Consider the Cauchy problem

\(\rm x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=u;\)

𝑢 = 𝑓(𝑡) on the initial curve Γ = (𝑡, 𝑡); 𝑡 > 0.

Consider the following statements:

𝑃: If 𝑓(𝑡) = 2𝑡 + 1, then there exists a unique solution to the Cauchy problem in a neighbourhood of Γ.

𝑄: If 𝑓(𝑡) = 2𝑡 − 1, then there exist infinitely many solutions to the Cauchy problem in a neighbourhood of Γ.

Then

Given -

Consider the Cauchy problem

\(\rm x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=u;\)

𝑢 = 𝑓(𝑡) on the initial curve Γ = (𝑡, 𝑡); 𝑡 > 0.

Concept -

If the PDE is the form of pP + qQ = R then 

(i) \(\frac{p(t)}{x'}= \frac{q(t)}{y'} = \frac{R}{u'(t)}\) ⇒ Infinite solutions 

(ii) \(\frac{p(t)}{x'} \neq \frac{q(t)}{y'} \neq \frac{R}{u'(t)}\) ⇒ Unique solution

(iii) \(\frac{p(t)}{x'} = \frac{q(t)}{y'} \neq \frac{R}{u'(t)}\) ⇒ No solution

where p = x, q = y and R = u  

Explanation -

we have the PDE is \(\rm x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=u;\) and 𝑢 = 𝑓(𝑡) on the initial curve Γ = (𝑡, 𝑡); 𝑡 > 0.

So x = t, y = t and u = f(t) 

Now for Statement (P) -

\(\frac{t}{1}= \frac{t}{1} \neq \frac{2t+1}{2}\)

Hence there is no solution. So Statement P is false.

Now for Statement (Q) -

\(\frac{t}{1}= \frac{t}{1} \neq \frac{2t-1}{2}\)

Hence there is no solution. So Statement Q is false.

Hence option (4) is true.