# Applying a substitution to a PDE

etotheipi
Homework Statement:
Please see below
Relevant Equations:
N/A
Problem: Consider the equation $$\frac{\partial v}{\partial t} = \frac{\partial^{2} v}{\partial x^2} + \frac{2v}{t+1}$$ where ##v(x,t)## is defined on ##0 \leq x \leq \pi## and is subject to the boundary conditions ##v(0,t) = 0##, ##v(\pi, t) = f(t)##, ##v(x,0) = h(x)## for some functions ##f(t)## and ##h(x)##. Using the substitution ##v=(t+1)^{2}u##, show that ##u## satisfies $$\frac{\partial u}{\partial t} = \frac{\partial^{2} u}{\partial x^2}$$ Attempt: I'm not sure if I'm doing the differentiation correctly. I did $$\frac{\partial v}{\partial t} = 2u(t+1)$$ $$\frac{\partial^{2} v}{\partial x^{2}} = \frac{\partial}{\partial x} \frac{\partial v}{\partial u} \frac{\partial u}{\partial x} = \frac{\partial}{\partial x} (t+1)^{2} \frac{\partial u}{\partial x} = (t+1)^{2} \frac{\partial^2 u}{\partial x ^2}$$ Plugging this in doesn't appear to give the result. My suspicion is that I was supposed to use the product rule for the first derivative, however I don't think that is right since I thought we were supposed to hold everything else constant during the differentiation? If I try this for the sake of it, I get $$\frac{\partial v}{\partial t} = 2u(t+1) + (t+1)^{2} \frac{\partial u}{\partial t}$$ $$\frac{\partial^{2} v}{\partial x^{2}} = \frac{\partial}{\partial x} \frac{\partial v}{\partial u} \frac{\partial u}{\partial x} = \frac{\partial}{\partial x} [(t+1)^{2} \frac{\partial u}{\partial x} + 2u(t+1)][\frac{\partial u}{\partial x}]$$ This seems even more wrong. So I wondered whether anyone could give me a pointer? Thanks!

## Answers and Replies

PeroK
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Homework Statement:: Please see below
Relevant Equations:: N/A

$$\frac{\partial v}{\partial t} = 2u(t+1)$$ $$\frac{\partial^{2} v}{\partial x^{2}} = \frac{\partial}{\partial x} \frac{\partial v}{\partial u} \frac{\partial u}{\partial x} = \frac{\partial}{\partial x} (t+1)^{2} \frac{\partial u}{\partial x} = (t+1)^{2} \frac{\partial^2 u}{\partial x ^2}$$
The second partial differentiation you could have simply written down, as ##t## and ##x## are independent variables.

For the first one you need to treat ##u## as a function of ##x## and ##t##.

• etotheipi
PeroK
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Using the substitution ##v=(t+1)^{2}u##

If you get stuck you should always write out functions with their arguments. In this case we have:
$$v(x, t) = (t+1)^2u(x, t)$$

• etotheipi
etotheipi
The second partial differentiation you could have simply written down, as ##t## and ##x## are independent variables.

For the first one you need to treat ##u## as a function of ##x## and ##t##.

Ah okay that works. Then I get

$$\frac{\partial v}{\partial t} = 2u(t+1) + (t+1)^{2}\frac{\partial u}{\partial t}$$ $$\frac{\partial^{2} v}{\partial x^{2}} = (t+1)^{2} \frac{\partial^{2} u}{\partial x^{2}}$$

and that checks out. The take away is that you only hold constant the variables which the one you're differentiating is a function of. The others you still have to treat with product rules etc.

If you get stuck you should always write out functions with their arguments. In this case we have:
$$v(x, t) = (t+1)^2u(x, t)$$

Yeah, I might start doing this. It makes it a little clearer. Thanks!

etotheipi
I thought I'd write up the solution I got for completeness, using the technique of guessing ##u(x,t) = A(x)B(t)## and then setting both sides equal to ##-\lambda##. One side turns out to be a "radioactive decay" one and the other is a "SHM" one: $$\frac{dB}{dt} = -\lambda B \implies B = Ce^{- \lambda t}$$ $$\frac{d^{2}A}{dx^{2}} = -\lambda A \implies A = D\cos{(\sqrt{\lambda} x + \phi)}$$ So I get ##v(x,t) = E(t+1)^{2}e^{- \lambda t}\cos{(\sqrt{\lambda} x + \phi)}##, and I guess I also need to substitute in the boundary conditions, though this should be fine since I have three unknowns and three conditions.

There are a few warning signs, since the question says that I "may find it helpful to use the substitution ##u(x,t) = w(x,t) + \gamma x## for a suitably chosen constant ##\gamma##". I've no idea what this is supposed to be implying...