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    #1
    Hello!!!

    I want to show that if $v_i(t,x_i), i=1, \dots, n$ are solutions of the equations $v_{i x_i x_i} -v_{it}=0, i=1, \dots,n$ respectively, then $v=v_1 v_2 \cdots v_n(v=v(t,x))$ is a solution of $\Delta v-v_t=0$.

    That's what I have done so far.

    $$v_x=(v_{1x} v_2 \cdots v_n)+(v_1 v_{2x} \cdots v_n)+ \dots+ (v_1 v_2 \cdots v_{nx})$$

    $$v_{xx}=(v_{1xx} v_2 \cdots v_n+ v_{1x} v_{2x} \cdots v_{n}+ \dots+ v_{1x} v_2 \cdots v_{nx})+ \dots+ (v_{1x} v_2 \cdots v_{nx}+ \dots+ v_1 v_2 \cdots v_{nxx} )$$

    So do we have to include now at $v_{xx}$ the sum $\sum_{i=1}^n v_{i x_i x_i}$?

    How can we use the fact that $v_{i x_i x_i} -v_{it}=0, i=1, \dots,n$ ?

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    #2
    Quote Originally Posted by evinda View Post
    Hello!!!

    I want to show that if $v_i(t,x_i), i=1, \dots, n$ are solutions of the equations $v_{i x_i x_i} -v_{it}=0, i=1, \dots,n$ respectively, then $v=v_1 v_2 \cdots v_n(v=v(t,x))$ is a solution of $\Delta v-v_t=0$.

    That's what I have done so far.

    $$v_x=(v_{1x} v_2 \cdots v_n)+(v_1 v_{2x} \cdots v_n)+ \dots+ (v_1 v_2 \cdots v_{nx})$$

    $$v_{xx}=(v_{1xx} v_2 \cdots v_n+ v_{1x} v_{2x} \cdots v_{n}+ \dots+ v_{1x} v_2 \cdots v_{nx})+ \dots+ (v_{1x} v_2 \cdots v_{nx}+ \dots+ v_1 v_2 \cdots v_{nxx} )$$

    So do we have to include now at $v_{xx}$ the sum $\sum_{i=1}^n v_{i x_i x_i}$?

    How can we use the fact that $v_{i x_i x_i} -v_{it}=0, i=1, \dots,n$ ?
    Hey evinda!!

    I think we have:
    $$v(t,\mathbf x) = v(t, x_1, x_2, ..., x_n) = v_1(t, x_1)v_2(t,x_2)...v_n(t,x_n)$$
    So:
    $$v_{x_1} = v_{1,x_1}v_2...v_n$$
    And:
    $$\nabla v = v_{\mathbf x} = (v_{1,x_1}v_2...v_n,\quad v_1v_{2,x_2}...v_n,\quad ...,\quad v_1v_2...v_{n,x_n})$$

    What would $\Delta v$ be?

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    #3 Thread Author
    Quote Originally Posted by I like Serena View Post
    Hey evinda!!

    I think we have:
    $$v(t,\mathbf x) = v(t, x_1, x_2, ..., x_n) = v_1(t, x_1)v_2(t,x_2)...v_n(t,x_n)$$
    So:
    $$v_{x_1} = v_{1,x_1}v_2...v_n$$
    And:
    $$\nabla v = v_{\mathbf x} = (v_{1,x_1}v_2...v_n,\quad v_1v_{2,x_2}...v_n,\quad ...,\quad v_1v_2...v_{n,x_n})$$

    What would $\Delta v$ be?
    A ok... $\Delta v$ is equal to $\nabla v \cdot \nabla v$, right?

    So does it hold that $\Delta v=(v_{1,x_1}v_2...v_n+v_{1,x_1}v_2...v_n,\quad v_1v_{2,x_2}...v_n+ v_1v_{2,x_2}...v_n,\quad ...,\quad v_1v_2...v_{n,x_n}+v_1v_2...v_{n,x_n})=(2v_{1,x_1}v_2...v_n,\quad 2v_1v_{2,x_2}...v_n,\quad ...,\quad 2v_1v_2...v_{n,x_n})$

    ?

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    #4
    Quote Originally Posted by evinda View Post
    $\Delta v$ is equal to $\nabla v \cdot \nabla v$, right?

    So does it hold that $\Delta v=(v_{1,x_1}v_2...v_n+v_{1,x_1}v_2...v_n,\quad v_1v_{2,x_2}...v_n+ v_1v_{2,x_2}...v_n,\quad ...,\quad v_1v_2...v_{n,x_n}+v_1v_2...v_{n,x_n})=(2v_{1,x_1}v_2...v_n,\quad 2v_1v_{2,x_2}...v_n,\quad ...,\quad 2v_1v_2...v_{n,x_n})$

    ?
    Not quite. $\Delta v = \nabla \cdot \nabla v$.
    That is:
    $$\nabla v = (v_{x_1}, ..., v_{x_n}) \\
    \Delta v = \nabla \cdot \nabla v = v_{x_1x_1} + ... + v_{x_nx_n}
    $$

    We can think of $\nabla$ as:
    $$\nabla = (\pd {}{x_1},\quad \pd {}{x_2},\quad ...,\quad \pd {}{x_n})$$

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    #5 Thread Author
    So in our case we have $$\Delta v= v_{1 x_1, x_1} v_2 \cdots v_n+ v_1 v_{2 x_2,x_2} \cdots v_n+ \dots+ v_1 v_2 \cdots v_{n x_n x_n}$$

    Right?

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    #6
    Quote Originally Posted by evinda View Post
    So in our case we have $$\Delta v= v_{1 x_1, x_1} v_2 \cdots v_n+ v_1 v_{2 x_2,x_2} \cdots v_n+ \dots+ v_1 v_2 \cdots v_{n x_n x_n}$$

    Right?
    Yep.

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    #7 Thread Author
    Nice... And then we have that

    $$\Delta v-v_t=(v_{1 x_1 x_1}-v_{1,t}) v_2 \cdots v_n+ \dots+ (v_{nx_n x_n}-v_{n,t}) v_1 \cdots v_{n-1}=0$$

    Right?
    Last edited by evinda; December 3rd, 2016 at 18:56.

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    #8
    If we put the $x_n$ indices lower, yes.

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    #9 Thread Author
    Quote Originally Posted by I like Serena View Post
    If we put the $x_n$ indices lower, yes.
    You are right, I edited my post... Thank you!!!!

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