# Thread: Application of chain rule

1. Hello!!!

Suppose that $u(t,x)$ is a solution of the heat equation $u_t-\Delta u=0$ in $(0,+\infty) \times \mathbb{R}^n$. I want to show that $u_k \equiv u(k^2 t, kx)$ is also a solution of the heat equation in $(0,+\infty) \times \mathbb{R}^n, \forall x \in \mathbb{R}^n$.

If we have a function $u(g,f)$ then the derivative in respect to $t$ is $\frac{\partial{u}}{\partial{g}} \frac{dg}{dt}+\frac{\partial{u}}{\partial{f}} \frac{df}{dt}$, right?

But then we would get that $\frac{\partial{u_k}}{\partial{t}}=k^2 \frac{\partial{u}}{\partial{k^2 t}}$.

But does the derivative $\frac{\partial{u}}{\partial{k^2 t}}$ make sense?

Or haven't I applied correctly the chain rule?

2. Originally Posted by evinda
Hello!!!

Suppose that $u(t,x)$ is a solution of the heat equation $u_t-\Delta u=0$ in $(0,+\infty) \times \mathbb{R}^n$. I want to show that $u_k \equiv u(k^2 t, kx)$ is also a solution of the heat equation in $(0,+\infty) \times \mathbb{R}^n, \forall x \in \mathbb{R}^n$.

If we have a function $u(g,f)$ then the derivative in respect to $t$ is $\frac{\partial{u}}{\partial{g}} \frac{dg}{dt}+\frac{\partial{u}}{\partial{f}} \frac{df}{dt}$, right?

But then we would get that $\frac{\partial{u_k}}{\partial{t}}=k^2 \frac{\partial{u}}{\partial{k^2 t}}$.

But does the derivative $\frac{\partial{u}}{\partial{k^2 t}}$ make sense?

Or haven't I applied correctly the chain rule?
Hey evinda!!

I'm afraid $t$ has 2 different meanings in this context, which is confusing.
Let's distinguish them by replacing one of them by $\tilde t$.
Oh, and $x$ is a vector in $\mathbb R^n$, so let's denote it by $\mathbf x$, and similarly denote $g$ by $\mathbf g$ to make sure we don't forget.
Also note that $u_{\mathbf x}(t,\mathbf x) = (u_{x_1}, u_{x_2}, ..., u_{x_n}) = \nabla u(t,\mathbf x)$.

So we have a function $u(f,\mathbf g)$ and we want the derivative with respect to $\tilde t$ of $u(f(\tilde t), \mathbf g(\tilde t))$.
Then we get:
$$\d u{\tilde t}=\pd uf \d f{\tilde t} + \pd u{\mathbf g} \cdot \d {\mathbf g}{\tilde t} = \pd {}f u(f({\tilde t}), {\mathbf g}({\tilde t})) \d {}{\tilde t} f({\tilde t}) + \pd {}{\mathbf g} u(f({\tilde t}), {\mathbf g}({\tilde t})) \cdot \d {}{\tilde t} {\mathbf g}({\tilde t})$$
In our case we have $u(t,\mathbf x)\equiv u(f,\mathbf g)$, so $f\equiv t$, and ${\mathbf g}\equiv {\mathbf x}$.
Substituting selectively (we can because there's no ambiguity in symbols any more), we get:
$$\d u{\tilde t}= \pd {}t u(f({\tilde t}), {\mathbf g}({\tilde t})) \d {}{\tilde t} f({\tilde t}) + \pd {}{\mathbf x}u(f({\tilde t}), {\mathbf g}({\tilde t})) \cdot \d {}{\tilde t} {\mathbf g}({\tilde t}) = u_t(f({\tilde t}),\mathbf g({\tilde t})) f_{\tilde t}(\tilde t) + u_{\mathbf x}(f({\tilde t}),\mathbf g({\tilde t})) \cdot \mathbf g_{\tilde t}(\tilde t)$$

3. Thread Author
I am a little confused now...

Don't we have in our case $f=k^2 t$ and $g=k \mathbf{x}$ ? Or am I wrong?

4. Originally Posted by evinda
I am a little confused now...

Don't we have in our case $f=k^2 t$ and $g=k \mathbf{x}$ ? Or am I wrong?
Properly distinguishing, we have $t = f=k^2 \tilde t$ and $\mathbf x = \mathbf g=k \mathbf{\tilde x}$.

5. Thread Author
Originally Posted by I like Serena
Properly distinguishing, we have $t = f=k^2 \tilde t$ and $\mathbf x = \mathbf g=k \mathbf{\tilde x}$.
And why do we find the derivative in respect to $\overline{t}$ and not in respect to $t$ ?

6. Originally Posted by evinda
And why do we find the derivative in respect to $\overline{t}$ and not in respect to $t$ ?
Because otherwise we're mixing up $t$ in $u(t,\mathbf x)$ and $\tilde t$ in $t=f(\tilde t)=k^2\tilde t$.

Since we introduced $t$ first as a parameter in $u(t,\mathbf x)$, I propose we don't introduce a new ambiguous $t$ that would have $t=f(t)$.

7. Thread Author
Originally Posted by I like Serena

So we have a function $u(f,\mathbf g)$ and we want the derivative with respect to $\tilde t$ of $u(f(\tilde t), \mathbf g(\tilde t))$.

In our case we have $u(t,\mathbf x)\equiv u(f,\mathbf g)$, so $f\equiv t$, and ${\mathbf g}\equiv {\mathbf x}$.
So we are given the function $u(t,\mathbf x)$. Don't we look for the derivate of a function of the form $u(\text{function of f =t}, \text{ function of } \mathbf{g}=\mathbf{x})$ ?

Could you explain it further to me?

8. Originally Posted by evinda
So we are given the function $u(t,\mathbf x)$. Don't we look for the derivate of a function of the form $u(\text{function of f =t}, \text{ function of } \mathbf{g}=\mathbf{x})$ ?

Could you explain it further to me?
We have:
$$u_k(\tilde t, \mathbf{\tilde x}) = u(k^2 \tilde t, k\mathbf {\tilde x})$$
From the chain rule:
$$u_{k,\tilde t}(\tilde t, \mathbf{\tilde x}) = u_t(k^2 \tilde t, k\mathbf {\tilde x}) k^2\\ u_{k,\mathbf{\tilde x}}(\tilde t, \mathbf{\tilde x}) = u_{\mathbf x}(k^2 \tilde t, k\mathbf {\tilde x})k = \nabla u(k^2 \tilde t, k\mathbf {\tilde x})k \\ u_{k,\mathbf{\tilde x}\mathbf{\tilde x}}(\tilde t, \mathbf{\tilde x}) = u_{\mathbf {xx}}(k^2 \tilde t, k\mathbf {\tilde x})k^2 = \Delta u(k^2 \tilde t, k\mathbf {\tilde x})k^2$$
So:
$$u_{k,\tilde t} - \Delta u_k = (u_{t} - \Delta u)k^2 = 0$$
Therefore $u_k$ is also a solution of the heat equation.

9. Thread Author
Originally Posted by I like Serena
We have:
$$u_k(\tilde t, \mathbf{\tilde x}) = u(k^2 \tilde t, k\mathbf {\tilde x})$$
From the chain rule:
$$u_{k,\tilde t}(\tilde t, \mathbf{\tilde x}) = u_t(k^2 \tilde t, k\mathbf {\tilde x}) k^2$$
Ah I see...

Originally Posted by I like Serena
$$u_{k,\mathbf{\tilde x}}(\tilde t, \mathbf{\tilde x}) = u_{\mathbf x}(k^2 \tilde t, k\mathbf {\tilde x})k = \nabla u(k^2 \tilde t, k\mathbf {\tilde x})k \\ u_{k,\mathbf{\tilde x}\mathbf{\tilde x}}(\tilde t, \mathbf{\tilde x}) = u_{\mathbf {xx}}(k^2 \tilde t, k\mathbf {\tilde x})k^2 = \Delta u(k^2 \tilde t, k\mathbf {\tilde x})k^2$$
How do we calculate $u_{k,\mathbf{\tilde x}}(\tilde t, \mathbf{\tilde x})$ given that $\mathbf{\tilde x}$ is a vector?

10. Originally Posted by evinda
How do we calculate $u_{k,\mathbf{\tilde x}}(\tilde t, \mathbf{\tilde x})$ given that $\mathbf{\tilde x}$ is a vector?
Generally, we have:
$$u_{\mathbf x} = \nabla u = (u_{x_1}, u_{x_2}, ..., u_{x_n})$$
So we should evaluate the derivative component for component.

In our case the derivative with respect to the first component is:
$$\pd {u_k}{\tilde x_1} = u_{k,\tilde x_1}(\tilde t, \tilde x_1, ..., \tilde x_n) = \pd {}{\tilde x_1}u(k^2\tilde t, k\tilde x_1, ..., k\tilde x_n) = u_{x_1}(k^2\tilde t, k\tilde x_1, ..., k\tilde x_n)\pd{}{\tilde x_1}(k\tilde x_1)$$

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•