Solving an initial value problem for the PDE $u_t + u_x = (x+t) \cos(xt)$

Question

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?

Answer 1

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 $b = 1$.

Answer 2

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!

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

Answer 3

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 $u(x,t)=f(x_0)+\int_0^s(2\sigma+x_0)\cos((\sigma+x_0)\sigma)~d\sigma=f(x-t)+\int_0^t(2\sigma+x-t)\cos((\sigma+x-t)\sigma)~d\sigma$

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

$f(x)=\sin x$

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