Solving PDE Problem: du/dt+du/dx=0 with Initial Condition u(x,0)=xe-x2

  • Thread starter Thread starter pentazoid
  • Start date Start date
  • Tags Tags
    Pde
pentazoid
Messages
142
Reaction score
0

Homework Statement



which solutions of du/dt+du/dx=0 is equal to xe-x2

Homework Equations


The Attempt at a Solution



u(x,0) = xe-x2
u(x,t)= (x-t)e(-x-t)2

what else do i need to do?
 
Physics news on Phys.org
pentazoid said:

Homework Statement



which solutions of du/dt+du/dx=0 is equal to xe-x2

Homework Equations





The Attempt at a Solution



u(x,0) = xe-x2
u(x,t)= (x-t)e(-x-t)2

what else do i need to do?
One way of solving a pde like that is to always try separation of variables , so that you assumeu(x,t)=X(x)T(t)where X is just a function of x and T just a function of t, plug this into your pde and see what u get!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top