Yes I think in that case it still would be zero. I think the main thing is the partial derivatives in the parenthesis. I'd imagine you would manipulate that somehow with the divergence theorem to make that parenthesis zero...That is my guess at least.
He said that the integral below equal to zero by an easy application of the divergence theorem.
\int_{\Omega }^{}f^{5}(f_{x}+2f_{y})=0
where f is some function
Can somebody explain how?
Hello, I'm back with another question.
I realize that I have no idea what the Duhamel principle is and what it is used for. Is there somebody who can clearly explain Duhamel's principle and maybe provide an example too?
Question, where did the n disappear to?
Ok so right now we have the left side of the inequality...somehow we need to convert the right to domain omega, and get constant a which depends on b. Does b/2 = a?
Ok well, I am going to come back to this tomorrow...it's already 3 am lol...I have another similar question with non-linear equations...but that won't take nearly as long.
good night!
I'm not sure if I did this right, but this is what I got so far and it looks like a mess and i did this assuming we're in R^3:
\int v(b_1\partial v_1/\partial x_1+b_2\partial v_2/\partial x_2+b_3\partial v_3/\partial x_3)=(vv_1b_1-\int v_1\partial v/\partial x)+(vv_2b_2-\int v_2\partial...
assuming that the above equation is true...I take equation A:
\int_{\Omega }^{}|\triangledown v|^{2}+v b \cdot \triangledown v+cv^{2} = 0
I then move things around and take the absolute value:
|\int_{\Omega }^{}b \cdot \triangledown vv| = \int_{\Omega }^{}|\triangledown v|^{2}+cv^{2}I plug it...
\int_{\Omega }^{}|\triangledown v|^{2}+b\triangledown vv+cv^{2} = 0
I tried to do integration by parts on b\triangledown vv
In hopes of trying to eliminate |\triangledown v|^{2}
But no luck...you mentioned the trace theorem. Is that necessary? I am not familiar with it.
ok, I'm a little confused of how to go from:
\int_{\Omega }^{}|\triangledown v|^{2}+b\triangledown vv+cv^{2} = 0
to
(c-\alpha)\int_{\Omega}^{}v^{2} \le -\int_{\Omega }^{}|\triangledown v|^{2}
since we're not exactly eliminating b anymore...I'm trying to work backworks and convert the...
Hmm. I thought that with the question saying that "If c is large enough compared to b", that was saying that c was so much larger, that we can just ignore the b part of the equation, which is why I set b = 0.
But now I think I see that I am considering only one case, and I need a more general...