha -Dnabla^2)c=0$ to the same equation (for $phi$) but with $alpha=0$. Since knowing $phi$ tells you everything about $c$, it suffices to study the $alpha=0$ case.

This is not what happens if $alpha=alpha(x)$ was a function of $x$; going through the same motions, if we set $phi(x,t) = c(x,t)e^{-a(x)t}$, we get instead $$e^{alpha(x)t}partial_t phi =(partial_t-alpha(x)) (phi e^{alpha(x)t}) = Dnabla^2(phi e^{alpha(x)t}) = e^{alpha(x)t}Dleft[nabla^2 phi + 2nablaalphacdotnabla phi + phi nabla^2 alpha + phi|nabla alpha|^2right]$$ which does not lead to a simpler equation.

Correct answer by Calvin Khor on December 3, 2020

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