- #1

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!