MHB Applying the Chain Rule to Derive Solutions of the Heat Equation

[evinda]

Hello! (Wave)

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? (Thinking)

 

[I like Serena]

Hello! (Wave)

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? (Thinking)

Hey evinda! (Smile)

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 $u_{\mathbf x}(t,\mathbf x) = (u_{x_1}, u_{x_2}, ..., u_{x_n}) = \nabla u(t,\mathbf x)$. (Nerd)

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:
$$
\d u{\tilde t}= \pd {}t u(f({\tilde t}), {\mathbf g}({\tilde t})) \d {}{\tilde t} f({\tilde t})
+ \pd {}{\mathbf x}u(f({\tilde t}), {\mathbf g}({\tilde t})) \cdot \d {}{\tilde t} {\mathbf g}({\tilde t})
= u_t(f({\tilde t}),\mathbf g({\tilde t})) f_{\tilde t}(\tilde t) + u_{\mathbf x}(f({\tilde t}),\mathbf g({\tilde t})) \cdot \mathbf g_{\tilde t}(\tilde t)
$$
(Thinking)

 

[evinda]

I am a little confused now... (Sweating)

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

 

[I like Serena]

I am a little confused now... (Sweating)

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 $\mathbf x = \mathbf g=k \mathbf{\tilde x}$. (Thinking)

 

[evinda]

Properly distinguishing, we have $t = f=k^2 \tilde t$ and $\mathbf x = \mathbf g=k \mathbf{\tilde x}$. (Thinking)

And why do we find the derivative in respect to $\overline{t}$ and not in respect to $t$ ? (Thinking)

 

[I like Serena]

And why do we find the derivative in respect to $\overline{t}$ and not in respect to $t$ ? (Thinking)

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 $t=f(t)$. (Worried)

 

[evinda]

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 $u(\text{function of f =t}, \text{ function of } \mathbf{g}=\mathbf{x})$ ? (Sweating)

Could you explain it further to me?

 

[I like Serena]

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

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. (Thinking)

 

[evinda]

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... (Smile)

$$u_{k,\mathbf{\tilde x}}(\tilde t, \mathbf{\tilde x}) = u_{\mathbf x}(k^2 \tilde t, k\mathbf {\tilde x})k = \nabla u(k^2 \tilde t, k\mathbf {\tilde x})k \\
u_{k,\mathbf{\tilde x}\mathbf{\tilde x}}(\tilde t, \mathbf{\tilde x}) = u_{\mathbf {xx}}(k^2 \tilde t, k\mathbf {\tilde x})k^2 = \Delta u(k^2 \tilde t, k\mathbf {\tilde x})k^2
$$

How do we calculate $u_{k,\mathbf{\tilde x}}(\tilde t, \mathbf{\tilde x})$ given that $\mathbf{\tilde x}$ is a vector? (Thinking)

 

[I like Serena]

How do we calculate $u_{k,\mathbf{\tilde x}}(\tilde t, \mathbf{\tilde x})$ given that $\mathbf{\tilde x}$ is a vector? (Thinking)

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. (Nerd)

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

 

[evinda]

A ok. And so
$$\frac{\partial^2{u_k}}{\partial{\tilde{x_1}}^2}=k \frac{\partial}{\partial{\tilde{x_1}}} (u_{x_1}(k^2 \tilde{t}, k \overline{x_1}, \dots, k \overline{x_n}))=k (u_{x_1 x_1})(k^2 \tilde{t}, k \overline{x_1}, \dots, k \overline{x_n}))=k^2 (u_{x_1 x_1})(k^2 \tilde{t}, k \overline{x_1}, \dots, k \overline{x_n}))$$

And thus $\Delta u_k=\sum_{i=1}^n \frac{\partial^2 u_k}{\partial{\tilde{x_i}^2}}=k^2 [u_{x_1 x_1} (k^2 \tilde{t}, k \mathbf{x})+ \dots+ u_{x_n x_n}(k^2 \tilde{t}, k \tilde{x})]=k^2 \Delta u (k^2 \tilde{t}, k \tilde{x})$Right? (Thinking)

 

[I like Serena]

Yes (Smile)
... if overline is the same as tilde, if we add a missing factor $k$ the third expression, and if we balance the parentheses. (Nerd)

 

[evinda]

Yes (Smile)
... if overline is the same as tilde, if we add a missing factor $k$ the third expression, and if we balance the parentheses. (Nerd)

Yes, I see... Thanks a lot! (Smile)