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Can you explain me how does the backward Euler method works? I have seen the formula and try to understand the method, but what I can't understand is why and how to use the Newton-Rapson method.

Do you have a link to a good tutorial? Something that can help me understand it graphically?

UPDATE

According to what Paul said, could you please tell me if what follows is correct?

The Cauchy problem is like this

$$u(t) = f\big ( u(t), t \big ) \quad \text{where} \quad f\big ( u(t), t \big ) = u'\big ( u(t), t \big )$$

then, according to the example given by Paul, I'll have something like:

$$u'(t)=\big [u(t) \big]^3$$

Now, the backward Euler method reads:

$$u_{t+1} = u_t + hf\big ( u_{t+1}, t+1 )~~~~~~~~~~\text{not really sure about this}$$

So for, $t = 0$ and $h = 1/2$ I should have the following

$$u_1 = u_0 + \frac{1}{2}(u_1)^3$$

For this equation I have to solve by means of Newton-Raphson before going to the next time $t = 1$:

$$u_1 - u_0 - \frac{1}{2}(u_1)^3 = 0$$

From here I don't know what to do, how to use Newton's method??

UPDATE

I think I finally got it. From backward Euler's method I have:

$$u_{t+1} = u_t + hf(u_{t+1}, t+1)$$

this can be seen as (just like everybody told me):

$$ F(u_{t+1})= u_{t+1} - u_t - hf(u_{t+1}, t+1) = 0$$

where $u_{t+1} = x$ so I have

$$ F(x)= x - u_t - hf(x, t+1) = 0$$

which now I see a familiar way to use Newton's (or whatever other method) to find the value of $x$

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I think your last formula is correct. –  Paul Jan 18 '13 at 22:31
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2 Answers 2

up vote 4 down vote accepted

You would use backward euler method to solve a differential equation of the form $u_t =f(u,t)$ where $f$ is not necessarily a linear function in u. When f is non-linear, then the backward euler method results in a set of non-linear equations that need to be solved for each time step. Ergo, Newton-raphson can be used to solve it.

For example, take

$$\frac{du}{dt}=u^3(t)$$

Backward euler results in

$$\frac{u_{t+1}-u_{t}}{\Delta t}=u_{t+1}^3$$

Since you can't solve for $u_{t+1}$ directly, you have to estimate it by a numerical scheme. This is where newton-raphson comes into play. Rearrange this equation so that the RHS=0, we get

$$\frac{u_{t+1}-u_{t}}{\Delta t}-u_{t+1}^3=0$$

Now, we can apply newton's method to find the root of this equation $u_{t+1}$, which in turn becomes the solution to the differential equation at the next timestep.

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Ok. Almost got it. First, when you say "of the form $u_t = f(u,t)$", this means $f(u,t) = u'\big ( u(t), t \big )$, I ask just because I don't want to mess up with the notation. Then, $du/dt = u^3(t) = f(u, t)$ right? Finally, when the RHS = 0, from the resulting equation $u_t$, $\Delta t$, are known but not $u_{t+1}$ so how do I evaluate that with Newton-Raphson? –  BRabbit27 Jan 18 '13 at 17:31
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Backwards Euler, as you state you know, is time advances using the equation: $U_{t+\Delta t} = U_{t}+ U'(t,U_{t+\Delta t}) \Delta t$.

You can not explicitly evaluate $U'_{t+\Delta t}$ since you don't know $U_{t+\Delta t}$. To make it more clear the equation is:

$X = U(t) + U'(t,X)\Delta t$

Therefore you must solve the equation for X (which is U at the next time step) using some methodology (this is where you use Newton-Raphson in your case).

As far as graphical interpretation, think about your standard Euler. The idea there is that you are taking the current value of $U$ and its derivative at the current time, and assuming that it is nearly linear for some small change in time ($\Delta t$). So you are drawing a straight line from $U(t)$ at a slope of $U'(t)$.

For backwards Euler, all you are doing is using the slope at the end of your line approximation rather than the start of it. As to why you would want to do this, it is a more complicated answer involving the stability of your solution.

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I appreciate your help, but still I can't see where does the Newton-Rapson comes to play... I mean, the equation of backward Euler I got it, and obviously I'm trying to compute the value of $U'(t, U_{t+\Delta t})$ which I can't. Newton-Rapson is used to find the zeros of a function isn't it? Sorry, I can't see how those two things fit... –  BRabbit27 Jan 18 '13 at 16:38
    
All this subindices confuse me A LOT. Because when I saw the backward Euler combined with Newton-Raphson, there appeared a superindex k, k+1... –  BRabbit27 Jan 18 '13 at 16:41
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I made a slight update to hopefully clarify for you. –  Godric Seer Jan 18 '13 at 21:01
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