Recent content by Runei

Whoops, you're absolutely right, I miswrote. The correct expression should be that we perturb U instead. So $$u(x,t,\alpha)=u(x,t,0)+\alpha\eta(x,t)$$

And I believe we have come full circle. What @Stephen Tashi said about making clear what arguments your function has, is particularly important here. A clear way to write the Euler-Lagrange equation would be $$\frac{\partial L}{\partial u}(x,y)-\frac{\partial}{\partial t}\left(\frac{\partial...

Digging further I have discovered where the problem lies, but I need a little help to understand the issue. It all comes from the derivation of the Euler-Lagrange equation. We define the Lagrangian as a function given by ##L(x,t,u,u_x,u_t,\alpha) = L(x,t,u,u_x,u_t)+\alpha\eta(x,t)##. We have...

Hi again, And thanks for you time digging into this. I'm unsure about how you have set up the different L functions, so let me try do re-iterate. The Lagrangian density ##L(x,t,u,u_x,u_t)## is some function dependent on those 5 arguments. As you say, ten there is another Lagrangian, L, which...

Some good points there. The reason for my thinking about all this, is that I am working on a problem in which we have a Lagrangian density given by ##L(x,t,u,u_t,u_x)##. The Euler-Lagrange function for this is $$\dfrac{\partial L}{\partial u} - \dfrac{\partial}{\partial t}\dfrac{\partial...

Hi, and thanks for your reply. So I am using it in the sense of taking the derivative with respect to its argument, as you show with the function ##f(x,y,z) = 2x^2+y+z## then ##\partial f/\partial x = 4x##. My question is then about the ways in which these arguments can be related, and how we...

Hi all, I was working on a problem using Euler-Lagrange equations, and I started wondering about the total and partial derivatives. After some fiddling around in equations, I feel like I have confused myself a bit. I'm not a mathematician by training, so there must exist some terminology which...

The funny thing is actually that a mixture of solvents will behave as a single solvent which either dissolves either or both of the polymers, or not. This can all be investigated using Hansen Solubility Parameters, which turns the problem into a geometric problem. The polymers will be...

There is also the possibility to use something called "solvent bonding", which does not rely on an adhesive (third material layer between the two polymers). For this approach, it is required to know a chemical that is a solvent for both polymers. When applying the solvent to one surface, and...

Thank you, so much for your answer !

Hi, My question is short and very simple: Is the loss of a leaving group primarily a random event? What is the actual mechanism that initiates that a specific leaving group.. leaves? Thanks in advance :)

When solving problems, particularly in optics, it is often that we represent the wave-function as a complex number, and then take the real part of it to be the final solution, after we do our analysis. u(\vec{r},t)=Re\{U(\vec{r},t)\}=\frac{1}{2}\left(U+U^*\right) Here U is the complex form of...

Well there are actually some more now that I think about it. One thing I realized is that the integral with ##t_2## should probably have a lower bound being ##t_1## instead...

Thanks for the replies! The equation is coupled with another equation namely $$ \int\limits_0^{t_1}\tau_1(t)\omega(t)dt+\int\limits_0^{t_2}\tau_2(t)\omega(t)dt+\int\limits_0^{t_1}\tau_{in}\omega(t)dt = 0 $$ The ##\tau_1(t)## and ##\tau_2(t)## are controllable functions - they can be chosen by...

Hello there! I'm currently doing some mathematical modelling at my work, and I have arrived at an interesting kind of circular relationship integral - and now I'm wondering about what to do. The integral looks very innocent at first glance: $$ \theta_s = \int\limits_0^{t_1} \omega (t) dt$$ So...