As I suspected. The assignment itself simply says to assume
\begin{align}
x_2 = h(x_1)
\end{align}
and then find ##h(x_1)##, which I have done with (4) by
\begin{align}
\dot{x}_2 = h'(x_1)\dot{x}_1 = -\frac{1}{2}x_1^2 \rightarrow \\
h'(x_1)x_2 = -\frac{1}{2}x_1^2 \rightarrow \\
h'(x_1)h(x_1) =...
So the theorem isn't applicable? The assignment suggests that if I were to use the theorem it would naturally lead me to a ##h## on the form ##x_1=h_c(x_2)##. However, I don't see how this theorem does that. Keep in mind, I'm only on an introductory course, so you might have to be a bit more...
I have to find the center manifold of the following system
\begin{align}
\dot{x}_1&=x_2 \\
\dot{x}_2&=-\frac{1}{2}x_1^2
\end{align}
which has a critical point at ##x_0=\begin{bmatrix}0 & 0\end{bmatrix}##. Its linearization at that point is
\begin{align}
D\mathbf {f}(\mathbf {x_0}) =...
For example having looked a solution sheet, I know that ##x(t)=\sin(t)## is not a solution for any system of the form ##\dot{x}(t)=f(x)##. I assume this is rather simple, but I simply cannot get my head around why it wouldn't be. I'm guessing it has to do with the dependence on ##x## rather then...
Ahhh my apologies! The coefficients in the picture is not the correct coefficients, but merely the coefficients not dependent on the step size. If you want to use those coefficients, then you have to multiple all of them by 1/h^2 in the case of the 5-point stencil, and 1/(6*h^2) for the 9-point...
Must have been a short lived link. I don't usually need to upload for public use, so I just took the first and best google suggestion.
What missing factor? Also keep in mind, that the Maple ark was merely a simple way, for me to check how those coefficients comes up. The actual implementations...
Well I don't derive it like you did. Instead wrote an equation, where the coefficients are unknown:
$$\nabla^{2}_{9}u =c_1U_{i+1,j+1}+c_2U_{i+1,j}+c_3U_{i+1,j-1}+c_4U_{i,j+1}+c_5U_{i,j}+c_6U_{i,j-1}+c_7U_{i-1,j+1}+c_8U_{i-1,j}+c_9U_{i-1,j-1}$$
Then taking the Taylor expansion of all the U's and...
I see. I wouldn't have thought of trying that route.
However! I succeed by substituting all the Taylor expansions, despite it being an extreme amount of grunt work, and then using method of undetermined characteristics. The key was to realize, that I wanted the operator to have a specific...
Hey there
I'm currently taking a course on numerical methods for solving differential equations, and atm we are working with the discrete laplacian operator. In particular the 9-point stencil:
However unlike the 5-point stencil, this one is getting to me. I have tried several things, in...
Yes I have made a correction. There should be a y in the cos function. ##C_{m,n}## shouldn't just look like it, they are constants, depending entirely on n and m. Also that is the problem I'm working on.
EDIT: If you wanted the original, problem here it is:
Interesting, as the assignment I'm doing atm, askes for en expression of for ##C_{m,n}##, before going on to ask for the use of initial conditions in a later sub assignment. If I had taking time into account. Then what do I do?
Hey there!
I am current taking an introductory course on PDE's, and our professor hasn't really emphasized last part of solutions from separation of variables. Now its not strictly going to be on the exam, however I remember doing this with ease a few years back, but for some reason now I...
Unforunately I only have another 4 days to complete this, so buying and reading a whole textbook on the subject isn't an option. I will keep it in mind though, for when I get some time off, as I find the subject very fascinating.
As for your proposed solution. I get where your going, but again...