Recent content by maggie56

Sorry, not sure i follow, what do you suggest i could set them as?

Homework Statement I have found the general solution to a second order pde to be U(x,t) = f(3x + t) + g(-x + t) where f and g are arbitrary functions I have initial conditions U(x,0) = sin(x) Du/dt (x,0) = cos (2x) The Attempt at a Solution I have found that U(x,0) = f(3x) +...

Homework Statement I have a pde, 16d2u/dxdy + du/dx + du/dy + au = 0 where a is constant. Homework Equations The Attempt at a Solution I have tried to solve this pde using the substitutions x=e^t and y=e^s so t=ln(x) and s=ln(y) then finding Du/dx= 1/x du/dt and du/dy= 1/y...

Homework Statement I have a PDE for which i have found the general solution to be u(x,y) = f1(3x + t) + f2(-x + t) where f1 and f2 are arbitrary functions. I have initial conditions u(x,0) = sin (x) and partial derivative du/dt (x,0) = cos (2x)Homework Equations u(x,y) = f1(3x + t) + f2(-x +...

It just says what is the particle path of the flow u= (-z + cos(at)) j + (y + sin(at)) k It is an example from a lecture, previously we had found the streamlines for the flow.

I don't understand how to find particle paths, for example i have a question that states; u= (-z + cos(at)) j + (y + sin(at)) k for the complementary function y' = -z x' = y so y''=-y therefore y = A cos t + B sin t and z = A sin t - B cos t Now for the particular integral, i...

thank you very much for your help

sorry I am completely stuck again, i think SLn(Fp) = |GLn(Fp)| / |Fp| which would give me \prod (p^{n(n-1)/2}(p-1)^n) / |Fp| ?

is it correct that for the determinant to be one in this upper triangular matrix that the diagonal entries must also be one? in this case will it be \prod (p ^ {i(i-1))/2}) for i = 1 to n

thank you for your patience! can i find the formula by multiplying the number of choices for each element in the matrix together, in which case i would have p^(n(n-1)/2)p^n choices for each matrix then the product of this over all matrices would be the formula. i hope I am not too wrong here...

only diagonal values can't be zero

so are there p-1 choices for each of them? which gives (p-1)^n choices along the diagonal?

they cannot be zero,

for an nxn matrix there will be n(n+1)/2 choices, so is it the product of i(i+1)/2 for i=1 to n?

sorry, all i can think is that you know the amount of zeros is \sum x! for x=1 to n totally irrelevant though. i really don't know how to do this?