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This is my first time to study PDE in grad level . I have to solve this IVP :

$$ u_t + u_x = (x+t) \cos(xt) \tag{*} $$ on $\mathbb{R} \times (0,\infty)$.

$$ u(x,0) = \sin(x) \tag{**} $$ on $(x,t) \in \mathbb{R} \times \{t=0\}$.

In my class note, I was given a formula to use , and I obtain:

$$ u(x,t) = \sin(x-tb) + \int\limits_0^t \left[x-(t-\sigma)b+\sigma\right] \cos\left(\left[x-(t-\sigma)b\right]\sigma\right) \, d\sigma $$

where $b$ is a constant.

I don't know what I need to do next. Is this one correct?

Peter
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    You remember that differentiating $\sin k$ gives $\cos k$? So the first thing to try is what happens if you put $u(x,t)=\sin xt$. That should lead you to a particular solution. The equation is linear, so if you find $u(x,t)$ such that $u_x+u_t=0$ you can add that to the particular solution. You need to juggle to satisfy the initial condition. – almagest Aug 31 '14 at 15:56
  • Should that be a $+\sigma$ in the cosine in the integrand? – ClassicStyle Aug 31 '14 at 16:03
  • The formular u(x,t) above I have is from what you said, it looks like a solution to me. – Peter Aug 31 '14 at 16:03

3 Answers3

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Why to learn that complicated formula? Will you even remember it when taking your exam? Why not using the method of characteristics, which I think it should be firstly taught when learning PDEs. In your case, it reads:

$$\frac{\mathrm{d}t}{1} = \frac{\mathrm{d}x}{1} = \frac{\mathrm{d}u}{(x+t) \cos{xt}}. $$

From the first equatlity we have: $\mathrm{d}t - \mathrm{d}x = 0, $ which after integrating yields $x-t=c$, where $c$ is a constant called characteristic. Take now the second equality and put $x$ as a function of $t$ to have:

$$\mathrm{d}t = \frac{\mathrm{du}}{(2t+c) \cos{[(t+c)t}]},$$

whichi is a separable differential equation. Solve the corresponding integral$^*$ and put the constant of integration as a function of $c = x-t$, solve for the given initial condition and you are done!

Cheers!

$^*$Note that the argument of the $\sin$ is a primitive of $2t+c$.

Dmoreno
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  • My professor taught that, so I just looked at the note. – Peter Aug 31 '14 at 16:31
  • @Peter, yeah, I'm aware. I was just wondering why your teacher did not explain to you this way to solve these problems! You may find your formula just came out of the blue. – Dmoreno Aug 31 '14 at 16:33
  • That solution did not fell out of the sky, there is a method for this. I have the name lying on my tongue. Heck what was it? – mvw Aug 31 '14 at 16:37
  • mmmm, nothing comes to my mind right now... It must be some characteristic or maybe specific method for this kind of problems... – Dmoreno Aug 31 '14 at 16:40
  • Hello again! I'm having trouble understanding the last part where the integration factor is a function of $c$. Why/how does this happen. I see how it works, but not really understanding it – ClassicStyle Aug 31 '14 at 16:41
  • I think the particular solution might be obtained using Duhamel's principle. – mvw Aug 31 '14 at 16:42
  • @TylerHG. Take a look at the example at: http://en.wikipedia.org/wiki/Method_of_characteristics#Example. You can always equate the Lagrange-Charpit equations to some $d\lambda$, where $\lambda$ is an auxiliar variable. – Dmoreno Aug 31 '14 at 16:51
  • @mvw : Yes Duhamel 's principle. My professor proved the method in class for general case, and gave out the formula. – Peter Aug 31 '14 at 16:56
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Just checking if your solution candidate fullfills the PDE $(*)$ and boundary condition $(**)$.

The condition $(**)$ holds.

For the PDE we need to calculate $u_x$ and $u_t$. For calculating $u_t$ we use the Leibniz rule. We get $$ \begin{align} u_x(x,t) &= \cos(x-tb) + \frac{\partial}{\partial x} \int\limits_0^t \left[x-(t-\sigma)b+\sigma\right] \cos\left(\left[x-(t-\sigma)b\right]\sigma\right) \, d\sigma \\ &= \cos(x-tb) + \int\limits_0^t \frac{\partial}{\partial x} \left[x-(t-\sigma)b+\sigma\right] \cos\left(\left[x-(t-\sigma)b\right]\sigma\right) \, d\sigma \\ &= \cos(x-tb) + \\ &\int\limits_0^t \cos\left(\left[x-(t-\sigma)b\right]\sigma\right) - \left[x-(t-\sigma)b+\sigma\right] \sin\left(\left[x-(t-\sigma)b\right]\sigma\right) \sigma \, d\sigma \end{align} $$ and $$ \begin{align} u_t(x,t) &= -b \cos(x-tb) + \frac{\partial}{\partial t} \int\limits_0^t \left[x-(t-\sigma)b+\sigma\right] \cos\left(\left[x-(t-\sigma)b\right]\sigma\right) \, d\sigma \\ &= -b \cos(x-tb) + \int\limits_0^t \frac{\partial}{\partial t} \left[x-(t-\sigma)b+\sigma\right] \cos\left(\left[x-(t-\sigma)b\right]\sigma\right) \, d\sigma + \\ & \left[x-(t-t)b+t\right] \cos\left(\left[x-(t-t)b\right]t\right) \frac{dt}{dt} \\ &= -b\cos(x-tb) + \\ &(-b) \int\limits_0^t \cos\left(\left[x-(t-\sigma)b\right]\sigma\right) - \left[x-(t-\sigma)b+\sigma\right] \sin\left(\left[x-(t-\sigma)b\right]\sigma\right) \sigma \, d\sigma + \\ & (x+t) \cos(xt) \\ &= -b \, u_x(x,t) + (x+t) \cos(xt) \end{align} $$ So, assuming I made no mistake, your candidate solves $$ u_t(x,t) + b \, u_x(x,t) = (x+t) \cos(xt) $$ which is your $(*)$ if $b = 1$.

mvw
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Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$

$\dfrac{dx}{ds}=1$ , letting $x(0)=x_0$ , we have $x=s+x_0=t+x_0$

$\dfrac{du}{ds}=(x+t)\cos xt=(2s+x_0)\cos((s+x_0)s)$ , letting $u(0)=f(x_0)$ , we have $u(x,t)=f(x_0)+\int_0^s(2\sigma+x_0)\cos((\sigma+x_0)\sigma)~d\sigma=f(x-t)+\int_0^t(2\sigma+x-t)\cos((\sigma+x-t)\sigma)~d\sigma$

$u(x,0)=\sin x$ :

$f(x)=\sin x$

$\therefore u(x,t)=\sin(x-t)+\int_0^t(2\sigma+x-t)\cos((\sigma+x-t)\sigma)~d\sigma$

doraemonpaul
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