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ymhiq
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Homework Statement
Hello, I have to demonstrate that multiplying a differential equation:
-d/dx[a(x)*d/dx{u(x)}]=f(x), 0<x<1 subject to u(0)=0 and u(1)=0.
by some function v(x) and integrating over an interval [0,1], I get a new equation that can be used in an optimisation problem, that equation is written as:
∫[a(x)*du/dx*dv/dy]dx=∫[f(x)*v(x)]dx, limits of integration of the integrals are the real numbers 0 and 1.
Homework Equations
The differential equation:
-d/dx[a(x)*d/dx{u(x)}]=f(x), 0<x<1 subject to u(0)=0 and u(1)=0.
∫[a(x)*du/dx*dv/dy]dx=∫[f(x)*v(x)]dx, limits of integration of the integrals are the real numbers 0 and 1.
The Attempt at a Solution
Firstly I did a little math:
-d[a(x)]/dx*d[u(x)]/dx-a(x)*d/dx[d{u(x)}/dx]=f(x)
Then I multiplied by v(x)
-v(x)*d[a(x)]/dx*d[u(x)]/dx-v(x)*a(x)*d/dx[d{u(x)}/dx]=v(x)*f(x)
Finally I integrated that over the limits of integration 0 and 1:
-∫v(x)*d[a(x)]/dx*d[u(x)]/dx-∫v(x)*a(x)*d/dx[d{u(x)}/dx]=∫v(x)*f(x)
RHS o this equation is equal to the RHS of the equation I have to demonstrate, however LHS is quite different. So I wrote to my teacher and he wrote me "you are almost there, the only thing you need is to apply the boundary conditions in order to nullify one of the terms of the LHS"; I don't know how. Besides nullifying the second term, the firs one is not exactly the same as in the equation I have to demonstrate.
Any help would be appreciated.
Thanks
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