- #1

Somefantastik

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Is there a pictorial way to expound on this for me?

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- #1

Somefantastik

- 230

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Is there a pictorial way to expound on this for me?

- #2

Defconist

- 7

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"Write the equation in the form Lu = 0,

where L is an operator. That is, if v is any function, Lv is a new function. For

instance,L = d/dx is the operator that takes v into its derivative dv/dx.

The definition we want for linearity is: L(u + v) = Lu + Lv and L(cu) = c*Lu

for any functions u, v and any constant c."

Is PDE u_x + u_yy = exp(x^2) linear ? Yes, because (u+v)_x + (u+v)_yy = u_x + v_x + u_yy + v_yy and what about (u_x)^2 = 0 ? No, it is not linear because (u_x + v_x)^2 is NOT equal to (u_x)^2 + (u_y)^2

First one of theese was not homogenous because the right side is 0 the other is though.

That homogenous part is some function of independent variables free from derivatives.

cheers,

Adam

- #3

Somefantastik

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I have Strauss's PDE book for next semester. I've looked through it and he looks pretty clear and concise.

Thanks again, it really helped.

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