# Thread: Show that the function is a solution of the equation

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 $v_{i x_i x_i} -v_{it}=0, i=1, \dots,n$ ?

2. Originally Posted by evinda
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?

Originally Posted by I like Serena
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 $\Delta v=(v_{1,x_1}v_2...v_n+v_{1,x_1}v_2...v_n,\quad v_1v_{2,x_2}...v_n+ v_1v_{2,x_2}...v_n,\quad ...,\quad v_1v_2...v_{n,x_n}+v_1v_2...v_{n,x_n})=(2v_{1,x_1}v_2...v_n,\quad 2v_1v_{2,x_2}...v_n,\quad ...,\quad 2v_1v_2...v_{n,x_n})$

?

4. Originally Posted by evinda
$\Delta v$ 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:
$$\nabla = (\pd {}{x_1},\quad \pd {}{x_2},\quad ...,\quad \pd {}{x_n})$$

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?

6. Originally Posted by evinda
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.

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?

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

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