Intgeration by parts function

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Mark Sc
Mark Sc 2021-1-31
回答： Teddy Negrete Peña 2021-2-2
Hi all,
I am new to the function of intgeration by parts, I would like to display and answer the following equation:
∫▒〖w (∂^2 u)/∂x+(∂^2 u)/∂y+(∂^2 u)/∂z dx〗
int(w(d^2u/dx+d^2u/dy+d^2u/dz)dx)=0
can anyone help me how to write this and use the function of intgeration by parts
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Walter Roberson
Walter Roberson 2021-1-31
MATLAB automatically does integration by parts when it can figure out the pattern using the content already present in expressions. It will not create new unknown function U to differentiate that just happens to have the magic propery that int(f,x) is 0. You have to supply that creativity.
You can simplify a bit by knowing that if w is a constant, then int(w*G(x,y,z)) = w * int(G(x,y,z)) and with the integral equalling 0 then int(G(x,y,z)) must equal 0, where G(x,y,z) = diff(u,x,x) + diff(u,y,y) + diff(u,z,z). So the w is a red herring, of no use at all... not unless I was right the first time and the w is only multiplying diff(u,x,x) rather than the whole sum.
Suppose U = c1*x^a1 + c2*y^a2 + c3*y^a3 then diff(U,x) = a1*c1*x^(a1-1), diff(U,x,x) = (a1-1)*a1*c1*x^(a1-2) and by symmetry diff(U,y,y) = (a2-1)*a2*c2*y^(a2-2) and diff(U,z,z) = (a3-1)*a3*c3*z^(a3-2) . Sum of those is
(a1-1)*a1*c1*x^(a1-2) + (a2-1)*a2*c2*y^(a2-2) + (a3-1)*a3*c3*z^(a3-2)
Integrate with respect to x:
a1*c1*x^(a1-1) + x*(a2-1)*a2*c2*y^(a2-2) + x*(a3-1)*a3*c3*z^(a3-2)
so
x*(a1*c1*x^(a1-2) + (a2-1)*a2*c2*y^(a2-2) + (a3-1)*a3*c3*z^(a3-2)) = 0
But that has to be true for all x, y, z, not just x = 0, so the the x^(a1-2), y^(a2-2), z^(a3-2) have to vanish. So let a1=2, a2=2, a3=2, getting
2*c1 + (2-1)*2*c2 + (2-1)*2*c3 = 0
c1 = -c2-c3
But this is only one of the possibilities, and does not take advantage of integration by parts.
Mark Sc
Mark Sc 2021-2-2
Hi,
thanks for your explanation

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回答（1 个）

Teddy Negrete Peña
Best regards Mark Sc,
Indeed you can use Matlab to check what you do analytically by hand, but before using Matlab functions, you must interpret that the "int" function can analytically solve an integral, but it must pose it one by one and use symbolic variables.
When you apply Integration by parts you must solve 2 integrals, the integral of dv and the integral vdu, then those integrals must be raised in the code.
Sincerely,
Teddy

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