Intgeration by parts function

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

[Walter Roberson]

In the top line of formula, the w seems to be only multiplying the first partial derivative, but in your int(), the w seems to be multiplying the sum of the partial derivatives. Which is correct?

[Mark Sc]

Hi,

Thanks for your comment,

The correct is:

w is multiplying the sum of the partial derivatives.

looking to hear your suggestion

[Walter Roberson]

syms u(x,y,z) w
f = w * (diff(u,x,x) + diff(u,y,y) + diff(u,z,z))

f(x, y, z) = 

int(f, x)

ans(x, y, z) = 

But you have to do the integration by parts manually.

[Mark Sc]

Hi,

Thanks for your reply.

I understand, but I thought this function solve integration by parts,

honsetly, I know integration by parts should be done manually, i thought, we could do that for verfications,

Looking forward to hearing your ideas!!.

Thank you,

[Walter Roberson]

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]

Hi,

thanks for your explanation

Answer 1

[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