Can we Use Partial Derivatives to Verify the Solution of PDE with Derivatives?

[mathmari]

Hey!

I want to verify that $$w(x,t)=\frac{1}{2c}\int_0^t\int_{c(t-\tau)-x}^{x+c(t-\tau)}f(y,\tau)dyd\tau$$ is the solution of the problem $$w_{tt}=c^2w_{xx}+f(x,t) , \ \ x>0, t>0 \\ w(x,0)=w_t(x,0)=0, \ \ x>0 \\ w(0,t)=0 , \ \ t\geq 0$$ For that we have to take the partial derivatives of $w$. But how can we do that in this case for example as for $t$ where we have at both integrals the $t$ ? Could you give me a hint? (Wondering)

 

[I like Serena]

Hey!

I want to verify that $$w(x,t)=\frac{1}{2c}\int_0^t\int_{c(t-\tau)-x}^{x+c(t-\tau)}f(y,\tau)dyd\tau$$ is the solution of the problem $$w_{tt}=c^2w_{xx}+f(x,t) , \ \ x>0, t>0 \\ w(x,0)=w_t(x,0)=0, \ \ x>0 \\ w(0,t)=0 , \ \ t\geq 0$$ For that we have to take the partial derivatives of $w$. But how can we do that in this case for example as for $t$ where we have at both integrals the $t$ ? Could you give me a hint? (Wondering)

Hey mathmari! (Wave)

How about defining $g(x,t,\tau) = \int_{c(t-\tau)-x}^{x+c(t-\tau)}f(y,\tau)dy$, and then differentiating one integral at a time using Leibniz's integral rule? (Wondering)

 

[mathmari]

How about defining $g(x,t,\tau) = \int_{c(t-\tau)-x}^{x+c(t-\tau)}f(y,\tau)dy$, and then differentiating one integral at a time using Leibniz's integral rule? (Wondering)

We have the following partial derivatives, or not?
\begin{align*}w_{t}&=\frac{\partial}{\partial{t}}\left [\frac{1}{2c}\int_0^tg(x,t,\tau )d\tau\right ]=\frac{1}{2c}g(x,t,t)+\frac{1}{2c}\int_0^tg_t(x,t,\tau )d\tau\\ & =\frac{1}{2c}\int_{-x}^xf(y,\tau)dy+\frac{1}{2c}\int_0^tg_t(x,t,\tau )d\tau\end{align*}
\begin{align*}w_{tt}&=\frac{1}{2c}\int_{-x}^xf_t(y,\tau)dy+\frac{1}{2c}\frac{\partial}{\partial{t}}\int_0^tg_t(x,t,\tau )d\tau\\ & =\frac{1}{2c}\int_{-x}^xf_t(y,\tau)dy+\frac{1}{2c}g_t(x,t,t)+\frac{1}{2c}\int_0^tg_{tt}(x,t,\tau )d\tau\end{align*}

\begin{align*}&w_x=\frac{\partial}{\partial{x}}\left [\frac{1}{2c}\int_0^tg(x,t,\tau )d\tau\right ]=\frac{1}{2c}\int_0^tg_x(x,t,\tau )d\tau \\ &w_{xx}=\frac{\partial}{\partial{x}}\left [\frac{1}{2c}\int_0^tg_x(x,t,\tau )d\tau\right ]=\frac{1}{2c}\int_0^tg_{xx}(x,t,\tau )d\tau \end{align*}

Is everythig correct so far? Now we have to calculate the derivaties of $g$, right?

 

[I like Serena]

I think it should be $f(y,t)$ and $f_t(y,t)$, shouldn't it?
Otherwise I believe it's all correct. (Happy)

 

[mathmari]

I think it should be $f(y,t)$ and $f_t(y,t)$, shouldn't it?
Otherwise I believe it's all correct. (Happy)

Ah ok! We have the following partial derivatives of $g$, or not?
\begin{align*}g_t&=\frac{\partial}{\partial{t}}\int_{c(t-\tau)-x}^{x+c(t-\tau)}f(y,\tau )dy\\ & =c\cdot f(x+c(t-\tau),\tau)-c\cdot f(c(t-\tau)-x, \tau)+\int_{c(t-\tau)-x}^{x+c(t-\tau)}f_t(y,\tau )dy\end{align*}

\begin{align*}g_{tt}&=c\cdot f_x(x+c(t-\tau),\tau)\cdot \frac{d}{dt}[x+c(t-\tau)]-c\cdot f_x(c(t-\tau)-x, \tau)\cdot \frac{d}{dt}[c(t-\tau)-x]+\frac{\partial}{\partial{t}}\int_{c(t-\tau)-x}^{x+c(t-\tau)}f_t(y,\tau )dy\\ & =c^2\cdot f_x(x+c(t-\tau),\tau)-c^2\cdot f_x(c(t-\tau)-x, \tau)+f_t(x+c(t-\tau),\tau )\cdot c-f(c(t-\tau)-x, \tau)\cdot c+\int_{c(t-\tau)-x}^{x+c(t-\tau)}f_t(y,\tau)dy\end{align*}

\begin{align*}g_x&=\frac{\partial}{\partial{x}}\int_{c(t-\tau)-x}^{x+c(t-\tau)}f(y,\tau )dy\\ & = f(x+c(t-\tau),\tau)-(-1)\cdot f(c(t-\tau)-x, \tau)+\int_{c(t-\tau)-x}^{x+c(t-\tau)}f_x(y,\tau )dy\\ & = f(x+c(t-\tau),\tau)+ f(c(t-\tau)-x, \tau)+\int_{c(t-\tau)-x}^{x+c(t-\tau)}f_x(y,\tau )dy\end{align*}

\begin{align*}g_{xx}&= f_x(x+c(t-\tau),\tau)+ f_x(c(t-\tau)-x, \tau)\cdot (-1)+\frac{\partial}{\partial{t}}\int_{c(t-\tau)-x}^{x+c(t-\tau)}f_x(y,\tau )dy\\ & = f_x(x+c(t-\tau),\tau)- f_x(c(t-\tau)-x, \tau)+f_x(x+c(t-\tau),\tau )-f_x(c(t-\tau)-x, \tau)+\int_{c(t-\tau)-x}^{x+c(t-\tau)}f_{xx}(y,\tau )dy \\ & = 2f_x(x+c(t-\tau),\tau)- 2f_x(c(t-\tau)-x, \tau)+\int_{c(t-\tau)-x}^{x+c(t-\tau)}f_{xx}(y,\tau )dy\end{align*}

(Wondering)

 

[mathmari]

I must have some mistakes...because if these derivative were correct, we would get

\begin{align*}w_{xx}&=\frac{1}{2c}\int_0^tg_{xx}(x,t,\tau)d\tau\\ & =\frac{1}{2c}\int_0^t\left [2f_x(x+c(t-\tau),\tau)- 2f_x(c(t-\tau)-x, \tau)+\int_{c(t-\tau)-x}^{x+c(t-\tau)}f_{xx}(y,\tau )dy\right ]d\tau\end{align*}

\begin{align*}w_{tt}&=\frac{1}{2c}\int_{-x}^xf_t(y,t)dy+\frac{1}{2c}g_t(x,t,t)+\frac{1}{2c}\int_0^tg_{tt}(x,t,\tau )d\tau \\ & = \frac{1}{2c}\int_{-x}^xf_t(y,t)dy+\frac{1}{2c}\left [c\cdot f(x,t)-c\cdot f(-x, t)+\int_{-x}^{x}f_t(y,t )dy\right ]+\frac{1}{2c}\int_0^t\left [c^2\cdot f_x(x+c(t-\tau),\tau)-c^2\cdot f_x(c(t-\tau)-x, \tau)+f_t(x+c(t-\tau),\tau )\cdot c-f(c(t-\tau)-x, \tau)\cdot c+\int_{c(t-\tau)-x}^{x+c(t-\tau)}f_t(y,\tau)dy\right ]d\tau \\ & = \frac{1}{c}\int_{-x}^xf_t(y,t)dy+\frac{f(x,t)-f(-x,t)}{2}+\frac{1}{2c}\int_0^t\left [c^2\cdot f_x(x+c(t-\tau),\tau)-c^2\cdot f_x(c(t-\tau)-x, \tau)+f_t(x+c(t-\tau),\tau )\cdot c-f(c(t-\tau)-x, \tau)\cdot c+\int_{c(t-\tau)-x}^{x+c(t-\tau)}f_t(y,\tau)dy\right ]d\tau\end{align*}

But these don't satisfy the problem, do they? (Wondering)

 

[mathmari]

I am trying it again.

I have done the following:

\begin{equation*}w(x,t)=\frac{1}{2c}\int_0^tg(x,t,\tau )d\tau \end{equation*}

\begin{align*}w_t&=\frac{1}{2c}g(x,t,t)+\frac{1}{2c}\int_0^tg_t(x,t,\tau )d\tau \\
& =\frac{1}{2c}g(x,t,t)+\frac{1}{2c}\int_0^t\frac{\partial}{\partial{t}}\int_{c(t-\tau)-x}^{x+c(t-\tau)}f(y,\tau )dyd\tau\\ & = \frac{1}{2c}\int_{-x}^xf(y,t)dy+\frac{1}{2c}\int_0^t\left [f(x+c(t-\tau ),\tau)\cdot c-f(c(t-\tau )-x,\tau)\cdot c+\int_{c(t-\tau)-x}^{x+c(t-\tau)}f_t(y,\tau )dy\right ]d\tau \\ & = \frac{1}{2c}\int_{-x}^xf(y,t)dy+\frac{1}{2}\int_0^tf(x+c(t-\tau ),\tau)d\tau-\frac{1}{2}\int_0^tf(c(t-\tau )-x,\tau)d\tau+\frac{1}{2c}\int_0^t\int_{c(t-\tau)-x}^{x+c(t-\tau)}f_t(y,\tau )dyd\tau\end{align*}

is the first derivative of $w$ as for $t$ correct? (Wondering)

 

[I like Serena]

I am trying it again.

I have done the following:

\begin{equation*}w(x,t)=\frac{1}{2c}\int_0^tg(x,t,\tau )d\tau \end{equation*}

\begin{align*}w_t&=\frac{1}{2c}g(x,t,t)+\frac{1}{2c}\int_0^tg_t(x,t,\tau )d\tau \\
& =\frac{1}{2c}g(x,t,t)+\frac{1}{2c}\int_0^t\frac{\partial}{\partial{t}}\int_{c(t-\tau)-x}^{x+c(t-\tau)}f(y,\tau )dyd\tau\\ & = \frac{1}{2c}\int_{-x}^xf(y,t)dy+\frac{1}{2c}\int_0^t\left [f(x+c(t-\tau ),\tau)\cdot c-f(c(t-\tau )-x,\tau)\cdot c+\int_{c(t-\tau)-x}^{x+c(t-\tau)}f_t(y,\tau )dy\right ]d\tau \\ & = \frac{1}{2c}\int_{-x}^xf(y,t)dy+\frac{1}{2}\int_0^tf(x+c(t-\tau ),\tau)d\tau-\frac{1}{2}\int_0^tf(c(t-\tau )-x,\tau)d\tau+\frac{1}{2c}\int_0^t\int_{c(t-\tau)-x}^{x+c(t-\tau)}f_t(y,\tau )dyd\tau\end{align*}

is the first derivative of $w$ as for $t$ correct? (Wondering)

I think this is correct yes. (Nod)

And we can simplify it bit more since $f(y,\tau)$ does not depend on $t$.
Therefore $\pd{}t f(y,\tau)=0$. (Thinking)

 

[mathmari]

I think this is correct yes. (Nod)

And we can simplify it bit more since $f(y,\tau)$ does not depend on $t$.
Therefore $\pd{}t f(y,\tau)=0$. (Thinking)

Ok! At the next step we have \begin{align*}w_{tt}&= \frac{1}{2c}\int_{-x}^xf_t(y,t)dy+\frac{1}{2}f(x,\tau)+\frac{1}{2}\int_0^t\frac{\partial}{\partial{t}}f(x+c(t-\tau ),\tau)d\tau-\frac{1}{2}f(-x,\tau)-\frac{1}{2}\int_0^t\frac{\partial}{\partial{t}}f(c(t-\tau )-x,\tau)d\tau\end{align*} Is the following correct? \begin{equation*}\frac{\partial}{\partial{t}}f(x+c(t-\tau ),\tau)=f_x(x+c(t-\tau ),\tau)\cdot \frac{d(x+c(t-\tau)}{dt}+f_t(x+c(t-\tau ),\tau)\cdot \frac{d\tau}{dt}\end{equation*} Or do we not use here the chain rule? (Wondering)

 

[mathmari]

If this is correct then we get \begin{align*}w_{tt}&= \frac{1}{2c}\int_{-x}^xf_t(y,t)dy+\frac{1}{2}f(x,t)+\frac{c}{2}\int_0^tf_x(x+c(t-\tau ),\tau)d\tau-\frac{1}{2}f(-x,t)-\frac{c}{2}\int_0^tf_x(c(t-\tau )-x,\tau)d\tau\end{align*}

The second derivative as for $x$ (if I have no mistakes) is \begin{align*}w_{xx}&=\frac{1}{2c}\int_0^tf_x(x+c(t-\tau ),\tau)d\tau-\frac{1}{2c}\int_0^tf_x(c(t-\tau )-x,\tau)d\tau\end{align*}

At the $w_{tt}$ is it correct that we have once $f(x,t)$ and once $f(-x,t)$ ? (Wondering)

 

[I like Serena]

Is the following correct? \begin{equation*}\frac{\partial}{\partial{t}}f(x+c(t-\tau ),\tau)=f_x(x+c(t-\tau ),\tau)\cdot \frac{d(x+c(t-\tau)}{dt}+f_t(x+c(t-\tau ),\tau)\cdot \frac{d\tau}{dt}\end{equation*} Or do we not use here the chain rule? (Wondering)

I believe that it should be:
\begin{align*}\frac{\partial}{\partial{t}}f(x+c(t-\tau ),\tau)&=f_x(x+c(t-\tau ),\tau)\cdot \frac{\partial(x+c(t-\tau))}{\partial t}+f_t(x+c(t-\tau ),\tau)\cdot \frac{\partial\tau}{\partial t} \\
&=f_x(x+c(t-\tau ),\tau)\cdot c
\end{align*}
shouldn't it? (Wondering)

 

[mathmari]

I believe that it should be:
\begin{align*}\frac{\partial}{\partial{t}}f(x+c(t-\tau ),\tau)&=f_x(x+c(t-\tau ),\tau)\cdot \frac{\partial(x+c(t-\tau))}{\partial t}+f_t(x+c(t-\tau ),\tau)\cdot \frac{\partial\tau}{\partial t} \\
&=f_x(x+c(t-\tau ),\tau)\cdot c
\end{align*}
shouldn't it? (Wondering)

Ah ok, with partial derivatives not just derivatives. But the result is the same as mine!

I am not really sure about the terms $f(x,t)$ and $f(-x,t)$ also about the term $\frac{1}{2c}\int_{-x}^xf_t(y,t)dy$.

If that integral would we $0$ and if we had twice $f(x,t)$ instead of once $f(x,t)$ and once $f(-x,t)$, then $w$ would satisfy the problem, wouldn't it?

Do we maybe use the fact that at the problem it is $x>0$ ?

(Wondering)

 

[mathmari]

Are maybe the limits of the limits wrong? These limits should describe the following set:

View attachment 8160

Are the limits that I used at the integral of $w(x,t)$ correct? (Wondering)

 

[I like Serena]

I believe you've already found that it should be:
$$w(x,t)=\frac{1}{2c}\int_0^{t-\frac{x}{c}}\int_{c(t-\tau)-x}^{x+c(t-\tau)}f(y,\tau)dyd\tau+\frac{1}{2c}\int_{t-\frac{x}{c}}^t\int_{x-c(t-\tau)}^{x+c(t-\tau)}f(y,\tau)dyd\tau$$
It also means that some of the steps are not correct yet. (Thinking)