1. ## Quick-ish DE question

Hey guys,

I'm doing some numerical lin.alg work that requires me to work by hand. I'm having trouble with the following question:-

Consider the IVP $\displaystyle dy/dx = U$
$\displaystyle x, y(0) = 1$
*U is some constant

We wish to nd the solution at a given, xed value of x > 0 using a forward-Euler method. To this end we divide the interval [0; x] into n equally spaced intervals of length h = x=n and use the forward-Euler scheme

$\displaystyle {Y}_{j+1}$ = $\displaystyle {Y}_{j}$ + $\displaystyle U*h*{Y}_{j+1}$

The question I'm struggling with is the following
:

Show that the forward Euler scheme implies that $\displaystyle {Y}_{j}$ = $\displaystyle (1+Uj)^n$ and hence,

$\displaystyle log{Y}_{n}$ = $\displaystyle nlog(1+Uj)$

Anyone got any tips, a start or something so that I can get through this question? I'm just completely lost with it.

2. Originally Posted by exphate
Hey guys,

I'm doing some numerical lin.alg work that requires me to work by hand. I'm having trouble with the following question:-

Consider the IVP $\displaystyle dy/dx = U$$\displaystyle x, y(0) = 1$
*U is some constant

We wish to nd the solution at a given, xed value of x > 0 using a forward-Euler method. To this end we divide the interval [0; x] into n equally spaced intervals of length h = x=n and use the forward-Euler scheme

$\displaystyle {Y}_{j+1}$ = $\displaystyle {Y}_{j}$ + $\displaystyle U*h*{Y}_{j+1}$

The question I'm struggling with is the following:

Show that the forward Euler scheme implies that $\displaystyle {Y}_{j}$ = $\displaystyle (1+Uj)^n$ and hence,

$\displaystyle log{Y}_{n}$ = $\displaystyle nlog(1+Uj)$

Anyone got any tips, a start or something so that I can get through this question? I'm just completely lost with it.

$\displaystyle {Y}_{j+1}$ = $\displaystyle {Y}_{j}$ + $\displaystyle U*h*{Y}_{j+1}$

or

$\displaystyle {Y}_{j+1}$ = $\displaystyle {Y}_{j}$ + $\displaystyle U*h*{Y}_{j}$

For the latter, it's a difference equation

$\displaystyle Y_{j+1} = (1 + h U) Y_j$

We look for a solution of the form $\displaystyle Y_j = c\rho^j.$ When you substitute you get

$\displaystyle c \rho^{j+1} = c (1 + hU)\rho^j$ giving $\displaystyle \rho = 1 + hU$

Thus, $\displaystyle Y_j = c (1 + hU)^j$. The initial condition (when $\displaystyle j = 0$) gives $\displaystyle c = 1$.