Originally Posted by

**angels_symphony** Hi, I just had a few questions on PDE's just to make sure I'm doing thigs right.

Question 1

Given a function f(x,y) is the integration with respect to one variable (say x) of zero c + g(y) where c is a constant?

Mr F says: I can't quite make sense of your question ... I'll answer what I think you're asking: If you integrate f(x, y) wrt x, y is treated as a constant. The arbitrary constant of integration will therefore be a function of y, g(y) say (because the partial derivative of g(y) wrt x is zero). The familiar 'C' is contained in g(y).

Eg. $\displaystyle f(x, y) = xy$. $\displaystyle \int f(x, y) \, dx = \int xy \, dx = \frac{yx^2}{2} + g(y)$. Depending on the boundary conditions, it may be that $\displaystyle g(y) = y^3 + y + 1$, say.

Question 2

How woul you solve a question like u_y +u_xy = 0? I was thinking maybe you could devide by u_y but that seems way to easy...

Mr F says: I'd re-write it as $\displaystyle \frac{\partial}{\partial y} \left( u + \frac{\partial u}{\partial x} \right) = 0$. Then, integrating wrt y (after having re-read the above), $\displaystyle u + \frac{\partial u}{\partial x} = g(x)$. So you're now, effectively, solving the ODE $\displaystyle \frac{du}{dx} + u = g(x)$ since u will be totally a function of x. btw You don't know what g(x) is but you might get it from the boundary conditions. (btw the integrating factor method can be used [in principle if g(x) is not known] to solve this ODE).

Question 3

After finding the u(x,t) = F(x+ct) where c is a constant is the solution to a PDE, how would you apply the initial conditions? Am I looking to find a value for c?

Mr F says: No. This is 'half' of the general solution to the wave equation. The wave equation will already have the value of c in it .... The initial conditions are used to get the functional form of F .... I strongly suggest you research **D'Alembert's solution of the wave equation**.

Thank you! Mr F says: You're welcome.